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[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 4.5 | If the three points $$\begin{aligned} & (1, a, b) \\ & (a, 2, b) \\ & (a, b, 3) \end{aligned}$$ are collinear (in 3-space), what is the value of $a+b$ ? | The first two points are distinct (otherwise we would have $a=1$ and $a=2$ simultaneously), and they both lie on the plane $z=b$, so the whole line is in this plane and $b=3$. Reasoning similarly with the last two points gives $a=1$, so $a+b=4$. | 4 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5.25 | Let $A B C$ be a triangle with incircle tangent to the perpendicular bisector of $B C$. If $B C=A E=$ 20, where $E$ is the point where the $A$-excircle touches $B C$, then compute the area of $\triangle A B C$. | Let the incircle and $B C$ touch at $D$, the incircle and perpendicular bisector touch at $X, Y$ be the point opposite $D$ on the incircle, and $M$ be the midpoint of $B C$. Recall that $A, Y$, and $E$ are collinear by homothety at $A$. Additionally, we have $M D=M X=M E$ so $\angle D X Y=\angle D X E=90^{\circ}$. Therefore $E, X$, and $Y$ are collinear. Since $M X \perp B C$, we have $\angle A E B=45^{\circ}$. The area of $A B C$ is $\frac{1}{2} B C \cdot A E \cdot \sin \angle A E B=100 \sqrt{2}$. | 100 \sqrt{2} | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 3.5 | On a spherical planet with diameter $10,000 \mathrm{~km}$, powerful explosives are placed at the north and south poles. The explosives are designed to vaporize all matter within $5,000 \mathrm{~km}$ of ground zero and leave anything beyond $5,000 \mathrm{~km}$ untouched. After the explosives are set off, what is the new surface area of the planet, in square kilometers? | The explosives have the same radius as the planet, so the surface area of the "cap" removed is the same as the new surface area revealed in the resulting "dimple." Thus the area is preserved by the explosion and remains $\pi \cdot(10,000)^{2}$. | 100,000,000 \pi | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5 | Find all ordered pairs of integers $(x, y)$ such that $3^{x} 4^{y}=2^{x+y}+2^{2(x+y)-1}$. | The right side is $2^{x+y}\left(1+2^{x+y-1}\right)$. If the second factor is odd, it needs to be a power of 3 , so the only options are $x+y=2$ and $x+y=4$. This leads to two solutions, namely $(1,1)$ and $(2,2)$. The second factor can also be even, if $x+y-1=0$. Then $x+y=1$ and $3^{x} 4^{y}=2+2$, giving $(0,1)$ as the only other solution. | (0,1), (1,1), (2,2) | HMMT_2 |
[
"Mathematics -> Number Theory -> Other"
] | 3.5 | How many of the integers $1,2, \ldots, 2004$ can be represented as $(m n+1) /(m+n)$ for positive integers $m$ and $n$ ? | For any positive integer $a$, we can let $m=a^{2}+a-1, n=a+1$ to see that every positive integer has this property, so the answer is 2004. | 2004 | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Function Theory -> Other",
"Mathematics -> Algebra -> Equations and Inequalities -> Other"
] | 7 | Let $R$ denote the set of all real numbers. Find all functions $f$ from $R$ to $R$ satisfying: (i) there are only finitely many $s$ in R such that $f(s)=0$, and (ii) $f\left(x^{4}+y\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in R. | The only such function is the identity function on $R$. Setting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find $f(y)=f(f(y))$ and thus $f(0)=f(f(0))=0$. It follows from (ii) that $f\left(x^{4}+y\right)=x^{3} f(x)+f(y)$ for all $x, y \in \mathbf{R}$. Set $y=0$ to obtain $f\left(x^{4}\right)=x^{3} f(x)$ for all $x \in \mathrm{R}$, and so $f\left(x^{4}+y\right)=f\left(x^{4}\right)+f(y)$ for all $x, y \in \mathbf{R}$. The functional equation suggests that $f$ is additive, that is, $f(a+b)=f(a)+f(b)$ for all $a, b \in \boldsymbol{R}$. We now show this. First assume that $a \geq 0$ and $b \in \boldsymbol{R}$. It follows that $f(a+b)=f\left(\left(a^{1 / 4}\right)^{4}+b\right)=f\left(\left(a^{1 / 4}\right)^{4}\right)+f(b)=f(a)+f(b)$. We next note that $f$ is an odd function, since $f(-x)=\frac{f\left(x^{4}\right)}{(-x)^{3}}=-f(x)$, $x \neq 0$. Since $f$ is odd, we have that, for $a<0$ and $b \in R$, $f(a+b)=-f((-a)+(-b))=-(f(-a)+f(-b))=f(a)+f(b)$. Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \in \mathbf{R}$. We now show that $\{s \in R \mid f(s)=0\}=\{0\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \in \mathrm{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \in\{s \in R \mid f(s)=0\}$ for all $n \in \mathrm{N}$. However, this is a contradiction to the given condition (i). It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f(b)$ for some $a, b \in \operatorname{R}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \in\{s \in R \mid f(s)=0\}=\{0\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \in \mathrm{R}$. This completes the proof. | \[
f(x) = x \quad \text{for all} \; x \in \mathbb{R}
\] | apmoapmo_sol |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 4 | Let $f(x)=x^{2}+x^{4}+x^{6}+x^{8}+\cdots$, for all real $x$ such that the sum converges. For how many real numbers $x$ does $f(x)=x$ ? | Clearly $x=0$ works. Otherwise, we want $x=x^{2} /\left(1-x^{2}\right)$, or $x^{2}+x-1=0$. Discard the negative root (since the sum doesn't converge there), but $(-1+\sqrt{5}) / 2$ works, for a total of 2 values. | 2 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5.25 | The Fibonacci numbers are defined by $F_{0}=0, F_{1}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 2$. There exist unique positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$ such that $\sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100} \sum_{i_{5}=0}^{100} F_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}}=F_{n_{1}}-5 F_{n_{2}}+10 F_{n_{3}}-10 F_{n_{4}}+5 F_{n_{5}}-F_{n_{6}}$. Find $n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}$. | We make use of the identity $\sum_{i=0}^{\ell} F_{i}=F_{\ell+2}-1$ (easily proven by induction) which implies $\sum_{i=k}^{\ell} F_{i}=F_{\ell+2}-F_{k+1}$. Applying this several times yields $\sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100} \sum_{i_{5}=0}^{100} F_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}} = \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100}\left(F_{i_{1}+i_{2}+i_{3}+i_{4}+102}-F_{i_{1}+i_{2}+i_{3}+i_{4}+1}\right) = \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100}\left(F_{i_{1}+i_{2}+i_{3}+204}-2 F_{i_{1}+i_{2}+i_{3}+103}+F_{i_{1}+i_{2}+i_{3}+2}\right) = \sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100}\left(F_{i_{1}+306}-3 F_{i_{1}+205}+3 F_{i_{1}+104}-F_{i_{1}+3}\right) = \sum_{i_{1}=0}^{100}\left(F_{i_{1}+408}-4 F_{i_{1}+307}+6 F_{i_{1}+206}-4 F_{i_{1}+105}+F_{i_{1}+4}\right) = F_{510}-5 F_{409}+10 F_{308}-10 F_{207}+5 F_{106}-F_{5}$. This representation is unique because the Fibonacci terms grow exponentially quickly, so e.g. the $F_{510}$ term dominates, forcing $n_{1}=510$ and similarly for the other terms. The final answer is $510+409+308+207+106+5=1545$. | 1545 | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Other",
"Mathematics -> Number Theory -> Other"
] | 4 | The Fibonacci numbers are defined by $F_{1}=F_{2}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 3$. If the number $$ \frac{F_{2003}}{F_{2002}}-\frac{F_{2004}}{F_{2003}} $$ is written as a fraction in lowest terms, what is the numerator? | Before reducing, the numerator is $F_{2003}^{2}-F_{2002} F_{2004}$. We claim $F_{n}^{2}-F_{n-1} F_{n+1}=$ $(-1)^{n+1}$, which will immediately imply that the answer is 1 (no reducing required). This claim is straightforward to prove by induction on $n$ : it holds for $n=2$, and if it holds for some $n$, then $$ F_{n+1}^{2}-F_{n} F_{n+2}=F_{n+1}\left(F_{n-1}+F_{n}\right)-F_{n}\left(F_{n}+F_{n+1}\right)=F_{n+1} F_{n-1}-F_{n}^{2}=-(-1)^{n+1}=(-1)^{n+2} $$ | 1 | HMMT_2 |
[
"Mathematics -> Number Theory -> Other"
] | 3.5 | Find the rightmost non-zero digit of the expansion of (20)(13!). | We can rewrite this as $(10 \times 2)(13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)=\left(10^{3}\right)(2 \times 13 \times 12 \times 11 \times 9 \times 8 \times 7 \times 6 \times 4 \times 3)$; multiplying together the units digits for the terms not equal to 10 reveals that the rightmost non-zero digit is 6. | 6 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers"
] | 4.5 | Let $A B C D$ be a quadrilateral inscribed in a unit circle with center $O$. Suppose that $\angle A O B=\angle C O D=135^{\circ}, B C=1$. Let $B^{\prime}$ and $C^{\prime}$ be the reflections of $A$ across $B O$ and $C O$ respectively. Let $H_{1}$ and $H_{2}$ be the orthocenters of $A B^{\prime} C^{\prime}$ and $B C D$, respectively. If $M$ is the midpoint of $O H_{1}$, and $O^{\prime}$ is the reflection of $O$ about the midpoint of $M H_{2}$, compute $O O^{\prime}$. | Put the diagram on the complex plane with $O$ at the origin and $A$ at 1. Let $B$ have coordinate $b$ and $C$ have coordinate $c$. We obtain easily that $B^{\prime}$ is $b^{2}, C^{\prime}$ is $c^{2}$, and $D$ is $b c$. Therefore, $H_{1}$ is $1+b^{2}+c^{2}$ and $H_{2}$ is $b+c+b c$ (we have used the fact that for triangles on the unit circle, their orthocenter is the sum of the vertices). Finally, we have that $M$ is $\frac{1}{2}\left(1+b^{2}+c^{2}\right)$, so the reflection of $O$ about the midpoint of $M H_{2}$ is $\frac{1}{2}\left(1+b^{2}+c^{2}+2 b+2 c+2 b c\right)=\frac{1}{2}(b+c+1)^{2}$, so we just seek $\frac{1}{2}|b+c+1|^{2}$. But we know that $b=\operatorname{cis} 135^{\circ}$ and $c=\operatorname{cis} 195^{\circ}$, so we obtain that this value is $\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})$. | \frac{1}{4}(8-\sqrt{6}-3 \sqrt{2}) | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5 | We say a point is contained in a square if it is in its interior or on its boundary. Three unit squares are given in the plane such that there is a point contained in all three. Furthermore, three points $A, B, C$, are given, each contained in at least one of the squares. Find the maximum area of triangle $A B C$. | Let $X$ be a point contained in all three squares. The distance from $X$ to any point in any of the three squares is at most $\sqrt{2}$, the length of the diagonal of the squares. Therefore, triangle $A B C$ is contained in a circle of radius $\sqrt{2}$, so its circumradius is at most $\sqrt{2}$. The triangle with greatest area that satisfies this property is the equilateral triangle in a circle of radius $\sqrt{2}$. (This can be proved, for example, by considering that the maximum altitude to any given side is obtained by putting the opposite vertex at the midpoint of its arc, and it follows that all the vertices are equidistant.) The equilateral triangle is also attainable, since making $X$ the circumcenter and positioning the squares such that $A X, B X$, and $C X$ are diagonals (of the three squares) and $A B C$ is equilateral, leads to such a triangle. This triangle has area $3 \sqrt{3} / 2$, which may be calculated, for example, using the sine formula for area applied to $A B X, A C X$, and $B C X$, to get $3 / 2(\sqrt{2})^{2} \sin 120^{\circ}$. | 3 \sqrt{3} / 2 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5 | Let $\Gamma$ denote the circumcircle of triangle $A B C$. Point $D$ is on $\overline{A B}$ such that $\overline{C D}$ bisects $\angle A C B$. Points $P$ and $Q$ are on $\Gamma$ such that $\overline{P Q}$ passes through $D$ and is perpendicular to $\overline{C D}$. Compute $P Q$, given that $B C=20, C A=80, A B=65$. | Suppose that $P$ lies between $A$ and $B$ and $Q$ lies between $A$ and $C$, and let line $P Q$ intersect lines $A C$ and $B C$ at $E$ and $F$ respectively. As usual, we write $a, b, c$ for the lengths of $B C, C A, A B$. By the angle bisector theorem, $A D / D B=A C / C B$ so that $A D=\frac{b c}{a+b}$ and $B D=\frac{a c}{a+b}$. Now by Stewart's theorem, $c \cdot C D^{2}+\left(\frac{a c}{a+b}\right)\left(\frac{b c}{a+b}\right) c=$ $\frac{a^{2} b c}{a+b}+\frac{a b^{2} c}{a+b}$ from which $C D^{2}=\frac{a b\left((a+b)^{2}-c^{2}\right)}{(a+b)^{2}}$. Now observe that triangles $C D E$ and $C D F$ are congruent, so $E D=D F$. By Menelaus' theorem, $\frac{C A}{A E} \frac{E D}{D F} \frac{F B}{B C}=1$ so that $\frac{C A}{B C}=\frac{A E}{F B}$. Since $C F=C E$ while $b>a$, it follows that $A E=\frac{b(b-a)}{a+b}$ so that $E C=\frac{2 a b}{a+b}$. Finally, $D E=\sqrt{C E^{2}-C D^{2}}=\frac{\sqrt{a b\left(c^{2}-(a-b)^{2}\right)}}{a+b}$. Plugging in $a=20, b=80, c=65$, we see that $A E=48, E C=32, D E=10$ as well as $A D=52, B D=13$. Now let $P D=x, Q E=y$. By power of a point about $D$ and $E$, we have $x(y+10)=676$ and $y(x+10)=1536$. Subtracting one from the other, we see that $y=x+86$. Therefore, $x^{2}+96 x-676=0$, from which $x=-48+2 \sqrt{745}$. Finally, $P Q=x+y+10=4 \sqrt{745}$. | 4 \sqrt{745} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Find the number of subsets $S$ of $\{1,2, \ldots 63\}$ the sum of whose elements is 2008. | Note that $1+2+\cdots+63=2016$. So the problem is equivalent to finding the number of subsets of $\{1,2, \cdots 63\}$ whose sum of elements is 8. We can count this by hand: $\{8\},\{1,7\},\{2,6\}$, $\{3,5\},\{1,2,5\},\{1,3,4\}$. | 66 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5 | Find the sum of squares of all distinct complex numbers $x$ satisfying the equation $0=4 x^{10}-7 x^{9}+5 x^{8}-8 x^{7}+12 x^{6}-12 x^{5}+12 x^{4}-8 x^{3}+5 x^{2}-7 x+4$ | For convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \Phi_{14}(x)$, which is suggestive. Indeed, consider $\omega$ a primitive 14 -th root of unity; since $\omega^{7}=-1$, we have $4 \omega^{10}=-4 \omega^{3},-7 \omega^{9}=7 \omega^{2}$, and so on, so that $P(\omega)=12\left(\omega^{6}-\omega^{5}+\cdots+1\right)=12 \Phi_{14}(\omega)=0$. Dividing, we find $P(x)=\Phi_{14}(x)\left(4 x^{4}-3 x^{3}-2 x^{2}-3 x+4\right)$. This second polynomial is symmetric; since 0 is clearly not a root, we have $4 x^{4}-3 x^{3}-2 x^{2}-3 x+4=0 \Longleftrightarrow 4\left(x+\frac{1}{x}\right)^{2}-3\left(x+\frac{1}{x}\right)-10=0$. Setting $y=x+1 / x$ and solving the quadratic gives $y=2$ and $y=-5 / 4$ as solutions; replacing $y$ with $x+1 / x$ and solving the two resulting quadratics give the double root $x=1$ and the roots $(-5 \pm i \sqrt{39}) / 8$ respectively. Together with the primitive fourteenth roots of unity, these are all the roots of our polynomial. Explicitly, the roots are $e^{\pi i / 7}, e^{3 \pi i / 7}, e^{5 \pi i / 7}, e^{9 \pi i / 7}, e^{11 \pi i / 7}, e^{13 \pi i / 7}, 1,(-5 \pm i \sqrt{39}) / 8$. The sum of squares of the roots of unity (including 1) is just 0 by symmetry (or a number of other methods). The sum of the squares of the final conjugate pair is $\frac{2\left(5^{2}-39\right)}{8^{2}}=-\frac{14}{32}=-\frac{7}{16}$. | -\frac{7}{16} | HMMT_2 |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions",
