problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
For integers $0\le a\le n$ , let $f(n,a)$ denote the number of coefficients in the expansion of $(x+1)^a(x+2)^{n-a}$ that is divisible by $3.$ For example, $(x+1)^3(x+2)^1=x^4+5x^3+9x^2+7x+2$ , so $f(4,3)=1$ . For each positive integer $n$ , let $F(n)$ be the minimum of $f(n,0),f(n,1),\ldots ,f(n,n)$ .
(1) Prove that there exist infinitely many positive integer $n$ such that $F(n)\ge \frac{n-1}{3}$ .
(2) Prove that for any positive integer $n$ , $F(n)\le \frac{n-1}{3}$ . | Let $g(n,a)$ be the number of coefficients in the expansion of $(x+1)^a (x-1)^{n-a}$ that are NOT divisible by 3, and let $G(n)=\max\limits_{a=0}^n g(n,a)$ . We know $g(n,a)=(n+1)-f(n,a)$ Part a) I will show there exists $a$ such that $f(n,a)=\frac{n-1}{3}$ $n=4,7$ both work.
Rephrasing the problem, $G(n)=\frac{2n+4}{3}$ works, then $G(3n+4)=2n+4$ , which solves the problem.
If $a\equiv 2(\bmod\; 3)$ , then we can compute $g(3n+4,a)$ as follows: $(x+1)^a (x-1)^{3n+4-a} = (x^2-1)^2 (x^3+1)^{\frac{a-2}{3}} (x^3-1)^{n-\frac{a-2}{3}}$ Note $(x^2-1)^2$ multiplies nicely with polynomial in $x^3$ , so $g(3n+4,a)=3g(n,\frac{a-2}{3})$ .
Since $g(4,1)=4$ , it follows that $g(16,5)=12, g(52,17)=36$ ,..., et cetera.
Part b) We need to show $G(n)\ge \frac{2n+4}{3}$ We proceed by strong induction. The base cases can be verified by hand. We establish a recursion in g. Call $(n,k)$ good if $g(n,k)\ge \frac{2n+4}{3}$ Case 1: $n\equiv 2(\bmod\; 3)$ .
Pick $k$ such that $g(n,k)\ge \frac{2n+4}{3}$ , which exists by our inductive hypothesis. $g(3n+2,3k)$ : $(x+1)^{3k} (x-1)^{3(n-k)} (x-1)^{2} = (x-1)^{2} (x^3+1)^k (x^3-1)^{n-k}$ so $g(3n+2,3k)= 3g(n,k)=2n+4 > \frac{6n+8}{3}$ Similarly, $g(3n+2,3k+2)=3g(n,k)$ as well.
Case 2: $n\equiv 1(\bmod\; 3)$ , which is discussed in part a).
Case 3: $3\mid n$ . This case actually depends on the previous two cases. While doing induction for the previous two cases, there is an inner induction for this particular case.
Replace $n$ with $3m+3$ . $g(3m+3,3k+1)$ : $(x+1)^{3k+1}(x-1)^{3(m-k)+2} \equiv (x+1)(x-1)^2 (x^3+1)^k (x^3-1)^{m-k} = (x^3-x^2-x+1)(x^3+1)^k (x^3-1)^{m-k} = (x^3+1)^{k+1}(x^3-1)^{m-k} - (x^2+x)(x^3+1)^k (x^3-1)^{m-k}$ Therefore, $g(3m+3,3k+1)=g(m+1,k+1)+2g(m,k)$ . Here is the inner induction for this particular case.
Similarly, $g(3m+3,3k+2)=g(3m+3, 3(m-k)+1)=g(m+1,m-k+1)+2g(m,m-k)=g(m+1,k)+2g(m,k)$ We need to show there exists $k$ such that $(m+1,k), (m,k)$ are both good.
Subcase 1: $m$ is 1 mod 3. Let $m=3l+1$ Then we have $g(3l+1,3k'+2)=3g(l-1,k')$ and $g(3l+2,3k'+2)=3g(l,k')$ , so this becomes the same subproblem.
Subcase 2: $m$ is 0 mod 3. Let $m=3l$ .
Then we note $g(3l+1,3k'+2)=3g(l-1,k')$ and $g(3l,3k'+2)=g(l,k')+2g(l-1,k')$ , reducing to a smaller subproblem.
Subcase 3: $m$ is 2 mod $3$ . Let $m=3l+2$ Then note $g(3l+2,3k'+2)=3g(l,k')$ and $g(3l+3,3k'+2)=g(l,k')+2g(l-1,k')$ , reducing to a smaller subproblem.
So this subproblem can be solved recursively as well. The end.
Note: This writeup can be improved to showing the following claim: for any $n$ there exists $k$ st $(n,k), (n+1,k)$ is good. I will fix this tomorrow. | [
"After trying serveral cases, I believe it should be (n-1)/3 for both parts, not (n+1)/3.",
"Sketch:\nPart a) If n satisfies, then 3n+4 also satisfies.\nPart b) Induction. We have f(3n+1,3m+2)=3f(n-1,m)+2, f(3n+2,3m+3)=3f(n,m+1) and f(3m+3,3n+1)=f(n+1,m+1)+2f(n,m), then we will prove a stronger statement: There... | [
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} |
Let $ABC$ be an acute triangle with $\angle ACB>2 \angle ABC$ . Let $I$ be the incenter of $ABC$ , $K$ is the reflection of $I$ in line $BC$ . Let line $BA$ and $KC$ intersect at $D$ . The line through $B$ parallel to $CI$ intersects the minor arc $BC$ on the circumcircle of $ABC$ at $E(E \neq B)$ . The line through $A$ parallel to $BC$ intersects the line $BE$ at $F$ .
Prove that if $BF=CE$ , then $FK=AD$ . | Let $CI$ meet circumcircle at $T$ .
Claim $: FTIB$ is parallelogram.
Proof $:$ Note that $TI || BF$ and $BF = CE = BT = TI$ (it's well-known that $TB = TI = TA$ ).
Claim $: TAF$ and $BFK$ are congruent.
Proof $:$ Note that $TF = BI = BK$ and $TA = TI = FB$ and $\angle FBK = \angle FBI + \angle IBK = \angle ITF + \angle IBK = \angle ITF + \angle ABC = \angle ITF + \angle ATB = \angle ATF$ .
Now Note that $FK = FA$ so we need to prove $FAD$ is isosceles. Note that $\angle TFA = \angle IBC = \frac{\angle B}{2}$ and $\angle FAB = \angle ABC$ so we need to prove $F,T,D$ are collinear. Lets assume not so $TF$ meets $BA$ at $S$ .
Claim $: SFBK$ is cyclic.
Proof $:$ Note that $\angle FSB = \angle TFA = \angle BKF$ cause $TAF$ and $BFK$ are congruent.
Now Note that $\angle SKB = \angle 180 - \angle SFB = \angle 180 - \frac{\angle B}{2} - \frac{\angle C}{2} = \angle 90 + \frac{\angle A}{2}$ and $\angle DKB = \angle CIB = \angle 90 + \frac{A}{2} \implies DKB = \angle SKB$ so $S$ and $D$ are same so $F,T,D$ are collinear as wanted so we're Done. | [
"<blockquote>It is confusing why they placed two easy geometry in a single test (P1 and P4). Maybe an inequality would have been better. Once you take out the midpoint of the arc, this is done easily.</blockquote>\n\nD'KBF is cyclic, not D'CBF",
"This problem is not that easy if you attend the test. All four stud... | [
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} |
Let $C=\{ z \in \mathbb{C} : |z|=1 \}$ be the unit circle on the complex plane. Let $z_1, z_2, \ldots, z_{240} \in C$ (not necessarily different) be $240$ complex numbers, satisfying the following two conditions:
(1) For any open arc $\Gamma$ of length $\pi$ on $C$ , there are at most $200$ of $j ~(1 \le j \le 240)$ such that $z_j \in \Gamma$ .
(2) For any open arc $\gamma$ of length $\pi/3$ on $C$ , there are at most $120$ of $j ~(1 \le j \le 240)$ such that $z_j \in \gamma$ .
Find the maximum of $|z_1+z_2+\ldots+z_{240}|$ . | Here is my <details><summary>solution</summary>Let the $240$ plurals be $e^{\theta_1 i},e^{\theta_2 i}, \cdots ,e^{\theta_{240}i},0 \leqslant \theta_1 \leqslant \theta_2 \leqslant \cdots \leqslant \theta_{240} \leqslant 2 \pi ,z_k=e^{\theta_k i}$ . And let $\omega_k=z_k+ $ $z_{k+40}+z_{k+80}+z_{k+120}+z_{k+160}+z_{k+200},(1\leqslant k \leqslant 40).$ For the complex number $\omega_k=a_{0 k}+a_{1 k}+a_{2 k}+a_{3 k}+a_{4 k}+a_{5 k},(1\leqslant k \leqslant 40),a_{i j}=z_{40 i+j}$ ,it should satisfy the following $2$ conditions:
(1) For any open arc $\Gamma$ of length $\pi$ on the unit circle, at most $5$ of $a_i (1 \leqslant i \leqslant 6)$ are on $\Gamma$ .
(2) For any open arc $\gamma$ of length $ \frac \pi 3$ on the unit circle, at most $3$ of $a_i (1 \leqslant i \leqslant 6)$ are on $\gamma$ .
So we have $|\omega|=|a_0+a_1+a_2+a_3+a_4+a_5| \leqslant |(a_0+a_5)+(a_1+a_4)+a_2+a_3|$ $\leqslant |2\cos(\frac \pi 2+\alpha)+2\cos(\frac \pi 6+\beta)+1+1|\leqslant 2+\sqrt3(\alpha,\beta\in[0,\frac\pi2])$ so $\displaystyle|\sum_{i=1}^{240}z_i|=|\sum_{i=1}^{40}\omega_i|\le40(2+\sqrt3).$ And when $z_1=z_2=\cdots=z_{40}=i,z_{41}=z_{42}=\cdots=z_{80}=-i,z_{81}=z_{82}=\cdots=$ $ z_{120}=\frac{\sqrt3}2+\frac 12i,$ $z_{121}=z_{122}=\cdots=z_{160}=\frac{\sqrt3}2-\frac 12i,z_{161}=z_{162}=\cdots=z_{240}=1$ ,we have $\displaystyle|\sum_{i=1}^{240}z_i|=80+40\sqrt3$ .</details> | [
"Just notice that $z_1+z_2+...+z_{240}=\\int_0^{2\\pi}F(\\theta)e^{i\\theta}d\\theta$ ,where F( $\\theta$ ) indicates the number of $z_i$ s in the arc $(\\theta-\\frac{\\pi}{6},\\theta+\\frac{\\pi}{6})$ This intergal equals $\\frac{1}{6} \\int_0^{2\\pi}(F(\\theta)e^{i\\theta}+F(\\theta+\\frac{\\pi}{3})e^{i(\\th... | [
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Let $m$ be a positive integer, and $A_1, A_2, \ldots, A_m$ (not necessarily different) be $m$ subsets of a finite set $A$ . It is known that for any nonempty subset $I$ of $\{1, 2 \ldots, m \}$ ,
\[ \Big| \bigcup_{i \in I} A_i \Big| \ge |I|+1. \]
Show that the elements of $A$ can be colored black and white, so that each of $A_1,A_2,\ldots,A_m$ contains both black and white elements. | <details><summary>Solution</summary>Consider a bipartite graph $G(S,T,E)$ with $S=\{A_1,\cdots,A_m\}$ representing the sets, $T=\bigcup_{j=1}^m A_j$ representing the elements and $E$ representing the edges. Connect an edge between $A_j$ and $b\in T$ if and only if $b\in A_j$ . It is equivalent to prove that there exists a coloring of $T$ in black and white such that each vertex in $S$ has at least one black neighbor and at least one white neighbor. The edges will be denoted by ordered pairs $(s,t)$ where $s\in S$ and $t\in T$ .
By Hall's marriage theorem, there exists an injection $M\colon S\to T$ such that $M(s)\in N_G(s)$ for all $s\in S$ . Let $E(M)$ denote the set of pairs $\{(s,M(s))\mid s\in S\}$ .
The main claim is as follows:**Claim.** Let $k$ be an integer with $0\le k\le m$ . Then there exists a subgraph $G_k(S,T,E_k)\subset G(S,T,E)$ and a subset $S_k\subset S$ such that the following conditions hold:
[list=1][*] $|S_k|=k$ ,
[*] $E(M)\subset E_k$ ,
[*] for a vertex $s\in S$ we have $\deg_{G_k}(s)=\begin{cases}2 & s\in S_k 1 & s\notin S_k\end{cases}$ ,
[*] $N_{G_k}(S_k)\cap M(S\setminus S_k)=\emptyset$ , and
[*] there exists a coloring of $N_{G_k}(S_k)$ in black and white such that for each vertex $s\in S_k$ , the two neighbors of $s$ in $G_k$ are of distinct colors.
[/list]**Proof.** We prove the claim by induction on $k$ . For the base case $k=0$ choose $E_0=E(M)$ , $S_0=\emptyset$ and the empty coloring. Now assume the claim is true for some $k<m$ and we will prove it for $k+1$ . Pick a subgraph $G_k(S,T,E_k)\subset G(S,T,E)$ , a subset $S_k\subset S$ and a coloring $c_k$ of $N_{G_k}(S_k)$ satisfying the induction hypothesis. Define $Y_k\doteqdot S\setminus S_k$ . By the problem statement we have $N_G(Y_k)\ge |Y_k| + 1$ . Hence there exist vertices $u_0\in Y_k$ and $v_0\in T\setminus M(Y_k)$ such that $e_0\doteqdot(u_0,v_0)\in E$ . Moreover we have $M(u_0)\notin N_{G_k}(S_k)$ by condition (4) in the induction hypothesis. Now define a subgraph $G_{k+1}(S,T,E_{k+1})\subset G(S,T,E)$ by $E_{k+1}\doteqdot E_k\cup\{e_0\}$ . Moreover define a subset $S_{k+1}\doteqdot S_k\cup\{u_0\}\subset S$ . It is easily seen that conditions (1)-(4) hold. It remains to provide a coloring of $N_{G_{k+1}}(S_{k+1})$ satisfying the induction step. We divide into cases.
*Case 1.* $v_0\in N_{G_k}(S_k)$ . Then use the coloring $c_k$ and color $M(u_0)$ with the color not presented in $v_0$ .
*Case 2.* $v_0\notin N_{G_k}(S_k)$ . Then use the coloring $c_k$ , color $v_0$ white and $M(u_0)$ black.
In both cases the new coloring satisfies condition (5). Hence the induction step is proved. $\square$ To finish the proof of the problem, substitute $k=m$ in the claim. Then we have a graph $G_m(S,T,E_m)\subset G(S,T,E)$ such that $\deg_{G_m}(s)=2$ for every $s\in S$ , and there exists a coloring of $N_{G_m}(S)$ such that for each vertex $s\in S$ , the two neighbors of $s$ in $G_m$ are of distinct colors. Now color the rest of vertices in $T$ arbitrarily (e.g color all of them white). This coloring satisfies the problem statement.</details>
Here is an informal and more intuitive version of the solution for better understanding.
<details><summary>Informal solution</summary>Let $X\doteqdot\bigcup_{i=1}^m A_i$ . By Hall's marriage solution, there exists distinct elements $a_1,\ldots,a_m\in X$ such that $a_i\in A_i$ for every $i\in[m]$ . By the conditions of the problem we have $|X|\ge m+1$ . Hence there exists an element $b_1\in X$ such that $b_1\notin\{a_1,\ldots,a_m\}$ . WLOG we have $b_1\in A_1$ . By the conditions of the problem again we have $\left|\bigcup_{i=2}^m A_i\right|\ge m$ . Hence there exists an element $b_2\in\bigcup_{i=2}^m A_i$ with $b_2\notin\{a_2,\ldots,a_m\}$ . We can assume WLOG that $b_2\in A_2$ . Continuing this way, after rearranging the $A_i$ we can assume that there exist $b_1,\ldots,b_n\in X$ such that $b_i\in A_i$ for every $i\in[m]$ , and moreover $b_i\ne a_j$ for every $i,j\in[m]$ with $i\le j$ . We will inductively color the pairs $\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_m,b_m\}$ such that at each step, if an element was colored previously then we will skip it. Moreover, we will make sure that for each $k\in[m]$ , each pair among $\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_k,b_k\}$ is colored with both colors. At the first step, note that $a_1\ne b_1$ so we can color $a_1$ black and $b_1$ white. For the induction step, assume $2\le k\le m$ and $\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_{k-1},b_{k-1}\}$ are already colored. Since $a_k\notin\bigcup_{i=1}^{k-1}\{a_i,b_i\}$ we can color it as we wish. If $b_k$ is already colored, color $a_k$ with the opposite color. Otherwise color $a_k$ black and $b_k$ white. Now we have obtained a partial coloring of $X$ such that each set $A_i$ consists of elements of both colors. Then we can color the rest of elements of $X$ arbitrarily (e.g. color all of them white). This finishes the solution.</details> | [
"<details><summary>I think this works?</summary>Consider a bipartite graph $(S,T)$ with $S=\\{A_1,\\cdots,A_m\\}$ representing the sets and $T=\\cup_{j=1}^m A_j$ representing the elements. Connect an edge between $A_j$ and $b_i$ if $b_i\\in A_j$ . WLOG, the graph is connected, since we can solve for each... | [
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In a cyclic convex hexagon $ABCDEF$ , $AB$ and $DC$ intersect at $G$ , $AF$ and $DE$ intersect at $H$ . Let $M, N$ be the circumcenters of $BCG$ and $EFH$ , respectively. Prove that the $BE$ , $CF$ and $MN$ are concurrent. | Let $T = BE \cap CF$ . Consider the inversion $\Phi$ at $T$ fixing the circle. Angle chase shows $\angle BH^*C = 180^\circ - \angle BGC$ (where $H^* = \Phi(H)$ ), i.e. $H \in \odot(BCG)$ . So $\Phi$ just swaps $\odot(EFH)$ and $\odot(BCG)$ , implying points $N,T,M$ are collinear. $\blacksquare$ [asy]
size(200);
pair A=dir(110),B=dir(50),C=dir(-20),D=dir(-60),E=dir(-140),F=dir(170),T=extension(F,C,E,B),G=extension(A,B,C,D),H=extension(A,F,E,D),N=circumcenter(F,H,E),M=circumcenter(G,B,C);
draw(unitcircle);
dot(" $A$ ",A,dir(A));
dot(" $B$ ",B,dir(B));
dot(" $C$ ",C,dir(C));
dot(" $D$ ",D,dir(D));
dot(" $E$ ",E,dir(E));
dot(" $F$ ",F,dir(F));
dot(" $G$ ",G,dir(G));
dot(" $H$ ",H,dir(H));
dot(" $T$ ",T,dir(-90));
dot(" $M$ ",M,dir(M));
dot(" $N$ ",N,dir(N));
draw(A--G--D--H--A,red);
draw(E--B^^F--C,blue);
draw(N--F--E--N^^M--B--C--M,brown);
draw(N--M,dotted);
[/asy]
Here's another different proof:**<span style="color:#f00">Claim 1:</span>** $\angle NFE + \angle MBC = \angle FTE$ .
