G285 2021 Summer Problem Set

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 $100$ .

$\textbf{(A)} \0 \qquad\textbf{(B)} \1 \qquad\textbf{(C)} \50 \qquad\textbf{(D)} \51 \qquad\textbf{(E)} \124$ (Error compiling LaTeX. Unknown error_msg)

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: 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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