"Mathematics -> Number Theory -> Prime Numbers"
] | 5 | Two positive rational numbers $x$ and $y$, when written in lowest terms, have the property that the sum of their numerators is 9 and the sum of their denominators is 10 . What is the largest possible value of $x+y$ ? | For fixed denominators $a<b$ (with sum 10), we maximize the sum of the fractions by giving the smaller denominator as large a numerator as possible: $8 / a+1 / b$. Then, if $a \geq 2$, this quantity is at most $8 / 2+1 / 1=5$, which is clearly smaller than the sum we get by setting $a=1$, namely $8 / 1+1 / 9=73 / 9$. So this is the answer. | 73 / 9 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5 | A tree grows in a rather peculiar manner. Lateral cross-sections of the trunk, leaves, branches, twigs, and so forth are circles. The trunk is 1 meter in diameter to a height of 1 meter, at which point it splits into two sections, each with diameter .5 meter. These sections are each one meter long, at which point they each split into two sections, each with diameter .25 meter. This continues indefinitely: every section of tree is 1 meter long and splits into two smaller sections, each with half the diameter of the previous. What is the total volume of the tree? | If we count the trunk as level 0, the two sections emerging from it as level 1, and so forth, then the $n$th level consists of $2^{n}$ sections each with diameter $1 / 2^{n}$, for a volume of $2^{n}(\pi / 4 \cdot 2^{-2 n})=(\pi / 4) \cdot 2^{-n}$. So the total volume is given by a simple infinite sum, $$ .25 \pi \cdot(1+1 / 2+1 / 4+\ldots)=.25 \pi \cdot 2=\pi / 2 $$ | \pi / 2 | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization"
] | 3.5 | I have chosen five of the numbers $\{1,2,3,4,5,6,7\}$. If I told you what their product was, that would not be enough information for you to figure out whether their sum was even or odd. What is their product? | Giving you the product of the five numbers is equivalent to telling you the product of the two numbers I didn't choose. The only possible products that are achieved by more than one pair of numbers are $12(\{3,4\}$ and $\{2,6\})$ and $6(\{1,6\}$ and $\{2,3\})$. But in the second case, you at least know that the two unchosen numbers have odd sum (and so the five chosen numbers have odd sum also). Therefore, the first case must hold, and the product of the five chosen numbers is $$ 1 \cdot 2 \cdot 5 \cdot 6 \cdot 7=1 \cdot 3 \cdot 4 \cdot 5 \cdot 7=420 $$ | 420 | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Field Theory"
] | 5 | Let $A$ be a set of integers such that for each integer $m$, there exists an integer $a \in A$ and positive integer $n$ such that $a^{n} \equiv m(\bmod 100)$. What is the smallest possible value of $|A|$? | Work in $R=\mathbb{Z} / 100 \mathbb{Z} \cong \mathbb{Z} / 4 \mathbb{Z} \times \mathbb{Z} / 25 \mathbb{Z}$. Call an element $r \in R$ type $(s, t)$ if $s=\nu_{2}(r) \leq 2$ and $t=\nu_{5}(r) \leq 2$. Also, define an element $r \in R$ to be coprime if it is of type $(0,0)$, powerful if it is of types $(0,2),(2,0)$, or $(2,2)$, and marginal otherwise. Then, note that if if $r \in R$ is marginal, then any power of $r$ is powerful. Therefore all marginal elements must be in $A$. We claim that all powerful elements are the cube of some marginal element. To show this take a powerful element $r$. In modulo 4 or 25, if $r$ is a unit, then since 3 is coprime to both the sizes of $(\mathbb{Z} / 4 \mathbb{Z})^{\times}$and $(\mathbb{Z} / 25 \mathbb{Z})^{\times}$, it is the cube of some element. Otherwise, if $r$ is zero then it is the cube of 2 or 5, respectively (since this case happens at least once this means that the constructed cube root is marginal). We now claim that 4 additional elements are needed to generate the coprime elements. To see this, note that $R^{\times} \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 20 \mathbb{Z}$ since there are primitive roots $\bmod 4$ and 25. Under this isomorphism, one can show that $(1,1),(1,2),(1,4)$, and $(0,1)$ generate anything, and that no element in $R^{\times}$has more than one of these as a multiple. To wrap up, note that there are $100-(20+1)(2+1)=37$ marginal elements, so 41 elements are needed in total. | 41 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 3.5 | Let $A B C D E$ be a convex pentagon such that $\angle A B C=\angle A C D=\angle A D E=90^{\circ}$ and $A B=B C=C D=D E=1$. Compute $A E$. | By Pythagoras, $A E^{2}=A D^{2}+1=A C^{2}+2=A B^{2}+3=4$ so $A E=2$. | 2 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5 | A regular dodecahedron is projected orthogonally onto a plane, and its image is an $n$-sided polygon. What is the smallest possible value of $n$ ? | We can achieve 6 by projecting onto a plane perpendicular to an edge of the dodecaheron. Indeed, if we imagine viewing the dodecahedron in such a direction, then 4 of the faces are projected to line segments (namely, the two faces adjacent to the edge and the two opposite faces), and of the remaining 8 faces, 4 appear on the front of the dodecahedron and the other 4 are on the back. Thus, the dodecahedron appears as shown. To see that we cannot do better, note that, by central symmetry, the number of edges of the projection must be even. So we just need to show that the answer cannot be 4. But if the projection had 4 sides, one of the vertices would give a projection forming an acute angle, which is not possible. So 6 is the answer. | 6 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 5 | There are eleven positive integers $n$ such that there exists a convex polygon with $n$ sides whose angles, in degrees, are unequal integers that are in arithmetic progression. Find the sum of these values of $n$. | The sum of the angles of an $n$-gon is $(n-2) 180$, so the average angle measure is $(n-2) 180 / n$. The common difference in this arithmetic progression is at least 1 , so the difference between the largest and smallest angles is at least $n-1$. So the largest angle is at least $(n-1) / 2+(n-2) 180 / n$. Since the polygon is convex, this quantity is no larger than 179: $(n-1) / 2-360 / n \leq-1$, so that $360 / n-n / 2 \geq 1 / 2$. Multiplying by $2 n$ gives $720-n^{2} \geq n$. So $n(n+1) \leq 720$, which forces $n \leq 26$. Of course, since the common difference is an integer, and the angle measures are integers, $(n-2) 180 / n$ must be an integer or a half integer, so $(n-2) 360 / n=360-720 / n$ is an integer, and then $720 / n$ must be an integer. This leaves only $n=3,4,5,6,8,9,10,12,15,16,18,20,24$ as possibilities. When $n$ is even, $(n-2) 180 / n$ is not an angle of the polygon, but the mean of the two middle angles. So the common difference is at least 2 when $(n-2) 180 / n$ is an integer. For $n=20$, the middle angle is 162 , so the largest angle is at least $162+38 / 2=181$, since 38 is no larger than the difference between the smallest and largest angles. For $n=24$, the middle angle is 165 , again leading to a contradiction. So no solution exists for $n=20,24$. All of the others possess solutions: \begin{tabular}{|c|l|} \hline$n$ & angles \\ \hline 3 & $59,60,61$ \\ 4 & $87,89,91,93$ \\ 5 & $106,107,108,109,110$ \\ 6 & $115,117,119,121,123,125$ \\ 8 & $128,130,132,134,136,138,140,142$ \\ 9 & $136, \ldots, 144$ \\ 10 & $135,137,139, \ldots, 153$ \\ 12 & $139,141,143, \ldots, 161$ \\ 15 & $149,150, \ldots, 163$ \\ 16 & $150,151, \ldots, 165$ \\ 18 & $143,145, \ldots, 177$ \\ \hline \end{tabular} (These solutions are quite easy to construct.) The desired value is then $3+4+5+6+$ $8+9+10+12+15+16+18=106$. | 106 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | You want to arrange the numbers $1,2,3, \ldots, 25$ in a sequence with the following property: if $n$ is divisible by $m$, then the $n$th number is divisible by the $m$ th number. How many such sequences are there? | Let the rearranged numbers be $a_{1}, \ldots, a_{25}$. The number of pairs $(n, m)$ with $n \mid m$ must equal the number of pairs with $a_{n} \mid a_{m}$, but since each pair of the former type is also of the latter type, the converse must be true as well. Thus, $n \mid m$ if and only if $a_{n} \mid a_{m}$. Now for each $n=1,2, \ldots, 6$, the number of values divisible by $n$ uniquely determines $n$, so $n=a_{n}$. Similarly, 7,8 must either be kept fixed by the rearrangement or interchanged, because they are the only values that divide exactly 2 other numbers in the sequence; since 7 is prime and 8 is not, we conclude they are kept fixed. Then we can easily check by induction that $n=a_{n}$ for all larger composite numbers $n \leq 25$ (by using $m=a_{m}$ for all proper factors $m$ of $n$ ) and $n=11$ (because it is the only prime that divides exactly 1 other number). So we have only the primes $n=13,17,19,23$ left to rearrange, and it is easily seen that these can be permuted arbitrarily, leaving 4 ! possible orderings altogether. | 24 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5.5 | An $11 \times 11$ grid is labeled with consecutive rows $0,1,2, \ldots, 10$ and columns $0,1,2, \ldots, 10$ so that it is filled with integers from 1 to $2^{10}$, inclusive, and the sum of all of the numbers in row $n$ and in column $n$ are both divisible by $2^{n}$. Find the number of possible distinct grids. | We begin by filling the 10 by 10 grid formed by rows and columns 1 through 10 with any values, which we can do in $\left(2^{10}\right)^{100}=2^{1000}$ ways. Then in column 0, there is at most 1 way to fill in the square in row 10, 2 ways for the square in row 9, down to $2^{10}$ ways in row 0. Similarly, there is 1 way to fill in the square in row 0 and column 10, 2 ways to fill in the square in row 0 and column 9, etc. Overall, the number of ways to fill out the squares in row or column 0 is $2^{1} \cdot 2^{2} \cdot 2^{3} \cdots 2^{9} \cdot 2^{10} \cdot 2^{9} \cdot 2^{8} \cdots 2^{1}=2^{100}$, so the number of possible distinct grids $2^{1000} \cdot 2^{100}=2^{1100}$. | 2^{1100} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5.25 | We have an $n$-gon, and each of its vertices is labeled with a number from the set $\{1, \ldots, 10\}$. We know that for any pair of distinct numbers from this set there is at least one side of the polygon whose endpoints have these two numbers. Find the smallest possible value of $n$. | Each number be paired with each of the 9 other numbers, but each vertex can be used in at most 2 different pairs, so each number must occur on at least $\lceil 9 / 2\rceil=5$ different vertices. Thus, we need at least $10 \cdot 5=50$ vertices, so $n \geq 50$. To see that $n=50$ is feasible, let the numbers $1, \ldots, 10$ be the vertices of a complete graph. Then each vertex has degree 9 , and there are $\binom{10}{2}=45$ edges. If we attach extra copies of the edges $1-2,3-4,5-6,7-8$, and $9-10$, then every vertex will have degree 10 . In particular, the graph has an Eulerian tour, so we can follow this tour, successively numbering vertices of the 50-gon according to the vertices of the graph we visit. Then, for each edge of the graph, there will be a corresponding edge of the polygon with the same two vertex labels on its endpoints. It follows that every pair of distinct numbers occurs at the endpoints of some edge of the polygon, and so $n=50$ is the answer. | 50 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 3.5 | $P$ is a point inside triangle $A B C$, and lines $A P, B P, C P$ intersect the opposite sides $B C, C A, A B$ in points $D, E, F$, respectively. It is given that $\angle A P B=90^{\circ}$, and that $A C=B C$ and $A B=B D$. We also know that $B F=1$, and that $B C=999$. Find $A F$. | Let $A C=B C=s, A B=B D=t$. Since $B P$ is the altitude in isosceles triangle $A B D$, it bisects angle $B$. So, the Angle Bisector Theorem in triangle $A B C$ given $A E / E C=A B / B C=t / s$. Meanwhile, $C D / D B=(s-t) / t$. Now Ceva's theorem gives us $$ \begin{gathered} \frac{A F}{F B}=\left(\frac{A E}{E C}\right) \cdot\left(\frac{C D}{D B}\right)=\frac{s-t}{s} \\ \Rightarrow \frac{A B}{F B}=1+\frac{s-t}{s}=\frac{2 s-t}{s} \Rightarrow F B=\frac{s t}{2 s-t} \end{gathered} $$ Now we know $s=999$, but we need to find $t$ given that $s t /(2 s-t)=F B=1$. So $s t=2 s-t \Rightarrow t=2 s /(s+1)$, and then $$ A F=F B \cdot \frac{A F}{F B}=1 \cdot \frac{s-t}{s}=\frac{\left(s^{2}-s\right) /(s+1)}{s}=\frac{s-1}{s+1}=\frac{499}{500} $$ | 499 / 500 | HMMT_2 |
[
"Mathematics -> Number Theory -> Congruences"
] | 5 | Positive integers $a, b$, and $c$ have the property that $a^{b}, b^{c}$, and $c^{a}$ end in 4, 2, and 9, respectively. Compute the minimum possible value of $a+b+c$. | This minimum is attained when $(a, b, c)=(2,2,13)$. To show that we cannot do better, observe that $a$ must be even, so $c$ ends in 3 or 7. If $c \geq 13$, since $a$ and $b$ are even, it's clear $(2,2,13)$ is optimal. Otherwise, $c=3$ or $c=7$, in which case $b^{c}$ can end in 2 only when $b$ ends in 8. However, no eighth power ends in 4, so we would need $b \geq 18$ (and $a \geq 2$), which makes the sum $2+18+3=23$ larger than 17. | 17 | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Group Theory"