*Proof:* Here by $\widehat{XY}$ we will mean angle subtended by minor arc $XY$ on circumference.
\begin{align*}
\angle FHE + \angle BGC &= \angle AHD + \angle AGD
&= 180^\circ - (\angle DAH + \angle ADH ) + 180^\circ - (\angle DAG + \angle ADG)
&= 180^\circ - (\widehat{DF} + \widehat{EA}) + 180^\circ - (\widehat{DB} + \widehat{AC})
&= 360^\circ - (180^\circ + \widehat{FE} + \widehat{BC})
&= 180^\circ - (\angle TBF + \angle TFB)
&= 180^\circ - \angle FTE
\end{align*}
Since $\angle NFE = 90^\circ - \angle FHE$ and $\angle MBC = 90^\circ - \angle BGC$ , so our Claim follows. $\square$ Now we can basically ignore points $A,D,H,G$ and only keep in mind that $N,M$ lie on perpendicular bisectors of segments $FE,BC$ (respectively) and our **Claim 1** is satisfied.
Let $K=FN \cap BM$ , $X=OM \cap FC$ , $Y = ON \cap BE$ and $O$ be circumcenter of $FBCE$ . Observe $X,Y \in \odot(BOF)$ as $$ \angle FYB = \angle FOB= \angle FXB = \widehat{FB} $$
[asy]
size(200);
pair F=dir(110),B=dir(70),C=dir(0),E=dir(-150),O=(0,0),T=extension(E,B,C,F),N=1.2*(E+F),M=extension(O,1/2*(B+C),N,T),K=extension(N,F,M,B),X=extension(O,M,F,C),Y=extension(O,N,E,B);
draw(unitcircle);
dot(" $E$ ",E,dir(E));
dot(" $F$ ",F,dir(F));
dot(" $B$ ",B,dir(B));
dot(" $C$ ",C,dir(C));
dot(" $O$ ",O,dir(-90));
dot(" $N$ ",N,dir(N));
dot(" $M$ ",M,dir(M));
dot(" $T$ ",T,dir(T));
dot(" $K$ ",K,dir(K));
dot(" $X$ ",X,dir(-90));
dot(" $Y$ ",Y,dir(-90));
draw(E--F--C--B--E^^F--Y^^B--X,red);
draw(B--O--N--K--M--O--F,brown);
draw(circumcircle(F,O,B),blue);
draw(N--M,dotted);
[/asy]
<span style="color:#f00">**Claim 2:**</span> $K \in \odot(BOF)$ .
*Proof:* This is just angle chase combined with **Claim 1**,
\begin{align*}
\angle FKB &= \angle NFT + \angle MBT - \angle FTB
&= (\angle NFE + \angle EFT) + (\angle MBC + \angle CBT) - \angle FTB
&= (\angle NFE + \angle MBC) + (\angle EFT + \angle CBT) - \angle FTB
&= \angle ETF + 2 \angle EFT - \angle FTB
&= \angle ETF + \angle EFT - \angle FET
&= 180^\circ - 2 \angle FET
&= 180^\circ - \angle FOB
\end{align*}
proving our Claim. $\square$ By Pascal on $FKBYOX$ w.r.t. $\odot(FOB)$ , we obtain points $N,T,M$ are collinear, as desired. $\blacksquare$ **Remark:** Angle chase shows that this problem is equivalent to [Polish Mathematical Olympiad finals 2021 P5](https://artofproblemsolving.com/community/c6h2537408p21604143) | [
"<details><summary>Solution</summary>Let $S$ be the point of intersection of $BE$ and $CF$ . We need to prove that $M,N,S$ are collinear.\nBy Pascal on $AFCDEF$ , we have that $G,H,S$ are collinear.\nNow an easy angle chase yields\n\\[\\angle AHN=90^\\circ-\\angle HEF=90^\\circ-\\angle DAF\\]\nso that $M... | [
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Let $p$ be a prime, $A$ is an infinite set of integers. Prove that there is a subset $B$ of $A$ with $2p-2$ elements, such that the arithmetic mean of any pairwise distinct $p$ elements in $B$ does not belong to $A$ . | Nice problem!
We will denote by $\mathbb{N}$ the set of positive integers. In which follows, we will assume for simplicity that $A \cap \mathbb{N}$ is infinite, since the same argues would work anyways. It is sufficient to find a subset $B \subseteq { (A\cap\mathbb{N}) }$ and we will replace $A$ by $A\cap \mathbb{N}$ .
Suppose for contradiction problem statement is false, that is, among any $2p-2$ elements in $A,$ exists $p$ with arithmetic mean in $A.$ **<span style="color:#f00">Claim 1:</span> For all $\alpha \in \mathbb{N}$ exists some constant $N_{\alpha} \in \mathbb{N}$ and some remainder $0<k_{\alpha} < p^{\alpha}$ such that $a \equiv k_{\alpha} \pmod {p^{\alpha}}, \forall a \in A, a\ge N_{\alpha}.$ Furthermore, the $N_{\alpha}'s$ are increaisng and $k_{\alpha+1} \equiv k_{\alpha} \pmod {p^{\alpha}}.$**
<details><summary>proof</summary>We proceed by induction.
First, denote by $T_a$ the set $\{ n \in A : n \equiv a \pmod {p} \}$ for all $a=0,\dots, p-1.$ If two different sets between those be infinite, say $T_a$ and $T_b$ then we can choose $2p-2$ distinct numbers $a_1,a_2, \dots a_{2p-1}$ such that $$ a_1 \equiv a_2 \equiv \dots \equiv a_{p-1} \equiv a \pmod {p} $$ $$ a_{p} \equiv a_{p+1} \equiv \dots \equiv a_{2p-2} \equiv b \pmod {p} $$ Then , modulo $p,$ every sum of $p$ terms in this sets will be equivalent to $$ ka+(p-k)b \equiv k(a-b) $$ where $1\le k \le p-1,$ since each residue appears $p-1$ times. Hence these sum can never be a multiple of $p$ (beacause both $k$ and $a-b$ are nonzero modulo $p$ ) and hence the arithmetic mean can never be an integer and cannot be in $A.$ Then, only one of the sets $T_i$ can be infinite and we can find $N_1.$ Now for the inductive step, we only need to consider consider the new set $A' = \left \{ \frac{a-k_{\alpha}}{p} : a \in A, a > N_\alpha \right \}.$ Clearly, the average of any $p$ elements in $A\cap (N_{\alpha}, \infty)$ is greater than $N_{\alpha}$ and equal to the sum of $p$ corresponding elements in $A'$ plus $k_\alpha.$ Then, for this average be on $A,$ it is necessary that the sum of the $p$ terms in $A'$ be a multiple of $p.$ Repeating the argument, if two of the sets $T'_a$ be infinite (these sets are defined the same way, but replacing $A$ by $A'$ ), we will find a contradiction. Then the inductive step is complete. $\square.$</details>
Now we will assume problem statement is false and get a contradiciton.
Pick from $A,$ positve integers $l_{(1,1)}<l_{(1,2)}\dots <l_{(1,p-1)} < l_{(2,1)} < \dots < l_{(p,p-1)}~(*)$ and consider the sets $L_a := \{l_{(a,i)} : 1\le i \le p-1 \}$ for every $1\le a \le p.$ Now, choose some number $\alpha$ sufficiently large, and consider an set $M \subseteq A$ with $p-1$ elements and such that $M$ does not contain one of the numbers selected above and $m> pN_{\alpha}, \forall m \in M ~(**).$ Consider for every $1 \le a \le p,$ the set $L_a \cup M.$ Then, if we assume for contradiction that problem statement is false, there will be two sets $L'_a \subseteq L_a$ and $M' \subseteq M,$ such that $|L'_a| + |M'| = p$ and $$ \frac{\sum_{i \in L'_a} i + \sum_{j \in M'} j}{p} \in A. $$
Note that $|L'_a|,|M'| \ge 1,$ beacause both sets has at most $p-1$ terms and the some of its cardinalities is $p.$ Then by $(**),$ it follows that the number above is at least $N_{\alpha}$ . Then, $$ \sum_{i \in L'_a} i + \sum_{j \in M'} j \equiv pk_{\alpha} \pmod {p^{\alpha +1}} $$ Again by $(*),$ it follows that the elements of $M$ are all equivalent to $ k_{\alpha} \pmod {p^{\alpha}}.$ Hence, looking at the above equivalence modulo $p^{\alpha},$ we will have, $$ \sum_{i \in L'_a} i \equiv (p-|M|)k_{\alpha} \pmod {p^{\alpha}}. $$ That is, every set $L_a$ has a non-empty subset $L'_a$ whose some of the elements is equivalent to $r_ak_{\alpha} \pmod {p^{\alpha}}$ for some $1\le r_a \le p-1.$ But since there are $p$ of these sets, it follows by pigeonhole that $$ \sum_{i \in L'_a} i \equiv \sum_{i \in L'_b} i \pmod {p^{\alpha}} $$ for some $a\ne b.$ But, by the choice of the numbers (we have marked this as $(*)$ ) we will have by an huge bound $$ 0< \left|\sum_{i \in L'_a} i - \sum_{i \in L'_b} i\right| < l_{(1,1)}+l_{(1,2)}\dots +l_{(1,p-1)} + l_{(2,1)} + \dots + l_{(p,p-1)} $$ which is a contradiction since $\alpha$ is any positive integer, and we could choose $\alpha$ such that $p^{\alpha}$ exceeds the right hand side of above. So we are done. | [
"Here\nhttps://artofproblemsolving.com/community/u800085h2807854p24762139",
"After given a hint from my classmate, I got a solution as follows.\n<details><summary>Click to expand</summary>Let $S_{i}( \\alpha )= \\{x \\in A : x \\equiv i( \\bmod p^\\alpha)\\}$ .If there exist $i,j$ and $ \\alpha $ such that ... | [
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"answer_score": 294,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808541.json"
} |
Let $a, b, c, p, q, r$ be positive integers with $p, q, r \ge 2$ . Denote
\[Q=\{(x, y, z)\in \mathbb{Z}^3 : 0 \le x \le a, 0 \le y \le b , 0 \le z \le c \}. \]
Initially, some pieces are put on the each point in $Q$ , with a total of $M$ pieces. Then, one can perform the following three types of operations repeatedly:
(1) Remove $p$ pieces on $(x, y, z)$ and place a piece on $(x-1, y, z)$ ;
(2) Remove $q$ pieces on $(x, y, z)$ and place a piece on $(x, y-1, z)$ ;
(3) Remove $r$ pieces on $(x, y, z)$ and place a piece on $(x, y, z-1)$ .
Find the smallest positive integer $M$ such that one can always perform a sequence of operations, making a piece placed on $(0,0,0)$ , no matter how the pieces are distributed initially. | Note: This seemingly easy problem is actually very hard. This is only a partial solution. I have yet to work out other details.
<details><summary>Solution if at least one of pqr is large...</summary>The answer is $p^aq^br^c$ . Let $f(x,y,z)$ be the number of pieces on the lattice point $(x,y,z)$ . For sets $X,Y,Z$ , let $X\times Y\times Z = \{ (m,n,t) | m\in X, n\in Y, t\in Z\}$ I first show $M\ge p^aq^br^c$ . Let $$ N=\sum\limits_{0\le x\le a, 0\le y\le b, 0\le z\le c} f(x,y,z)p^{-x}q^{-y}r^{-z} $$ Note $N$ is an invariant. In the end, $f(0,0,0)\ge 1$ . Therefore, if $M<p^aq^br^c$ , setting $f(a,b,c)=M$ gives $N<1$ , contradiction.
Now I show $p^aq^br^c$ always works. We proceed by strong induction, and we will use the inductive hypothesis very strongly. The key intuition is that almost all the pieces must be on $(a,b,c)$ while there must be a piece on $x=0$ . Combining these two gives a way to place a cell in $(0,0,0)$ .
Case 1: there are no pieces with $x=0$ . We group the $p^aq^br^c$ pieces into $p$ random groups of $p^{a-1}q^br^c$ pieces. Each move uses pieces in the same group only. By inductive hypothesis on $(a-1,b,c)$ on $\{1,\cdots,a\} \times \{0,\cdots,b\} \times \{0,\cdots, c\}$ , each group of $p^{a-1}q^br^c$ piece can result in one piece on $(1,0,0)$ . Now there are $p$ pieces on $(1,0,0)$ , we can use these to get one piece on $(0,0,0)$ .
Case 2: There is a piece with $x=0$ , a piece with $y=0$ and a piece with $z=0$ .
Claim: There are at most $(b+1)(c+1)-2$ pieces in $\{0,\cdots,a-1\} \times \{0,\cdots,b\} \times \{0,\cdots,c\}$ .
Proof: We can move $\lfloor \frac{f(a,n,t)}{p} \rfloor$ pieces from $(a,n,t)$ to $(a-1,n,t)$ . After transferring all cells from $(a,n,t)$ to $(a-1,n,t)$ for $0\le n\le b, 0\le t\le c$ .
Let $A$ be the number of pieces in $\{0,\cdots,a-1\} \times \{0,\cdots,b\} \times \{0,\cdots,c\}$ . If $A+\sum\limits_{0\le n\le b, 0\le t\le c} \lfloor \frac{f(a,n,t)}{p} \rfloor \ge p^{a-1}q^br^c$ , by inductive hypothesis on $(a-1,b,c)$ , we can place a cell on $(0,0,0)$ . Therefore, it suffices to handle the case where $A+\sum\limits_{0\le n\le b, 0\le t\le c} \lfloor \frac{f(a,n,t)}{p} \rfloor \le p^{a-1}q^br^c-1$ (1)
We also have $A+\sum\limits_{0\le n\le b, 0\le t\le c} f(a,n,t) =p^aq^br^c$ (2) because there are a total of $p^aq^br^c$ pieces on the board. $p(1)-(2)$ gives $(p-1)A - \sum\limits_{0\le n\le b, 0\le t\le c} R(f(a,n,t),p) \le -p$ It follows that $(p-1)A - \sum\limits_{0\le n\le b, 0\le t\le c} (p-1) \le -p$ so $A\le (b+1)(c+1)-2$ .
Using similar arguments, we can see that the number of stones in $\{0,\cdots,a\} \times \{1,\cdots,b\} \times \{0,\cdots,c\}$ is at most $(a+1)(c+1)-2$ and the number of stones in $\{0,\cdots,a\} \times \{0,\cdots,b\} \times \{1,\cdots,c\}$ is at most $(a+1)(b+1)-2$ . This means the number of stones not in $(a,b,c)$ is at most $(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)-6$ , so the number of stones in $(a,b,c)$ is at least $p^aq^br^c -(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1) + 6$ .
Let $X=(0,k,l)$ be a square with at least one cell. If I can transfer $q^kr^l-1$ stones to $X$ , I win by inductive hypothesis on $a=0, b=k, c=l$ . To transfer $q^kr^l-1$ stones to $X$ , I need $(q^kr^l-1)(p^aq^{b-k}r^{c-l}) = p^aq^br^c - p^aq^{b-k}r^{c-l} \le p^aq^br^c-p^a$ . Therefore, if $p^a\ge (a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)-6$ I am done. Similarly, if $q^b, r^c\ge (a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)-6$ I am also done.
