problem stringlengths 10 5.15k | answer stringlengths 0 1.22k | solution stringlengths 0 11.1k | difficulty float64 0.75 2.02k | difficulty_raw listlengths 3 8 |
|---|---|---|---|---|
Find the least positive integer $n$ such that
$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$ | 1 | Let S be the sum of the sequence. We begin the same as in Solution 1 to get $S\sin(1)=\cot(45)-\cot(46)+\cot(47)-\cot(48)+...+\cot(133)-\cot(134)$. Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence $\frac{1}{\sin(46)\sin(47)}+\frac{1}{\sin(48)\sin(49)}+...+\frac{1}{\si... | 6.75 | [
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Find the sum of all positive two-digit integers that are divisible by each of their digits. | 630 | Using casework, we can list out all of these numbers: \[11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.\] | 3 | [
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A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$, and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$. Find the mean of $\mathcal{S}$. | 651 | Since this is a weighted average problem, the mean of $S$ is $\frac{13}{27}$ as far from $1$ as it is from $2001$. Thus, the mean of $S$ is $1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}$.
~ pi_is_3.14 | 4.375 | [
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Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$, given that there are no multiple roots. | 500 | We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$. | 5.625 | [
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In triangle $ABC$, angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24$. The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square ... | 291 | After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$, meaning $\triangle TAC$ is an isosceles triangle and $AC=24$.
Using law of sines on $\triangle ABC$, we can create the following equation:
$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$
$\angle ABC=45^{\circ}$ and $\angle BAC=6... | 6.625 | [
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An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 937 | Solving for $y$ in terms of $x$ gives $y=\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$, which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\sqrt{4-x^2}/2)$ and $(0,1)$, so by the distance formula we have
\[2x=\sqrt{x^2+(1-\sqrt{4-x... | 6.25 | [
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A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 79 | Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$. In the diagram below, the lowest $y$-coordinate at each of $a$, $b$, $c$, and $d$ corresponds to the value of the roll.
The red path corresponds to the sequence of rolls $2, 3, 5, 5$. This establishes a bijection between val... | 4 | [
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Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive... | 923 | More directly than Solution 2, we have \[DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.\] | 5.875 | [
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Call a positive integer $N$ a 7-10 double if the digits of the base-$7$ representation of $N$ form a base-$10$ number that is twice $N$. For example, $51$ is a 7-10 double because its base-$7$ representation is $102$. What is the largest 7-10 double? | 315 | Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be \[abc\] in base 7. Then the number in expanded form is \[49a+7b+c\] in base 7 and \[100a+10b+c\] in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. \[98b+14b+2c=... | 5.625 | [
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In triangle $ABC$, $AB=13$, $BC=15$ and $CA=17$. Point $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, and $F$ is on $\overline{CA}$. Let $AD=p\cdot AB$, $BE=q\cdot BC$, and $CF=r\cdot CA$, where $p$, $q$, and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$. The ratio of the area of triangle $DEF$ ... | 61 | By the barycentric area formula, our desired ratio is equal to \begin{align*} \begin{vmatrix} 1-p & p & 0 \\ 0 & 1-q & q \\ r & 0 & 1-r \notag \end{vmatrix} &=1-p-q-r+pq+qr+pr\\ &=1-(p+q+r)+\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\ &=1-\frac{2}{3}+\frac{\frac{4}{9}-\frac{2}{5}}{2}\\ &=\frac{16}{45} \end{align*}, so the answe... | 6.25 | [
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Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime ... | 200 | The distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore,
For $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities.
For $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(0,2)$, $(2,0)$... | 4.75 | [
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In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so tha... | 149 | Let each point $P_i$ be in column $c_i$. The numberings for $P_i$ can now be defined as follows. \begin{align*}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{align*}
We can now convert the five given equalities. \begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 ... | 6.875 | [
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A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 5 | Note that the center is obviously at some point $(r,r,r)$ since it must be equidistant from the x-y, x-z, and y-z planes. Now note that the point-plane distance formula is $\widehat{n}\cdot \vec{d}$ (the direction vector of the normal to the plane dotted with the distance between a point on the plane and the point we'r... | 6.125 | [
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In a certain circle, the chord of a $d$-degree arc is $22$ centimeters long, and the chord of a $2d$-degree arc is $20$ centimeters longer than the chord of a $3d$-degree arc, where $d < 120.$ The length of the chord of a $3d$-degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m... | 174 | Let $z=\frac{d}{2}$, $R$ be the circumradius, and $a$ be the length of 3d degree chord. Using the extended sine law, we obtain: \[22=2R\sin(z)\] \[20+a=2R\sin(2z)\] \[a=2R\sin(3z)\] Dividing the second from the first we get $\cos(z)=\frac{20+a}{44}$ By the triple angle formula we can manipulate the third equation as fo... | 6.75 | [
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A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | 351 | Let $a_n$ be the number of ways if the first house has mail, and let $b_n$ be the number of ways if the first house does not get mail.