] | 5 | Find the value of $$ \binom{2003}{1}+\binom{2003}{4}+\binom{2003}{7}+\cdots+\binom{2003}{2002} $$ | Let $\omega=-1 / 2+i \sqrt{3} / 2$ be a complex cube root of unity. Then, by the binomial theorem, we have $$ \begin{aligned} \omega^{2}(\omega+1)^{2003} & =\binom{2003}{0} \omega^{2}+\binom{2003}{1} \omega^{3}+\binom{2003}{2} \omega^{4}+\cdots+\binom{2003}{2003} \omega^{2005} \\ 2^{2003} & =\binom{2003}{0}+\binom{2003}{1}+\binom{2003}{2}+\cdots+\binom{2003}{2003} \\ \omega^{-2}\left(\omega^{-1}+1\right)^{2003} & =\binom{003}{0} \omega^{-2}+\binom{2003}{1} \omega^{-3}+\binom{2003}{2} \omega^{-4}+\cdots+\binom{2003}{2003} \omega^{-2005} \end{aligned} $$ If we add these together, then the terms $\binom{2003}{n}$ for $n \equiv 1(\bmod 3)$ appear with coefficient 3 , while the remaining terms appear with coefficient $1+\omega+\omega^{2}=0$. Thus the desired sum is just $\left(\omega^{2}(\omega+1)^{2003}+2^{2003}+\omega^{-2}\left(\omega^{-1}+1\right)^{2003}\right) / 3$. Simplifying using $\omega+1=-\omega^{2}$ and $\omega^{-1}+1=-\omega$ gives $\left(-1+2^{2003}+-1\right) / 3=\left(2^{2003}-2\right) / 3$. | \left(2^{2003}-2\right) / 3 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 4 | A regular hexagon has one side along the diameter of a semicircle, and the two opposite vertices on the semicircle. Find the area of the hexagon if the diameter of the semicircle is 1. | The midpoint of the side of the hexagon on the diameter is the center of the circle. Draw the segment from this center to a vertex of the hexagon on the circle. This segment, whose length is $1 / 2$, is the hypotenuse of a right triangle whose legs have lengths $a / 2$ and $a \sqrt{3}$, where $a$ is a side of the hexagon. So $1 / 4=a^{2}(1 / 4+3)$, so $a^{2}=1 / 13$. The hexagon consists of 6 equilateral triangles of side length $a$, so the area of the hexagon is $3 a^{2} \sqrt{3} / 2=3 \sqrt{3} / 26$. | 3 \sqrt{3} / 26 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 4.5 | Equilateral triangles $A B F$ and $B C G$ are constructed outside regular pentagon $A B C D E$. Compute $\angle F E G$. | We have $\angle F E G=\angle A E G-\angle A E F$. Since $E G$ bisects $\angle A E D$, we get $\angle A E G=54^{\circ}$. Now, $\angle E A F=108^{\circ}+60^{\circ}=168^{\circ}$. Since triangle $E A F$ is isosceles, this means $\angle A E F=6^{\circ}$, so the answer is $54^{\circ}-6^{\circ}=48^{\circ}$. | 48^{\circ} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Probability -> Other"
] | 4 | Mark has a cursed six-sided die that never rolls the same number twice in a row, and all other outcomes are equally likely. Compute the expected number of rolls it takes for Mark to roll every number at least once. | Suppose Mark has already rolled $n$ unique numbers, where $1 \leq n \leq 5$. On the next roll, there are 5 possible numbers he could get, with $6-n$ of them being new. Therefore, the probability of getting another unique number is $\frac{6-n}{5}$, so the expected number of rolls before getting another unique number is $\frac{5}{6-n}$. Since it always takes 1 roll to get the first number, the expected total number of rolls is $1+\frac{5}{5}+\frac{5}{4}+\frac{5}{3}+\frac{5}{2}+\frac{5}{1}=\frac{149}{12}$. | \frac{149}{12} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations",
"Mathematics -> Discrete Mathematics -> Graph Theory"
] | 5.25 | The game of Penta is played with teams of five players each, and there are five roles the players can play. Each of the five players chooses two of five roles they wish to play. If each player chooses their roles randomly, what is the probability that each role will have exactly two players? | Consider a graph with five vertices corresponding to the roles, and draw an edge between two vertices if a player picks both roles. Thus there are exactly 5 edges in the graph, and we want to find the probability that each vertex has degree 2. In particular, we want to find the probability that the graph is composed entirely of cycles. Thus there are two cases. The first case is when the graph is itself a 5-cycle. There are 4! ways to choose such a directed cycle (pick an arbitrary vertex $A$ and consider a vertex it connects to, etc.), and thus $\frac{4!}{2}=12$ ways for the undirected graph to be a 5-cycle. Now, there are 5! ways to assign the edges in this cycle to people, giving a total contribution of $12 \cdot 5$!. The second case is when the graph is composed of a 2-cycle and a 3-cycle, which only requires choosing the two vertices to be the 2-cycle, and so there are $\binom{5}{2}=10$ ways. To assign the players to edges, there are $\binom{5}{2}=10$ ways to assign the players to the 2-cycle. For the 3-cycle, any of the $3!=6$ permutations of the remaining 3 players work. The total contribution is $10 \cdot 10 \cdot 6$. Therefore, our answer is $\frac{12 \cdot 120+10 \cdot 10 \cdot 6}{10^{5}}=\frac{51}{2500}$. | \frac{51}{2500} | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization"
] | 5 | Find the largest integer $n$ such that $3^{512}-1$ is divisible by $2^{n}$. | Write $$ \begin{aligned} 3^{512}-1 & =\left(3^{256}+1\right)\left(3^{256}-1\right)=\left(3^{256}+1\right)\left(3^{128}+1\right)\left(3^{128}-1\right) \\ & =\cdots=\left(3^{256}+1\right)\left(3^{128}+1\right) \cdots(3+1)(3-1) \end{aligned} $$ Now each factor $3^{2^{k}}+1, k \geq 1$, is divisible by just one factor of 2 , since $3^{2^{k}}+1=$ $\left(3^{2}\right)^{2^{k-1}}+1 \equiv 1^{2^{k-1}}+1=2(\bmod 4)$. Thus we get 8 factors of 2 here, and the remaining terms $(3+1)(3-1)=8$ give us 3 more factors of 2 , for a total of 11. | 11 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> Volume",
"Mathematics -> Calculus -> Integral Calculus -> Techniques of Integration -> Multi-variable",
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 5 | A wealthy king has his blacksmith fashion him a large cup, whose inside is a cone of height 9 inches and base diameter 6 inches. At one of his many feasts, he orders the mug to be filled to the brim with cranberry juice. For each positive integer $n$, the king stirs his drink vigorously and takes a sip such that the height of fluid left in his cup after the sip goes down by $\frac{1}{n^{2}}$ inches. Shortly afterwards, while the king is distracted, the court jester adds pure Soylent to the cup until it's once again full. The king takes sips precisely every minute, and his first sip is exactly one minute after the feast begins. As time progresses, the amount of juice consumed by the king (in cubic inches) approaches a number $r$. Find $r$. | First, we find the total amount of juice consumed. We can simply subtract the amount of juice remaining at infinity from the initial amount of juice in the cup, which of course is simply the volume of the cup; we'll denote this value by $V$. Since volume in the cup varies as the cube of height, the amount of juice remaining in the cup after $m$ minutes is $V \cdot \prod_{n=1}^{m}\left(\frac{9-\frac{1}{n^{2}}}{9}\right)^{3}=V \cdot\left(\prod_{n=1}^{m}\left(1-\frac{1}{9 n^{2}}\right)\right)^{3}$. We can now factor the term inside the product to find $V\left(\prod_{n=1}^{m} \frac{(3 n+1)(3 n-1)}{9 n^{2}}\right)^{3}=V\left(\frac{(3 m+1)!}{3^{3 m}(m!)^{3}}\right)^{3}$. If remains to evaluate the limit of this expression as $m$ goes to infinity. However, by Stirling's approximation, we have $\lim _{m \rightarrow \infty} \frac{(3 m+1)!}{3^{3 m}(m!)^{3}} =\lim _{m \rightarrow \infty} \frac{\left(\frac{3 n+1}{e}\right)^{3 n+1} \cdot \sqrt{2 \pi(3 n+1)}}{\left(\frac{3 n}{e}\right)^{3 n} \sqrt{(2 \pi n)^{3}}} =\lim _{m \rightarrow \infty} \frac{(3 n+1) \sqrt{3}}{2 \pi n e}\left(\frac{3 n+1}{3 n}\right)^{3 n} =\frac{3 \sqrt{3}}{2 \pi}$. Therefore the total amount of juice the king consumes is $V-V\left(\frac{3 \sqrt{3}}{2 \pi}\right)^{3}=\left(\frac{3^{2} \cdot \pi \cdot 9}{3}\right)\left(\frac{8 \pi^{3}-81 \sqrt{3}}{8 \pi^{3}}\right)=\frac{216 \pi^{3}-2187 \sqrt{3}}{8 \pi^{2}}$. | \frac{216 \pi^{3}-2187 \sqrt{3}}{8 \pi^{2}} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 4.5 | Five points are chosen uniformly at random on a segment of length 1. What is the expected distance between the closest pair of points? | Choose five points arbitrarily at $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ in increasing order. Then the intervals $\left(a_{2}-x, a_{2}\right),\left(a_{3}-x, a_{3}\right),\left(a_{4}-x, a_{4}\right),\left(a_{5}-x, a_{5}\right)$ must all be unoccupied. The probability that this happens is the same as doing the process in reverse: first defining these intervals, then choosing five random points none of which lie in the four intervals. This transformed process clearly has a $(1-4x)^{5}$ probability of success. It follows that the desired probability is $\int_{0}^{1/4}(1-4x)^{5} dx=\frac{1}{24}$. | \frac{1}{24} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4 | A binary string of length $n$ is a sequence of $n$ digits, each of which is 0 or 1 . The distance between two binary strings of the same length is the number of positions in which they disagree; for example, the distance between the strings 01101011 and 00101110 is 3 since they differ in the second, sixth, and eighth positions. Find as many binary strings of length 8 as you can, such that the distance between any two of them is at least 3 . You get one point per string. | The maximum possible number of such strings is 20 . An example of a set attaining this bound is \begin{tabular}{ll} 00000000 & 00110101 \\ 11001010 & 10011110 \\ 11100001 & 01101011 \\ 11010100 & 01100110 \\ 10111001 & 10010011 \\ 01111100 & 11001101 \\ 00111010 & 10101100 \\ 01010111 & 11110010 \\ 00001111 & 01011001 \\ 10100111 & 11111111 \end{tabular} This example is taken from page 57 of F. J. MacWilliams and N. J. A. Sloane, The Theory of Error Correcting Codes (New York: Elsevier Publishing, 1977). The proof that 20 is the best possible is elementary but too long to reproduce here; see pages $537-541$ of MacWilliams and Sloane for details. In general, a set of $M$ strings of length $n$ such that any two have a distance of at least $d$ is called an $(n, M, d)$-code. These objects are of basic importance in coding theory, which studies how to transmit information through a channel with a known error rate. For example, since the code given above has minimum distance 3, I can transmit to you a message consisting of strings in this code, and even if there is a possible error rate of one digit in each string, you will still be able to determine the intended message uniquely. | \begin{tabular}{ll} 00000000 & 00110101 \ 11001010 & 10011110 \ 11100001 & 01101011 \ 11010100 & 01100110 \ 10111001 & 10010011 \ 01111100 & 11001101 \ 00111010 & 10101100 \ 01010111 & 11110010 \ 00001111 & 01011001 \ 10100111 & 11111111 \ \end{tabular} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5 | Let $f(x)=x^{2}-2 x$. How many distinct real numbers $c$ satisfy $f(f(f(f(c))))=3$ ? | We see the size of the set $f^{-1}\left(f^{-1}\left(f^{-1}\left(f^{-1}(3)\right)\right)\right)$. Note that $f(x)=(x-1)^{2}-1=3$ has two solutions: $x=3$ and $x=-1$, and that the fixed points $f(x)=x$ are $x=3$ and $x=0$. Therefore, the number of real solutions is equal to the number of distinct real numbers $c$ such that $c=3, c=-1, f(c)=-1$ or $f(f(c))=-1$, or $f(f(f(c)))=-1$. The equation $f(x)=-1$ has exactly one root $x=1$. Thus, the last three equations are equivalent to $c=1, f(c)=1$, and $f(f(c))=1$. $f(c)=1$ has two solutions, $c=1 \pm \sqrt{2}$, and for each of these two values $c$ there are two preimages. It follows that the answer is $1+1+1+2+4=9$. | 9 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5.25 | Alice, Bob, and Charlie are playing a game with 6 cards numbered 1 through 6. Each player is dealt 2 cards uniformly at random. On each player's turn, they play one of their cards, and the winner is the person who plays the median of the three cards played. Charlie goes last, so Alice and Bob decide to tell their cards to each other, trying to prevent him from winning whenever possible. Compute the probability that Charlie wins regardless. | If Alice has a card that is adjacent to one of Bob's, then Alice and Bob will play those cards as one of them is guaranteed to win. If Alice and Bob do not have any adjacent cards, since Charlie goes last, Charlie can always choose a card that will win. Let $A$ denote a card that is held by Alice and $B$ denote a card that is held by Bob. We will consider the ascending order of which Alice and Bob's cards are held. If the ascending order in which Alice and Bob's cards are held are $A B A B$ or $B A B A$, then Charlie cannot win. In these 2 cases, there will always be 2 consecutive cards where one is held by Alice and the other is held by Bob. Therefore, the only cases we need to consider are the ascending orders $A A B B$, $A B B A$, and their symmetric cases. In the case $A A B B$, we must make sure that the larger card Alice holds and the smaller card Bob holds are not consecutive. Alice can thus have $\{1,2\},\{2,3\}$, or $\{1,3\}$. Casework on what Bob can have yields 5 different combinations of pairs of cards Alice and Bob can hold. Since this applies to the symmetric case $B B A A$ as well, we get 10 different combinations. In the case $A B B A$, we see that Alice's cards must be $\{1,6\}$ and Bob's cards must be $\{3,4\}$. Considering the symmetric case $B A A B$ as well, this gives us 2 more combinations. Thus, there are 12 total possible combinations of Alice's and Bob's cards such that Charlie will win regardless. The total number of ways to choose Alice's and Bob's cards is given by $\binom{6}{2}\binom{4}{2}=90$, so the probability that Charlie is guaranteed to win is $\frac{12}{90}=\frac{2}{15}$. | \frac{2}{15} | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5 | Find all positive integer solutions $(m, n)$ to the following equation: $$ m^{2}=1!+2!+\cdots+n! $$ | A square must end in the digit $0,1,4,5,6$, or 9 . If $n \geq 4$, then $1!+2!+\cdots+n$ ! ends in the digit 3 , so cannot be a square. A simple check for the remaining cases reveals that the only solutions are $(1,1)$ and $(3,3)$. | (1,1), (3,3) | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Compute the number of quadruples $(a, b, c, d)$ of positive integers satisfying $12a+21b+28c+84d=2024$. | Looking at the equation $\bmod 7$ gives $a \equiv 3(\bmod 7)$, so let $a=7a^{\prime}+3$. Then mod 4 gives $b \equiv 0(\bmod 4)$, so let $b=4b^{\prime}$. Finally, $\bmod 3$ gives $c \equiv 2(\bmod 3)$, so let $c=3c^{\prime}+2$. Now our equation yields $$84a^{\prime}+84b^{\prime}+84c^{\prime}+84d=2024-3 \cdot 12-2 \cdot 28=1932 \Longrightarrow a^{\prime}+b^{\prime}+c^{\prime}+d=23$$ Since $a, b, c, d$ are positive integers, we have $a^{\prime}$ and $c^{\prime}$ are nonnegative and $b^{\prime}$ and $d$ are positive. Thus, let $b^{\prime\prime}=b^{\prime}+1$ and $d^{\prime}=d+1$, so $a^{\prime}, b^{\prime\prime}, c^{\prime}, d^{\prime}$ are nonnegative integers summing to 21. By stars and bars, there are $\binom{24}{3}=2024$ such solutions. | 2024 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations",
