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Difference between revisions of "G285 2021 Summer Problem Set"

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Geometry285 is playing the game "Guess And Choose". In this game, Geometry285 selects a subset of not necessarily distinct integers <math>P=\{a,b,c \cdots \}</math> from the set <math>S=\{1,2,3,4 \cdots k-1,k \}</math> such that the sum of all elements in <math>P</math> is <math>k</math>. Each distinct is selected chronologically and placed in <math>P</math>, such that <math>1 \le a \le k</math>, <math>1 \le b \le a</math>, <math>1 \le c \le b</math>, and so on. Then, the elements are randomly arranged. Suppose <math>S_{p,k}</math> represents the total number of outcomes that a subset <math>P</math> containing <math>p</math> integers sums to <math>k</math>. If distinct permutations of the same set <math>P</math> are considered unique, find the remainder when <cmath>\sum_{p=1}^{1000}\sum_{k=1}^{1000} S_{a,k}</cmath> is divided by <math>100</math>.
 
Geometry285 is playing the game "Guess And Choose". In this game, Geometry285 selects a subset of not necessarily distinct integers <math>P=\{a,b,c \cdots \}</math> from the set <math>S=\{1,2,3,4 \cdots k-1,k \}</math> such that the sum of all elements in <math>P</math> is <math>k</math>. Each distinct is selected chronologically and placed in <math>P</math>, such that <math>1 \le a \le k</math>, <math>1 \le b \le a</math>, <math>1 \le c \le b</math>, and so on. Then, the elements are randomly arranged. Suppose <math>S_{p,k}</math> represents the total number of outcomes that a subset <math>P</math> containing <math>p</math> integers sums to <math>k</math>. If distinct permutations of the same set <math>P</math> are considered unique, find the remainder when <cmath>\sum_{p=1}^{1000}\sum_{k=1}^{1000} S_{a,k}</cmath> is divided by <math>100</math>.
  
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<math>\textbf{(A)} \0 \qquad\textbf{(B)} \1 \qquad\textbf{(C)} \50 \qquad\textbf{(D)} \51 \qquad\textbf{(E)} \124</math>
+
<math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 50 \qquad\textbf{(D)}\ 51 \qquad\textbf{(E)}\ 124</math>
  
 
[[G285 Summer Problem Set Problem 7|Solution]]
 
[[G285 Summer Problem Set Problem 7|Solution]]
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<math>\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 69 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 71 \qquad\textbf{(E)}\ 72</math>
 
<math>\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 69 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 71 \qquad\textbf{(E)}\ 72</math>
 +
 +
[[G285 Summer Problem Set Problem 8|Solution]]
  
 
Solution: The first summation is simply <math>9 \cdot 9 = 81</math> by Vieta's. The second summation is <math>-9+r_{12}</math>. The minimum possible value is <math>72+r_{12}</math>, so we need to minimize <math>r_{12}</math>. If we do bounding, when <math>x=0</math> we have <math>p(x)=1</math>, and when <math>x=-1</math> we have <math>p(x)=-230</math>. The shift implies there is a root <math>r_k</math> where <math>0 \le k \le 12</math> such that <math>-1 < r_k < 0</math>. However, <math>x=-1.5</math> seems very close to <math>0</math>, and <math>x<-1.5</math> approaches infinity, so there is another root <math>-2 < r_k < -1</math>. Therefore, we have the smallest root <math>r_{12}</math> must be <math>72+(-2+\{r_{12} \})</math>, where <math>\{x \}</math> is the fractional part. The answer is <math>\boxed{70}</math>
 
Solution: The first summation is simply <math>9 \cdot 9 = 81</math> by Vieta's. The second summation is <math>-9+r_{12}</math>. The minimum possible value is <math>72+r_{12}</math>, so we need to minimize <math>r_{12}</math>. If we do bounding, when <math>x=0</math> we have <math>p(x)=1</math>, and when <math>x=-1</math> we have <math>p(x)=-230</math>. The shift implies there is a root <math>r_k</math> where <math>0 \le k \le 12</math> such that <math>-1 < r_k < 0</math>. However, <math>x=-1.5</math> seems very close to <math>0</math>, and <math>x<-1.5</math> approaches infinity, so there is another root <math>-2 < r_k < -1</math>. Therefore, we have the smallest root <math>r_{12}</math> must be <math>72+(-2+\{r_{12} \})</math>, where <math>\{x \}</math> is the fractional part. The answer is <math>\boxed{70}</math>

Revision as of 11:49, 24 June 2021

Welcome to the Birthday Problem Set! In this set, there are multiple choice AND free-response questions. Feel free to look at the solutions if you are stuck:

Problem 1

Find $\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil$

$\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46$

Solution

Problem 2

Let \[f(x,y) = \begin{cases}x^y & \text{ if } x^2>y \text{ and } |x|<y\\f(f(\sqrt{|x|},y),y) & \text{ otherwise} \end{cases}\] If $y$ is a positive integer, find the sum of all values of $x$ such that $f(x,y) \neq k$ for some constant $k$.