$a_n=a_{n-2}+a_{n-3}$ because if the first house gets mail, the next house that gets mail must either be the third or fourth house.
$b_n=a_{n-1}+a_{n-2}$ because if the first house does... | 5.75 | [
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The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are... | 85 | [asy] import three; draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1)); draw((0,0,0)--(1,0,0)); draw((0,0,1)--(1,0,1)); for(int i = 0; i < 2; ++i) { for(int j = 0; j < 2; ++j) { for(int k = 0; k < 2; ++k) { dot((i,j,k)); } } } // dot((0,... | 6.375 | [
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Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$? | 816 | The two-digit perfect squares are $16, 25, 36, 49, 64, 81$. We try making a sequence starting with each one:
$16 - 64 - 49$. This terminates since none of them end in a $9$, giving us $1649$.
$25$.
$36 - 64 - 49$, $3649$.
$49$.
$64 - 49$, $649$.
$81 - 16 - 64 - 49$, $81649$.
The largest is $81649$, so our answer is $\... | 4.5 | [
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] |
Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who cou... | 298 | Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cup F$ be the number of students who study both. Then $\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700$, and $\left\lceil 30\% \cdot 2001 \right\rceil... | 5.875 | [
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Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$. | 898 | Calculate the first few terms:
\[211,375,420,523,267,-211,-375,-420,-523,\dots\]
At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously $211+420+267=\boxed{898}$
~Dhillonr25
~ pi_is_3.14 | 4.75 | [
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Let $R = (8,6)$. The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 67 | [asy] pointpen = black; pathpen = black+linewidth(0.7); pair R = (8,6), P = (32,60)/7, Q= (80,24)/7; D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6)); D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6)); D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP("R",R... | 4.5 | [
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A set of positive numbers has the triangle property if it has three distinct elements that are the lengths of the sides of a triangle whose area is positive. Consider sets $\{4, 5, 6, \ldots, n\}$ of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible... | 253 | Out of all ten-element subsets with distinct elements that do not possess the triangle property, we want to find the one with the smallest maximum element. Call this subset $\mathcal{S}$. Without loss of generality, consider any $a, b, c \,\in \mathcal{S}$ with $a < b < c$. $\,\mathcal{S}$ does not possess the triangle... | 6.375 | [
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Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n... | 251 | Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$. Let $DF$ = $b$ and diameter $HI$ go through $J$ the midpoint of $EF$. Since a diamet... | 6 | [
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Let $\triangle{PQR}$ be a right triangle with $PQ = 90$, $PR = 120$, and $QR = 150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with... | 725 | [asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0... | 6.75 | [
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A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$, and that $f(x) = 1-|x-2|$ for $1\le x \le 3$. Find the smallest $x$ for which $f(x) = f(2001)$. | 429 | Iterating the condition $f(3x) = 3f(x)$, we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$. We know the definition of $f(x)$ from $1 \le x \le 3$, so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$. Indeed,
\[f(2001... | 6.75 | [
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Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 929 | We consider how many ways we can have 2*2 grid
$(1)$: All the girds are red--$1$ case
$(2)$: One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases
$(3)$: Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cas... | 5.5 | [
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How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$, where $i$ and $j$ are integers and $0\leq i < j \leq 99$? | 784 | Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$; therefore, the order of 10 modulo $7,11$, and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 ... | 6.125 | [
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Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$. The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$, where $m$ and $n... | 341 | At first, it wins $6$ games, only one way
Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways
Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways
Fourthly, it wins $3$ games, this time, i... | 5.5 | [
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Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i + 1}$, replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular t... | 101 | On the first construction, $P_1$, four new tetrahedra will be constructed with side lengths $\frac 12$ of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume $\left(\frac 12\right)^3 = ... | 7.125 | [
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In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB = 8$, $BD = 10$, and $BC = 6$. The length $CD$ may be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 69 | Draw a line from $B$, parallel to $\overline{AD}$, and let it meet $\overline{CD}$ at $M$. Note that $\triangle{DAB}$ is similar to $\triangle{BMC}$ by AA similarity, since $\angle{ABD}=\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\angle{BMC}=\angle{ADM}=\angle{DAB}$. Now since $ADMB$ is an isosceles trapezoid... | 5.625 | [
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There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$. These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$, where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldo... | 840 | $z$ can be written in the form $\text{cis\,}\theta$. Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$
Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\the... | 7.5 | [
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Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$, be the points on $\overline{EF}$, $\overline{EH}$, and $\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel... | 417 | [asy] import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l)... | 7.125 | [
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Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit ... | 59 | Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is \[\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}\] thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{059}.$
... | 4.25 | [
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The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where ... | 154 | Let the radius of the circles be $r$. The longer dimension of the rectangle can be written as $14r$, and by the Pythagorean Theorem, we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$.
Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\rig... | 5.75 | [
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Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit number and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive intege... | 25 | Let Jane's age $n$ years from now be $10a+b$, and let Dick's age be $10b+a$. If $10b+a>10a+b$, then $b>a$. The possible pairs of $a,b$ are:
$(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \dots , (8,9)$
That makes 36. But $10a+b>25$, so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5),... | 5.5 | [
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Consider the sequence defined by $a_k =\dfrac{1}{k^2+k}$ for $k\geq 1$. Given that $a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}$, for positive integers $m$ and $n$ with $m<n$, find $m+n$. | 840 | Note that $a_1 + a_2 + \cdots + a_i = \dfrac{i}{i+1}$. This can be proven by induction. Thus, $\sum\limits_{i=m}^{n-1} a_i = \sum\limits_{i=1}^{n-1} a_i - \sum\limits_{i=1}^{m-1} a_i = \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29$. Cross-multiplying yields $29n - 29m - mn = 0$, and adding $29^2$ to both sid... | 5.25 | [
5,
5,
5,
5,
5,
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6,
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] |
Let $A_1,A_2,A_3,\cdots,A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\cdots,A_{12}\} ?$ | 183 | Proceed as above to initially get 198 squares (with overcounting). Then note that any square with all four vertices on the dodecagon has to have three sides "between" each vertex, giving us a total of three squares. However, we counted these squares with all four of their sides plus both of their diagonals, meaning we ... | 5.625 | [
5,
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6,
6,
6,
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5,
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] |
The solutions to the system of equations
$\log_{225}x+\log_{64}y=4$
$\log_{x}225-\log_{y}64=1$
are $(x_1,y_1)$ and $(x_2,y_2)$. Find $\log_{30}\left(x_1y_1x_2y_2\right)$. | 12 | Let $A=\log_{225}x$ and let $B=\log_{64}y$.
From the first equation: $A+B=4 \Rightarrow B = 4-A$.
Plugging this into the second equation yields $\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}$ and thus, $B=1\pm\sqrt{5}$.
So, $\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sq... | 6.625 | [
7,
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6,
7,
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6,
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] |
The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $x,y$ and $r$ with $|x|>|y|$,
\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]
What are the first three digits to the right of the decimal point in the decimal representation... | 428 | $1^n$ will always be 1, so we can ignore those terms, and using the definition ($2002 / 7 = 286$):
\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]
Since the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{286... | 5.5 | [
6,
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5,
5,
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] |
Find the smallest integer $k$ for which the conditions
(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers
(2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$
(3) $a_9=k$
are satisfied by more than one sequence. | 748 | From $(2)$, $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$
Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$.
Since $\gcd(13,21)=1$, the next smallest possible value for $a_1$ that yield... | 5.5 | [
6,
5,
5,
5,
5,
6,
6,
6
] |
Harold, Tanya, and Ulysses paint a very long picket fence.
Harold starts with the first picket and paints every $h$ th picket;
Tanya starts with the second picket and paints every $t$ th picket; and
Ulysses starts with the third picket and paints every $u$ th picket.
Call the positive integer $100h+10t+u$ paintable wh... | 757 | The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$. Note that the smallest number, $min \{ h,t,u \},$ divides the other $2$, and the next smallest divide the largest number, otherwise there is a common solution by... | 5.5 | [
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In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10$. Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCF... | 148 | By the Pythagorean Theorem, $BC=35$. By the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$. We can find the coordinates of F, and use that the find the equation of line EF. Then, we can find the coordinates of G. Triangle $ADC = 1110/7$ and we can find the area triangle $AGF$ with the shoelace theorem, so subtract... | 5.75 | [
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] |
Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$. A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$, which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$. The beam continues to be reflected off the faces of the cube. The length of the light path from the time ... | 230 | When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the l... | 6.375 | [
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6
] |
Let $F(z)=\dfrac{z+i}{z-i}$ for all complex numbers $z\neq i$, and let $z_n=F(z_{n-1})$ for all positive integers $n$. Given that $z_0=\dfrac{1}{137}+i$ and $z_{2002}=a+bi$, where $a$ and $b$ are real numbers, find $a+b$. | 275 | Iterating $F$ we get:
\begin{align*} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac... | 6.5 | [
6,
6,
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7,
7
] |
In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$, respectively, and $AB=24$. Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$. The area of triangle $AFB$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any ... | 63 | Use the same diagram as in Solution 1. Call the centroid $P$. It should be clear that $PE=9$, and likewise $AP=12$, $AE=12$. Then, $\sin \angle AEP = \frac{\sqrt{55}}{8}$. Power of a Point on $E$ gives $FE=\frac{16}{3}$, and the area of $AFB$ is $AE * EF* \sin \angle AEP$, which is twice the area of $AEF$ or $FEB$ (the... | 6.875 | [
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7
] |
A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer. Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is t... | 30 | Let the sum of the integers in $\mathcal{S}$ be $N$, and let the size of $|\mathcal{S}|$ be $n+1$. After any element $x$ is removed, we are given that $n|N-x$, so $x\equiv N\pmod{n}$. Since $1\in\mathcal{S}$, $N\equiv1\pmod{n}$, and all elements are congruent to 1 mod $n$. Since they are positive integers, the largest ... | 6.125 | [
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] |
Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given th... | 163 | We let $A$ be the origin, or $(0,0,0)$, $B = (0,0,12)$, and $D = (12,0,0)$. Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \frac{12-6}2 = 3$, so $G = (a,b,3)$, where a and b are variables.