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 5 | Let $A B C$ be a triangle whose incircle has center $I$ and is tangent to $\overline{B C}, \overline{C A}, \overline{A B}$, at $D, E, F$. Denote by $X$ the midpoint of major arc $\widehat{B A C}$ of the circumcircle of $A B C$. Suppose $P$ is a point on line $X I$ such that $\overline{D P} \perp \overline{E F}$. Given that $A B=14, A C=15$, and $B C=13$, compute $D P$. | Let $H$ be the orthocenter of triangle $D E F$. We claim that $P$ is the midpoint of $\overline{D H}$. Indeed, consider an inversion at the incircle of $A B C$, denoting the inverse of a point with an asterik. It maps $A B C$ to the nine-point circle of $\triangle D E F$. According to $\angle I A X=90^{\circ}$, we have $\angle A^{*} X^{*} I=90^{\circ}$. Hence line $X I$ passes through the point diametrically opposite to $A^{*}$, which is the midpoint of $\overline{D H}$, as claimed. The rest is a straightforward computation. The inradius of $\triangle A B C$ is $r=4$. The length of $E F$ is given by $E F=2 \frac{A F \cdot r}{A I}=\frac{16}{\sqrt{5}}$. Then, $D P^{2}=\left(\frac{1}{2} D H\right)^{2}=\frac{1}{4}\left(4 r^{2}-E F^{2}\right)=4^{2}-\frac{64}{5}=\frac{16}{5}$. Hence $D P=\frac{4 \sqrt{5}}{5}$. | \frac{4 \sqrt{5}}{5} | HMMT_2 |
[
"Mathematics -> Algebra -> Sequences and Series -> Other"
] | 4.5 | If $a_{1}=1, a_{2}=0$, and $a_{n+1}=a_{n}+\frac{a_{n+2}}{2}$ for all $n \geq 1$, compute $a_{2004}$. | By writing out the first few terms, we find that $a_{n+4}=-4 a_{n}$. Indeed, $$ a_{n+4}=2\left(a_{n+3}-a_{n+2}\right)=2\left(a_{n+2}-2 a_{n+1}\right)=2\left(-2 a_{n}\right)=-4 a_{n} $$ Then, by induction, we get $a_{4 k}=(-4)^{k}$ for all positive integers $k$, and setting $k=501$ gives the answer. | -2^{1002} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5.25 | Let $D$ be a regular ten-sided polygon with edges of length 1. A triangle $T$ is defined by choosing three vertices of $D$ and connecting them with edges. How many different (non-congruent) triangles $T$ can be formed? | The problem is equivalent to finding the number of ways to partition 10 into a sum of three (unordered) positive integers. These can be computed by hand to be $(1,1,8),(1,2,7),(1,3,6),(1,4,5),(2,2,6),(2,3,5),(2,4,4),(3,3,4)$ | 8 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Number Theory -> Congruences"
] | 5.25 | Let $a \star b=ab-2$. Compute the remainder when $(((579 \star 569) \star 559) \star \cdots \star 19) \star 9$ is divided by 100. | Note that $$(10a+9) \star (10b+9)=(100ab+90a+90b+81)-2 \equiv 90(a+b)+79 \pmod{100}$$ so throughout our process all numbers will end in 9, so we will just track the tens digit. Then the "new operation" is $$a \dagger b \equiv -(a+b)+7 \bmod 10$$ where $a$ and $b$ track the tens digits. Now $$(a \dagger b) \dagger c \equiv (-(a+b)+7) \dagger c \equiv a+b-c \pmod{10}$$ Thus, our expression has tens digit congruent to $$-0+1-2+3-\cdots-54+55-56-57+7 \equiv 2 \bmod 10$$ making the answer 29. | 29 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations",
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 5.5 | In $\triangle A B C, \omega$ is the circumcircle, $I$ is the incenter and $I_{A}$ is the $A$-excenter. Let $M$ be the midpoint of arc $\widehat{B A C}$ on $\omega$, and suppose that $X, Y$ are the projections of $I$ onto $M I_{A}$ and $I_{A}$ onto $M I$, respectively. If $\triangle X Y I_{A}$ is an equilateral triangle with side length 1, compute the area of $\triangle A B C$. | Using Fact 5, we know that $I I_{A}$ intersects the circle $(A B C)$ at $M_{A}$, which is the center of $(I I_{A} B C X Y)$. Let $R$ be the radius of the latter circle. We have $R=\frac{1}{\sqrt{3}}$. We have $\angle A I M=\angle Y I I_{A}=\angle Y I X=\frac{\pi}{3}$. Also, $\angle I I_{A} M=\angle I M I_{A}$ by calculating the angles from the equilateral triangle. Using $90-60-30$ triangles, we have: $A I=\frac{1}{2} M I=\frac{1}{2} I I_{A}=R$, $A M=\frac{\sqrt{3}}{2} M I=\sqrt{3} R$, $M M_{A}^{2}=A M^{2}+A M_{A}^{2}=7 R^{2}$. Now, let $J$ and $N$ be the feet of the altitudes from $A$ and $B$ respectively on $M M_{A}$. Note that as $M$ is an arc midpoint of $B C, N$ is actually the midpoint of $B C$. $M_{A} J=\frac{A M_{A}^{2}}{M M_{A}}=\frac{4}{\sqrt{7}} R$, $M_{A} N=\frac{B M_{A}^{2}}{M M_{A}}=\frac{1}{\sqrt{7}} R$. Thus $J N=\frac{3}{\sqrt{7}} R$. Also, we have, $B N^{2}=M_{A} N \cdot M N=\frac{6}{7} R^{2}$. Now, $[A B C]=\frac{1}{2} J N \cdot B C=J N \cdot B N=\frac{3 \sqrt{6}}{7} R^{2}=\frac{\sqrt{6}}{7}$. | \frac{\sqrt{6}}{7} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5 | Find the number of ordered pairs of positive integers $(x, y)$ with $x, y \leq 2020$ such that $3 x^{2}+10 x y+3 y^{2}$ is the power of some prime. | We can factor as $(3 x+y)(x+3 y)$. If $x \geq y$, we need $\frac{3 x+y}{x+3 y} \in\{1,2\}$ to be an integer. So we get the case where $x=y$, in which we need both to be a power of 2, or the case $x=5 y$, in which case we need $y$ to be a power of 2. This gives us $11+9+9=29$ solutions, where we account for $y=5 x$ as well. | 29 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | The numbers $1,2, \ldots, 20$ are put into a hat. Claire draws two numbers from the hat uniformly at random, $a<b$, and then puts them back into the hat. Then, William draws two numbers from the hat uniformly at random, $c<d$. Let $N$ denote the number of integers $n$ that satisfy exactly one of $a \leq n \leq b$ and $c \leq n \leq d$. Compute the probability $N$ is even. | The number of integers that satisfy exactly one of the two inequalities is equal to the number of integers that satisfy the first one, plus the number of integers that satisfy the second one, minus twice the number of integers that satisfy both. Parity-wise, this is just the number of integers that satisfy the first one, plus the number of integers that satisfy the second one. The number of integers that satisfy the first one is $b-a+1$. The probability this is even is $\frac{10}{19}$, and odd is $\frac{9}{19}$. This means the answer is $$\frac{10^{2}+9^{2}}{19^{2}}=\frac{181}{361}$$. | \frac{181}{361} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Let $A_{1}, A_{2}, \ldots, A_{m}$ be finite sets of size 2012 and let $B_{1}, B_{2}, \ldots, B_{m}$ be finite sets of size 2013 such that $A_{i} \cap B_{j}=\emptyset$ if and only if $i=j$. Find the maximum value of $m$. | In general, we will show that if each of the sets $A_{i}$ contain $a$ elements and if each of the sets $B_{j}$ contain $b$ elements, then the maximum value for $m$ is $\binom{a+b}{a}$. Let $U$ denote the union of all the sets $A_{i}$ and $B_{j}$ and let $|U|=n$. Consider the $n$ ! orderings of the elements of $U$. Note that for any specific ordering, there is at most one value of $i$ such that all the elements in $A_{i}$ come before all the elements in $B_{i}$ in this ordering; this follows since $A_{j}$ shares at least one element with $B_{i}$ and $B_{j}$ shares at least one element with $A_{i}$ for any other $j \neq i$. On the other hand, the number of ways to permute the $(a+b)$ elements in $A_{i} \cup B_{i}$ so that all the elements in $A_{i}$ come first is equal to $a!b!$. Therefore, the number of permutations of $U$ where all the elements in $A_{i}$ come before all the elements in $B_{i}$ is equal to: $$n!\cdot \frac{a!b!}{(a+b)!}=\frac{n!}{\binom{a+b}{a}}$$ Summing over all $m$ values of $i$, the total number of orderings where, for some $i$, the elements in $A_{i}$ come before $B_{i}$ is equal to $$\frac{n!m}{\binom{a+b}{a}}$$ But there are at most $u$ ! such orderings, since there are $u$ ! total orderings, so it follows that $m \leq\binom{ a+b}{a}$. Equality is attained by taking $U$ to be a set containing $(a+b)$ elements, letting $A_{i}$ range over all $a$-element subsets of $U$, and letting $B_{i}=U \backslash A_{i}$ for each $i$. | \binom{4025}{2012} | HMMT_2 |
[
"Mathematics -> Precalculus -> Trigonometric Functions"
] | 5 | Let $w, x, y$, and $z$ be positive real numbers such that $0 \neq \cos w \cos x \cos y \cos z$, $2 \pi =w+x+y+z$, $3 \tan w =k(1+\sec w)$, $4 \tan x =k(1+\sec x)$, $5 \tan y =k(1+\sec y)$, $6 \tan z =k(1+\sec z)$. Find $k$. | From the identity $\tan \frac{u}{2}=\frac{\sin u}{1+\cos u}$, the conditions work out to $3 \tan \frac{w}{2}=4 \tan \frac{x}{2}=5 \tan \frac{y}{2}=6 \tan \frac{z}{2}=k$. Let $a=\tan \frac{w}{2}, b=\tan \frac{x}{2}, c=\tan \frac{y}{2}$, and $d=\tan \frac{z}{2}$. Using the identity $\tan (M+N)=\frac{\tan M+\tan N}{1-\tan M \tan N}$, we obtain $\tan \left(\frac{w+x}{2}+\frac{y+z}{2}\right) =\frac{\tan \left(\frac{w+x}{2}\right)+\tan \left(\frac{y+z}{2}\right)}{1-\tan \left(\frac{w+x}{2}\right) \tan \left(\frac{y+z}{2}\right)} =\frac{\frac{a+b}{1-a b}+\frac{c+d}{1-c d}}{1-\left(\frac{a+b}{1-a b}\right)\left(\frac{c+d}{1-c d}\right)} =\frac{a+b+c+d-a b c-a b d-b c d-a c d}{1+a b c d-a b-a c-a d-b c-b d-c d}$. Because $x+y+z+w=\pi$, we get that $\tan \left(\frac{x+y+z+w}{2}\right)=0$ and thus $a+b+c+d=a b c+a b d+b c d+a c d$. Substituting $a, b, c, d$ corresponding to the variable $k$, we obtain that $k^{3}-19 k=0$. Therefore, $k$ can be only $0, \sqrt{19},-\sqrt{19}$. However, $k=0$ is impossible as $w, x, y, z$ will all be 0. Also, $k=-\sqrt{19}$ is impossible as $w, x, y, z$ will exceed $\pi$. Therefore, $k=\sqrt{19}$. | \sqrt{19} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Circles",
"Mathematics -> Calculus -> Differential Calculus -> Related Rates"
] | 5 | A circle is tangent to both branches of the hyperbola $x^{2}-20y^{2}=24$ as well as the $x$-axis. Compute the area of this circle. | Invert about the unit circle centered at the origin. $\omega$ turns into a horizontal line, and the hyperbola turns into the following: $$\begin{aligned} \frac{x^{2}}{\left(x^{2}+y^{2}\right)^{2}}-\frac{20y^{2}}{\left(x^{2}+y^{2}\right)^{2}}=24 & \Longrightarrow x^{2}-20y^{2}=24\left(x^{2}+y^{2}\right)^{2} \\ & \Longrightarrow 24x^{4}+\left(48y^{2}-1\right)x^{2}+24y^{4}+20y^{2}=0 \\ & \Longrightarrow\left(48y^{2}-1\right)^{2} \geq 4(24)\left(24y^{4}+20y^{2}\right) \\ & \Longrightarrow 1-96y^{2} \geq 1920y^{2} \\ & \Longrightarrow y \leq \sqrt{1/2016} \end{aligned}$$ This means that the horizontal line in question is $y=\sqrt{1/2016}$. This means that the diameter of the circle is the reciprocal of the distance between the point and line, which is $\sqrt{2016}$, so the radius is $\sqrt{504}$, and the answer is $504\pi$. | 504\pi | HMMT_2 |
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 5.5 | Compute the number of even positive integers $n \leq 2024$ such that $1,2, \ldots, n$ can be split into $\frac{n}{2}$ pairs, and the sum of the numbers in each pair is a multiple of 3. | There have to be an even number of multiples of 3 at most $n$, so this means that $n \equiv 0,2 \pmod{6}$. We claim that all these work. We know there are an even number of multiples of 3, so we can pair them; then we can pair $3k+1$ and $3k+2$ for all $k$. This means the answer is $\frac{2022}{3}+1=675$. | 675 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 5 | Jerry and Neil have a 3-sided die that rolls the numbers 1, 2, and 3, each with probability $\frac{1}{3}$. Jerry rolls first, then Neil rolls the die repeatedly until his number is at least as large as Jerry's. Compute the probability that Neil's final number is 3. | If Jerry rolls $k$, then there is a $\frac{1}{4-k}$ probability that Neil's number is 3, since Neil has an equal chance of rolling any of the $4-k$ integers not less than $k$. Thus, the answer is $$\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{18}$$. | \frac{11}{18} | HMMT_2 |
[