It remains to handle small cases?</details> | [
"Very nice problem!\nThere is my solution,i hope that there are no mistakes in it.\n<details><summary>my solution</summary>Answer is $p^{a}q^{b}r^{c}$ ,to the lowerbound place $p^{a}q^{b}r^{c}-1$ pieces in (a;b;c).\nNow we will proof that if we have $p^{a}q^{b}r^{c}$ pieces we can make a piece placed in (0;0;0... | [
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"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808544.json"
} |
Find all pairs of positive integers $(m, n)$ , such that in a $m \times n$ table (with $m+1$ horizontal lines and $n+1$ vertical lines), a diagonal can be drawn in some unit squares (some unit squares may have no diagonals drawn, but two diagonals cannot be both drawn in a unit square), so that the obtained graph has an Eulerian cycle. | The answer is $m=n$ only. Making the degree of all vertices even is the equivalent and sufficient condition to create an eulerian path.
Construction: let S be the set of points on an edge but not a corner. Connect each pair with a line of slope 1.
Proof of optimality: we prove a critical claim: let S be the set of points on an edge but not a corner. Then there exists a matching $(a_1, b_1), \cdots, (a_k,b_k)$ in S such that $S=\{ a_1, a_2, \cdots, a_k, b_1, b_2, \cdots, b_k\}$ such that there is a path from $a_i$ to $b_i$ with only new edges.
Proof: focus on $a_i$ . If I draw a diagonal containing $a_i$ it must affect the parity of degree of another vertex call $c$ . Repeat the argument until $c\in S$ , in which case $a_i$ is paired with $c$ . If it goes into an already paired vertex the degree of that vertex will be forever odd.
The paths connecting $a_i,b_i$ and $a_j,b_j$ must not be in the same square. Therefore, the only way they can intersect (geometrically) is in a 2x2 square, a path goes from top left corner to center to bottom right corner, and another goes from bottom left to center to top right. We can "adjust" the paths such that the two paths do not cross each other i.e. the edges of one side are always to one side of the other path.
Now, it may be intuitive to think of $S$ as points of a circle, where two edges do not cross each other (i.e. one edge stays on one side of any other edge)
We can color all vertices black and white with the checkerboard coloring. Note a vertex can only be paired of another of the same color. If a vertex connects to a vertex on the same row or on the opposite row, we can show one arc it cuts into has an odd number of points if $m\ne n$ , contradicting continuity! | [
"<details><summary>My sol (BEWARE: AWFUL WORDING AND NO LATEX)</summary>CLAIM: m = n are the only integers such it is possible for the table to have an Eulerian cycle\nIt is well known that it is equivalent for every point to have an even degree for it to have an Eulerian cycle. \nAt m = n, we have a construction (... | [
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"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2811451.json"
} |
Given a non-right triangle $ABC$ with $BC>AC>AB$ . Two points $P_1 \neq P_2$ on the plane satisfy that, for $i=1,2$ , if $AP_i, BP_i$ and $CP_i$ intersect the circumcircle of the triangle $ABC$ at $D_i, E_i$ , and $F_i$ , respectively, then $D_iE_i \perp D_iF_i$ and $D_iE_i = D_iF_i \neq 0$ . Let the line $P_1P_2$ intersects the circumcircle of $ABC$ at $Q_1$ and $Q_2$ . The Simson lines of $Q_1$ , $Q_2$ with respect to $ABC$ intersect at $W$ .
Prove that $W$ lies on the nine-point circle of $ABC$ . | Great problem! My solution uses angle chasing to find interesting carectization of the points, and then the solution is quite natural!
*<span style="color:#00f">Key Lemma:</span> Let $K$ be intersection of the tangents from $B$ and $C$ to $(ABC)$ and then let $\omega$ be the circle with center $K$ passing through $B.$ Further, let $\gamma$ be the apolloniam circle of $A$ with respect to the segment $BC.$ Then, points $P_1$ and $P_2$ are the intersections of $\omega$ and $\gamma.$*
<details><summary>proof</summary>This is just angle chasing. First let $X_1,Y_1,Z_1$ be the projections of $P_1$ with respect to $BC,AC,AB,$ respectively. It is well known that $\triangle X_1Y_1Z_1 \sim \triangle D_1E_1F_1$ (I know it as Lemma 2.5 from *Geometry of Conics, A. V. Akopyan, A. A. Zaslavsky;* I hope this is well known; it is just angle chasing anyways ).
[asy]
import graph;
import geometry;
import olympiad;
size(200);
pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L;
A=dir(126.791);
B=dir(200.87);
C=dir(-20.87);
O=(0,0);
M=B*0.5+C*0.5;
Ob=2*B-O;
Oc=2*C-O;
Oba=bisectorpoint(Ob,O);
Oca=bisectorpoint(Oc,O);
K=extension(B,Oba,C,Oca);
Oa=2*A-O;
Oaa=bisectorpoint(O,Oa);
L=extension(A,Oaa,B,C);
path u,p,q,r;
u=Circle(O,1,400);
p=Circle(K,distance(K,B),400);
q=Circle(L,distance(L,A),400);
r=circumcircle(B,O,C);
pair Px=intersectionpoints(p,q)[0];
pair Py=intersectionpoints(p,q)[1];
path l =Px--Py;
pair Qx=intersectionpoints(l,u)[0];
pair Qy=2*O-Qx;
pair Da,Ea,Fa;
Da=foot(Px,B,C);
Ea=foot(Px,A,C);
Fa=foot(Px,A,B);
fill(A--Fa--Px--Ea--cycle, 0.9*white+ 0.1*cyan);
fill(B--Da--Px--Fa--cycle, 0.9*white+0.1*green);
fill(C--Da--Px--Ea--cycle,0.9*white+0.1*magenta);
draw(A--B--C--cycle);
draw(Ea--Da--Fa--cycle);
draw(A--Px--C);
draw(Px--B);
draw(Ea--Px--Fa);
draw(Px--Da);
dot(" $A$ ",A,dir(A));
dot(" $B$ ",B,dir(B));
dot(" $C$ ",C,dir(C));
dot(" $P_1$ ",Px,dir(140));
dot(" $X_1$ ",Da,dir(Da));
dot(" $Y_1$ ",Ea,dir(Ea));
dot(" $Z_1$ ",Fa,dir(Fa));
[/asy]
By problem condition we must have that $$ \measuredangle ABP_1 + \measuredangle P_1CA = \frac{\pi}{2} $$ and $$ \measuredangle CBP_1 + \measuredangle CAP_1 = \measuredangle P_1CB + \measuredangle BAP_1 $$ We can easily manipulate this to get $$ \measuredangle BP_1C= \measuredangle A + \frac{\pi}{2} $$ so it follows easily that $K$ is indeed the circumcenter of $\triangle BP_1C.$ Finally, we have that by law of sines at $(BX_1P_1Z_1)$ and $(CX_1P_1Y_1),$ $$ \frac{P_1X_1}{X_1Y_1}=\frac{P_1X_1}{X_1Z_1} \iff \frac{\sin \angle P_1BC}{\sin \angle ABC} = \frac{\sin \angle P_1CB}{\sin \angle ACB} $$ so it follows that $P_1$ lies on $\gamma$ . $\square.$</details>
[asy]
import graph;
import geometry;
import olympiad;
size(300);
pair A,B,C,O,M,Ob,Oc,Oba,Oca,K,Oa,Oaa,L;
A=dir(126.791);
B=dir(200.87);
C=dir(-20.87);
O=(0,0);
M=B*0.5+C*0.5;
Ob=2*B-O;
Oc=2*C-O;
Oba=bisectorpoint(Ob,O);
Oca=bisectorpoint(Oc,O);
K=extension(B,Oba,C,Oca);
Oa=2*A-O;
Oaa=bisectorpoint(O,Oa);
L=extension(A,Oaa,B,C);
path u,p,q,r;
u=Circle(O,1);
p=Circle(K,distance(K,B));
q=Circle(L,distance(L,A));
r=circumcircle(B,O,C);
pair Px=intersectionpoints(p,q)[0];
pair Py=intersectionpoints(p,q)[1];
path l =Px--Py;
pair Qx=intersectionpoints(l,u)[0];
pair Qy=2*O-Qx;
pair Ota=2*B-Oba;
draw(u,red);
draw(p,green);
draw(q,blue);
draw(A--B--C--cycle);
draw(Px--Py--Qx--Qy--cycle,dotted+magenta);
draw(K--Ota,gray);
draw(K--Oca,gray);
dot(" $A$ ",A,dir(90));
dot(" $B$ ",B,dir(150));
dot(" $C$ ",C,dir(30));
dot(" $K$ ",K,dir(K));
dot(" $O$ ",O,dir(90));
dot(" $P_1$ ",Px,dir(150));
dot(" $P_2$ ",Py,dir(Py));
dot(" $Q_1$ ",Qx,dir(-90));
dot(" $Q_2$ ",Qy,dir(Qy));
[/asy]
By the *Key Lemma,* we must have that line $P_1P_2$ is the radical axes of $\omega$ and $\gamma.$ Hence, since $(ABC)$ is ortogonal to both $\omega$ and $\gamma,$ we must have that $O,$ the circumcenter of $\triangle ABC,$ must lie on $P_1P_2.$
For finish, let $H$ be the orthocenter of $\triangle ABC,$ $U$ be the intersection of the Simson lines of $Q_1$ and $Q_2,$ and $T$ be the reflection of $H$ across $U.$ By above, we must have that $Q_1,Q_2$ are diametrically oppositive and it follows easily from angle chasing that their Simson lines are perpendicular $(i)$ . It is well known that the Simson line of $Q_1$ (respectively $Q_2$ ) bissects segment $HQ_1$ (respectively $HQ_2$ ); so it follows that $TQ_1, TQ_2$ is parallel to the Simson line of $Q_1,Q_2,$ repsectively; then it follows from conclusion $(i)$ that $\angle Q_1TQ_2 = {\pi}{/2}.$ But since points $Q_1$ and $Q_2$ are diametrically oppositive, this implies that $T$ lies on $(ABC)$ and so $U$ must lie on the nine-point circle (beacause the homothety with ratio ${1}{/2}$ centered at $H$ sends $(ABC)$ to the nine-point circle). So done. | [
"This is not hard. Notice that O,P1and P2 are collinear and OP1*OP2=R^2. Then it is easy by orthopole.",
"<blockquote>Interesting but easy problem.\nWe will use the following steps to finish the problem.\n<details><summary>Lemma 1</summary>**Lemma 1.** Let $P$ be a point lie on $(ABC)$ , $l$ be the Simson li... | [
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"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2811452.json"
} |
Let $a_1, a_2, \ldots, a_n$ be $n$ positive integers that are not divisible by each other, i.e. for any $i \neq j$ , $a_i$ is not divisible by $a_j$ . Show that
\[ a_1+a_2+\cdots+a_n \ge 1.1n^2-2n. \]
*Note:* A proof of the inequality when $n$ is sufficient large will be awarded points depending on your results. | Partial solution with Desmos and Justanaccount that proves if the statement is true for all $n<C$ then it must be true for all $n$ by strong induction on $n$ . Because my solution seems a bit sketchy, if there are any mistakes in my solution, please point it out. Thank you!
<details><summary>Solution</summary>Let $A$ be the set of $x_i$ dividing $x_j$ for some $j\ne i$ . By inductive hypothesis, the contribution of elements in A is at least $2\cdot 1.1|A|^2$ because they get double contribution. (i.e. they get at least $1.2|A|^2$ EXTRA contribution).
Let $\lambda=2$ (for now, this will change) and $k$ be the number of $a_i$ 's less than $\lambda n$ . Furthermore, let $x_i=\frac{a_i}{2^{\nu_2(a_i)}}$ . We know $x_i$ is pairwise distinct. Consider a directed graph with vertices $\{1,\cdots,n\}$ such that $i\to j$ iff $x_i\mid x_j$ . We can see that having $\nu_2(a_i)$ be the length of the longest possible directed path from $i$ is necessary (and works).
Let $B=( \{ x_1,\cdots,x_n \} \cap \{1, \cdots , \lambda n\} ) \backslash A$ By considering $\frac{x_i}{3^{\nu_3(x_i)}}$ we can see that there are at most $\frac{\lambda n}{3}$ values of $x_i$ . Therefore, the set of $i$ with outdegree 0 is at most $\frac{\lambda n}{3}$ since they form an anticlique. Thus there are at least $k-\frac{\lambda n}{3}$ elements in $A$ .
To count the total contribution of stuff in $2\sum_{x\in A} x + \sum_{x\in B} x \ge 1.1 |A|^2 + (|A|+|B|)^2 \ge 1.2(k-\frac{\lambda n}{3})^2 + k^2$ where the last step comes from the inductive hypothesis.
For the $n-k$ numbers greater than $\lambda n$ , we can get they are at least $\lambda n + 1, \lambda n +3$ , ..., which is $\lambda n (n-k) + (n-k)^2$ . (We can apply this if we prove that $k<n$ is forced, otherwise we can't induct this way). After this we can change $\lambda$ accordingly.
We can also combine with the bound $1.1k^2+(1-k)(1+\lambda-k)$ , which should work, giving us 1.101.</details>
I may have typoed on desmos... I'll check again tomorrow. | [
"This isn't a complete solution. Its a very stupid bounding idea. It proves the bound $n^2-2n$ . Idk if you can optimise it in the current form.\n\nOrder the numbers in increasing order. Fix $k \\geqslant 2$ . If $a_k \\leqslant 2(k-2)$ then by the Piegon-Hole Principle some two among $a_1, a_2, \\cdots, a_{k-... | [
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"boxed": false,
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"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2811457.json"
} |
Given a positive integer $n$ , find all $n$ -tuples of real number $(x_1,x_2,\ldots,x_n)$ such that
\[ f(x_1,x_2,\cdots,x_n)=\sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \big| k_1x_1+k_2x_2+\cdots+k_nx_n-1 \big| \]
attains its minimum. | Good problem.
We claim that the minimum happens only when $x_k=\frac 1{n+1}$ for all $1\leq k\leq n$ . For the rest of the proof, assume $(x_1,x_2,\ldots,x_n)$ is a triple achieving the minimum.
We get rid of the case $n=1$ , where $f(x)=1+|x-1|+|2x-1|$ . Writing it out as a piece wise function, we easily see that the minimum is at $\frac 12$ . For the rest, we assume $n\geq 2$ .
First, we reduce the problem to having all of the $x_k$ equal. Indeed, we know
\begin{align*}
f(x_1,x_2,\cdots,x_n)&=\sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \big| k_1x_1+k_2x_2+\cdots+k_nx_n-1 \big|
&= \frac 12\sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \left(\big| k_1x_1+k_2x_2+\cdots+k_nx_n-1 \big|+ \big| k_2x_1+k_1x_2+\cdots+k_nx_n-1 \big|\right)
&\geq \sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \left| k_1\cdot\frac{x_1+x_2}2+ k_1\cdot\frac{x_1+x_2}2+k_3x_3\cdots+k_nx_n-1 \right|\quad(1)
&=f\left(\frac{x_1+x_2}2, \frac{x_1+x_2}2,x_3,x_n\right)
\end{align*}
so if $nx=\sum\limits_{k=1}^n x_k$ , then \[f(x_1,x_2,\ldots,x_n)\geq f(x,x,\ldots,x)\tag{2}\]
Now, assuming all of the $x_k$ are equal, this problem is equivalent to minimizing
\[\sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \left| x\sum_{j=1}^n k_j-1 \right|\]
For $0\leq j\leq 2n$ , let $t_j$ be the number of ordered $n$ -tuples $(k_1,\ldots,k_n)\in\{0,1,2\}^n$ with
\[\sum_{m=1}^n k_m=j\]
Note that by definition,
\[\sum t_jx^j=(1+x+x^2)^n\]
Now, to find the minimum, we can note that $f$ is a piece wise function of $x$ , and each part is linear. In addition, the slope is clearly increasing (as a single term $t_j|jx-1|$ goes from $-t_jjx$ to $t_jjx$ ). Thus, we look for where the slope changes from negative to positive (say at $j$ ), and then $x=\frac 1j$ is our minimum.
We claim $j=n+1$ . This requires us to show that
\[\sum_{m=0}^nmt_m<\sum_{m=n+1}^{2n}mt_m\qquad \sum_{m=0}^{n+1}mt_m>\sum_{m=n+2}^{2n}mt_m\tag{3}\]
First, we note $mt_m$ are the coefficients of $x^{m-1}$ in $\frac{\mathrm d}{\mathrm dx}(1+x+x^2)$ , which is $n$ times
\[g_n(x)=(2x+1)(1+x+x^2)^{n-1}\]
We shall prove the following claim:**<span style="color:#ff0000">Claim.</span>** Let $u_{n,j}$ be the coefficient of $x^j$ in $g(x)$ . Then
\[u_{n,j}<u_{n,2n-1-j}\qquad 0\leq j\leq n-1\]
and
\[u_{n,j}>u_{n,2n-j}\qquad 1\leq j\leq n\]**Proof.** We proceed by induction on $n$ :
*Base Case.* $n=2$ . Then $g_2(x)=2+6x+6x^2+4x^3$ which satisfies our claim.
*Induction Hypothesis.* Assume the claim is true for some $n=m\geq 1$ . We show it for $n=m+1$ .
*Induction Step.* We know $g_{m+1}(x)=(1+x+x^2)g_m(x)$ , so
\[u_{m+1,j}=u_{m,j-2}+u_{m,j-1}+u_{m,j}\]
Thus, the first inequality is true as
\begin{align*}
u_{m+1,j}&=u_{m,j-2}+u_{m,j-1}+u_{m,j}
&< u_{m,2m-j-1}+u_{m,2m-j}+u_{m,2m-j+1}=u_{m+1,2(m+1)-j-1}
\end{align*}
which follows by just applying the induction hypothesis and checking bounds to make sure that at least one of the equalities fail. The similar idea works for the second. $\square$ $(3)$ follows by adding our claim until it is satisfied. Thus, we must have equality at $x=\frac 1{n+1}$ .