We can now calculate the coordinates of E. Drawing the ... | 7 | [
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] |
Given that
\begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ &(3)& z=|x-y|. \end{eqnarray*}
How many distinct values of $z$ are possible? | 9 | We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$. From this, we have \begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ \end{eqnarray*} Because $a$ and $c$ are digits, and $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\boxed{009}$ possible values (since all digits ... | 4.125 | [
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] |
Three vertices of a cube are $P=(7,12,10)$, $Q=(8,8,1)$, and $R=(11,3,9)$. What is the surface area of the cube? | 294 | $PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}$
$PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}$
$QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}$
So, $PQR$ is an equilateral triangle. Let the side of the cube be $a$.
$a\sqrt{2}=\sqrt{98}$
So, $a=7$, and hence the surface area is $6a^2=\boxed{294}$. | 4.375 | [
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4,
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] |
It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$ | 111 | $abc=6^6$. Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$. $b=\sqrt[3]{abc}=6^2=36$.
Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$. Out of these, the only value of $a$ that works is $a=27$, from which we ... | 5.375 | [
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6
] |
Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$.
If $n=202$, then the area of the garden enclosed by the path, not including the path itself, is $m\left(\sqrt3/... | 803 | When $n>1$, the path of blocks has $6(n-1)$ blocks total in it. When $n=1$, there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is
\[(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A\],
where $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of the first... | 6.375 | [
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7,
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7,
6,
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] |
Find the sum of all positive integers $a=2^n3^m$ where $n$ and $m$ are non-negative integers, for which $a^6$ is not a divisor of $6^a$. | 42 | Notice that the condition is equivalent to saying
\[v_2(a^6) \geq v_2(6^a) \implies 6n \geq a\] \[v_3(a^6) \geq v_3(6^a) \implies 6m \geq a.\]
Notice that we cannot have both expressions to be equality state, as that would result in $a^6 = 6^a.$ Testing, we see the possible pairs $(n, m)$ are $(1, 0), (2, 0), (3, 0), ... | 6.375 | [
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] |
Find the integer that is closest to $1000\sum_{n=3}^{10000}\frac1{n^2-4}$. | 521 | We know that $\frac{1}{n^2 - 4} = \frac{1}{(n+2)(n-2)}$. We can use the process of fractional decomposition to split this into two fractions: $\frac{1}{(n+2)(n-2)} = \frac{A}{(n+2)} + \frac{B}{(n-2)}$ for some A and B.
Solving for A and B gives $1 = (n-2)A + (n+2)B$ or $1 = n(A+B)+ 2(B-A)$. Since there is no n term on... | 5.75 | [
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] |
It is known that, for all positive integers $k$,
$1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6$.
Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$. | 112 | $\frac{k(k+1)(2k+1)}{6}$ is a multiple of $200$ if $k(k+1)(2k+1)$ is a multiple of $1200 = 2^4 \cdot 3 \cdot 5^2$. So $16,3,25|k(k+1)(2k+1)$.
Since $2k+1$ is always odd, and only one of $k$ and $k+1$ is even, either $k, k+1 \equiv 0 \pmod{16}$.
Thus, $k \equiv 0, 15 \pmod{16}$.