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 4.5 | Compute the smallest positive integer such that, no matter how you rearrange its digits (in base ten), the resulting number is a multiple of 63. | First, the number must be a multiple of 9 and 7. The first is easy to check and holds for all permutations. Note that when two adjacent digits $a$ and $b$ are swapped, the number changes by $9(a-b) \cdot 10^{k}$ (we disregard sign), so $9(a-b)$ must also be a multiple of 63 for all digits $a$ and $b$. In particular, this is sufficient, since a permutation can be represented as a series of transpositions. This means that $a-b$ must be a multiple of 7 for all digits $a$ and $b$, so either all digits are equal or they are in $\{0,7\},\{1,8\}$, or $\{2,9\}$. We find the minimum for each case separately. We first provide the following useful fact: the first repunit (numbers $1,11,111, \ldots$) that is a multiple of 7 is 111111. This is because $10 \bmod 7=3$, and 3 is a generator modulo 7 (of course, you can just compute the powers of 3 by hand, and it will not take much longer). If a number $k \cdot 1 \ldots 1$ is a multiple of 63, then either $k$ or $1 \ldots 1$ is a multiple of 7; if it is $k$, then it's clear that we need 777777777 to make the sum a multiple of 9. If $1 \ldots 1$ is a multiple of 7, then it is at least 111111, then to make a multiple of 9, we need 333333. If the only digits are 7 and 0, then we need at least nine sevens to make the digit sum a multiple of nine, which has more digits than 333333. If the only digits are 8 and 1, then we can note that since 8 and 1 are both $1(\bmod 7)$, these numbers are equivalent to the repunits modulo 7, so such numbers have at least six digits. The best such six-digit number with digits summing to a multiple of 9 is 111888, which is our new candidate. If the only digits are 9 and 2, then by analogous logic such numbers have at least six digits. But the smallest such number is 999999, which is not better. So our best answer is 111888. It works. | 111888 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Graph Theory"
] | 4.5 | You would like to provide airline service to the 10 cities in the nation of Schizophrenia, by instituting a certain number of two-way routes between cities. Unfortunately, the government is about to divide Schizophrenia into two warring countries of five cities each, and you don't know which cities will be in each new country. All airplane service between the two new countries will be discontinued. However, you want to make sure that you set up your routes so that, for any two cities in the same new country, it will be possible to get from one city to the other (without leaving the country). What is the minimum number of routes you must set up to be assured of doing this, no matter how the government divides up the country? | Each city $C$ must be directly connected to at least 6 other cities, since otherwise the government could put $C$ in one country and all its connecting cities in the other country, and there would be no way out of $C$. This means that we have 6 routes for each of 10 cities, counted twice (since each route has two endpoints) $\Rightarrow 6 \cdot 10 / 2=30$ routes. On the other hand, this is enough: picture the cities arranged around a circle, and each city connected to its 3 closest neighbors in either direction. Then if $C$ and $D$ are in the same country but mutually inaccessible, this means that on each arc of the circle between $C$ and $D$, there must be (at least) three consecutive cities in the other country. Then this second country would have 6 cities, which is impossible. So our arrangement achieves the goal with 30 routes. | 30 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Kelvin the frog currently sits at $(0,0)$ in the coordinate plane. If Kelvin is at $(x, y)$, either he can walk to any of $(x, y+1),(x+1, y)$, or $(x+1, y+1)$, or he can jump to any of $(x, y+2),(x+2, y)$ or $(x+1, y+1)$. Walking and jumping from $(x, y)$ to $(x+1, y+1)$ are considered distinct actions. Compute the number of ways Kelvin can reach $(6,8)$. | Observe there are $\binom{14}{6}=3003$ up-right paths from $(0,0)$ to $(6,8)$, each of which are 14 steps long. Any two of these steps can be combined into one: $UU, RR$, and $RU$ as jumps, and $UR$ as walking from $(x, y)$ to $(x+1, y+1)$. The number of ways to combine steps is the number of ways to group 14 actions into singles and consecutive pairs, which is $F_{15}=610$. Every path Kelvin can take can be represented this way, so the answer is $610 \cdot 3003=1831830$. | 1831830 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Other"
] | 5.25 | Yang has the sequence of integers $1,2, \ldots, 2017$. He makes 2016 swaps in order, where a swap changes the positions of two integers in the sequence. His goal is to end with $2,3, \ldots, 2017,1$. How many different sequences of swaps can Yang do to achieve his goal? | Let $n=2017$. The problem is asking to write a cycle permutation of $n$ integers as the product of $n-1$ transpositions. Say that the transpositions Yang uses are $\left(a_{i}, b_{i}\right)$ (i.e. swapping the $a_{i}$-th integer in the sequence with the $b_{i}$-th integer in the sequence). Draw the graph with edges $\left(a_{i}, b_{i}\right)$. One can show that the result is a cycle if and only if the resulting graph is acyclic, so it must be a tree. There are $n^{n-2}$ trees by Cayley's formula, and for each tree, it can be made in $(n-1)$! ways (any ordering of the edges). So the total number of ways to end with a cycle is $n^{n-2} \cdot(n-1)$!. By symmetry, each cycle can be made in the same number of ways, so in particular the cycle $2,3, \ldots, n, 1$ can be made in $\frac{n^{n-2} \cdot(n-1)!}{(n-1)!}=n^{n-2}$ ways. | 2017^{2015} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 4 | In triangle $A B C$, points $M$ and $N$ are the midpoints of $A B$ and $A C$, respectively, and points $P$ and $Q$ trisect $B C$. Given that $A, M, N, P$, and $Q$ lie on a circle and $B C=1$, compute the area of triangle $A B C$. | Note that $M P \parallel A Q$, so $A M P Q$ is an isosceles trapezoid. In particular, we have $A M=M B=B P=P Q=\frac{1}{3}$, so $A B=\frac{2}{3}$. Thus $A B C$ is isosceles with base 1 and legs $\frac{2}{3}$, and the height from $A$ to $B C$ is $\frac{\sqrt{7}}{6}$, so the area is $\frac{\sqrt{7}}{12}$. | \frac{\sqrt{7}}{12} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 3.5 | Suppose $A B C$ is a triangle such that $A B=13, B C=15$, and $C A=14$. Say $D$ is the midpoint of $\overline{B C}, E$ is the midpoint of $\overline{A D}, F$ is the midpoint of $\overline{B E}$, and $G$ is the midpoint of $\overline{D F}$. Compute the area of triangle $E F G$. | By Heron's formula, $[A B C]=\sqrt{21(21-15)(21-14)(21-13)}=84$. Now, unwinding the midpoint conditions yields $[E F G]=\frac{[D E F]}{2}=\frac{[B D E]}{4}=\frac{[A B D]}{8}=\frac{[A B C]}{16}=\frac{84}{16}=\frac{21}{4}$. | \frac{21}{4} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 5 | Barry picks infinitely many points inside a unit circle, each independently and uniformly at random, $P_{1}, P_{2}, \ldots$ Compute the expected value of $N$, where $N$ is the smallest integer such that $P_{N+1}$ is inside the convex hull formed by the points $P_{1}, P_{2}, \ldots, P_{N}$. Submit a positive real number $E$. If the correct answer is $A$, you will receive $\lfloor 100 \cdot \max (0.2099-|E-A|, 0)\rfloor$ points. | Clearly, $N \geq 3$, and let's scale the circle to have area 1. We can see that the probability to not reach $N=4$ is equal to the probability that the fourth point is inside the convex hull of the past three points. That is, the probability is just one minus the expected area of those $N$ points. The area of this turns out to be really small, and is around 0.074, and so $(1-0.074)$ of all sequences of points make it to $N=4$. The probability to reach to the fifth point from there should be around $(1-0.074)(1-0.074 \cdot 2)$, as any four points in convex configuration can be covered with 2 triangles. Similarly, the chance of reaching $N=6$ should be around $(1-0.074)(1-0.074 \cdot 2)(1-0.074 \cdot 3)$, and so on. Noting that our terms eventually decay to zero around term $1/0.074=13$, our answer should be an underestimate. In particular, we get $$3+(1-0.074)(1+(1-0.074 \cdot 2)(1+(1-0.074 \cdot 3)(1+\cdots))) \approx 6.3$$ Guessing anything slightly above this lower bound should give a positive score. | 6.54 | HMMT_2 |
[
"Mathematics -> Number Theory -> Prime Numbers"
] | 5 | Given that the 32-digit integer 64312311692944269609355712372657 is the product of 6 consecutive primes, compute the sum of these 6 primes. | Because the product is approximately $64 \cdot 10^{30}$, we know the primes are all around 200000. Say they are $200000+x_{i}$ for $i=1, \ldots, 6$. By expanding $\prod_{i=1}^{6}\left(200000+x_{i}\right)$ as a polynomial in 200000, we see that $$31231 \cdot 10^{25}=200000^{5}\left(x_{1}+\cdots+x_{6}\right)$$ plus the carry from the other terms. Note that $31231=975 \cdot 32+31$, so $x_{1}+\cdots+x_{6} \leq 975$. Thus, $$16\left(x_{1}x_{2}+x_{1}x_{3}+\cdots+x_{5}x_{6}\right) \leq 16 \cdot \frac{5}{12}\left(x_{1}+\cdots+x_{6}\right)^{2}<\frac{20}{3} \cdot 1000^{2}<67 \cdot 10^{5}$$ so the carry term from $200000^{4}\left(x_{1}x_{2}+\cdots+x_{5}x_{6}\right)$ is at most $67 \cdot 10^{25}$. The other terms have negligible carry, so it is pretty clear $x_{1}+\cdots+x_{6}>972$, otherwise the carry term would have to be at least $$31231 \cdot 10^{25}-200000^{5}(972)=127 \cdot 10^{25}$$ It follows that $x_{1}+\cdots+x_{6}$ lies in [973, 975], so the sum of the primes, $6 \cdot 200000+\left(x_{1}+\cdots+x_{6}\right)$, lies in $[1200973,1200975]$. As these primes are all greater than 2, they are all odd, so their sum is even. Thus it must be 1200974. | 1200974 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5.5 | Compute $\sqrt[4]{5508^{3}+5625^{3}+5742^{3}}$, given that it is an integer. | Let $a=5625=75^{2}$ and $b=117$. Then we have $5508^{3}+5265^{3}+5742^{3}=(a-b)^{3}+a^{3}+(a+b)^{3}=3a^{3}+6ab^{2}=3a(a^{2}+2b^{2})$. We have $3a=3^{3} \cdot 5^{4}$, so $a^{2}+2b^{2}=3^{4} \cdot(625^{2}+2 \cdot 19^{2})$ should be 3 times a fourth power. This means $625^{2}+2 \cdot 19^{2}=3x^{4}$ for some integer $x$. By parity, $x$ must be odd, and also $x^{2} \sqrt{3} \approx 625$. Approximating $\sqrt{3}$ even as 2, we get $x$ should be around 19. Then $x=17$ is clearly too small, and $x=21$ is too big. (You can also check mod 7 for this latter one.) Thus, $x=19$. The final answer is then $3^{2} \cdot 5 \cdot 19=855$. | 855 | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 4.5 | A pebble is shaped as the intersection of a cube of side length 1 with the solid sphere tangent to all of the cube's edges. What is the surface area of this pebble? | Imagine drawing the sphere and the cube. Take a cross section, with a plane parallel to two of the cube's faces, passing through the sphere's center. In this cross section, the sphere looks like a circle, and the cube looks like a square (of side length 1) inscribed in that circle. We can now calculate that the sphere has diameter $d:=\sqrt{2}$ and surface area $S:=\pi d^{2}=2 \pi$, and that the sphere protrudes a distance of $x:=\frac{\sqrt{2}-1}{2}$ out from any given face of the cube. It is known that the surface area chopped off from a sphere by any plane is proportional to the perpendicular distance thus chopped off. Thus, each face of the cube chops of a fraction $\frac{x}{d}$ of the sphere's surface. The surface area of the pebble contributed by the sphere is thus $S \cdot\left(1-6 \cdot \frac{x}{d}\right)$, whereas the cube contributes 6 circles of radius $\frac{1}{2}$, with total area $6 \cdot \pi\left(\frac{1}{2}\right)^{2}=\frac{3}{2} \pi$. The pebble's surface area is therefore $$S \cdot\left(1-6 \cdot \frac{x}{d}\right)+\frac{3}{2} \pi=2 \pi \cdot\left(1-6 \cdot \frac{\sqrt{2}-1}{2 \sqrt{2}}\right)+\frac{3}{2} \pi=\frac{6 \sqrt{2}-5}{2} \pi$$ | \frac{6 \sqrt{2}-5}{2} \pi | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 5 | Point $P$ is inside a square $A B C D$ such that $\angle A P B=135^{\circ}, P C=12$, and $P D=15$. Compute the area of this square. | Let $x=A P$ and $y=B P$. Rotate $\triangle B A P$ by $90^{\circ}$ around $B$ to get $\triangle B C Q$. Then, $\triangle B P Q$ is rightisosceles, and from $\angle B Q C=135^{\circ}$, we get $\angle P Q C=90^{\circ}$. Therefore, by Pythagorean's theorem, $P C^{2}=x^{2}+2y^{2}$. Similarly, $P D^{2}=y^{2}+2x^{2}$. Thus, $y^{2}=\frac{2P C^{2}-P D^{2}}{3}=21$, and similarly $x^{2}=102 \Longrightarrow xy=3\sqrt{238}$. Thus, by the Law of Cosines, the area of the square is $$\begin{aligned} A B^{2} & =A P^{2}+B P^{2}-2 \cos \left(135^{\circ}\right)(A P)(B P) \\ & =x^{2}+y^{2}+\sqrt{2}xy \\ & =123+6\sqrt{119} \end{aligned}$$ | 123+6\sqrt{119} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4.5 | An $n$-string is a string of digits formed by writing the numbers $1,2, \ldots, n$ in some order (in base ten). For example, one possible 10-string is $$35728910461$$ What is the smallest $n>1$ such that there exists a palindromic $n$-string? | The following is such a string for $n=19$ : $$ 9|18| 7|16| 5|14| 3|12| 1|10| 11|2| 13|4| 15|6| 17|8| 19 $$ where the vertical bars indicate breaks between the numbers. On the other hand, to see that $n=19$ is the minimum, notice that only one digit can occur an odd number of times in a palindromic $n$-string (namely the center digit). If $n \leq 9$, then (say) the digits 1,2 each appear once in any $n$-string, so we cannot have a palindrome. If $10 \leq n \leq 18$, then 0,9 each appear once, and we again cannot have a palindrome. So 19 is the smallest possible $n$. | 19 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 3.5 | Mr. Canada chooses a positive real $a$ uniformly at random from $(0,1]$, chooses a positive real $b$ uniformly at random from $(0,1]$, and then sets $c=a /(a+b)$. What is the probability that $c$ lies between $1 / 4$ and $3 / 4$ ? | From $c \geq 1 / 4$ we get $$ \frac{a}{a+b} \geq \frac{1}{4} \Longleftrightarrow b \leq 3 a $$ and similarly $c \leq 3 / 4$ gives $$ \frac{a}{a+b} \leq \frac{3}{4} \Longleftrightarrow a \leq 3 b $$ Choosing $a$ and $b$ randomly from $[0,1]$ is equivalent to choosing a single point uniformly and randomly from the unit square, with $a$ on the horizontal axis and $b$ on the vertical axis. To find the probability that $b \leq 3 a$ and $a \leq 3 b$, we need to find the area of the shaded region of the square. The area of each of the triangles on the side is $(1 / 2)(1)(1 / 3)=1 / 6$, and so the area of the shaded region is $1-2(1 / 6)=2 / 3$. | 2 / 3 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 5 | Let $r_{1}, \ldots, r_{n}$ be the distinct real zeroes of the equation $x^{8}-14 x^{4}-8 x^{3}-x^{2}+1=0$. Evaluate $r_{1}^{2}+\cdots+r_{n}^{2}$ | Observe that $x^{8}-14 x^{4}-8 x^{3}-x^{2}+1 =\left(x^{8}+2 x^{4}+1\right)-\left(16 x^{4}+8 x^{3}+x^{2}\right) =\left(x^{4}+4 x^{2}+x+1\right)\left(x^{4}-4 x^{2}-x+1\right)$. The polynomial $x^{4}+4 x^{2}+x+1=x^{4}+\frac{15}{4} x^{2}+\left(\frac{x}{2}+1\right)^{2}$ has no real roots. On the other hand, let $P(x)=x^{4}-4 x^{2}-x+1$. Observe that $P(-\infty)=+\infty>0, P(-1)=-1<0, P(0)=1>0$, $P(1)=-3<0, P(+\infty)=+\infty>0$, so by the intermediate value theorem, $P(x)=0$ has four distinct real roots, which are precisely the real roots of the original degree 8 equation. By Vieta's formula on $P(x)$, $r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2} =\left(r_{1}+r_{2}+r_{3}+r_{4}\right)^{2}-2 \cdot\left(\sum_{i<j} r_{i} r_{j}\right) =0^{2}-2(-4)=8$ | 8 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations",