Now, this means that all the numbers sum to $\frac n{n+1}$ . This means if $x_1\leq x_2\leq\cdots\leq x_n$ , we have $x_1\leq \frac 1{n+1}$ .
First, suppose $n=2$ . Then because $(1)$ needs to hold, we know $x_1+2x_2-1$ and $x_2+2x_1-1$ have the same sign, but they sum to $0$ , so they are $0$ , as desired.
We now write down the chain of equalities in the form of $(2)$ in the form of
\begin{align*}
f(x_1,\ldots,x_n)&\geq f\left(x_1,\ldots,x_{n-2},x,\frac 1{n+1}\right)
&\geq f\left(x_1,\ldots,x_{n-3},x,\frac 1{n+1}, \frac 1{n+1}\right)
&\vdots
&\geq f\left(x_1,x,\frac 1{n+1}, \ldots,\frac 1{n+1}\right)
&\geq f\left(\frac 1{n+1},\ldots,\frac 1{n+1}\right)
\end{align*}
where $x$ is arbitrary to make the average $\frac 1{n+1}$ . We know all of these are equalities, and thus must the last one. However, we know that this means all the inequalities in the form $(1)$ are true, so for all $(k_1,\ldots,k_n)\in\{0,1,2\}^n$ ,
\[k_1x_1+k_2x_2+\cdots+k_nx_n\qquad k_2x_1+k_1x_2+\cdots+k_nx_n\]
have the same sign. Taking $k_2=\cdots k_{n-1}=1$ and $k_1=k_n=2$ gives the numbers $\frac 1{n+1}-x_1,\frac 1{n+1}-x$ , but as these sum to 0, they must be 0. Thus $x_1$ is the average, so all numbers are the average
This completes the proof. | [
"<details><summary>Sketch</summary>The answer is $x_1=\\cdots=x_n=\\frac{1}{n+1}$ only. \n\nClaim: minimum holds when and only when $x_i$ 's are equal.\n\nProof: Let $y=\\frac{x_1+x_2}{2}$ . I will prove that $f(x_1,x_2,x_3,\\cdots,x_n)\\ge f(y,y,x_3,\\cdots,x_n)$ Note for any $k,m,c\\in \\mathbb{R}$ , $|kx_... | [
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Given a positive integer $n$ , let $D$ is the set of positive divisors of $n$ , and let $f: D \to \mathbb{Z}$ be a function. Prove that the following are equivalent:
(a) For any positive divisor $m$ of $n$ ,
\[ n ~\Big|~ \sum_{d|m} f(d) \binom{n/d}{m/d}. \]
(b) For any positive divisor $k$ of $n$ ,
\[ k ~\Big|~ \sum_{d|k} f(d). \] | Nice problem.
We first claim the following:**[color=#ff0000]Claim.** For any positive integers $m$ and $k$ , we have
\[m\left|~\sum_{d\mid m}\mu\left(\frac md\right)\binom{kd-1}{d-1}\right.\]
where $\mu(j)$ is the classic mobs function.**Proof.** Consider any prime $p\mid m$ , and suppose $\nu_p(m)=\alpha\geq 1$ . Note that the only $d$ that show up in the sum are ones with $\nu_p(d)\in\{\alpha,\alpha-1\}$ , as otherwise $p^2\mid m/d$ and thus $\mu(m/d)=0$ . If we let $m=p^\alpha n$ , the above sum is
\[\sum_{d\mid n}\mu\left(\frac nd\right)\left(\binom{kdp^\alpha-1}{dp^\alpha-1}-\binom{kdp^{\alpha-1}-1}{dp^{\alpha-1}-1}\right)\]
Thus it suffices to show
\[\binom{kdp^\alpha-1}{dp^\alpha-1}\equiv\binom{kdp^{\alpha-1}-1}{dp^{\alpha-1}-1}\pmod{p^\alpha}\]
Note that
\begin{align*}
\binom{kdp^\alpha-1}{dp^\alpha-1}&=\prod_{k=1}^{dp^\alpha-1}\frac{kdp^\alpha-1-k}{dp^\alpha-1-k}
&=\prod_{k=1}^{dp^{\alpha-1}-1}\frac{kdp^\alpha-1-kp}{dp^\alpha-1-kp}\cdot\prod_{k=0}^{dp^{\alpha-1}-1}\prod_{\ell=1}^{p-1}\frac{kdp^\alpha-1-kp-\ell}{dp^\alpha-1-kp-\ell}
&\equiv \prod_{k=1}^{dp^{\alpha-1}-1}\frac{kdp^{\alpha-1}-1-k}{dp^{\alpha-1}-1-k}\cdot\prod_{k=0}^{dp^{\alpha-1}-1}\prod_{\ell=1}^{p-1}1
&=\binom{kdp^{\alpha-1}-1}{dp^{\alpha-1}-1}\pmod{p^\alpha}
\end{align*}
completing the proof. $\square$ Now, we let
\[g(k)=\sum_{d\mid k}f(k)\]
and by the Mobius Inversion Formula, we get
\[f(k)=\sum_{d\mid k}\mu\left(\frac kd\right)g(k)\]
Thus, we know
\begin{align*}
\sum_{d\mid m}f(d)\binom{n/d}{m/d}&=\sum_{d\mid m}\binom{n/d}{m/d}\sum_{k\mid d}\mu\left(\frac dk\right)g(k)
&=\sum_{k\mid m}g(k)\sum_{d\mid m/k}\mu(d)\binom{n/dk}{m/dk}
&=\frac nm\sum_{k\mid m}g(k)\sum_{d\mid m/k}\mu(d)\binom{n/dk-1}{m/dk-1}
\end{align*}
Note that by the claim, the second sum is divisible by $m/k$ . Now, if (b) holds, then literal substitution shows
\[\frac nm\cdot g(k)\cdot \sum_{d\mid m/k}\mu(d)\binom{n/dk-1}{m/dk-1}\equiv 0\pmod n\]
as $k\mid g(k)$ and $m/k$ divides the sum. On the contrary, if (a) holds, assume (b) doesn't and let $m$ be the smallest divisor of $n$ that doesn't satisfy (b). Then, we know
\[n\mid\frac nm\sum_{k\mid m}g(k)\sum_{d\mid m/k}\mu(d)\binom{n/dk-1}{m/dk-1}\]
All other than the last term are divisible by $n$ for the same reason as above, and the last term is
\[\frac nmg(m)\]
Thus, we must have $m\mid g(m)$ , a contradiction.
This completes the proof. | [
"<details><summary>Solution</summary>For convenience, write $g(d)=\\sum_{e \\mid d} f(e)$ so that $f(d)=\\sum_{e \\mid d} \\mu\\left(\\frac{d}{e}\\right) g(e)$ by Möbius Inversion. Then we need to prove that $k \\mid g(k)$ for all $k$ iff\n\\[n \\mid \\sum_{d \\mid m} \\binom{n/d}{m/d} \\sum_{e \\mid d} \\m... | [
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Let $m,n$ be two positive integers with $m \ge n \ge 2022$ . Let $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be $2n$ real numbers. Prove that the numbers of ordered pairs $(i,j) ~(1 \le i,j \le n)$ such that
\[ |a_i+b_j-ij| \le m \]
does not exceed $3n\sqrt{m \log n}$ . | <blockquote>Let $m,n$ be two positive integers with $m \ge n \ge 2022$ . Let $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be $2n$ real numbers. Prove that the numbers of ordered pairs $(i,j) ~(1 \le i,j \le n)$ such that
\[ |a_i+b_j-ij| \le m \]does not exceed $3n\sqrt{m \log n}$ .</blockquote>
Consider the bipartite graph $\mathcal{G}(a_i,b_i)$ , where we join two vertices iff $|a_i + b_j - ij| \leq m$ , Let $E$ denote the total number of edges in the graph, let $\mathcal{F}$ denote the number of walks of length $2$ in the graph ( $a_i\rightarrow b_j \rightarrow a_k$ ), evidently by jensen we have $\mathcal{F}=\sum \binom{d_i}{2} \geq n \binom{\frac{E}{n}}{2}$ .**Claim 1:** There are no 4-cycles $a_i \rightarrow b_j \rightarrow a_k \rightarrow b_l \rightarrow a_i$ such that $(k-i)(j-l) \geq 4m$ .
<details><summary>Proof</summary>Assume FTSOC that there exists a 4-cycle, say $a_i \rightarrow b_j \rightarrow a_k \rightarrow b_l \rightarrow a_i $ so we have the inequalities $$ ij-m \leq a_i + b_j \leq ij+m...(1) $$ $$ il-m \leq a_i + b_l \leq il +m ...(2) $$ $$ kj-m \leq a_k+b_j \leq kj+m ... (3) $$ $$ kl-m \leq a_k + b_l \leq kl+m ...(4) $$ By adding $(1)$ and $(4)$ , we obtain the inequality $$ ij+kl-2m \leq a_i+b_j+a_k + b_l \leq ij+kl+2m $$ and by adding $(2)$ and $(3)$ we obtain $$ il+kj-2m \leq a_i + b_j + a_k +b_l \leq il+kj+2m $$ but these two together yield that $$ 4m \geq (k-i)(j-l) $$</details>
By Claim 1, for any two $a_i,a_j$ they have at most $\frac{4m}{j-k}$ possible common neighbours, summing this up and using jensen yields the desired inequality. $\blacksquare$ | [
"does anyone have any ideas?",
"Consider a bipartite graph $(U,V)$ with vertices $( \\{u_1,\\cdots,u_n\\}, \\{v_1,\\cdots,v_n\\})$ such that there is an edge between $u_i$ and $v_j$ if $|a_i+b_j-ij|\\le m$ .\n\nLet $E$ be the total number of edges, and $X$ be the number of 2-step walks $v_i\\leftrig... | [
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Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$ . Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$ , if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$ , where $A,C,D,B$ lie on $\ell$ in this order, then $\angle APC=\angle BPD$ . | Let $P$ be a point such that $\omega_1,\omega_2$ , and the point circle $P$ are coaxial. Let the radical axis of $\omega_1,\omega_2$ meet $\ell$ at $Q$ . Then the circle centered at $Q$ with radius $\sqrt{QA\cdot QB}=\sqrt{QC\cdot QD}=QP$ is an Appolonian circle with respect to $A,B$ and $C,D$ , so the angle bisectors of $\angle APB$ and $\angle CPD$ meet at the same point on $\ell$ , or are the same, as desired. | [
"Let $P$ be a point such that after inversion wrt $P$ $\\omega_1^*$ and $\\omega_2^*$ are concentric, which is well known to exist (let $O$ be the new common center).\nNow, the angle condition gets simply translated to $\\angle A^*PC^*=\\angle B^*PD^*$ , where $(A^*B^*C^*D^*P)$ is cyclic. But this cycl... | [
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Two positive real numbers $\alpha, \beta$ satisfies that for any positive integers $k_1,k_2$ , it holds that $\lfloor k_1 \alpha \rfloor \neq \lfloor k_2 \beta \rfloor$ , where $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$ . Prove that there exist positive integers $m_1,m_2$ such that $\frac{m_1}{\alpha}+\frac{m_2}{\beta}=1$ . | Okay here is a complete solution. Thanks to Justanaccount for helping out!
Assume $\alpha,\beta \notin \mathbb{Q}$ Note the number of $k_1$ satisfying $\lfloor k_1 \alpha \rfloor < X$ is precisely $\lfloor \frac{X}{\alpha} \rfloor$ .
Let $a=\frac{1}{\alpha}$ and $b=\frac{1}{\beta}$ . From $\lfloor k_1\alpha \rfloor \ne \lfloor k_2\beta\rfloor$ we can see if $\lfloor Xa \rfloor > \lfloor (X-1)a \rfloor$ then $\lfloor Xb \rfloor = \lfloor (X-1)b \rfloor$ , for otherwise there exists a solution $(k_1,k_2)$ to $\lfloor k_1\alpha \rfloor = \lfloor k_2\beta \rfloor = X$ In other words, if $\{xa\} > 1-a$ then $\{xb\} < 1-b$ .**Claim**
There exists integers $m,n,k$ such that $ma+nb=k$ .**Proof of Claim**
We first prove there exists $n$ such that $||na||, ||nb||<\epsilon$ where $||na||=\min\{1-\{na\}, \{na\} \}$ .
We can easily find $||na||<\epsilon^3$ , and we note $||na||, ||2na||, \cdots, ||\lceil \frac{1}{\epsilon} \rceil na||$ are all smaller than $\epsilon$ . We can prove with pigeonhole principle at least one of $||tnb||$ where $1\le t\le \lceil \frac{1}{\epsilon} \rceil$ works.
Suppose $\{na\},\{nb\} < \epsilon$ . Let $x=\{na\}, y=\{nb\}$ . Assume for the sake of contradiction $\frac xy$ is not rational.
If there exists integers $M,A,B$ such that $Mx\in (A,A+a)$ and $My\in (B,B+b)$ , then I am done, because $\{Mna\}=\{(Mn-1)a\} + 1$ and $\{Mnb\} = \{(Mn-1)b\} + 1$ .
We take $M=\lfloor \frac{A+a}{x} \rfloor$ . This guarantees $x(\frac{A+a}{x}-1)=A+a-x \le Mx \le x\frac{A+a}{x} = A+a$ . Since $a>x$ , $Mx\in (A,A+a)$ We have $My \in ((A+a)\frac yx - y, (A+a)\frac yx)$ . We need to show for some $A$ , there exists an integer $B$ such that $(A+a)\frac yx - y \ge B$ and $(A+a)\frac yx \le B+b$ . If $\frac yx$ is irrational, there exists $A$ such that $y\le \{(A+a)\frac yx\} \le b$ . Therefore, $\frac yx$ must be rational.
Now, let $na=C+x, nb=D+y$ then $x=qy$ for some $q\in \mathbb{Q}$ . Therefore, $\frac{na-C}{nb-D}=q \rightarrow (na-C)=q(nb-D)$
Let $q=\frac uv$ where $u,v\in \mathbb{Z}$ then $v(na-C)=u(nb-D) \rightarrow (vn)a-(un)b=vC-uD$ **Back to the problem** if $m,n$ are of opposite signs, say $ma\equiv nb(\bmod\; 1)$ , then pick $||tb||=\epsilon$ .
If $\{tb\}=\epsilon < \frac{ab}{mn}$ then $\{tmb\}=m\epsilon < \frac{ab}{n} < b$ and $\{tma\} = \{tnb\} < \frac{ab}{m} < a$ . It follows that $\{tma\} < \{(tm-1)a\}$ and $\{tmb\} < \{(tm-1)b\}$ , contradiction.
If $\{tb\}=1-\epsilon>1-\frac{ab}{mn}$ then $\{tmb\}=1-m\epsilon > 1- \frac{ab}{n} > 1-b$ and $\{tma\} = \{tnb\} > 1- \frac{ab}{m} > 1- a$ . It follows that $\{tma\} > \{(tm+1)a\}$ and $\{tmb\} > \{(tm+1)b\}$ , contradiction.
Therefore, $m,n$ are of the same sign. WLOG $m,n,k$ are all positive and $k$ is minimal.
Assume for the sake of contradiction $k\ge 2$ . Thus, either $ma\ge 1$ or $nb\ge 1$ . WLOG $nb\ge 1$ , so it suffices to take $\{Xa\}<a$ and $\{Xb\} < \frac 1n < b$ WLOG $\gcd(k,m,n)=1$ . Let $\{Xa\}=Z+\epsilon$ , then I want $Xk-Zm\equiv 1(\bmod\; n)$ , which wins because $\{\frac{X(k-ma)}{n} \} = \{ \frac{Xk}{n} - Xa \frac mn\} = \{ \frac{Xk-(Z+\epsilon)m}{n} \}$ But this is doable by density; $(X, Xa (\bmod\; n))$ are dense in $\mathbb{Z}_n \times [0,n)$ so we can clearly take $Xk\equiv Zm+1 (\bmod\; n)$ and $\{Xa\} < \epsilon$ for any $\epsilon > 0$ (another way to think about it is that $\{\frac{Xa}{n}\}$ is dense in $[0,1)$ , and fixing $X$ mod $n$ preserves the density ) | [
"UPD: See solns at PM #4, #7\n\nSome ideas: assume $\\alpha,\\beta \\notin \\mathbb{Q}$ Note the number of $k_1$ satisfying $\\lfloor k_1 \\alpha \\rfloor < X$ is precisely $\\lfloor \\frac{X}{\\alpha} \\rfloor$ .\n\nLet $a=\\frac{1}{\\alpha}$ and $b=\\frac{1}{\\beta}$ . From $\\lfloor k_1\\alpha \\rfloor... | [
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Given a positive integer $n \ge 2$ . Find all $n$ -tuples of positive integers $(a_1,a_2,\ldots,a_n)$ , such that $1<a_1 \le a_2 \le a_3 \le \cdots \le a_n$ , $a_1$ is odd, and
(1) $M=\frac{1}{2^n}(a_1-1)a_2 a_3 \cdots a_n$ is a positive integer;
(2) One can pick $n$ -tuples of integers $(k_{i,1},k_{i,2},\ldots,k_{i,n})$ for $i=1,2,\ldots,M$ such that for any $1 \le i_1 <i_2 \le M$ , there exists $j \in \{1,2,\ldots,n\}$ such that $k_{i_1,j}-k_{i_2,j} \not\equiv 0, \pm 1 \pmod{a_j}$ . | <details><summary>Easy for P3, easiest problem on T3 for me</summary>Let $a$ be the number of odd terms in $a_i$ . The answer is $2^a|a_1-1$ only.