Problem 3

Let circles $\omega_1$ and $\omega_2$ with centers $Q$ and $L$ concur at points $A$ and $B$ such that $AQ=20$, $AL=28$. Suppose a point $P$ on the extension of $AB$ is formed such that $PQ=29$ and lines $PQ$ and $PL$ intersect $\omega_1$ and $\omega_2$ at $C$ and $D$ respectively. If $DC=\frac{16\sqrt{37}}{\sqrt{145}}$, the value of $\sin^2(\angle LAQ)$ can be represented as $\frac{m \sqrt{n}}{r}$, where $m$ and $r$ are relatively prime positive integers, and $n$ is square free. Find $2m+3n+4r$

$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 39 \qquad\textbf{(D)}\ 45 \qquad\textbf{(E)}\ 54$

Problem 4

Let $ABCD$ be a rectangle with $BC=6$ and $AB=8$. Let points $M$ and $N$ lie on $ABCD$ such that $M$ is the midpoint of $BC$ and $N$ lies on $AD$. Let point $Q$ be the center of the circumcircle of quadrilateral $MNOP$ such that $O$ and $P$ lie on the circumcircle of $\triangle MNP$ and $\triangle MNO$ respectively, along with $OD \perp QO$ and $MP \perp BP$. If the shortest distance between $Q$ and $AB$ is $3$, $\triangle AOQ$ and $\triangle QBP$ are degenerate, and $BP=AO$, find $25 \cdot OD \cdot PC$

$\textbf{(A)}\ 209 \qquad\textbf{(B)}\ 228 \qquad\textbf{(C)}\ 54\sqrt{57} \qquad\textbf{(D)}\ 90\sqrt{19} \qquad\textbf{(E)}\ 72\sqrt{57}$

Problem 5

Suppose $\triangle ABC$ is an equilateral triangle. Let points $D$ and $E$ lie on the extensions of $AB$ and $AC$ respectively such that $\angle AED=60^o$ and $DE=14$. If there exists a point $P$ outside of $\triangle ADE$ such that $AP=PD=28$, and there exists a point $O$ outside outside of $CBDE$ such that $OE=OA$, the area $2APEO$ can be represented as $m\sqrt{n}+o\sqrt{p}$, where $n$ and $p$ are squarefree,. Find $m+n+o+p$

$\textbf{(A)}\ 152 \qquad\textbf{(B)}\ 162 \qquad\textbf{(C)}\ 164 \qquad\textbf{(D)}\ 214\qquad\textbf{(E)}\ 224$

Problem 6

$16$ people are attending a hotel conference, $8$ of which are executives, and $8$ of which are speakers. Each person is designated a seat at one of $4$ round tables, each containing $4$ seats. If executives must sit at least one speaker and executive, there are $N$ ways the people can be seated. Find $\left \lfloor \sqrt{N} \right \rfloor$. Assume seats, people, and table rotations are distinguishable.

$\textbf{(A)}\ 720 \qquad\textbf{(B)}\ 1440 \qquad\textbf{(C)}\ 2520 \qquad\textbf{(D)}\ 3456\qquad\textbf{(E)}\ 5760$

Problem 7

Geometry285 is playing the game "Guess And Choose". In this game, Geometry285 selects a subset of not necessarily distinct integers $P=\{a,b,c \cdots \}$ from the set $S=\{1,2,3,4 \cdots k-1,k \}$ such that the sum of all elements in $P$ is $k$. Each distinct is selected chronologically and placed in $P$, such that $1 \le a \le k$, $1 \le b \le a$, $1 \le c \le b$, and so on. Then, the elements are randomly arranged. Suppose $S_{p,k}$ represents the total number of outcomes that a subset $P$ containing $p$ integers sums to $k$. If distinct permutations of the same set $P$ are considered unique, find the remainder when \[\sum_{p=1}^{1000}\sum_{k=1}^{1000} S_{a,k}\] is divided by $100$.

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 50 \qquad\textbf{(D)}\ 51 \qquad\textbf{(E)}\ 124$

Solution

Problem 8

Let $p(x)=x^{12}-9x^{11}+16x^{10}+256x^5+1$, Let $r_1, r_2, r_3, r_4, r_5, r_6, ..., r_{12}$ be the twelve roots that satisfies $p(x)=0$, find the least possible value of \[\left \lfloor \sum_{n=1}^{12}\sum_{k=1}^{12} r_nr_k-\sum_{s=1}^{11} r_s \right \rfloor\]

$\textbf{(A)}\ 67 \qquad\textbf{(B)}\ 69 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 71 \qquad\textbf{(E)}\ 72$

Solution

Solution: The first summation is simply $9 \cdot 9 = 81$ by Vieta's. The second summation is $-9+r_{12}$. The minimum possible value is $72+r_{12}$, so we need to minimize $r_{12}$. If we do bounding, when $x=0$ we have $p(x)=1$, and when $x=-1$ we have $p(x)=-230$. The shift implies there is a root $r_k$ where $0 \le k \le 12$ such that $-1 < r_k < 0$. However, $x=-1.5$ seems very close to $0$, and $x<-1.5$ approaches infinity, so there is another root $-2 < r_k < -1$. Therefore, we have the smallest root $r_{12}$ must be $72+(-2+\{r_{12} \})$, where $\{x \}$ is the fractional part. The answer is $\boxed{70}$

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