If $k \equiv 0 \pmod{3}$, then $3|k$. ... | 5.875 | [
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] |
Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$. (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$.) | 49 | Rewriting the given information and simplifying it a bit, we have \begin{align*} k \le \frac{2002}{n} < k+1 &\implies \frac{1}{k} \ge \frac{n}{2002} > \frac{1}{k+1}. \\ &\implies \frac{2002}{k} \ge n > \frac{2002}{k+1}. \end{align*}
Now note that in order for there to be no integer solutions to $n,$ we must have $\lef... | 4 | [
5,
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4,
3,
4,
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4,
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] |
Let $\mathcal{S}$ be the set $\lbrace1,2,3,\ldots,10\rbrace$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$. | 501 | Let the two disjoint subsets be $A$ and $B$, and let $C = S-(A+B)$. For each $i \in S$, either $i \in A$, $i \in B$, or $i \in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$.
However, there are $2^{10}$ ways to organize the elements of $S$ such that $A = \emptyset$ and $... | 5.75 | [
5,
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] |
While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $x$ for which the sine of $x$ degrees is the same as the sine of $x$ radians are $\frac{m\pi}{n-\pi}... | 900 | Note that $x$ degrees is equal to $\frac{\pi x}{180}$ radians. Also, for $\alpha \in \left[0 , \frac{\pi}{2} \right]$, the two least positive angles $\theta > \alpha$ such that $\sin{\theta} = \sin{\alpha}$ are $\theta = \pi-\alpha$, and $\theta = 2\pi + \alpha$.
Clearly $x > \frac{\pi x}{180}$ for positive real value... | 6.75 | [
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7,
7,
7,
7,
7
] |
Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$, and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$, where $m$, $n$, and $p$ are positive integers and $m$ is not divisible by the square of an... | 518 | Let the second term of each series be $x$. Then, the common ratio is $\frac{1}{8x}$, and the first term is $8x^2$.
So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$. Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$.
The only solution in the appropriate form is $x = \frac... | 6 | [
6,
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6,
6,
6,
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] |
A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p... | 660 | We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$, and a missed shot is represented by $(1,0)$. Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$. We can find the number of s... | 6.875 | [
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] |
In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{P... | 901 | [asy] size(10cm); pair A,B,C,D,E,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,W... | 6.125 | [
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7
] |
The perimeter of triangle $APM$ is $152$, and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$. Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$. | 98 | Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have \[\frac{19}{AM} = \frac{152-2AM-19+19}{15... | 6.125 | [
6,
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] |
Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$, and the product of the radii is $68$. The x-axis and the line $y = mx$, where $m > 0$, are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$, where $a$, $b$, and $c$ are positive integ... | 282 | Let the smaller angle between the $x$-axis and the line $y=mx$ be $\theta$. Note that the centers of the two circles lie on the angle bisector of the angle between the $x$-axis and the line $y=mx$. Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$. Let $\tan{\frac{\t... | 7.625 | [
8,
7,
7,
7,
8,
8,
8,
8
] |
Given that
$\frac{((3!)!)!}{3!} = k \cdot n!,$
where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$ | 839 | Note that\[{{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.\]Because $120\cdot719!<720!$, conclude that $n$ must be less than 720, so the maximum value of $n$ is 719. The requested value of $k+n$ is therefore $120+719=\boxed{839}$.
~yofro | 5.875 | [
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] |
One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to t... | 301 | We want to find $\frac{\sum\limits_{n=1}^{50} (4n-1)\pi}{10000\pi}=\frac{\sum\limits_{n=1}^{50} (4n-1)}{10000}=\frac{(\sum\limits_{n=1}^{50} (4n) )-50}{10000}=\frac{101}{200} \rightarrow 101+200=\boxed{301}$ | 5.375 | [
7,
5,
5,
5,
5,
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7,
5
] |
Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list. | 484 | Thinking of this problem algorithmically, one can "sort" the array to give: \[{1, 2, 3, 5, 8, 13, 21, 34}\]
Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole $j = i + 1$ shebang. Then, we see that if we ... | 4 | [
5,
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Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$ | 12 | Using the properties of logarithms, we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$. Therefore, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]
Now, manipulate the second equation. \begin{align*} \log_{10} (\sin x + \cos x) &= \frac{1}{2}(\log_{10} n - \log_{10} ... | 4.625 | [
5,
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5,
5,
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] |
Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$ | 505 | [asy] size(220); import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6)); draw(box((0,-.1,0),(0.4,0.6,0.3))); draw(box((-.1,0,0),(0.5,0.5,0.3))); draw(box((0,0,-.1),(0.4,0.5,0.4))); draw(box((0,0,0),(0.4,0.5,0.3)),linewidth(1.2)+linetype("1")); [/asy]
The set can be broken into... | 6.375 | [
7,
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] |
The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers. Find $m + n + p.$ | 348 | [asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue); [/asy]
[asy] size(120); import three; unitsize(1cm); draw(... | 6.375 | [
6,
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7,
7,
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] |
Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s.$ | 380 | [asy] size(220); pointpen = black; pathpen = black + linewidth(0.7); pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle); D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7)); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2); [/asy]