Call points $(x_1, \cdots, x_n)$ .
Claim: The maximum number of points that can be placed in $\times_{j=1}^n \mathbb{Z}_{a_j}$ is at most $\frac{1}{2^n} (a_1-1) \prod\limits_{j=2}^n a_j$ Proof: we induct on $n$ .
Base Case: $n=1$ , clear
Inductive Step: For $i=0,\cdots, a_n-1$ , let $X_i$ denote the union of planes $x_n=2i, x_n=2i+1$ . By inductive hypothesis, we can put at most $2^{1-n} (a_1-1)a_2 \cdots a_{n-1}$ points in $X_i$ (since which row you put don't matter). Summing $X_i$ over $0\le i\le n-1$ counts everything twice, and dividing by 2 yields the desired result.
Therefore this problem is about characterising all equality cases.
Claim: it suffices to solve the problem when $a_i$ is odd for all $i$ .
Proof: if $a_i$ is even then we can split into $X_0, X_1, \cdots, X_{a_i/2-1}$ and use induction to complete our construction. Alternatively, if $(a_1,\cdots,a_{i-1}, a_{i+1},\cdots, a_n)$ fails, then $(a_1,\cdots,a_n)$ fails by mimicking the inductive argument.
At this point M must be an integer since it is an upper bound, so $2^n|a_1-1$ is necessary.
Now I claim this is also sufficient. We first solve the problem for $a_1=\cdots=a_n=k2^n+1$ . On the line $x_2=c_2, x_3=c_3, \cdots, x_n=c_n$ for fixed $c_2,\cdots,c_n$ place points with $x_n\in \{0,2,\cdots, 2(k-1)\} + 2k\sum\limits_{j=1}^{n-1} 2^{j-1}c_{j+1}$ .
We will prove there is no conflict between $x_2=c_2, \cdots, x_{n}=c_{n}$ and $x_2=c_1+e_2, \cdots, x_{n-1}=c_{n-1}+e_{n}$ for $e_j\in \{-1,0,1\}$ for $2\le j\le n$ and $e_j$ are not all zero.
Note the set of points on $x_2=c_2+e_1, \cdots, x_{n}=c_{n}+e_{n-1}$ is precisely the set of points on $x_2=c_2, \cdots, x_{n}=c_{n}$ shifted by $2k \sum\limits_{j=1}^{n-1} 2^{j-1}e_{j+1}$ . Note $0<|\sum\limits_{j=1}^{n-1} 2^{j-1}e_{j+1}|\le 2^{n-1}-1$ . For simplicity let $Y=\sum\limits_{j=1}^{n-1} 2^{j-1}e_{j+1}$ Let $[X,X+2k-2]$ be the interval of points on the line $x_2=c_2, \cdots, x_{n}=c_{n}$ , then $[X+2kY, X+2kY+2k-2]$ is the interval of points on the line $x_2=c_2, \cdots, x_n=c_n$ . WLOG $Y$ is positive, and we can easily check there is no conflict when $1\le Y\le 2^{n-1}-1$ by only checking $Y=1, Y=2^{n-1}-1$ .
Now we extend to $a_2,\cdots,a_n$ being odd integers at least $a_1$ . To go from $(a_1,\cdots,a_{i-1}, a_i, a_{i+1}, \cdots,a_n)$ to $(a_1,\cdots,a_{i-1}, a_i+2, a_{i+1}, \cdots,a_n)$ , we place the same set of points on $x_i=a_i-1$ on the plane $x_i=a_i+1$ and the same set of points on $x_i=a_i$ to $x_i=a_i+2$ . This way, there is no conflict with a fixed $x_i$ , no conflict between planes $x_i=a_i-1, x_i=a_i$ implies no conflict between planes $x_i=a_i, x_i=a_i+1$ and no conflicts between $x_i=a_i+1, x_i=a_i+2$ , and no conflict between planes $x_i=1, x_i=a_i$ implies no conflict between planes $x_i=1, x_i=a_i+2$ , completing the extension step.
At each step, for a fixed value of $x_2,\cdots,x_n$ there are exactly $k=\frac{a_1-1}{2^n}$ points, so this construction works.</details> | [] | [
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Find all positive integer $k$ such that one can find a number of triangles in the Cartesian plane, the centroid of each triangle is a lattice point, the union of these triangles is a square of side length $k$ (the sides of the square are not necessarily parallel to the axis, the vertices of the square are not necessarily lattice points), and the intersection of any two triangles is an empty-set, a common point or a common edge. | Claim: the only such numbers are the multiples of $3$ .
If $k=3t$ pick an axis and grid aligned $k\times k$ square divided in $t\times t$ square each of side $3\times 3$ , and divide these in two triangles. In this way it is straight forward enough to see that the barycenters must have integer coordinates.
Lemma: given triangles with integer centroids (calls such triangles good) $ABC$ and $A'BC$ , the vertices $A$ and $A'$ differ by three times an integer vector
Proof: $A'-A=3(\frac{A'+B+C}{3}-\frac{A+B+C}{3})$ . Since the two centroids have integer coordinates, their differences is an integer vector, so we are done.
Now, using the lemma, divide the vertices of the triangulation in ( $\leq$ ) $3$ equivalence classes where the equivalence relation is differing by three times an integer vector [from any vertex we can reach a vertex of a fixed triangle, by multiple steps of the form $A\to A'$ if there is a segment $BC$ such that $ABC,A'BC$ both belong to the triangulation].
Since the square has $4>3$ vertices, at least two will be in the same of the three equivalence classes. Let $\Delta x$ be the difference between the $x$ coordinates of these two vertices, and $\Delta y$ that of $y$ coordinates. If the two vertices are opposite wrt the square we have $\Delta x^2+\Delta y^2=2k^2$ . Since the two vertices are equivalent, $3|\Delta x,\Delta y\implies 3|\Delta x^2+\Delta y^2=2k^2\implies 3|k$ . Similarly if the two vertices are adjacent wrt the square, $3|\Delta x^2+\Delta y^2=k^2\implies 3|k$ . In both cases, $k$ is divisible by $3$ , so we are done. | [
"easy for China TST\n\n<details><summary>solution</summary>The answer is all $3|k$ .\n\nConstruction: \nIf $3|k$ then take two triangles. One with $(0, 0), (0, k), (k, 0)$ and one with $(k, k), (0, k), (k, 0)$ .\nTheir centroids are both lattice points.\n\nProof that others won't work:\nWe would prove that in... | [
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Show that there exist constants $c$ and $\alpha > \frac{1}{2}$ , such that for any positive integer $n$ , there is a subset $A$ of $\{1,2,\ldots,n\}$ with cardinality $|A| \ge c \cdot n^\alpha$ , and for any $x,y \in A$ with $x \neq y$ , the difference $x-y$ is not a perfect square. | <blockquote>The main idea is **Trivial solutions --> Generalize**.
I'll present two similar approaches.
<details><summary>Solution 1)</summary>Denote $A$ as the set of numbers which contain only digits of $1,4,7,9,12$ under hexadecimal expression.
Obviously, we have chosen at least $0.0001n^{\log_{16}5}$ numbers in $[n]$ .
For any two distinct numbers in $A$ , assume in the hexadecimal system, the smallest bit weight is $16^t$ on their different bits. Then their difference can be rewritten as $x\times 16^t$ , where $x\equiv 2,3,5,6,8,10,11,13,14\pmod {16}$ , which satisfies the "not a square number" condition.
Therefore, pick $C=0.0001, \alpha = \log_{16}5$ . We are done. $\blacksquare$ *Solution by Yiran ZHANG, Cheng JIANG, students in China National Team (2022), from Shanghai Middle School*</details>
<details><summary>Solution 2)</summary>Basically this is induction.
We consider $16k, 16k+2, 16k+5, 16k+8, 16k+10, 16k+13$ . Here $1+k\in \{1,2,\cdots, \frac{n}{16}\}$ .
Since $x^2\equiv 0,1,4,9\pmod {16}$ , we know that if $i\not\equiv j\pmod {16}$ , then $i-j$ isn't a square.
On the other hand, if $i\equiv j\pmod {16}$ , then $(16k+c)-(16l+c)$ is square $\Rightarrow$ $k-l$ is square. By inductive hypothesis, we can pick $6\times C\times \left(\frac{n}{16}\right)^\alpha = \frac{6}{16^\alpha}\cdot C\cdot n^{\alpha}$ .
Therefore, we want to pick $16^{\alpha}$ where $\alpha > \frac 12$ . In fact, $\alpha = \log_{16}\frac{401}{100}$ , $C = \frac{1}{10^{2022}}$ gives us the desired. $\blacksquare$</details>
I believe these two are more succinct than any one above. :D</blockquote>
I solve this using also hexademical expression,but i choose $2,5,7,13,15$ | [
"<blockquote>Show that there exist constants $c$ and $\\alpha > \\frac{1}{2}$ , such that for any positive integer $n$ , there is a subset $A$ of $\\{1,2,\\ldots,n\\}$ with cardinality $|A| \\ge c \\cdot n^\\alpha$ , and for any $x,y \\in A$ with $x \\neq y$ , the difference $x-y$ is not a perfect squ... | [
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(1) Prove that, on the complex plane, the area of the convex hull of all complex roots of $z^{20}+63z+22=0$ is greater than $\pi$ .
(2) Let $a_1,a_2,\ldots,a_n$ be complex numbers with sum $1$ , and $k_1<k_2<\cdots<k_n$ be odd positive integers. Let $\omega$ be a complex number with norm at least $1$ . Prove that the equation
\[ a_1 z^{k_1}+a_2 z^{k_2}+\cdots+a_n z^{k_n}=w \]
has at least one complex root with norm at most $3n|\omega|$ .
| I'll give a proof for Gauss Lucas Thm Here.
<blockquote>For any complex polynomial $P(z)$ , the roots of the derivative $P'(z)$ lies in the convex hull of the roots of $P(z).$ </blockquote>
Proof. let $P(z)=\lambda \prod_{k=1}^n(z-z_k)^{\alpha_k},$ then if $P'(Z)=0,$ $$ \frac{P'(Z)}{P(Z)}=\sum_{k=1}^n\frac{\alpha_k}{Z-z_k}=0. $$ $$ \Longrightarrow\sum_{k=1}^n\frac{\alpha_k(\overline Z-\overline{z_k})}{|Z-z_k|^2}=0 $$ $$ \Longrightarrow \sum_{k=1}^n\frac{\alpha_k}{|Z-z_k|^2}Z=\sum_{k=1}^n\frac{\alpha_kz_k}{|Z-z_k|^2} $$ Therefore $Z$ can be written as a convex linear combination of $z_1,\ldots ,z_n,$ done $.\Box$ ------
Part 1. By Gauss-Lucas Thm the convex hull of all complex roots of $z^{20}+63z+22=0$ is subset of the convex hull of all complex roots of $20z^{19}+63=0,$ which is a regular 19-gon with length $(63/20)^{1/19}.$ Therefore the radius of incircle of the 19-gon is $$ \left(\frac{63}{20}\right)^{1/19}\cos\frac{\pi}{19}>1 $$ after some "simple" estimations, so the total area is greater than $\pi.\Box$ ------
Part 2. Let $P(z)= a_1 z^{k_1}+a_2 z^{k_2}+\cdots+a_n z^{k_n}-w.$ To use Gauss-Lucas Thm, we look at
\[Q(z)=R_0(z):=-z^{k_n}P\left(\frac 1z\right)=wz^{k_n}-a_1z^{k_n-k_1}-\cdots -a_{n-1}z^{k_n-k_{n-1}}-a_n.\]
Take derivative and delete trivial zeros $z=0,$ \[Q'(z)=wk_nz^{k_n-1}-a_1(k_n-k_1)z^{k_n-k_1-1}-\cdots -a_{n-1}(k_n-k_{n-1})z^{k_n-k_{n-1}-1},\]
\[R_1(z):=R_0'(z)\cdot z^{-(k_n-k_{n-1}-1)}=wk_nz^{k_{n-1}}-a_1(k_n-k_1)z^{k_{n-1}-k_1-1}-\cdots -a_{n-1}(k_n-k_{n-1}).\]
Keep doing the above operation(taking derivative and delete $z=0$ 's), define
\[R_t(z)=R_{t-1}'(z)\cdot z^{-(k_{n-t+1}-k_{n-t}-1)}=wk_n\cdots k_{n-t+1}z^{k_{n-t}}-a_1(k_n-k_1)\cdots (k_{n-t+1}-k_1)z^{k_{n-t}-k_1-1}-\cdots -a_{n-t}(k_n-k_{n-t})\cdots (k_{n-t+1}-k_{n-t}).\]
By Gauss-Lucas Thm, let $S(P)$ be the set of zeros for polynomial $P,$ then $$ \text{conv}{(S(R_t))}\subseteq\text{conv}{(S(R_{t-1}))}\subseteq\cdots \subseteq\text{conv}{(S(R_{0}))}. $$ By Vieta Thm, the product of $k_{n-t}$ roots for $R_t$ is \(\frac{a_{n-t}(k_n-k_{n-t})\cdots (k_{n-t+1}-k_{n-t})}{wk_n\cdots k_{n-t+1}},\) so maximum norm in $S(R_t),$ say $M(S(R_t)),$ $$ \ge \sqrt[k_{n-t}]{\frac{|a_{n-t}|(k_n-k_{n-t})\cdots (k_{n-t+1}-k_{n-t})}{|w|k_n\cdots k_{n-t+1}}}. $$ Therefore we only need to find a $1\le t\le n$ such that
\[M(S(R_{n-t}))\ge\sqrt[k_{t}]{\frac{|a_{t}|(k_n-k_{t})\cdots (k_{t+1}-k_{t})}{|w|k_n\cdots k_{t+1}}} \ge \frac 1{3n|w|}.\]
\[\iff \sqrt[k_{t}]{\left(1+\frac{k_t}{k_n-k_t}\right)\cdots \left(1+\frac{k_t}{k_{t+1}-k_t}\right)} \le {3n|w|}\sqrt[k_{t}]{\frac{|a_t|}{|w|}}.\]
For all $s<t\le n,$ by Taylor's Series
\[\sqrt[k_{t}]{1+\frac{k_t}{k_s-k_t}}\le 1+\frac{1}{k_s-k_t}\le 1+\frac{1}{s-t}.\]
We only need
\[|a_t|^{1/k_t}\ge\frac 1{3n}\prod_{t<s\le n}\frac{s-t+1}{s-t}=\frac{n-t+1}{3n},\]
which is just trivial since $|a_1|+\cdots +|a_n|\ge |a_1+\cdots +a_n|=1.\Box$ | [
"<details><summary>hint</summary>Use the fact that the roots of $P'(x)$ lies in the convex hall of P(x).\nFor (2),consider $Q_0(z)=P(\\frac{1}{z})$ and recursively define $Q_{i+1}(z)$ be $Q'_{i}(z)$ with all the zero roots deleted.\nEstimate each $a_i$ by the constant factor of $Q_{n-i}(z)$ .</details>",... | [
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Initially, each unit square of an $n \times n$ grid is colored red, yellow or blue. In each round, perform the following operation for every unit square simultaneously:
- For a red square, if there is a yellow square that has a common edge with it, then color it yellow.
- For a yellow square, if there is a blue square that has a common edge with it, then color it blue.
- For a blue square, if there is a red square that has a common edge with it, then color it red.
It is known that after several rounds, all unit squares of this $n \times n$ grid have the same color. Prove that the grid has became monochromatic no later than the end of the $(2n-2)$ -th round. | The crux of this problem is the following claim.**<span style="color:#07f">Claim:</span>** There exists a cell that never changes color.
*Proof:* By the condition, the color of every cell must eventually become constant. Consider a cell $\mathcal{A}$ that first becomes a constant color, and assume WLOG that the color is red. Assume FTSOC that $\mathcal{A}$ was not always red. Then, it must have a red neighbor $\mathcal{B}$ before it changed to red. However, $\mathcal{B}$ cannot become yellow, or $\mathcal{A}$ would become yellow, a contradiction. Thus, the color of $\mathcal{B}$ becomes constant before $\mathcal{A}$ does, contradicting the assumption that $\mathcal{A}$ is the first square to become constant. $\square$ Thus, there exists a cell $\mathcal{A}$ that never changes color. Consider the graph with vertices representing cells and edges connecting orthogonally adjacent cells. Then, the cells adjacent to $\mathcal{A}$ must be red or blue, so they must become red after $1$ round. Similarly, the cells of distance $2$ away from $\mathcal{A}$ must be red or blue after $1$ round, so they must become red after $2$ rounds. We can continue this argument to show that all cells a distance $k$ away from $\mathcal{A}$ must become red after $k$ rounds. Since all cells are of distance at most $2n-2$ from $\mathcal{A}$ , they must all become red in $2n-2$ rounds. $\square$ | [
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Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABC$ and $\triangle ADC$ are $I,J$ , respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$ . The line perpendicular to $BD$ through $P$ intersects with the outer angle bisector of $\angle BAD$ and the outer angle bisector $\angle BCD$ at $E,F$ , respectively. Show that $PE=PF$ . | <blockquote>Solved with **BarisKoyuncu**, **SerdarBozdag** and **sevket12**.