"Mathematics -> Number Theory -> Congruences",
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 7 | For each prime $p$, a polynomial $P(x)$ with rational coefficients is called $p$-good if and only if there exist three integers $a, b$, and $c$ such that $0 \leq a<b<c<\frac{p}{3}$ and $p$ divides all the numerators of $P(a)$, $P(b)$, and $P(c)$, when written in simplest form. Compute the number of ordered pairs $(r, s)$ of rational numbers such that the polynomial $x^{3}+10x^{2}+rx+s$ is $p$-good for infinitely many primes $p$. | By Vieta, the sum of the roots is $-10(\bmod p)$. However, since the three roots are less than $p/3$, it follows that the roots are $\left(p-a^{\prime}\right)/3,\left(p-b^{\prime}\right)/3,\left(p-c^{\prime}\right)/3$, where there are finitely many choices $a^{\prime}<b^{\prime}<c^{\prime}$. By pigeonhole, one choice, say $(u, v, w)$ must occur for infinitely many $p$. We then get that the roots of $P$ are $-u/3,-v/3$, and $-w/3$. Moreover, we must have that $u, v, w$ are all $1(\bmod 3)$ or all $2(\bmod 3)$, and by Vieta, we have $u+v+w=30$. The polynomial is then uniquely determined by $u, v, w$. Thus, it suffices to count triples $u<v<w$ of positive integers such that $u, v, w$ are all $1(\bmod 3)$ or all $2(\bmod 3)$ and that $u+v+w=30$. It's not very hard to list them all now. When $u, v, w \equiv 1(\bmod 3)$, there are 7 triples: $(1,4,25),(1,7,22),(1,10,19),(1,13,16),(4,7,19)$, $(4,10,16)$, and $(7,10,13)$. When $u, v, w \equiv 2(\bmod 3)$, there are 5 triples: $(2,5,23),(2,8,20),(2,11,17),(5,8,17)$, and $(5,11,14)$. Hence, the answer is $7+5=12$. | 12 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 5 | Let $A_{1} A_{2} \ldots A_{19}$ be a regular nonadecagon. Lines $A_{1} A_{5}$ and $A_{3} A_{4}$ meet at $X$. Compute $\angle A_{7} X A_{5}$. | Inscribing the nondecagon in a circle, note that $$\angle A_{3} X A_{5}=\frac{1}{2}(\widehat{A_{1} A_{3}}-\widehat{A_{4} A_{5}})=\frac{1}{2} \widehat{A_{5} A_{3} A_{4}}=\angle A_{5} A_{3} X$$ Thus $A_{5} X=A_{5} A_{3}=A_{5} A_{7}$, so $$\begin{aligned} \angle A_{7} X A_{5} & =90^{\circ}-\frac{1}{2} \angle X A_{5} A_{7}=\frac{1}{2} \angle A_{1} A_{5} A_{7} \\ & =\frac{1}{4} \widehat{A_{1} A_{8} A_{7}}=\frac{1}{4} \cdot \frac{13}{19} \cdot 360^{\circ}=\frac{1170^{\circ}}{19} \end{aligned}$$ | \frac{1170^{\circ}}{19} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Area"
] | 5 | Let $A B C D$ be a rectangle such that $A B=20$ and $A D=24$. Point $P$ lies inside $A B C D$ such that triangles $P A C$ and $P B D$ have areas 20 and 24, respectively. Compute all possible areas of triangle $P A B$. | There are four possible locations of $P$ as shown in the diagram. Let $O$ be the center. Then, $[P A O]=10$ and $[P B O]=12$. Thus, $[P A B]=[A O B] \pm[P A O] \pm[P B O]=120 \pm 10 \pm 12$, giving the four values $98,118,122$, and 142. | 98, 118, 122, 142 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 4.5 | Given $\triangle A B C$ with $A B<A C$, the altitude $A D$, angle bisector $A E$, and median $A F$ are drawn from $A$, with $D, E, F$ all lying on \overline{B C}$. If \measuredangle B A D=2 \measuredangle D A E=2 \measuredangle E A F=\measuredangle F A C$, what are all possible values of \measuredangle A C B$ ? | Let $H$ and $O$ be the orthocenter and circumcenter of $A B C$, respectively: it is well-known (and not difficult to check) that \measuredangle B A H=\measuredangle C A O$. However, note that \measuredangle B A H=\measuredangle B A D=\measuredangle C A F$, so \measuredangle C A F=\measuredangle C A O$, that is, $O$ lies on median $A F$, and since $A B<A C$, it follows that $F=O$. Therefore, \measuredangle B A C=90^{\circ}$. Now, we compute \measuredangle A C B=\measuredangle B A D=\frac{2}{6} \measuredangle B A C=30^{\circ}$. | 30^{\circ} \text{ or } \pi / 6 \text{ radians} | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization"
] | 5 | Compute the sum of all positive integers $n$ such that $n^{2}-3000$ is a perfect square. | Suppose $n^{2}-3000=x^{2}$, so $n^{2}-x^{2}=3000$. This factors as $(n-x)(n+x)=3000$. Thus, we have $n-x=2a$ and $n+x=2b$ for some positive integers $a, b$ such that $ab=750$ and $a<b$. Therefore, we have $n=a+b$, so the sum will be just the sum of divisors of $750=2 \cdot 3 \cdot 5^{3}$, which is $$(1+2)(1+3)(1+5+25+125)=1872$$. | 1872 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5.25 | Kelvin the Frog is hopping on a number line (extending to infinity in both directions). Kelvin starts at 0. Every minute, he has a $\frac{1}{3}$ chance of moving 1 unit left, a $\frac{1}{3}$ chance of moving 1 unit right and $\frac{1}{3}$ chance of getting eaten. Find the expected number of times Kelvin returns to 0 (not including the start) before getting eaten. | First we compute probability that the mouse returns to 0 before being eaten. Then probability that it is at 0 in $2n$ minutes without being eaten is given by $\frac{1}{3^{2n}}\binom{2n}{n}$. Therefore, the overall expectation is given by $\sum_{n \geq 1}\binom{2n}{n} 9^{-n}=-1+\sum_{n \geq 0}\binom{2n}{n} 9^{-n}=-1+\frac{1}{\sqrt{1-4/9}}=-1+\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}-5}{5}$ where we use the well known fact that $\sum_{n \geq 0}\binom{2n}{n} x^{n}=\frac{1}{\sqrt{1-4x}}$ for $x=\frac{1}{9}$. | \frac{3\sqrt{5}-5}{5} | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 3.5 | Let rectangle $A B C D$ have lengths $A B=20$ and $B C=12$. Extend ray $B C$ to $Z$ such that $C Z=18$. Let $E$ be the point in the interior of $A B C D$ such that the perpendicular distance from $E$ to \overline{A B}$ is 6 and the perpendicular distance from $E$ to \overline{A D}$ is 6 . Let line $E Z$ intersect $A B$ at $X$ and $C D$ at $Y$. Find the area of quadrilateral $A X Y D$. | Draw the line parallel to \overline{A D}$ through $E$, intersecting \overline{A B}$ at $F$ and \overline{C D}$ at $G$. It is clear that $X F E$ and $Y G E$ are congruent, so the area of $A X Y D$ is equal to that of $A F G D$. But $A F G D$ is simply a 12 by 6 rectangle, so the answer must be 72 . (Note: It is also possible to directly compute the values of $A X$ and $D Y$, then use the formula for the area of a trapezoid.) | 72 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Plane Geometry -> Circles"
] | 5 | $A B C D$ is a cyclic quadrilateral with sides $A B=10, B C=8, C D=25$, and $D A=12$. A circle $\omega$ is tangent to segments $D A, A B$, and $B C$. Find the radius of $\omega$. | Denote $E$ an intersection point of $A D$ and $B C$. Let $x=E A$ and $y=E B$. Because $A B C D$ is a cyclic quadrilateral, $\triangle E A B$ is similar to $\triangle E C D$. Therefore, $\frac{y+8}{x}=\frac{25}{10}$ and $\frac{x+12}{y}=\frac{25}{10}$. We get $x=\frac{128}{21}$ and $y=\frac{152}{21}$. Note that $\omega$ is the $E$-excircle of $\triangle E A B$, so we may finish by standard calculations. Indeed, first we compute the semiperimeter $s=\frac{E A+A B+B E}{2}=\frac{x+y+10}{2}=\frac{35}{3}$. Now the radius of $\omega$ is (by Heron's formula for area) $r_{E}=\frac{[E A B]}{s-A B}=\sqrt{\frac{s(s-x)(s-y)}{s-10}}=\sqrt{\frac{1209}{7}}=\frac{\sqrt{8463}}{7}$ | \sqrt{\frac{1209}{7}} \text{ OR } \frac{\sqrt{8463}}{7} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers"
] | 5 | $S$ is a set of complex numbers such that if $u, v \in S$, then $u v \in S$ and $u^{2}+v^{2} \in S$. Suppose that the number $N$ of elements of $S$ with absolute value at most 1 is finite. What is the largest possible value of $N$ ? | First, if $S$ contained some $u \neq 0$ with absolute value $<1$, then (by the first condition) every power of $u$ would be in $S$, and $S$ would contain infinitely many different numbers of absolute value $<1$. This is a contradiction. Now suppose $S$ contains some number $u$ of absolute value 1 and argument $\theta$. If $\theta$ is not an integer multiple of $\pi / 6$, then $u$ has some power $v$ whose argument lies strictly between $\theta+\pi / 3$ and $\theta+\pi / 2$. Then $u^{2}+v^{2}=u^{2}\left(1+(v / u)^{2}\right)$ has absolute value between 0 and 1 , since $(v / u)^{2}$ lies on the unit circle with angle strictly between $2 \pi / 3$ and $\pi$. But $u^{2}+v^{2} \in S$, so this is a contradiction. This shows that the only possible elements of $S$ with absolute value \leq 1 are 0 and the points on the unit circle whose arguments are multiples of $\pi / 6$, giving $N \leq 1+12=13$. To show that $N=13$ is attainable, we need to show that there exists a possible set $S$ containing all these points. Let $T$ be the set of all numbers of the form $a+b \omega$, where $a, b$ are integers are $\omega$ is a complex cube root of 1 . Since $\omega^{2}=-1-\omega, T$ is closed under multiplication and addition. Then, if we let $S$ be the set of numbers $u$ such that $u^{2} \in T, S$ has the required properties, and it contains the 13 complex numbers specified, so we're in business. | 13 | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions"
] | 5 | Let $\{a_{i}\}_{i \geq 0}$ be a sequence of real numbers defined by $a_{n+1}=a_{n}^{2}-\frac{1}{2^{2020 \cdot 2^{n}-1}}$ for $n \geq 0$. Determine the largest value for $a_{0}$ such that $\{a_{i}\}_{i \geq 0}$ is bounded. | Let $a_{0}=\frac{1}{\sqrt{2}^{2020}}\left(t+\frac{1}{t}\right)$, with $t \geq 1$. (If $a_{0}<\frac{1}{\sqrt{2}^{2018}}$ then no real $t$ exists, but we ignore these values because $a_{0}$ is smaller.) Then, we can prove by induction that $a_{n}=\frac{1}{\sqrt{2}^{2020 \cdot 2^{n}}}\left(t^{2^{n}}+\frac{1}{t^{2^{n}}}\right)$. For this to be bounded, it is easy to see that we just need $\frac{t^{2^{n}}}{\sqrt{2}^{2020 \cdot 2^{n}}}=\left(\frac{t}{\sqrt{2}^{2020}}\right)^{2^{n}}$ to be bounded, since the second term approaches 0. We see that this is is equivalent to $t \leq 2^{2020 / 2}$, which means $a_{0} \leq \frac{1}{\sqrt{2}^{2020}}\left(\sqrt{2}^{2020}+\left(\frac{1}{\sqrt{2}}\right)^{2020}\right)=1+\frac{1}{2^{2020}}$. | 1+\frac{1}{2^{2020}} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 5 | Let $P$ be a polynomial such that $P(x)=P(0)+P(1) x+P(2) x^{2}$ and $P(-1)=1$. Compute $P(3)$. | Plugging in $x=-1,1,2$ results in the trio of equations $1=P(-1)=P(0)-P(1)+P(2)$, $P(1)=P(0)+P(1)+P(2) \Rightarrow P(1)+P(2)=0$, and $P(2)=P(0)+2 P(1)+4 P(2)$. Solving these as a system of equations in $P(0), P(1), P(2)$ gives $P(0)=-1, P(1)=-1, P(2)=1$. Consequently, $P(x)=x^{2}-x-1 \Rightarrow P(3)=5$. | 5 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 4.5 | Knot is ready to face Gammadorf in a card game. In this game, there is a deck with twenty cards numbered from 1 to 20. Each player starts with a five card hand drawn from this deck. In each round, Gammadorf plays a card in his hand, then Knot plays a card in his hand. Whoever played a card with greater value gets a point. At the end of five rounds, the player with the most points wins. If Gammadorf starts with a hand of $1,5,10,15,20$, how many five-card hands of the fifteen remaining cards can Knot draw which always let Knot win (assuming he plays optimally)? | Knot can only lose if all of his cards are lower than 10; if not he can win by playing the lowest card that beats Gammadorf's card, or if this is not possible, his lowest card, each turn. There are $\binom{7}{5}=21$ losing hands, so he has $\binom{15}{5}-\binom{7}{5}$ possible winning hands. | 2982 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4.5 | Find the number of solutions in positive integers $(k ; a_{1}, a_{2}, \ldots, a_{k} ; b_{1}, b_{2}, \ldots, b_{k})$ to the equation $$a_{1}(b_{1})+a_{2}(b_{1}+b_{2})+\cdots+a_{k}(b_{1}+b_{2}+\cdots+b_{k})=7$$ | Let $k, a_{1}, \ldots, a_{k}, b_{1}, \ldots, b_{k}$ be a solution. Then $b_{1}, b_{1}+b_{2}, \ldots, b_{1}+\cdots+b_{k}$ is just some increasing sequence of positive integers. Considering the $a_{i}$ as multiplicities, the $a_{i}$ 's and $b_{i}$ 's uniquely determine a partition of 7. Likewise, we can determine $a_{i}$ 's and $b_{i}$ 's from any partition of 7, so the number of solutions is $p(7)=15$. | 15 | HMMT_2 |
[
"Mathematics -> Algebra -> Abstract Algebra -> Group Theory"
] | 5.25 | Find the smallest integer $n \geq 5$ for which there exists a set of $n$ distinct pairs $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ of positive integers with $1 \leq x_{i}, y_{i} \leq 4$ for $i=1,2, \ldots, n$, such that for any indices $r, s \in\{1,2, \ldots, n\}$ (not necessarily distinct), there exists an index $t \in\{1,2, \ldots, n\}$ such that 4 divides $x_{r}+x_{s}-x_{t}$ and $y_{r}+y_{s}-y_{t}$. | In other words, we have a set $S$ of $n$ pairs in $(\mathbb{Z} / 4 \mathbb{Z})^{2}$ closed under addition. Since $1+1+1+1 \equiv 0(\bmod 4)$ and $1+1+1 \equiv-1(\bmod 4),(0,0) \in S$ and $S$ is closed under (additive) inverses. Thus $S$ forms a group under addition (a subgroup of $(\mathbb{Z} / 4 \mathbb{Z})^{2}$ ). By Lagrange's theorem (from basic group theory), $n \mid 4^{2}$, so $n \geq 8$. To achieve this bound, one possible construction is $\{1,2,3,4\} \times\{2,4\}$ | 8 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations"