Denote the height of $\t... | 5.75 | [
6,
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5,
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6,
6,
6
] |
In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by $30$. Find the sum of the four terms. | 129 | The sequence is of the form $a-d,$ $a,$ $a+d,$ $\frac{(a+d)^2}{a}$. Since the first and last terms differ by 30, we have \[\frac{(a+d)^2}{a}-a+d=30\] \[d^2+3ad=30a\] \[d^2+3ad-30a=0\] \[d=\frac{-3a + \sqrt{9a^2+120a}}{2}.\] Let $9a^2+120a=x^2$, where $x$ is an integer. This yields the following: \[9a^2+120a-x^2=0\] \[a... | 5.75 | [
6,
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6,
6,
5,
6,
5
] |
An integer between $1000$ and $9999$, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there? | 615 | If the common sum of the first two and last two digits is $n$, such that $1 \leq n \leq 9$, there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit). This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers. If the common sum of the first tw... | 4.25 | [
5,
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5,
4,
4
] |
Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$
[asy] pointpen = black; pathpen = black+linewidth(0.7); size(220); /* We will WLOG AB = 2 to draw ... | 83 | Without loss of generality, let $AC = BC = 1$. Then, using the Law of Sines in triangle $AMC$, we get $\frac {1}{\sin 150} = \frac {MC}{\sin 7}$, and using the sine addition formula to evaluate $\sin 150 = \sin (90 + 60)$, we get $MC = 2 \sin 7$.
Then, using the Law of Cosines in triangle $MCB$, we get $MB^2 = 4\sin^2... | 5.375 | [
5,
5,
6,
5,
5,
6,
6,
5
] |
An angle $x$ is chosen at random from the interval $0^\circ < x < 90^\circ.$ Let $p$ be the probability that the numbers $\sin^2 x, \cos^2 x,$ and $\sin x \cos x$ are not the lengths of the sides of a triangle. Given that $p = d/n,$ where $d$ is the number of degrees in $\text{arctan}$ $m$ and $m$ and $n$ are positive ... | 92 | Note that the three expressions are symmetric with respect to interchanging $\sin$ and $\cos$, and so the probability is symmetric around $45^\circ$. Thus, take $0 < x < 45$ so that $\sin x < \cos x$. Then $\cos^2 x$ is the largest of the three given expressions and those three lengths not forming a triangle is equival... | 5.75 | [
6,
5,
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6,
6,
6,
7,
5
] |
In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$) | 777 | [asy] real x = 1.60; /* arbitrary */ pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C+D)/2,... | 6.5 | [
7,
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7,
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6,
7
] |
Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base-$2$ representation has more $1$'s than $0$'s. Find the remainder when $N$ is divided by $1000$. | 155 | We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need m... | 6.375 | [
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The decimal representation of $m/n,$ where $m$ and $n$ are relatively prime positive integers and $m < n,$ contains the digits $2, 5$, and $1$ consecutively, and in that order. Find the smallest value of $n$ for which this is possible. | 127 | To find the smallest value of $n$, we consider when the first three digits after the decimal point are $0.251\ldots$.
Otherwise, suppose the number is in the form of $\frac{m}{n} = 0.X251 \ldots$, where $X$ is a string of $k$ digits and $n$ is small as possible. Then $10^k \cdot \frac{m}{n} - X = \frac{10^k m - nX}{n}... | 6.125 | [
6,
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In $\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\overline{CA},$ and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}.$ Suppose that $\overline{DF}$ meets $\overl... | 289 | Firstly, angle bisector theorem yields $\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}$. We're given that $AM=MC$. Therefore, the cross ratio
\[(A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120}\]
We need a fourth point for this cross ratio to be useful, so reflect point $F$ over angle bisector $BD$ to a point $F'... | 6.875 | [
7,
7,
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7,
7,
7,
7,
7
] |
The product $N$ of three positive integers is $6$ times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $N$. | 336 | Let the three integers be $a, b, c$. $N = abc = 6(a + b + c)$ and $c = a + b$. Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$. Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so $N$ is one of $12\cdot 13 = 156, 12\cdot 8 = 96, 12\cdo... | 4.125 | [
4,
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4,
4,
4,
5,
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4
] |
Let $N$ be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when $N$ is divided by 1000? | 120 | We want a number with no digits repeating, so we can only use the digits $0-9$ once in constructing our number. To make the greatest number, we want the greatest digit to occupy the leftmost side and the least digit to occupy the rightmost side. Therefore, the last three digits of the greatest number should be an arran... | 4 | [
5,
4,
4,
4,
4,
3,
4,
4
] |
Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$, $B$, and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$, $B$ is never immediately followed by $C$, and $C$ is never immediately followed by $A$. How many seven-... | 192 | There are three choices for the first letter and two choices for each subsequent letter, so there are $3\cdot2^{n-1}\ n$-letter good words. Substitute $n=7$ to find there are $3\cdot2^6=\boxed{192}$ seven-letter good words. ~ aopsav (Credit to AoPS Alcumus) | 4 | [
4,
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4,
4,
4,
4,
4,
4
] |
In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 28 | Let the large tetrahedron be $ABCD$, and the small tetrahedron be $WXYZ$, with $W$ on $ABC$, $X$ on $BCD$, $Y$ on $ACD$, and $Z$ on $ABD$. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let $AB=1$, for our convenience. Dropping an altit... | 5.625 | [
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6
] |
A cylindrical log has diameter $12$ inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a $45^\circ$ angle with the plane of the first cut. The intersection of these two planes has e... | 216 | The volume of the wedge is half the volume of a cylinder with height $12$ and radius $6$. (Imagine taking another identical wedge and sticking it to the existing one). Thus, $V=\dfrac{6^2\cdot 12\pi}{2}=216\pi$, so $n=\boxed{216}$.