Let $AB \cap CD = X , AD \cap BC = Y , BI \cap DJ = M$ . Applying Desargues on triangles $\triangle AIB$ and $\triangle CJD$ , we obtain that the points $X , M , G = AI \cap CJ$ are collinear. Now note that since $AI$ is the exterior bisector of $\angle XAC$ and $CJ$ is the bisector of $\angle XCA$ , we get that $G$ is the excenter of $\triangle XAC$ . So, since $G,X,M$ are collinear, this means $M$ lies on the bisector of $\angle (AB,CD)$ , so $d(M,AB)=d(M,CD)$ . Since $M$ also lies on $BI$ and $DJ$ , we similarly obtain that $d(M,AB)=d(M,BC)=d(M,CD)=d(M,AD)$ , implying that the quadrilateral $ABCD$ has an excircle with center $M$ .
Now we will show that $\angle ADE = \angle CDF$ . Note that if we show this, then $A,F$ would be isogonal to $\angle EDC$ , and since $E,C$ are obviously isogonal to $\angle EDC$ , by isogonality lemma we would obtain $P , M$ are isogonal wrt $\angle EDC$ , so this would mean $\angle ADP = \angle JDF$ , and so $\angle EDP = \angle FDP$ , and from here we would be done. So it indeed suffices to show $\angle ADE = \angle CDF$ .
Let $H$ and $N$ be the foots from $E$ and $F$ to $DA$ and $DC$ respectively. Since $EHPD$ and $DPFN$ are cyclic, to show $\angle ADE = \angle CDF$ , it suffices to show $H,P,N$ are collinear. By Desargues theorem on $\triangle AEH$ and $\triangle CFN$ , this collinearity is equivalent to the points $ D , M , O = EH \cap FN$ being collinear. Let $S$ be the intersection of the perpendicular from $D$ to $DA$ with $MA$ and let $T$ be the intersection of the perpendicular from $D$ to $DC$ with $MC$ . By Thales, it suffices to show $\frac{MS}{ME} = \frac{MT}{MF}$ which is equivalent to $ST \parallel EF$ , or $BD \perp ST$ . With this, we get rid of points $E,F$ .
Let $K,L$ be the foots from $M$ to $CD,AD$ respectively. Let $Z$ be the foot from $M$ to $BD$ . We will show that $\triangle KZL \sim \triangle SDT$ , and this would imply that $\angle DTS = \angle ZLK = \angle ZDK = 90^\circ - \angle ZDT \implies BD \perp ST$ .
Also note that $\angle KZL = \angle ADK = 90^\circ - \angle KDS = \angle SDT$ , so it suffices to show $\frac{DS}{DT} = \frac{ZK}{ZL}$ .
Now let $\theta = \angle XAM = \angle MAD$ and $\gamma = \angle YCM = \angle MCD$ . We clearly have $DS = DA \cdot \tan{\theta}$ and $DT = DC \cdot \tan{\gamma}$ , so $\frac{DS}{DT} = \frac{DA}{DC} \cdot \frac{\tan{\theta}}{\tan{\gamma}}$ .
By LOS and ratio lemma, we get $\frac{ZK}{ZL} = \frac{\sin{\angle PDC}}{\sin{\angle PDA}} = \frac{PC}{PA} \cdot \frac{DA}{DC}$ , so it suffices to show $\frac{PC}{PA} = \frac{\tan{\theta}}{\tan{\gamma}}$ .
Note that $\frac{PC}{PA} = \frac{\text{area}(BAD)}{\text{area}(BCD)} = \frac{DA \cdot AB \cdot \sin{2\theta}}{DC \cdot CB \cdot \sin{2\gamma}} = \frac{\tan{\theta}}{\tan{\gamma}} \iff DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$ .
By law of cosines in $\triangle ABD$ and $\triangle CBD$ , we have $DA^2 + AB^2 + 2 \cdot DA \cdot AB \cdot \cos{2\theta} = DC^2 + CB^2 + 2 \cdot DC \cdot CB \cdot \cos{2\gamma}$ .
Now using the fact that $\cos{2\theta} = 2\cos^2{\theta} - 1$ , this gives that $(DA - AB)^2 + 4\cdot DA \cdot AB \cdot \cos^2{\theta} = (DC - CB)^2 + 4\cdot DC \cdot CB \cdot \cos^2{\gamma}$ . Since $ABCD$ has an excircle, $DA - AB = DC - CB$ and so we get $DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$ , and we are done. $\square$ </blockquote> $F$ | [
"Is the problem right?\nCould you please check the original problem?\nI think there are some typos somewhere.\n\nHere is a proof that states : $AC,BD,IJ$ can never concurrent.\n\nLet $AC,BD$ intersect at $K$ Let incenter of triangle $ABK, CDK$ be $S,T$ .\nThen it is trivial that $S$ lies on segment $BI$ ... | [
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Find all functions $f: \mathbb R \to \mathbb R$ such that for any $x,y \in \mathbb R$ , the multiset $\{(f(xf(y)+1),f(yf(x)-1)\}$ is identical to the multiset $\{xf(f(y))+1,yf(f(x))-1\}$ .
*Note:* The multiset $\{a,b\}$ is identical to the multiset $\{c,d\}$ if and only if $a=c,b=d$ or $a=d,b=c$ . | I might have written a bit complicated :blush:
\[\{f(xf(y)+1),f(yf(x)-1)\}=\{xf(f(y))+1,yf(f(x))-1\}\]
Answers are $f(x)=x$ and $f(x)=-x$ which clearly fit. Let $P(x,y)$ be the assertion. Note that $P(0,0)$ gives $\{f(1),f(-1)\}=\{1,-1\}$ . If $f(1)=1$ , then $P(1,1)$ and $P(-1,1)$ imply $f(0)=0$ . If $f(1)=-1$ , then $P(1,1)$ and $P(1,-1)$ give $f(0)=0$ again.**<span style="color:#00f">Claim: </span>** $f$ is surjective.**<span style="color:#f00">Proof: </span>**If $f(f(x))\not \equiv 0$ , then $yf(f(x))-1$ is surjective. Assume that $f(f(x))=0$ for all $x$ . Then, $f(xf(y)+1)=\pm 1$ however $f\not \equiv 0$ so $xf(y)+1$ is surjective but $f(0)=0$ gives a contradiction. $\square$ **<span style="color:#00f">Claim:</span>** $f$ is injective at $0$ .**<span style="color:#f00">Proof:</span>** Suppose that $f(r)=0$ and $r\neq 0$ . **<span style="color:#0f0">Case I:</span>** If $f(1)=1$ , then $P(x,r)$ yields $\{1,f(rf(x)-1)\}=\{1,rf(f(x))-1\}$ hence $f(rf(x)-1)=rf(f(x))-1$ and $P(r,x)$ yields $f(rf(x)+1)=rf(f(x))+1$ . Since $f$ is surjective, we have $f(rx-1)=rf(x)-1$ and $f(rx+1)=rf(x)+1$ . Their difference implies $f(x+2)=f(x)+2$ . Notice that $f(r^2+1)=1$ . Now, $P(x,1)$ and $P(x,r^2+1)$ imply $f(x+1)\in \{x+1,f(f(x))-1\}$ and $f(x+1)\in \{x+1,(r^2+1)f(f(x))-1\}$ . This requires either $f(x+1)=x+1$ or $f(f(x))=0$ . We have $f(r^2+2k+1)=2k+1\neq r^2+2k+1$ and $f(f(r^2+2k))=f(f(r^2)+2k)=f(f(r^2))+2k\neq 0$ for sufficiently large $k$ which results in a contradiction.**<span style="color:#0f0">Case II: </span>**If $f(1)=-1$ , then $P(x,r)$ gives $\{-1,f(rf(x)-1)\}=\{1,rf(f(x))-1\}$ hence $f(rf(x)-1)=1$ and since $f$ is surjective $rf(x)-1$ is surjective however $f(x)=-1$ for all $x$ is impossible. $\square$ **<span style="color:#00f">Claim: </span>** $f$ is injective.**<span style="color:#f00">Proof:</span>** Assume that $f(a)=c=f(b)$ where $c\neq 0$ . $P(x,a)$ and $P(x,b)$ yield $f(cx+1)\in \{xf(c)+1,af(f(x))-1\}$ and $f(cx+1)\in \{xf(c)+1,bf(f(x))-1\}$ . Notice that $af(f(x))-1\neq bf(f(x))-1$ thus, $f(cx+1)=xf(c)+1$ . However $x=\frac{a-1}{c}$ and $x=\frac{b-1}{c}$ gives a contradiction since $f(c)\neq 0$ . $\square$ Now we split into two cases.**<span style="color:#0f0">First Case:</span>** If $f(1)=1$ , $P(m,1)$ gives $f(m+1)=m+1$ by induction hence $f(n)=n$ for positive integers. $P(f(x),y)$ and $P(f(y),x)$ imply $f(f(x)f(y)+1)\in \{f(x)f(f(y))+1,yf(f(f(x)))-1\}$ and $f(f(x)f(y)+1)\in \{f(y)f(f(x))+1,xf(f(f(y)))-1\}$ . Plug $y=n$ in order to see that $f(nf(x)+1)\in \{nf(x)+1,nf^3(x)-1\}$ and $f(nf(x)+1)\in \{nf^2(x)+1,nx-1\}$ . For sufficiently large $n$ , we have $nf(x)+1\neq nx-1$ and $nf^3(x)-1\neq nf^2(x)+1$ for fixed $x$ . Thus, either $nf(x)+1=nf^2(x)+1$ or $nf^3(x)-1=nx-1$ . Both of them yield $f^3(x)=x$ .
Suppose that $f(t)\neq t$ . Plug $P(f(t),1/t)$ to get $0\in \{f(f(t)f(\frac{1}{t})+1),f(\frac{f(f(t))}{t}-1)\}$ or $0\in \{f(t)f(\frac{1}{t})+1,\frac{f(f(t))}{t}-1\}$ . We see that $f(f(t))\neq t$ because otherwise $t=f(f(f(t)))=f(t)$ would contradict. Thus, $f(t)f(1/t)=-1$ . Plug $P(1/t,f(t))$ to observe (by injectivity) $2\in \{\frac{f(f(t))}{t}+1,f(t)f(1/t)-1\}=\{\frac{f(f(t))}{t}+1,-2\}$ and this implies $f(f(t))=t$ or $f(t)=t$ again. Hence $f(x)=x$ for all reals. $\square$ **<span style="color:#0f0">Second Case: </span>**If $f(1)=-1$ , by similar induction $f(n)=-n$ . Notice that for $f(p)=p$ , plugging $P(1/p,p)$ implies $-2\in \{2,pf(1/p)-1\}$ so $f(1/p)=-1/p$ . Pick $P(p,1/p)$ to get $0=pf(-1/p)+1$ hence $f(-1/p)=-1/p=f(p)$ but this requires $p^2=1$ which is impossible. Thus, $f(p)=p$ iff $p=0$ . Now plug $P(x,1/f(f(x))$ which implies $0\in \{xf(1/f(f(x)))+1,f(x)/f(f(x))-1\}$ and since $f(f(x))\neq f(x)$ we observe $f(\frac{1}{f(f(x))})=-\frac{1}{x}$ . Hence $f(f(-\frac{1}{f(x)}))=\frac{1}{x}$ Similarily $P(-1/f(f(x)),x)$ yields $f(-\frac{1}{f(f(x))})=\frac{1}{x}=f(f(-\frac{1}{f(x)}))$ thus (by using the surjectivity of $f$ ), $\frac{-1}{f(x)}=f(\frac{-1}{x})$ . We get $f(f(f(x))).\frac{1}{x}=f(f(f(x)))f(f(-1/f(x)))=-1$ thus, $f(f(f(x)))=-x$ . This gives $f$ is odd.
Suppose that $f(1+t)=f(1-t)$ . $P(t/f(y),y)$ implies $1-t\in \{t.\frac{f(f(y))}{f(y)}+1,yf(f(\frac{t}{f(y)}))-1\}$ . Note that $\{f(x+1),-f(f(x)+1)\}=\{1-x,-f(f(x))-1\}$ and $\{f(1-x),f(f(x)-1)\}=\{x+1,f(f(x))-1\}$ . We obtain $f(1-t)\neq 1+t$ so $f(1-t)=f(f(t))-1$ and $f(f(t)-1)=t+1$ and $f(1+t)=1-t$ .
\[1-f(t)=f(f(f(f(t)-1)))=f(f(t))-1\]
If $f(2-t)\neq 2-t$ , then $y=2-t$ gives $1=f(f(\frac{t}{f(2-t)}))$ or $f(2-t)=t$ . This requires $t-2=f(f(f(2-t)))=f(f(t))=2-f(t)$ or $f(t)=4-t$ . However, $P(1,f(t))$ gives $f(1+t)=1-t\in \{f(5-t),f(1-t)\}$ which is impossible. Hence $f(2-t)=t-2$ . Since $\{f(2-x),f(-f(1-x)-1)\}=\{x,-f(f(1-x))-1\}$ we see that $t-2=-f(f(1-t))-1$ or $f(f(1-t))=t-1$ and this implies $f(t-1)=t-1=f(-t-1)$ and this contradicts. Thus, there is no $t$ where $f(1+t)=1-t$ .
Notice that $f(x+1)\neq 1-x$ gives $f(x+1)=-f(f(x))-1$ and $f(-f(x)-1)=1-x$ . Plug $P(x,-n)$ to observe $\{f(nx+1),f(-nf(x)-1)\}=\{-nx+1,-nf(f(x))-1\}$ . Since $f(nx+1)\neq 1-nx$ we get $f(-nf(x)-1)=-nx+1$ or $f(-nf(\frac{x}{n})-1)=1-x=f(-f(x)-1)$ thus, $f(x/n)=f(x)/n$ .
Also $f(1-x)=f(f(x))-1$ and $f(-1-x)=f(f(x))+1$ hence $f(x+2)=f(x)-2$ . Furthermore, $2f(x+1)=f(2x+2)=f(2x)-2=2f(x)-2$ or $f(x+1)=f(x)-1$ . In conclusion, $f(x)-1=f(x+1)=-f(f(x))-1$ gives $f(f(x))=-f(x)$ and since $f$ is surjective, $f(x)=-x$ for all reals as desired. $\blacksquare$ | [
"NOTE: THE PREVIOUS SOLUTION I POSTED HAS A SIGN ERROR, hopefully this one is correct.\n\nNote 2: This solves for f(1)=1 only \n\n<details><summary>Solution to f(1)=1 only</summary>The answer is $f(x)\\equiv x, f(x)\\equiv -x$ only. They both work. \n\nLet $P(x,y)$ denote the assertion that the multiset $\\{(f... | [
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Find all positive integers $a,b,c$ and prime $p$ satisfying that
\[ 2^a p^b=(p+2)^c+1.\] | My 300th post!!!**Solution**
Notice that $2\mid 2^ap^b,$ we have $p\equiv 1\pmod 2.$ Let $c=2^{\theta}r$ where $2\nmid r,$ then $(p+2)^{2^{\theta}}+1\mid 2^ap^b.$ Lemma $:$ $(p+2)^{2^{\theta}}+1$ is divisible by $p.$ If it is not true $,$ $(p+2)^{2^{\theta}}+1=2^{\beta},$ then $\beta >2.$ Therefore $(p+2)^{2^{\theta}}+1\equiv 0\pmod 4,\theta =0,c=r,$ then $p+3=2^{\beta},$ therefore $p\equiv 5\pmod 8,$ so that $\left(\frac{-2}p\right)=-1.$ However $2^{r+1}\equiv -2\pmod p,$ contradiction! Therefore $(p+2)^{2^{\theta}}+1$ is divisible by $p.$ If $r\geq 3,$ let $N=\frac{(p+2)^{2^{\theta}r}+1}{(p+2)^{2^{\theta}}+1},$ then $N=\sum\limits_{j=0}^{r-1}(p+2)^{j\cdot 2^{\theta}}(-1)^j\equiv r\equiv 1\pmod 2,$ therefore $2\nmid N.$ From $N\mid 2^ap^b,$ $N=p^k,$ where $k\in\mathbb N.$ Using LTE Lemma $,$ $v_p(N)=v_p(r),$ therefore $N\leq r.$ Let $x=(p+2)^{2^{\theta}},$ then $x\geq 5,$ therefore $N=\frac{x^r+1}{x+1}\geqslant x^{r-1}-x^{r-2}\geqslant\frac 12x^{r-1}>r,$ contradiction!
Now $r=1.$ If $\theta =0,c=1,$ then $p+3=2^ap^b\equiv 0\pmod p,$ so $p=3,a=b=1,(a,b,c,p)=(1,1,1,3).$ Consider $\theta\geq 1,$ then $2^ap^b=(p+2)^{2^{\theta}}+1\equiv 2\pmod 4,$ therefore $a=1,2p^b=(p+2)^{2^{\theta}}+1\cdots (*).$ If $\theta =1,$ $2^ap^b=p^2+4p+5,$ therefore $p=5,$ we can get $b=2,(a,b,c,p)=(1,2,2,5).$ Now let $\theta\geq 2,$ then $2^{2^{\theta}}\equiv -1\pmod p,2^{2^{\theta +1}}\equiv 1\pmod p,$ so $\delta _p(2)=2^{\theta +1},2^{\theta+1}\mid (p-1).$ Therefore $p\equiv 1\pmod {16}.$ From $(*)$ we have $2(-1)^b\equiv 2\pmod {p+1},$ so $2\mid b.$ We can write $(*)$ as $2(p^b-1)=(p+2)^{2^{\theta}}-1.$ Using LTE Lemma $,$ $v_2\left((p+2)^{2^{\theta}}-1\right)=v_2\left((p+2)^2-1\right) +\theta -1=\theta +2.$ However $,$ $\theta +1=v_2(p^b+1)\geqslant v_2\left(p^2-1\right)\geqslant v_2(p-1)+v_2(p+1)\geqslant\theta +2,$ contradiction!