] | 5 | Two 18-24-30 triangles in the plane share the same circumcircle as well as the same incircle. What's the area of the region common to both the triangles? | Notice, first of all, that $18-24-30$ is 6 times $3-4-5$, so the triangles are right. Thus, the midpoint of the hypotenuse of each is the center of their common circumcircle, and the inradius is $\frac{1}{2}(18+24-30)=6$. Let one of the triangles be $A B C$, where $\angle A<\angle B<\angle C=90^{\circ}$. Now the line $\ell$ joining the midpoints of sides $A B$ and $A C$ is tangent to the incircle, because it is the right distance (12) from line $B C$. So, the hypotenuse of the other triangle lies along $\ell$. We may formulate this thus: The hypotenuse of each triangle is parallel to the shorter leg, and therefore perpendicular to the longer leg, of the other. Now it is not hard to see, as a result of these parallel and perpendicularisms, that the other triangle "cuts off" at each vertex of $\triangle A B C$ a smaller, similar right triangle. If we compute the dimensions of these smaller triangles, we find that they are as follows: 9-12-15 at $A, 6-8-10$ at $B$, and 3-4-5 at $C$. The total area chopped off of $\triangle A B C$ is thus $$\frac{9 \cdot 12}{2}+\frac{6 \cdot 8}{2}+\frac{3 \cdot 4}{2}=54+24+6=84$$ The area of $\triangle A B C$ is $18 \cdot 24 / 2=216$. The area of the region common to both the original triangles is thus $216-84=132$. | 132 | HMMT_2 |
[
"Mathematics -> Number Theory -> Congruences"
] | 5 | Let \(a_{1}, a_{2}, \ldots\) be an infinite sequence of integers such that \(a_{i}\) divides \(a_{i+1}\) for all \(i \geq 1\), and let \(b_{i}\) be the remainder when \(a_{i}\) is divided by 210. What is the maximal number of distinct terms in the sequence \(b_{1}, b_{2}, \ldots\)? | It is clear that the sequence \(\{a_{i}\}\) will be a concatenation of sequences of the form \(\{v_{i}\}_{i=1}^{N_{0}},\{w_{i} \cdot p_{1}\}_{i=1}^{N_{1}},\{x_{i} \cdot p_{1} p_{2}\}_{i=1}^{N_{2}},\{y_{i} \cdot p_{1} p_{2} p_{3}\}_{i=1}^{N_{3}}\), and \(\{z_{i} \cdot p_{1} p_{2} p_{3} p_{4}\}_{i=1}^{N_{4}}\), for some permutation \((p_{1}, p_{2}, p_{3}, p_{4})\) of \((2,3,5,7)\) and some sequences of integers \(\{v_{i}\} \cdot\{w_{i}\} \cdot\{x_{i}\} \cdot\{y_{i}\} \cdot\{z_{i}\}\), each coprime with 210. In \(\{v_{i}\}_{i=1}^{N_{0}}\), there are a maximum of \(\phi(210)\) distinct terms \(\bmod 210\). In \(\{w_{i} \cdot p_{1}\}_{i=1}^{N_{1}}\), there are a maximum of \(\phi\left(\frac{210}{p_{1}}\right)\) distinct terms mod 210. In \(\{x_{i} \cdot p_{1} p_{2}\}_{i=1}^{N_{2}}\), there are a maximum of \(\phi\left(\frac{210}{p_{1} p_{2}}\right)\) distinct terms \(\bmod 210\). In \(\{y_{i} \cdot p_{1} p_{2} p_{3}\}_{i=1}^{N_{3}}\), there are a maximum of \(\phi\left(\frac{210}{p_{1} p_{2} p_{3}}\right)\) distinct terms \(\bmod 210\). In \(\{z_{i} \cdot p_{1} p_{2} p_{3} p_{4}\}_{i=1}^{N_{4}}\), there can only be one distinct term \(\bmod 210\). Therefore we wish to maximize \(\phi(210)+\phi\left(\frac{210}{p_{1}}\right)+\phi\left(\frac{210}{p_{1} p_{2}}\right)+\phi\left(\frac{210}{p_{1} p_{2} p_{3}}\right)+1\) over all permutations \((p_{1}, p_{2}, p_{3}, p_{4})\) of \((2,3,5,7)\). It's easy to see that the maximum occurs when we take \(p_{1}=2, p_{2}=3, p_{3}=5, p_{4}=7\) for an answer of \(\phi(210)+\phi(105)+\phi(35)+\phi(7)+1=127\). This upper bound is clearly attainable by having the \(v_{i}\)'s cycle through the \(\phi(210)\) integers less than 210 coprime to 210, the \(w_{i}\)'s cycle through the \(\phi\left(\frac{210}{p_{1}}\right)\) integers less than \(\frac{210}{p_{1}}\) coprime to \(\frac{210}{p_{1}}\), etc. | 127 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons",
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 4 | A plane $P$ slices through a cube of volume 1 with a cross-section in the shape of a regular hexagon. This cube also has an inscribed sphere, whose intersection with $P$ is a circle. What is the area of the region inside the regular hexagon but outside the circle? | One can show that the hexagon must have as its vertices the midpoints of six edges of the cube, as illustrated; for example, this readily follows from the fact that opposite sides of the hexagons and the medians between them are parallel. We then conclude that the side of the hexagon is $\sqrt{2} / 2$ (since it cuts off an isosceles triangle of leg $1 / 2$ from each face), so the area is $(3 / 2)(\sqrt{2} / 2)^{2}(\sqrt{3})=3 \sqrt{3} / 4$. Also, the plane passes through the center of the sphere by symmetry, so it cuts out a cross section of radius $1 / 2$, whose area (which is contained entirely inside the hexagon) is then $\pi / 4$. The sought area is thus $(3 \sqrt{3}-\pi) / 4$. | (3 \sqrt{3}-\pi) / 4 | HMMT_2 |
[
"Mathematics -> Number Theory -> Factorization",
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 5 | Niffy's favorite number is a positive integer, and Stebbysaurus is trying to guess what it is. Niffy tells her that when expressed in decimal without any leading zeros, her favorite number satisfies the following: - Adding 1 to the number results in an integer divisible by 210 . - The sum of the digits of the number is twice its number of digits. - The number has no more than 12 digits. - The number alternates in even and odd digits. Given this information, what are all possible values of Niffy's favorite number? | Note that Niffy's favorite number must end in 9, since adding 1 makes it divisible by 10. Also, the sum of the digits of Niffy's favorite number must be even (because it is equal to twice the number of digits) and congruent to 2 modulo 3 (because adding 1 gives a multiple of 3 ). Furthermore, the sum of digits can be at most 24 , because there at most 12 digits in Niffy's favorite number, and must be at least 9 , because the last digit is 9 . This gives the possible sums of digits 14 and 20. However, if the sum of the digits of the integer is 20 , there are 10 digits, exactly 5 of which are odd, giving an odd sum of digits, which is impossible. Thus, Niffy's favorite number is a 7 digit number with sum of digits 14 . The integers which we seek must be of the form $\overline{A B C D E F 9}$, where $A, C, E$ are odd, $B, D, F$ are even, and $A+B+C+D+E+F=5$. Now, note that $\{A, C, E\}=\{1,1,1\}$ or $\{1,1,3\}$, and these correspond to $\{B, D, F\}=\{0,0,2\}$ and $\{0,0,0\}$, respectively. It suffices to determine which of these six integers are congruent to $-1(\bmod 7)$, and we see that Niffy's favorite number must be 1010309. | 1010309 | HMMT_2 |
[
"Mathematics -> Geometry -> Plane Geometry -> Triangulations",
"Mathematics -> Algebra -> Intermediate Algebra -> Other"
] | 5 | Suppose $A B C$ is a triangle with circumcenter $O$ and orthocenter $H$ such that $A, B, C, O$, and $H$ are all on distinct points with integer coordinates. What is the second smallest possible value of the circumradius of $A B C$ ? | Assume without loss of generality that the circumcenter is at the origin. By well known properties of the Euler line, the centroid $G$ is such that $O, G$, and $H$ are collinear, with $G$ in between $O$ and $H$, such that $G H=2 G O$. Thus, since $G=\frac{1}{3}(A+B+C)$, and we are assuming $O$ is the origin, we have $H=A+B+C$. This means that as long as $A, B$, and $C$ are integer points, $H$ will be as well. However, since $H$ needs to be distinct from $A, B$, and $C$, we must have \triangle A B C$ not be a right triangle, since in right triangles, the orthocenter is the vertex where the right angle is. Now, if a circle centered at the origin has any integer points, it will have at least four integer points. (If it has a point of the form $(a, 0)$, then it will also have $(-a, 0),(0, a)$, and $(0,-a)$. If it has a point of the form $(a, b)$, with $a, b \neq 0$, it will have each point of the form $( \pm a, \pm b)$.) But in any of these cases where there are only four points, any triangle which can be made from those points is a right triangle. Thus we need the circumcircle to contain at least eight lattice points. The smallest radius this occurs at is \sqrt{1^{2}+2^{2}}=\sqrt{5}$, which contains the eight points $( \pm 1, \pm 2)$ and $( \pm 2, \pm 1)$. We get at least one valid triangle with this circumradius: $$ A=(-1,2), B=(1,2), C=(2,1) $$ The next valid circumradius is \sqrt{1^{2}+3^{2}}=\sqrt{10}$ which has the valid triangle $$ A=(-1,3), B=(1,3), C=(3,1) $$ | \sqrt{10} | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Luna has an infinite supply of red, blue, orange, and green socks. She wants to arrange 2012 socks in a line such that no red sock is adjacent to a blue sock and no orange sock is adjacent to a green sock. How many ways can she do this? | Luna has 4 choices for the first sock. After that, she has 3 choices for each of 2011 remaining socks for a total of $4 \cdot 3^{2011}$. | 4 \cdot 3^{2011} | HMMT_2 |
[
"Mathematics -> Geometry -> Solid Geometry -> 3D Shapes"
] | 5 | A tetrahedron has all its faces triangles with sides $13,14,15$. What is its volume? | Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. Let $A D, B E$ be altitudes. Then $B D=5, C D=9$. (If you don't already know this, it can be deduced from the Pythagorean Theorem: $C D^{2}-B D^{2}=\left(C D^{2}+A D^{2}\right)-\left(B D^{2}+A D^{2}\right)=A C^{2}-A B^{2}=56$, while $C D+B D=B C=14$, giving $C D-B D=56 / 14=4$, and now solve the linear system.) Also, $A D=\sqrt{A B^{2}-B D^{2}}=12$. Similar reasoning gives $A E=33 / 5$, $E C=42 / 5$. Now let $F$ be the point on $B C$ such that $C F=B D=5$, and let $G$ be on $A C$ such that $C G=A E=33 / 5$. Imagine placing face $A B C$ flat on the table, and letting $X$ be a point in space with $C X=13, B X=14$. By mentally rotating triangle $B C X$ about line $B C$, we can see that $X$ lies on the plane perpendicular to $B C$ through $F$. In particular, this holds if $X$ is the fourth vertex of our tetrahedron $A B C X$. Similarly, $X$ lies on the plane perpendicular to $A C$ through $G$. Let the mutual intersection of these two planes and plane $A B C$ be $H$. Then $X H$ is the altitude of the tetrahedron. To find $X H$, extend $F H$ to meet $A C$ at $I$. Then $\triangle C F I \sim \triangle C D A$, a 3-4-5 triangle, so $F I=C F \cdot 4 / 3=20 / 3$, and $C I=C F \cdot 5 / 3=25 / 3$. Then $I G=C I-C G=26 / 15$, and $H I=I G \cdot 5 / 4=13 / 6$. This leads to $H F=F I-H I=9 / 2$, and finally $X H=\sqrt{X F^{2}-H F^{2}}=\sqrt{A D^{2}-H F^{2}}=3 \sqrt{55} / 2$. Now $X A B C$ is a tetrahedron whose base $\triangle A B C$ has area $A D \cdot B C / 2=12 \cdot 14 / 2=84$, and whose height $X H$ is $3 \sqrt{55} / 2$, so its volume is $(84)(3 \sqrt{55} / 2) / 3=42 \sqrt{55}$. | 42 \sqrt{55} | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 4 | Let $b(x)=x^{2}+x+1$. The polynomial $x^{2015}+x^{2014}+\cdots+x+1$ has a unique "base $b(x)$ " representation $x^{2015}+x^{2014}+\cdots+x+1=\sum_{k=0}^{N} a_{k}(x) b(x)^{k}$ where each "digit" $a_{k}(x)$ is either the zero polynomial or a nonzero polynomial of degree less than $\operatorname{deg} b=2$; and the "leading digit $a_{N}(x)$ " is nonzero. Find $a_{N}(0)$. | Comparing degrees easily gives $N=1007$. By ignoring terms of degree at most 2013, we see $a_{N}(x)\left(x^{2}+x+1\right)^{1007} \in x^{2015}+x^{2014}+O\left(x^{2013}\right)$. Write $a_{N}(x)=u x+v$, so $a_{N}(x)\left(x^{2}+x+1\right)^{1007} \in(u x+v)\left(x^{2014}+1007 x^{2013}+O\left(x^{2012}\right)\right) \subseteq u x^{2015}+(v+1007 u) x^{2014}+O\left(x^{2013}\right)$. Finally, matching terms gives $u=1$ and $v+1007 u=1$, so $v=1-1007=-1006$. | -1006 | HMMT_2 |
[
"Mathematics -> Algebra -> Linear Algebra -> Matrices"
] | 4.5 | $M$ is an $8 \times 8$ matrix. For $1 \leq i \leq 8$, all entries in row $i$ are at least $i$, and all entries on column $i$ are at least $i$. What is the minimum possible sum of the entries of $M$ ? | Let $s_{n}$ be the minimum possible sum for an $n$ by $n$ matrix. Then, we note that increasing it by adding row $n+1$ and column $n+1$ gives $2 n+1$ additional entries, each of which has minimal size at least $n+1$. Consequently, we obtain $s_{n+1}=s_{n}+(2 n+1)(n+1)=s_{n}+2 n^{2}+3 n+1$. Since $s_{0}=0$, we get that $s_{8}=2\left(7^{2}+\ldots+0^{2}\right)+3(7+\ldots+0)+8=372$. | 372 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5 | Consider the cube whose vertices are the eight points $(x, y, z)$ for which each of $x, y$, and $z$ is either 0 or 1 . How many ways are there to color its vertices black or white such that, for any vertex, if all of its neighbors are the same color then it is also that color? Two vertices are neighbors if they are the two endpoints of some edge of the cube. | Divide the 8 vertices of the cube into two sets $A$ and $B$ such that each set contains 4 vertices, any two of which are diagonally adjacent across a face of the cube. We do casework based on the number of vertices of each color in set $A$. - Case 1: 4 black. Then all the vertices in $B$ must be black, for 1 possible coloring. - Case 2: 3 black, 1 white. Then there are 4 ways to assign the white vertex. The vertex in $B$ surrounded by the black vertices must also be black. Meanwhile, the three remaining vertices in $B$ may be any configuration except all black, for a total of $4\left(2^{3}-1\right)=28$ possible colorings. - Case 3: 2 black, 2 white. Then, there are 6 ways to assign the 2 white vertices. The 4 vertices of $B$ cannot all be the same color. Additionally, we cannot have 3 black vertices of $B$ surround a white vertex of $A$ with the other vertex of $B$ white, and vice-versa, so we have a total of $6\left(2^{4}-2-4\right)=60$ possible colorings. - Case 4: 1 black, 3 white. As in case 2, there are 28 possible colorings. - Case 5: 5 white. As in case 1, there is 1 possible coloring. So there is a total of $1+28+60+28+1=118$ possible colorings. | 118 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 4.5 | It can be shown that there exists a unique polynomial $P$ in two variables such that for all positive integers $m$ and $n$, $$P(m, n)=\sum_{i=1}^{m} \sum_{j=1}^{n}(i+j)^{7}$$ Compute $P(3,-3)$. | Note that for integers $m>0, n>1$, $$P(m, n)-P(m, n-1)=\sum_{i=1}^{m}(i+n)^{7}=(n+1)^{7}+(n+2)^{7}+(n+3)^{7}$$ for all real $n$. Moreover, $P(3,1)-P(3,0)=P(3,1) \Longrightarrow P(3,0)=0$. Then $$\begin{aligned} P(3,-3) & =P(3,0)-\left(1^{7}+2^{7}+3^{7}\right)-\left(0^{7}+1^{7}+2^{7}\right)-\left((-1)^{7}+0^{7}+1^{7}\right) \\ & =-3^{7}-2 \cdot 2^{7}-2=-2445 \end{aligned}$$ | -2445 | HMMT_2 |