Part 1:
Part 2:
- AMBRIGGS | 5 | [
4,
5,
6,
6,
4,
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4,
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] |
In triangle $ABC,$ $AB = 13,$ $BC = 14,$ $AC = 15,$ and point $G$ is the intersection of the medians. Points $A',$ $B',$ and $C',$ are the images of $A,$ $B,$ and $C,$ respectively, after a $180^\circ$ rotation about $G.$ What is the area of the union of the two regions enclosed by the triangles $ABC$ and $A'B'C'?$ | 112 | Since a $13-14-15$ triangle is a $5-12-13$ triangle and a $9-12-15$ triangle "glued" together on the $12$ side, $[ABC]=\frac{1}{2}\cdot12\cdot14=84$.
There are six points of intersection between $\Delta ABC$ and $\Delta A'B'C'$. Connect each of these points to $G$.
[asy] size(8cm); pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C... | 6 | [
6,
5,
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6,
6,
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6,
7
] |
Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$, respectively. | 400 | Let $\theta=\angle BDA$. Let $AB=BC=CD=x$. By the extended law of sines, \[\frac{x}{\sin\theta}=25\] Since $AC\perp BD$, $\angle CAD=90-\theta$, so \[\frac{x}{\sin(90-\theta)=\cos\theta}=50\] Hence $x=25\sin\theta=50\cos\theta$. Solving $\tan\theta=2$, $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}$. Thu... | 6.25 | [
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6,
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6
] |
Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences. | 348 | If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$, $f(2)=1716$, and $f(3)=1848$. Plugging in the values for x gives us a system of three equations:
$a+b+c=1440$
$4a+2b+c=1716$
$9a+3b+c=1848$
Solvin... | 5.375 | [
4,
5,
6,
5,
6,
6,
6,
5
] |
Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$ | 6 | Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]
Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+... | 6.125 | [
6,
6,
6,
6,
7,
6,
6,
6
] |
Two positive integers differ by $60$. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers? | 156 | Call the two integers $b$ and $b+60$, so we have $\sqrt{b}+\sqrt{b+60}=\sqrt{c}$. Square both sides to get $2b+60+2\sqrt{b^2+60b}=c$. Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$, and $(b+n+30)(b-n+30)=900$. The sum of these two factors is $2b+60$, so they must both be even. To maximize $b$, we want to ma... | 5.625 | [
6,
6,
5,
5,
5,
6,
6,
6
] |
Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may be expressed as $\frac {m\sqrt {n}}{p},$ where $m,$ $n,$ and $p$ are positive integers... | 578 | We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$
Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$, $MN=AM-AN$, and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$
From the third equation, we get $CN=\frac{168} {25}.$
By the Pythagorean Theorem in $\D... | 6.625 | [
7,
7,
6,
6,
7,
6,
7,
7
] |
The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of th... | 134 | Let there be $N$ members of the committee. Suppose candidate $n$ gets $a_n$ votes. Then $a_n$ as a percentage out of $N$ is $100\frac{a_n}{N}$. Setting up the inequality $a_n \geq 1 + 100\frac{a_n}{N}$ and simplifying, $a_n \geq \lceil(\frac{N}{N - 100})\rceil$ (the ceiling function is there because $a_n$ is an integer... | 6.125 | [
6,
5,
6,
6,
6,
6,
7,
7
] |
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are... | 683 | This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to $341.33/1024$. We can either round up or down. If we round down, we see $341/10... | 5.625 | [
6,
5,
6,
6,
5,
5,
6,
6
] |
Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct e... | 51 | The y-coordinate of $F$ must be $4$. All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting $F = (f,4)$, and knowing that $\angle FAB = 120^\circ$, we can use rewrite $F$ using complex numbers: $f + 4 i = (b + 2 i)\left(e^{i(2 \pi / 3)}\right) = (b + 2 i)\left(-1/2 + ... | 6.625 | [
7,
7,
7,
6,
7,
6,
7,
6
] |
Let \[P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).\] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let
$\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$
where $m, n,$ and $p$ ar... | 15 | As in Solution 1, we find that the roots of $P(x)$ we care about are the 24th roots of unity except $1$. Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for $1$, which is only counted once.