Therefore $\boxed{(a,b,c,p)=(1,1,1,3),(1,2,2,5)}.\blacksquare$ | [
"If $c=0$ then obviously $a=1$ and $b=0$ Let $c>0$ $mod2$ gives $a>1$ If $p=2$ then $RHS\\equiv 1\\not\\equiv 0\\equiv LHS (mod 2)$ contradiction.So $p>=3$ If $b=0$ then we have:\nIf $c=1$ we have the solutionw on the form $(a,b,c,p):(a,0,1,2^a-3)$ with $p=prime$ If $c=odd>1$ then $2^a=(p+3)... | [
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Let $n$ be a positive integer, $x_1,x_2,\ldots,x_{2n}$ be non-negative real numbers with sum $4$ . Prove that there exist integer $p$ and $q$ , with $0 \le q \le n-1$ , such that
\[ \sum_{i=1}^q x_{p+2i-1} \le 1 \mbox{ and } \sum_{i=q+1}^{n-1} x_{p+2i} \le 1, \]
where the indices are take modulo $2n$ .
*Note:* If $q=0$ , then $\sum_{i=1}^q x_{p+2i-1}=0$ ; if $q=n-1$ , then $\sum_{i=q+1}^{n-1} x_{p+2i}=0$ . | Here is another version of my proof above, which is simpler in my opinion. Moreover the gas station lemma is proved for completeness.
<details><summary>Solution</summary>Let us write $[n]\doteqdot\{1,2,\ldots,n\}$ for brevity.
*Lemma.* (gas station lemma) Let $n$ be a positive integer and let $a_1,\ldots,a_n$ be real numbers such that $\sum_{i=1}^n a_i=0$ . Then there exists an integer $p$ such that $\sum_{i=1}^k a_{p+i-1}\ge 0$ for all positive integers $k\in[n]$ , where the indices are taken modulo $n$ .
<details><summary>Proof</summary>Pick an integer $c$ such that $0\le c\le n-1$ and $\sum_{i=1}^c a_i$ is minimal, where the empty sum is $0$ . We claim that $p\doteqdot c+1$ works. To prove this, let $k\in[n]$ be a positive integer. We wish to prove that $\sum_{i=1}^k a_{p+i-1}\ge 0$ . Equivalently:
\begin{align*}
\iff&\sum_{i=c+1}^{c+k} a_{i}\ge 0
\iff&\sum_{i=1}^{c+k}a_i\ge\sum_{i=1}^{c}a_i.
\end{align*}
We distinguish between two cases.
- *Case 1:* $c+k<n$ . By minimality we obtain $\sum_{i=1}^{c+k}a_i\ge\sum_{i=1}^{c}a_i$ .
- *Case 2:* $c+k\ge n$ . Note that $c+k\le (n-1)+n\le 2n-1$ . Hence $0\le c+k-n\le n-1$ . Since $\sum_{i=1}^n a_i=0$ and by minimality we obtain $$ \sum_{i=1}^{c+k}a_i=\sum_{i=1}^{c+k-n}a_i\ge\sum_{i=1}^{c}a_i. $$
Since we have exhausted all cases, the lemma is proved.</details>
Now define real numbers $a_1,b_1,a_2,b_2,\ldots,a_n,b_n$ by $a_i\doteqdot x_{2i-1}$ and $b_i\doteqdot x_{2i}$ for all $i\in[n]$ . It suffices to prove that there exist integers $p$ and $q$ such that $0\le q\le n-1$ and $$ \sum_{i=q+2}^{n} a_{p+i}\le 1\mbox{ and }\sum_{i=1}^q b_{p+i}\le 1, $$ where indices are taken modulo $n$ . The assignment $(p,q)\mapsto (2p+1,q)$ solves the original problem.
Define $A\doteqdot\sum_{i=1}^n a_i$ and $B\doteqdot\sum_{i=1}^n b_i$ . If $A\le 1$ then we can choose $(p,q)=(0,0)$ . Similarly, if $B\le 1$ then we can choose $(p,q)=(0,n-1)$ . So assume $A>1$ and $B>1$ . Define real numbers $c_1,\ldots,c_n$ by $c_i\doteqdot Ba_i-Ab_i$ for all $i\in[n]$ . Note that $$ \sum_{i=1}^n c_i=B\sum_{i=1}^n a_i-A\sum_{i=1}^n b_i=BA-AB=0. $$ By the gas station lemma, there exists an integer $p$ such that $\sum_{i=1}^k c_{p+i}\ge 0$ for all integers $k\in[n]$ , where indices are taken modulo $n$ . Let $q$ be the maximal integer $q$ such that $0\le q\le n$ and $\sum_{i=1}^q b_{p+i}\le 1$ . Since we assume $B>1$ then $q\le n-1$ . Assume by contradiction that $\sum_{i=q+2}^n a_{p+i}>1$ . By maximality we obtain $\sum_{i=1}^{q+1} b_{p+i}>1$ . Hence $$ 0\le\sum_{i=1}^{q+1}c_{p+i}=B\sum_{i=1}^{q+1} a_{p+i}-A\sum_{i=1}^{q+1} b_{p+i}<B\sum_{i=1}^{q+1} a_{p+i}-A. $$ Therefore $\sum_{i=1}^{q+1} a_{p+i}>\frac{A}{B}$ . Now we have $$ \sum_{i=q+2}^n c_{p+i}=\sum_{i=1}^n c_{p+i}-\sum_{i=1}^{q+1} c_{p+i}=-\sum_{i=1}^{q+1} c_{p+i}\le 0. $$ Hence $$ 0\ge\sum_{i=q+2}^n c_{p+i}=B\sum_{i=q+2}^n a_{p+i}-A\sum_{i=q+2}^n b_{p+i}>B-A\sum_{i=q+2}^n b_{p+i}. $$ Therefore $\sum_{i=q+2}^n b_{p+i}>\frac{B}{A}$ . Now by AM-GM
\begin{align*}\sum_{i=1}^n(a_{p+i}+b_{p+i})&=\sum_{i=1}^{q+1} a_{p+i}+\sum_{i=1}^{q+1} b_{p+i}+\sum_{i=q+2}^n a_{p+i}+\sum_{i=q+2}^n b_{p+i}
&>\frac{A}{B}+1+1+\frac{B}{A}
&>4,
\end{align*}
contradiction.</details> | [
"Sketch and unLaTeXed since I am on mobile:\nLet S be the sum of x_i where i is even. Similarly let T be the sum of x_i where i is odd. WLOG S>=T. Consider a function f:1,...,n->R such that f(i)=S*x_(2i+1)-T*x_(2i). Then sum f(i)=0 so by gas station lemma, there exists an index a such that f(a),f(a+1),...,f(a+n-1) ... | [
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Given a positive integer $n$ , let $D$ be the set of all positive divisors of $n$ . The subsets $A,B$ of $D$ satisfies that for any $a \in A$ and $b \in B$ , it holds that $a \nmid b$ and $b \nmid a$ . Show that
\[ \sqrt{|A|}+\sqrt{|B|} \le \sqrt{|D|}. \] | Solved with Luke Robitaille, Justin Lee, and Espen Slettnes.
Let \[V=\bigcup_{a\in A}\{\text{divisors of }a\} \quad\text{and}\quad W=\bigcup_{a\in A}\{\text{multiples of }a\}\cap D\]
<span style="color:red">**Claim:**</span> \(|V|\cdot|W|\ge|A|\cdot|D|\).
*Proof.* We induct on the number of prime divisors of \(n\), with base case \(n=1\) trivial.
For the inductive step, take a prime \(p\mid n\). For any set \(S\), let \(S_i=\{s\in S\mid \nu_p(s)=i\}\). Then \(|V_0|\ge|V_1|\ge\cdots\ge|V_n|\) and \(|W_0|\le|W_1|\le\cdots\le|W_n|\). By Chebyshev's inequality, \begin{align*} |V|\cdot|W|=\left(\sum_{i=0}^n|V_i|\right)\left(\sum_{i=0}^n|W_i|\right) &\ge(n+1)\sum_{i=0}^n|V_i|\cdot|W_i| &\ge(n+1)\sum_{i=0}^n|A_i|\cdot|D_i| =|D|\sum_{i=0}^n|A_i|=|A|\cdot|D|. \end{align*} \(\blacksquare\)
We may assume that for any \(d\in D\), if \(d\) has a divisor and a multiple in \(A\), then \(d\in A\). Then \(V\cap W=A\) and \(|V\cup W|=|D|-|B|\), so \[|D|-|B|+|A|=|V|+|W|\ge2\sqrt{V|\cdot|W|}\ge2\sqrt{|A|\cdot|D|},\] which rearranges to the desired.
| [
"Oops this is pretty much the same as above\n\nLet $A_{down} = \\{ x \\text{ st } x|a \\text{ for some } a\\in A\\}$ and $A_{up} = \\{ \\text{ st } a|x|n \\text{ for some } a\\in A\\}$ . Define $B_{up}, B_{down}$ similarly.\n\nMain Claim: $|A| |D| \\le |A_{up}||A_{down}|$ Proof: We induct on $\\omega(n)$ . L... | [
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Let $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ be two squares such that the boundaries of $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ does not contain any line segment. Construct 16 line segments $A_iB_j$ for each possible $i,j \in \{1,2,3,4\}$ . What is the maximum number of line segments that don't intersect the edges of $A_1A_2A_3A_4$ or $B_1B_2B_3B_4$ ? (intersection with a vertex is not counted).
*Proposed by Allen Zheng* | There are two feasible interpretations of this problem, so I'll provide a solution for both. The construction for both of them is two concentric squares with sides parallel.
<details><summary>Solution with one interpretation</summary>If the problem means that an intersection is not counted if and only if the intersection point is a vertex of **both segments**, then the answer is $4^2=\boxed{16},$ because none of the segments intersect the *interior* of any one of the segments $A_1A_2, A_2, A_3, A_3A_4, A_4A_1, B_1B_2, B_2B_3,$ $B_4B_4,$ or $B_4B_1.$ $\square$</details>
<details><summary>Solution with another interpretation</summary>If the problem means that an intersection is not counted if and only if the intersection point is a vertex of **at least one of the segments**, then the answer is $\boxed{12},$ because, unlike in the first interpretation, if we line up the squares such that the subscripts of the vertices correspond, the segments $A_1B_3, A_2B_4, A_3B_1,$ and $A_4B_2$ wouldn't be counted. $\square$</details>
<details><summary>Note</summary>The wording of this question is ambiguous, so I think the CMIMC staff did the right thing in accepting both of these answers.</details> | [
"<details><summary>My Protest with 16</summary>[asy]\nlabel(\" $A_1$ \",(0,0),W);\nlabel(\" $A_2$ \",(4,0),E);\nlabel(\" $A_3$ \",(4,4),E);\nlabel(\" $A_4$ \",(0,4),W);\ndraw((0,0)--(4,0));\ndraw((4,0)--(4,4));\ndraw((4,4)--(0,4));\ndraw((0,0)--(0,4));\ndraw((0,0)--(4,4));\ndraw((0,4)--(4,0));\nlabel(\" $B_1$ \",(1... | [
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Find the smallest positive integer $n$ for which $315^2-n^2$ evenly divides $315^3-n^3$ .
*Proposed by Kyle Lee* | <details><summary>Solution</summary>First, we have
\begin{align*}
\left(315-n^2\right) \mid \left(315^3-n^3\right) &\implies (315-n)(315+n) \mid (315-n)(315^2+315n+n^2)
&\implies (315+n) \mid (315^2+315n+n^2)
&\implies (315+n) \mid (315^2+315n+n^2-n(315+n))
&\implies (315+n) \mid(315^2+315n+n^2-315n-n^2)
&\implies (315+n) \mid 315^2.
\end{align*}
Then, it is not difficult to see that the smallest divisor of $315^2$ larger than $315$ is $405,$ so the answer is $$ 405-315=\boxed{90}. $$ $\square$</details> | [
"<details><summary>Euclidean Algorithm</summary>$$ 315^2-n^2=(315+n)(315-n) $$ $$ (315-n)(315^2+315n+n^2) $$ $$ (315+n)|(315^2+315n+n^2) $$ $$ (315+n)|(315^2) $$ $$ n=\\boxed{90} $$</details>"
] | [
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"path": "Contest Collections/2022 Contests/2022 CMIMC/2791419.json"
} |
Let $ABCD$ be a rectangle with $AB=10$ and $AD=5.$ Suppose points $P$ and $Q$ are on segments $CD$ and $BC,$ respectively, such that the following conditions hold:
- $BD \parallel PQ$
- $\angle APQ=90^{\circ}.$
What is the area of $\triangle CPQ?$ *Proposed by Kyle Lee* | <details><summary>Solution</summary>Since $BD \parallel PQ$ and $AP \perp PQ,$ it follows that $AP \perp BD.$ Now, let us place rectangle $ABCD$ into the Cartesian Plane such that $$ A=(0,5), B=(10,5), C=(10,0), D=(0,0). $$ Since the slope of line $BD$ is
\begin{align*}
\frac{5-0}{10-0} &= \frac{5}{10}
&= \frac{1}{2},
\end{align*}
it follows that the slope of line $AP$ is $-2,$ so its equation is $y=-2x+5.$ The equation of line $CD$ is obviously just $y=0,$ which implies that the intersection of $AP$ and $CD$ is $$ P = \left(\frac{5}{2},0\right). $$ Now, let $Q = (10, k),$ where $k \in [0,5].$ Then, we have
\begin{align*}
\frac{k-0}{10-\frac{5}{2}} = \frac{1}{2} &\implies \frac{k}{\frac{15}{2}} = \frac{1}{2}
&\implies k = \frac{15}{4}.
\end{align*}
Now, we know all the coordinates of the vertices of $\triangle{CPQ},$ so we can deduce that $CP=\tfrac{15}{2}$ and $CQ=\tfrac{15}{4},$ which implies that
\begin{align*}
[CPQ] &= \frac{CP \cdot CQ}{2}
&= \frac{\frac{15}{2} \cdot \frac{15}{4}}{2}
&= \frac{\frac{225}{8}}{2}
&= \boxed{\frac{225}{16}}.
\end{align*} $\square$</details> | [
"<details><summary>Solution</summary>Slope of $PQ$ is $\\frac{1}{2}$ thus, the slope of $AP$ is $-2.$ Therefore $PC=\\frac{15}{2}$ and $CQ=\\frac{15}{4}$ so our answer is $\\boxed{\\frac{225}{16}}$</details>"
] | [
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Let $\triangle ABC$ be equilateral with integer side length. Point $X$ lies on $\overline{BC}$ strictly between $B$ and $C$ such that $BX<CX$ . Let $C'$ denote the reflection of $C$ over the midpoint of $\overline{AX}$ . If $BC'=30$ , find the sum of all possible side lengths of $\triangle ABC$ .
*Proposed by Connor Gordon* | <details><summary>Solution</summary>[asy]
pair C = dir(60), A = dir(180), B = dir(300);
draw(A--B--C--cycle);
pair X = (2B+C)/3;
draw(A--X);
pair Cp = A+X-C;
draw(A--Cp--X,red);
draw(Cp--B,blue);
pair P = extension(Cp,X,A,B);
dot(A^^B^^C^^X^^Cp^^P);
label(" $A$ ",A,dir(180));
label(" $B$ ",B,dir(300));
label(" $C$ ",C,dir(60));
label(" $X$ ",X,dir(0));
label(" $C'$ ",Cp,dir(-120));
label(" $P$ ",P,dir(-90));
[/asy]
Note that $CAC'X$ is a parallelogram, so $\overline{BX}||\overline{AC'}$ and thus $XAC'B$ is a trapezoid. Let $\overline{AB}$ and $\overline{XC'}$ intersect at $P$ . A simple angle chase shows that $\triangle APC'$ and $\triangle XPB$ are equilateral, so it follows that $AXBC'$ is an isosceles trapezoid and $AX=BC'=30$ .
To finish, note that the length $\ell$ of an internal cevian of an equilateral triangle with side length $s$ satisfies $\tfrac{s\sqrt{3}}{2}\leq \ell < s$ , so it follows that $30<s\leq 30\cdot\tfrac{2}{\sqrt{3}}=\sqrt{1200}$ , where one can easily note that $34<\sqrt{1200}<35$ . It follows that the possible values for $s$ are $31$ , $32$ , $33$ , and $34$ , for a sum of $\boxed{130}$ .</details> | [] | [
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For any integer $a$ , let $f(a) = |a^4 - 36a^2 + 96a - 64|$ . What is the sum of all values of $f(a)$ that are prime?