[
"Mathematics -> Number Theory -> Greatest Common Divisors (GCD)"
] | 5 | An ordered pair $(a, b)$ of positive integers is called spicy if $\operatorname{gcd}(a+b, ab+1)=1$. Compute the probability that both $(99, n)$ and $(101, n)$ are spicy when $n$ is chosen from $\{1,2, \ldots, 2024\}$ uniformly at random. | We claim that $(a, b)$ is spicy if and only if both $\operatorname{gcd}(a+1, b-1)=1$ and $\operatorname{gcd}(a-1, b+1)=1$. To prove the claim, we note that $$\operatorname{gcd}(a+b, ab+1)=\operatorname{gcd}(a+b, b(-b)+1)=\operatorname{gcd}(a+b, b^{2}-1)$$ Hence, we have $$\begin{aligned} \operatorname{gcd}(a+b, ab+1)=1 & \Longleftrightarrow \operatorname{gcd}(a+b, b^{2}-1)=1 \\ & \Longleftrightarrow \operatorname{gcd}(a+b, b-1)=1 \text{ and } \operatorname{gcd}(a+b, b+1)=1 \\ & \Longleftrightarrow \operatorname{gcd}(a+1, b-1)=1 \text{ and } \operatorname{gcd}(a-1, b+1)=1 \end{aligned}$$ proving the claim. Thus, $n$ works if and only if all following four conditions hold: - $\operatorname{gcd}(n+1,98)=1$, or equivalently, $n$ is neither $-1(\bmod 2)$ nor $-1(\bmod 7)$; - $\operatorname{gcd}(n-1,100)=1$, or equivalently, $n$ is neither $1(\bmod 2)$ nor $1(\bmod 5)$; - $\operatorname{gcd}(n+1,100)=1$, or equivalently, $n$ is neither $-1(\bmod 2)$ nor $-1(\bmod 5)$; and - $\operatorname{gcd}(n-1,102)=1$, or equivalently, $n$ is neither $1(\bmod 2), 1(\bmod 3)$, nor $1(\bmod 17)$. Thus, there are $1,2,3,6,17$ possible residues modulo $2,3,5,7$, and 17, respectively. The residues are uniformly distributed within $\{1,2, \ldots, 2024!\}$. Hence, the answer is $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{6}{7} \cdot \frac{16}{17}=\frac{96}{595}$. | \frac{96}{595} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Inequalities"
] | 3.5 | Find the smallest integer $n$ such that $\sqrt{n+99}-\sqrt{n}<1$. | This is equivalent to $$\begin{aligned} \sqrt{n+99} & <\sqrt{n}+1 \\ n+99 & <n+1+2 \sqrt{n} \\ 49 & <\sqrt{n} \end{aligned}$$ So the smallest integer $n$ with this property is $49^{2}+1=2402$. | 2402 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other",
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5.25 | You have a twig of length 1. You repeatedly do the following: select two points on the twig independently and uniformly at random, make cuts on these two points, and keep only the largest piece. After 2012 repetitions, what is the expected length of the remaining piece? | First let $p(x)$ be the probability density of $x$ being the longest length. Let $a_{n}$ be the expected length after $n$ cuts. $a_{n}=\int_{0}^{1} p(x) \cdot\left(x a_{n-1}\right) d x=a_{n-1} \int_{0}^{1} x p(x) d x=a_{1} a_{n-1}$. It follows that $a_{n}=a_{1}^{n}$, so our answer is \left(a_{1}\right)^{2012}$. We now calculate $a_{1}$. Let $P(z)$ be the probability that the longest section is \leq z$. Clearly $P(z)=0$ for $z \leq \frac{1}{3}$. To simulate making two cuts we pick two random numbers $x, y$ from $[0,1]$, and assume without loss of generality that $x \leq y$. Then picking two such points is equivalent to picking a point in the top left triangle half of the unit square. This figure has area \frac{1}{2}$ so our $P(z)$ will be double the area where $x \leq z, y \geq 1-z$ and $y-x \leq z$. For \frac{1}{3} \leq z \leq \frac{1}{2}$ the probability is double the area bounded by $x=z, 1-z=y, y-x=z$. This is $2\left(\frac{1}{2}(3 z-1)^{2}\right)=(3 z-1)^{2}$. For \frac{1}{2} \leq z \leq 1$ this value is double the hexagon bounded by $x=0, y=1-z, y=x, x=z, y=1$, $y=x+z$. The complement of this set, however, is three triangles of area \frac{(1-z)^{2}}{2}$, so $P(z)=1-3(1-z)^{2}$ for \frac{1}{2} \leq z \leq 1$. Now note that $P^{\prime}(z)=p(z)$. Therefore by integration by parts $a_{1}=\int_{0}^{1} z p(z) d z=\int_{0}^{1} z P^{\prime}(z) d z=$ ${ }_{0}^{1} z P(z)-\int_{0}^{1} P(z) d z$. This equals $$ \begin{gathered} 1-\int_{\frac{1}{3}}^{\frac{1}{2}}(3 z-1)^{2} d z-\int_{\frac{1}{2}}^{1} 1-3(1-z)^{2} d z \\ =1-\left[\frac{1}{\frac{1}{3}} \frac{(3 z-1)^{3}}{9}-\frac{1}{2}+\left[\left[_{\frac{1}{2}}^{1}(z-1)^{3}\right.\right.\right. \\ =1-\frac{1}{72}-\frac{1}{2}+\frac{1}{8}=\frac{1}{2}+\frac{1}{9} \\ =\frac{11}{18} \end{gathered} $$ So the answer is \left(\frac{11}{18}\right)^{2012}$. | \left(\frac{11}{18}\right)^{2012} | HMMT_2 |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Complex Numbers",
"Mathematics -> Calculus -> Differential Calculus -> Applications of Derivatives"
] | 5.25 | Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk has radius 5. Each succeeding disk has its center attached to a point on the circumference of the previous disk and has a radius equal to $2 / 3$ of the previous disk. Each disk spins around its center (relative to the disk it is attached to) at a rate of \pi / 6$ radians per second. Initially, at $t=0$, the centers of the disks are aligned on a single line, going outward. Alice is sitting at the limit point of all these disks. After 12 seconds, what is the length of the trajectory that Alice has traced out? | Suppose the center of the largest teacup is at the origin in the complex plane, and let $z=\frac{2}{3} e^{\pi i t / 6}$. The center of the second disk is at $5 e^{\pi i t / 6}$ at time $t$; that is, \frac{15}{2} z$. Then the center of the third disk relative to the center of the second disk is at \frac{15}{2} z^{2}$, and so on. Summing up a geometric series, we get that Alice's position is $$ \begin{aligned} \frac{15}{2}\left(z+z^{2}+z^{3}+\cdots\right) & =\frac{15}{2}\left(1+z^{2}+z^{3}+\cdots\right)-\frac{15}{2} \\ & =\frac{15}{2}\left(\frac{1}{1-z}\right)-\frac{15}{2} \end{aligned} $$ Now, after 12 seconds, $z$ has made a full circle in the complex plane centered at 0 and of radius $2 / 3$. Thus $1-z$ is a circle centered at 1 of radius $2 / 3$. So $1-z$ traces a circle, and now we need to find the path that \frac{1}{1-z}$ traces. In the complex plane, taking the reciprocal corresponds to a reflection about the real axis followed by a geometric inversion about the unit circle centered at 0 . It is well known that geometric inversion maps circles not passing through the center of the inversion to circles. Now, the circle traced by $1-z$ contains the points $1-2 / 3=1 / 3$, and $1+2 / 3=5 / 3$. Therefore the circle \frac{1}{1-z}$ contains the points 3 and $3 / 5$, with the center lying halfway between. So the radius of the circle is $$ \frac{1}{2}\left(3-\frac{3}{5}\right)=\frac{6}{5} $$ and so the perimeter is $2 \pi(6 / 5)=12 \pi / 5$. Scaling by \frac{15}{2}$ gives an answer of $$ \frac{15}{2}\left(\frac{12 \pi}{5}\right)=18 \pi $$ | 18 \pi | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 5 | Let $a, b$ be integers chosen independently and uniformly at random from the set $\{0,1,2, \ldots, 80\}$. Compute the expected value of the remainder when the binomial coefficient $\binom{a}{b}=\frac{a!}{b!(a-b)!}$ is divided by 3. | By Lucas' Theorem we're looking at $\prod_{i=1}^{4}\binom{a_{i}}{b_{i}}$ where the $a_{i}$ and $b_{i}$ are the digits of $a$ and $b$ in base 3. If any $a_{i}<b_{i}$, then the product is zero modulo 3. Otherwise, the potential residues are $\binom{2}{0}=1,\binom{2}{1}=2,\binom{2}{2}=1,\binom{1}{0}=1,\binom{1}{1}=1,\binom{0}{0}=1$. So each term in the product has a $\frac{1}{3}$ chance of being zero; given that everything is nonzero, each term has a $\frac{1}{6}$ chance of being 2 and a $\frac{5}{6}$ chance of being 1. The probability that an even number of terms are 1 given that none are zero is then given by the roots of unity filter $\frac{\left(\frac{5}{6}+\frac{1}{6} \cdot(1)\right)^{4}+\left(\frac{5}{6}+\frac{1}{6} \cdot(-1)\right)^{4}}{2}=\frac{81+16}{162}=\frac{97}{162}$. Thus the expected value is $\left(\frac{2}{3}\right)^{4}\left(2-\frac{97}{162}\right)=\frac{1816}{6561}$ | \frac{1816}{6561} | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 5.25 | Let $S$ be the set \{1,2, \ldots, 2012\}. A perfectutation is a bijective function $h$ from $S$ to itself such that there exists an $a \in S$ such that $h(a) \neq a$, and that for any pair of integers $a \in S$ and $b \in S$ such that $h(a) \neq a, h(b) \neq b$, there exists a positive integer $k$ such that $h^{k}(a)=b$. Let $n$ be the number of ordered pairs of perfectutations $(f, g)$ such that $f(g(i))=g(f(i))$ for all $i \in S$, but $f \neq g$. Find the remainder when $n$ is divided by 2011 . | Note that both $f$ and $g$, when written in cycle notation, must contain exactly one cycle that contains more than 1 element. Assume $f$ has $k$ fixed points, and that the other $2012-k$ elements form a cycle, (of which there are (2011 - $k$ )! ways). Then note that if $f$ fixes $a$ then $f(g(a))=g(f(a))=g(a)$ implies $f$ fixes $g(a)$ So $g$ must send fixed points of $f$ to fixed points of $f$. It must, therefore send non-fixed points to non-fixed points. This partitions $S$ into two sets, at least one of which must be fixed by $g$, since $g$ is a perfectutation. If $g$ fixes all of the the non-fixed points of $f$, then, since any function commutes with the identity, $g$ fixes some $m$ of the fixed points and cycles the rest in $(k-m-1)$ ! ways. So there are \sum_{m=0}^{k-2}\binom{k}{m}(k-m-1)$ ! choices, which is \sum_{m=0}^{k-2} \frac{k!}{(k-m) m!}$. If $g$ fixes all of the fixed points of $f$, then order the non-fixed points of $f a_{1}, a_{2}, \ldots, a_{2012-k}$ such that $f\left(a_{i}\right)=a_{i+1}$. If $g\left(a_{i}\right)=a_{j}$ then $f\left(g\left(a_{i}\right)\right)=a_{j+1}$ thus $g\left(a_{i+1}\right)=a_{j+1}$. Therefore the choice of $g\left(a_{1}\right)$ uniquely determines $g\left(a_{i}\right)$ for the rest of the $i$, and $g\left(a_{m}\right)=a_{m+j-i}$. But $g$ has to be a perfectutation, so $g$ cycles through all the non-fixed points of $f$, which happens if and only if $j-i$ is relatively prime to $2012-k$. So there are \phi(2012-k)$ choices. Therefore for any $f$ there are \sum_{m=0}^{k-2} \frac{k!}{(k-m) m!}+\phi(2012-k)$ choices of $g$, but one of them will be $g=f$, which we cannot have by the problem statement. So there are $-1+\sum_{m=0}^{k-2} \frac{k!}{(k-m) m!}+\phi(2012-k)$ options. Now note that a permutation can not fix all but one element. So $n=\sum_{k=0}^{2010}\binom{2012}{k}(2011-k)!(-1+$ $\left.\sum_{m=0}^{k-2} \frac{k!}{(k-m) m!}+\phi(2012-k)\right)$ Modulo 2011 (which is prime), note that all terms in the summand except the one where $k=1$ vanish. Thus, $n \equiv(2010)!(-1+(-1)) \equiv 2(\bmod 2011)$ by Wilson's Theorem. | 2 | HMMT_2 |
[
"Mathematics -> Discrete Mathematics -> Combinatorics"
] | 4.5 | How many elements are in the set obtained by transforming $\{(0,0),(2,0)\} 14$ times? | Transforming it $k \geq 1$ times yields the diamond $\{(n, m):|n-1|+|m| \leq k+1\}$ with the points $(1, k),(1, k+1),(1,-k),(1,-k-1)$ removed (this can be seen inductively). So we get $(k+1)^{2}+k^{2}-4$ lattice points, making the answer 477. | 477 | HMMT_2 |
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions",
"Mathematics -> Algebra -> Algebra -> Equations and Inequalities"
] | 5 | Let $f(x)=x^{2}+a x+b$ and $g(x)=x^{2}+c x+d$ be two distinct real polynomials such that the $x$-coordinate of the vertex of $f$ is a root of $g$, the $x$-coordinate of the vertex of $g$ is a root of $f$ and both $f$ and $g$ have the same minimum value. If the graphs of the two polynomials intersect at the point (2012, - 2012), what is the value of $a+c$ ? | It is clear, by symmetry, that 2012 is the equidistant from the vertices of the two quadratics. Then it is clear that reflecting $f$ about the line $x=2012$ yields $g$ and vice versa. Thus the average of each pair of roots is 2012 . Thus the sum of the four roots of $f$ and $g$ is 8048 , so $a+c=-8048$. | -8048 | HMMT_2 |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other"
] | 5 | Franklin has four bags, numbered 1 through 4. Initially, the first bag contains fifteen balls, numbered 1 through 15 , and the other bags are empty. Franklin randomly pulls a pair of balls out of the first bag, throws away the ball with the lower number, and moves the ball with the higher number into the second bag. He does this until there is only one ball left in the first bag. He then repeats this process in the second and third bag until there is exactly one ball in each bag. What is the probability that ball 14 is in one of the bags at the end? | Pretend there is a 16 th ball numbered 16. This process is equivalent to randomly drawing a tournament bracket for the 16 balls, and playing a tournament where the higher ranked ball always wins. The probability that a ball is left in a bag at the end is the probability that it loses to ball 16. Of the three balls $14,15,16$, there is a \frac{1}{3}$ chance 14 plays 15 first, a \frac{1}{3}$ chance 14 plays 16 first, and a \frac{1}{3}$ chance 15 plays 16 first. In the first case, 14 does not lose to 16 , and instead loses to 15 ; otherwise 14 loses to 16 , and ends up in a bag. So the answer is \frac{2}{3}$. | \frac{2}{3} | HMMT_2 |
Subsets and Splits