The possible imaginary parts of the 12th root... | 8.125 | [
8,
8,
8,
8,
8,
9,
8,
8
] |
The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$? | 217 | A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}}$$= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$, for $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace$.... | 4.875 | [
4,
5,
5,
5,
5,
5,
5,
5
] |
Set $A$ consists of $m$ consecutive integers whose sum is $2m$, and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$. Find $m.$ | 201 | Note that since set $A$ has $m$ consecutive integers that sum to $2m$, the middle integer (i.e., the median) must be $2$. Therefore, the largest element in $A$ is $2 + \frac{m-1}{2}$.
Further, we see that the median of set $B$ is $0.5$, which means that the "middle two" integers of set $B$ are $0$ and $1$. Therefore, ... | 5.5 | [
6,
6,
6,
5,
6,
5,
6,
4
] |
A convex polyhedron $P$ has $26$ vertices, $60$ edges, and $36$ faces, $24$ of which are triangular and $12$ of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have? | 241 | Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal. We have ${26 \choose 2} = \frac{26\cdot25}2 = 325$ total line segments determined by the vertices. Of these, $60$ are edges. Each triangular face has $0$ face diagonals and each quadrilateral face has $2$, so there ... | 6.375 | [
6,
6,
6,
6,
7,
6,
7,
7
] |
Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$. | 86 | Without loss of generality, let $(0,0)$, $(2,0)$, $(0,2)$, and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$. Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$. Because the segment has length 2, $x^... | 6 | [
6,
6,
6,
6,
5,
7,
7,
5
] |
Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not atte... | 849 | Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2.
These inequalities follow: \[\frac{a}{q} < \frac{160}{300} = \frac{8}{15}\] \[\frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}\] Solving for a and b and adding the two inequalities:... | 6.5 | [
6,
6,
6,
7,
6,
7,
7,
7
] |
An integer is called snakelike if its decimal representation $a_1a_2a_3\cdots a_k$ satisfies $a_i<a_{i+1}$ if $i$ is odd and $a_i>a_{i+1}$ if $i$ is even. How many snakelike integers between 1000 and 9999 have four distinct digits? | 882 | Let's create the snakelike number from digits $a < b < c < d$, and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of $5\cdot{10 \choose 4}$ But, this over-counts since it counts numbers like 0213. We can corre... | 5.875 | [
6,
6,
4,
7,
6,
6,
7,
5
] |
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$ | 588 | Let our polynomial be $P(x)$.
It is clear that the coefficient of $x$ in $P(x)$ is $-1 + 2 - 3 + \ldots + 14 - 15 = -8$, so $P(x) = 1 -8x + Cx^2 + Q(x)$, where $Q(x)$ is some polynomial divisible by $x^3$.
Then $P(-x) = 1 + 8x + Cx^2 + Q(-x)$ and so $P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)$, where $R(x)$ is some po... | 6.125 | [
6,
6,
6,
7,
6,
6,
6,
6
] |
Define a regular $n$-pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that
the points $P_1, P_2,\ldots, P_n$ are coplanar and no three of them are collinear,
each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint,
all of t... | 199 | We use the Principle of Inclusion-Exclusion (PIE).
If we join the adjacent vertices of the regular $n$-star, we get a regular $n$-gon. We number the vertices of this $n$-gon in a counterclockwise direction: $0, 1, 2, 3, \ldots, n-1.$
A regular $n$-star will be formed if we choose a vertex number $m$, where $0 \le m \... | 6.625 | [
7,
6,
7,
7,
7,
6,
6,
7
] |
Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is s... | 35 | We let $AB=3, AC=4, DE=6, DG=7$ for the purpose of labeling. Clearly, the dividing segment in $DEFG$ must go through one of its vertices, without loss of generality $D$. The other endpoint ($D'$) of the segment can either lie on $\overline{EF}$ or $\overline{FG}$. $V_2$ is a trapezoid with a right angle then, from whic... | 5.875 | [
6,
5,
6,
6,
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6,
6,
6
] |
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