*Proposed by Alexander Wang* | <details><summary>Solution</summary>First, we can factor the given function as $$ f(a) = |(a-2)(a-4)(a^2+6a-8)|. $$ For this to be prime, exactly two of $(a-2), (a-4),$ and $(a^2+6a-8)$ must be $\pm 1$ and the third must be $\pm p$ for a prime $p.$ We now split into cases:**a-2 is 1:** Then, we have $a=3$ which yields $f(a)=19,$ which is prime.**a-2 is -1:** Then, we have $a=1$ which yields $f(a)=3,$ which is also prime.**a-2 is not 1 or -1:** Then, it follows that $a-2$ must be of the form $\pm p$ for a prime $p$ which means that both of $(a-4)$ and $(a^2+6a-8)$ must be $\pm 1.$ If $a-4=1,$ then $a=5$ which gives $f(a)=141,$ which is not prime. If $a-4=-1,$ then $a=3,$ which we have already considered.
Hence, putting the results from these three cases together, the answer is $$ 19+3=\boxed{22}. $$ $\square$</details> | [
"With some playing around we have $$ f(a)=|(a-2)g(a)| $$ for some cubic $g(a)$ . In order for this to be prime we must either have $a=1$ or $a=3$ . The answer is $\\boxed{22}$ @below inspection shows that if $g(a) = \\pm 1$ then $a$ is not an integer",
"technically $g(a)=\\pm 1$ could be possible @... | [
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There are $9$ points arranged in a $3\times 3$ square grid. Let two points be adjacent if the distance between them is half the side length of the grid. (There should be $12$ pairs of adjacent points). Suppose that we wanted to connect $8$ pairs of adjacent points, such that all points are connected to each other. In how many ways is this possible?
*Proposed by Kevin You* | [] | [
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A $3\times2\times2$ right rectangular prism has one of its edges with length $3$ replaced with an edge of length $5$ parallel to the original edge. The other $11$ edges remain the same length, and the $6$ vertices that are not endpoints of the replaced edge remain in place. The resulting convex solid has $8$ faces, as shown below.
Find the volume of the solid.
*Proposed by Justin Hsieh*
**Attachments:**
 | [] | [
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} | |
There are 36 contestants in the CMU Puyo-Puyo Tournament, each with distinct skill levels.
The tournament works as follows:
First, all $\binom{36}{2}$ pairings of players are written down on slips of paper and are placed in a hat.
Next, a slip of paper is drawn from the hat, and those two players play a match. It is guaranteed that the player with a higher skill level will always win the match.
We continue drawing slips (without replacement) and playing matches until the results of the match completely determine the order of skill levels of all 36 contestants (i.e. there is only one possible ordering of skill levels consistent with the match results), at which point the tournament immediately finishes.
What is the expected value of the number of matches played before the stopping point is reached?
*Proposed by Dilhan Salgado* | pretty nice, solved this during my bio class
Let $1, 2, 3, \ldots, 36$ be the players, where a higher number represents a higher skill level. I'll use the notation $i-j$ to denote the match where $i$ plays $j$ . The key is to notice that only the matches of the form $k-(k+1)$ matter (ie, $1-2$ , $2-3$ , etc). This is because if we play all matches (excluding $1-2$ ) involving $1$ and $2$ , we still won't know who is better between the two players.
The rest is just computation. Obviously, the minimum number of slips needed is $35$ and the maximum is ${36\choose 2} = 630$ . If we need $n$ slips to determine the order, then we need the $35$ different $k-(k+1)$ slips to be drawn as well as $n-35$ "extra" slips. There are ${595\choose {n-35}}$ ways to choose these extra slips. Next, we fix the last slip to be drawn as one of the $k-(k+1)$ slips. There are $35$ ways to do this. Finally, we simply rearrange the remaining $34$ $k-(k+1)$ slips and the $n-35$ extra slips. There are $(n-1)!$ ways to rearrange these slips. In all, there are $35{595\choose {n-35}}(n-1)!$ ways to rearrange/pick the slips in order for $n$ slips to determine the order. Each one orderings of these has a $$ \frac 1{630}\cdot \frac{1}{629}\cdots \frac{1}{631-n} $$ probability of occurring, so the probability that $n$ slips determine the order is $$ \frac1{630}\cdot \frac{1}{629}\cdots \frac{1}{631-n} \cdot 35{595\choose {n-35}}(n-1)! $$ .
The desired expected value is simply the sum: \begin{align*}
&\sum_{n = 35}^{630} \left(n\cdot \frac1{630}\cdot \frac{1}{629}\cdots \frac{1}{631-n} \cdot 35{595\choose {n-35}}(n-1)!\right)
&=35\sum_{n = 35}^{630}\frac{n!{595\choose {n-35}}}{630\cdot 629\cdots \cdot (631-n)}
&=35\sum_{n = 35}^{630}\frac{\frac{n!595!}{(n-35)!(630-n)!}}{630\cdot 629\cdots \cdot (631-n)}
&=35\sum_{n = 35}^{630}\frac{\frac{n!595!}{(n-35)!(630-n)!}}{\frac{630!}{(630-n)!}}
&=35\sum_{n = 35}^{630}\frac{\frac{n!595!}{(n-35)!(630-n)!}}{\frac{630!}{(630-n)!}}
&=35\sum_{n = 35}^{630}\frac{n!595!}{630!(n-35)!}
&=35\sum_{n = 35}^{630}\frac{{n\choose 35}}{{630\choose 35}}
\end{align*}.
Now, we can just apply the hockey stick identity to find that our answer is $$ 35\cdot \frac{{631\choose 36}}{{630\choose 35}} = \boxed{\frac{22085}{36}}. $$ $\blacksquare$ | [] | [
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For natural numbers $n$ , let $r(n)$ be the number formed by reversing the digits of $n$ , and take $f(n)$ to be the maximum value of $\frac{r(k)}k$ across all $n$ -digit positive integers $k$ .
If we define $g(n)=\left\lfloor\frac1{10-f(n)}\right\rfloor$ , what is the value of $g(20)$ ?
*Proposed by Adam Bertelli* | [] | [
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Adam places down cards one at a time from a standard 52 card deck (without replacement) in a pile. Each time he places a card, he gets points equal to the number of cards in a row immediately before his current card that are all the same suit as the current card. For instance, if there are currently two hearts on the top of the pile (and the third card in the pile is not hearts), then placing a heart would be worth 2 points, and placing a card of any other suit would be worth 0 points. What is the expected number of points Adam will have after placing all 52 cards?
*Proposed by Adam Bertelli* | really fun problem to solve
Number the cards as follows for convenience: $\{A1, A2, \ldots, A13\}, \{B1, B2, \ldots, B13\}, \{C1, C2, \ldots, C13\}, \{D1, D2, \ldots, D13\}$ . The letter corresponds to the suit of the card. Suppose we place each card in a row, where cards to the left are placed earlier (ie, the leftmost card is the first card placed down). This is obviously equivalent to the original problem.
Now, here's the cool part: Assume we have the following "placement" of cards (this is just an example): $$ B1, A1, A2, A3, A4, \ldots $$ The score of card $A2$ is obviously $1$ , the score of card $A3$ is $2$ , and the score of card $A4$ is $3$ (this comes from the given scoring system). However, we can also interpret the score of an individual card as the number of "runs" of same-suit cards (of any length) that a particular card "ends". For example, the card $A4$ "ends" exactly $3$ runs - the run $A3, A4$ , the run $A2, A3, A4$ , and the run $A1, A2, A3, A4$ . This is obviously true for any arbitrary card.
Using this new definition, we see that the final score is simply the number of runs (of any length) in the entire row. This is now quite easy to count via casework and linearity of expectation.
We wish to find the expected number of runs of length $n$ in the entire row ( $n$ will obviously range from $2$ to $13$ in our final answer). There are $53-n$ possible "locations" where the run could be. At each one of these locations, there is a $\frac {12}{51}\cdot \frac{11}{50}\cdots \frac{14-n}{53-n}$ probability of a run of length $n$ occuring. Thus, the expected number of runs of length $n$ in the entire row is just $$ \left(\frac {12}{51}\cdot \frac{11}{50}\cdots \frac{14-n}{53-n}\right)\cdot (53-n) $$ by linearity. The expression is equal to $$ \cfrac{\cfrac{12!}{(13-n)!}}{\cfrac{51!}{(52-n)!}}\cdot (53-n) = \frac{12!(52-n)!\cdot (53-n)}{51!(13-n)!} = \frac{12!(53-n)!}{51!(13-n)!}. $$ Summing over all possible "run-lengths", we see that the expected score that we will get is $$ \sum^{13}_{n = 2} \frac{12!(53-n)!}{51!(13-n)!} = \sum^{13}_{n = 2} \frac{12!40!(53-n)!}{51!40!(13-n)!} = \sum^{13}_{n = 2} \frac{12!40!}{51!}\cdot {{53-n}\choose 40}=\frac{12!40!}{51!}\sum^{13}_{n= 2}{53-n\choose 40}. $$ Finally, we can apply the Hockey Stick identity to get that $$ \sum^{13}_{n= 2}{53-n\choose 40} = {52\choose 41} $$ , so the final answer is $$ \frac{12!40!}{51!}\cdot {52\choose 41} = \boxed{\frac{624}{41}} $$ . $\blacksquare$ | [] | [
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} |
Let $\{\varepsilon_i\}_{i\ge 1}, \{\theta_i\}_{i\ge 0}$ be two infinite sequences of real numbers, such that $\varepsilon_i \in \{-1,1\}$ for all $i$ , and the numbers $\theta_i$ obey $$ \tan \theta_{n+1} = \tan \theta_{n}+\varepsilon_n \sec(\theta_{n})-\tan \theta_{n-1} , \qquad n \ge 1 $$ and $\theta_0 = \frac{\pi}{4}, \theta_1 = \frac{2\pi}{3}$ . Compute the sum of all possible values of $$ \lim_{m \to \infty} \left(\sum_{n=1}^m \frac{1}{\tan \theta_{n+1} + \tan \theta_{n-1}} + \tan \theta_m - \tan \theta_{m+1}\right) $$ *Proposed by Grant Yu* | [] | [
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Let $ABCD$ be a cyclic quadrilateral with $AB=3, BC=2, CD=6, DA=8,$ and circumcircle $\Gamma.$ The tangents to $\Gamma$ at $A$ and $C$ intersect at $P$ and the tangents to $\Gamma$ at $B$ and $D$ intersect at $Q.$ Suppose lines $PB$ and $PD$ intersect $\Gamma$ at points $W \neq B$ and $X \neq D,$ respectively. Similarly, suppose lines $QA$ and $QC$ intersect $\Gamma$ at points $Y \neq A$ and $Z \neq C,$ respectively. What is the value of $\frac{{WX}^2}{{YZ}^2}?$ *Proposed by Kyle Lee* | [] | [
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Let $F_n$ denote the $n$ th Fibonacci number, with $F_0=0, F_1=1$ and $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 2$ . There exists a unique two digit prime $p$ such that for all $n$ , $p | F_{n+100} + F_n$ . Find $p$ .
*Proposed by Sam Rosenstrauch* | [] | [
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Let a tree on $mn + 1$ vertices be $(m,n)$ -nice if the following conditions hold:
- $m + 1$ colors are assigned to the nodes of the tree
- for the first $m$ colors, there will be exactly $n$ nodes of color $i$ $(1\le i \le m)$
- the root node of the tree will be the unique node of color $m+1$ . \item the $(m,n)$ -nice trees must also satisfy the condition that for any two non-root nodes $i, j$ , if the color of $i$ equals the color of $j$ , then the color of the parent of $i$ equals the color of the parent of $j$ .
- Nodes of the same color are considered indistinguishable (swapping any two of them results in the same tree).
Let $N(u,v,l)$ denote the number of $(u,v)$ -nice trees with $l$ leaves. Note that $N(2,2,2) = 2, N(2,2,3) = 4, N(2,2,4) = 6$ . Compute the remainder when $\sum_{l = 123}^{789} N(8,101,l)$ is divided by $101$ .
Definition: Any rooted, ordered tree consists of some set of nodes, each of which has a (possibly empty) ordered list of children. Each node is the child of exactly one other node, with the exception of the root, which has not parent. There also cannot be any cycles of nodes which are all linearly children of each other.
*Proposed by Advait Nene* | [] | [
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Let $ABC$ be a triangle with $AB = 5, BC = 13,$ and $AC = 12$ . Let $D$ be a point on minor arc $AC$ of the circumcircle of $ABC$ (endpoints excluded) and $P$ on $\overline{BC}$ . Let $B_1, C_1$ be the feet of perpendiculars from $P$ onto $CD, AB$ respectively and let $BB_1, CC_1$ hit $(ABC)$ again at $B_2, C_2$ respectively. Suppose that $D$ is chosen uniformly at random and $AD, BC, B_2C_2$ concur at a single point. Compute the expected value of $BP/PC$ .
*Proposed by Grant Yu* | [] | [
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Alice and Bob live on the same road. At time $t$ , they both decide to walk to each other's houses at constant speed. However, they were busy thinking about math so that they didn't realize passing each other. Alice arrived at Bob's house at $3:19\text{pm}$ , and Bob arrived at Alice's house at $3:29\text{pm}$ . Charlie, who was driving by, noted that Alice and Bob passed each other at $3:11\text{pm}$ . Find the difference in minutes between the time Alice and Bob left their own houses and noon on that day.
*Proposed by Kevin You* | <details><summary>Solution</summary>First of all, clearly, Alice is faster than Bob. Now, let the ratio of Alice's speed to Bob's speed be $r.$ Note that, after passing each other, Alice has $r$ times less distance to get to Bob's house than Bob has to get to Alice's house and is also $r$ times faster. Therefore, it follows that the time it would take Alice to reach Bob's house after passing Bob is $r^2$ times less than the time it would take Bob to reach Alice's house after passing Alice.
We thus get the equation
\begin{align*}
\frac{18}{r^2} = 8 &\implies r^2 = \frac{18}{8} = \frac{9}{4}
&\implies r = \frac{3}{2}.
\end{align*}
Therefore, when Alice and Bob pass each other, Alice has $\tfrac{2}{5}$ of the way left to travel. Since this takes $8$ minutes, it follows that the first part of the trip, which is $\tfrac{3}{5}$ of the total distance, took $12$ minutes. Since they passed each other at $3:11\text{ pm},$ this tells us that they both left their homes $12$ minutes earlier, at $2:59\text{ pm}.$ Hence, the answer is
\begin{align*}
2\cdot60+59 &= 120 + 59
&= \boxed{179}.
\end{align*} $\square$</details> | [
"ok this one was harder than #5\n",
"<details><summary>Solution</summary>Since they past each other at $3:11$ we know that it took 8 minutes for Alice to reach Bob's house after they met and 18 minutes for Bob to reach Alice's house after they met. Now let us say that they drove $x$ minutes before meeting eac... | [
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Arthur, Bob, and Carla each choose a three-digit number. They each multiply the digits of their own numbers. Arthur gets 64, Bob gets 35, and Carla gets 81. Then, they add corresponding digits of their numbers together. The total of the hundreds place is 24, that of the tens place is 12, and that of the ones place is 6. What is the difference between the largest and smallest of the three original numbers?
*Proposed by Jacob Weiner* | <details><summary>Solution</summary>Let the numbers chosen by Arthur, Bob, and Carla be $\overline{a_1a_2a_3}, \overline{b_1b_2b_3},$ and $\overline{c_1c_2c_3}$ respectively, where $$ x_i \in \{1,2,\ldots,9\} \ \forall \ x \in \{a,b,c\}, i \in \{1,2,3\}. $$ It is not difficult to see that $$ \max(a_1) = 8, \max(b_1)=7, \max(c_1)=9. $$ Since the sum of these is $24,$ it follows that $$ (a_1, b_1, c_1) = (8,7,9). $$ Additionally, $$ \min(a_3)=\min(b_3)=\min(c_3)=1, $$ so it follows that $$ b_3 \leq 4. $$ Hence, it follows that $$ b_3=1, $$ which implies that $$ b_2=5. $$ Now, we have $$ a_2a_3=8, c_2c_3=9, a_2+c_2=7, a_1+c_1=5. $$ If $c_2$ is $1$ or $9,$ then $a_2$ would have to be $6$ and $-2$ respectively, both of which aren't possible, so $$ c_2=3. $$ From here, it is easy to deduce that $$ c_3=3, a_2=4, a_3=2. $$ Hence, collecting all of the variables we have found through logical reasoning, we can conclude that $$ \begin{bmatrix} a_1 & a_2 & a_3 b_1 & b_2 & b_3 c_1 & c_2 & c_3 \end{bmatrix} = \begin{bmatrix} 8 & 4 & 2 7 & 5 & 1 9 & 3 & 3 \end{bmatrix}, $$ so the answer is $$ 933-751 = \boxed{182}. $$ $\square$</details> | [
"the first numbers i tested without reading the third sentence worked haha\n\n<details><summary>Solution</summary>Let the hundreds place of the three numbers be $a$ , $b$ , and $c$ , respectively. Note that $\\operatorname{max}(a)=8$ , $\\operatorname{max}(b)=7$ , and $\\operatorname{max}(c)=9$ . We have that... | [
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How many 4-digit numbers have exactly $9$ divisors from the set $\{1,2,3,4,5,6,7,8,9,10\}$ ?
*Proposed by Ethan Gu* | <details><summary>Solution</summary>It is not difficult to see that the excluded number must be $7,8,$ or $9.$ We now consider each case separately:**7:** Note that the number must be divisible by
\begin{align*}
\frac{\text{lcm} (1,2,3,\ldots,10)}{7} &= \frac{2520}{7}
&= 360.
\end{align*}
If we let the number be $360k,$ then the $4$ -digit condition implies that $$ 3 \leq k \leq 27, $$ which has $25$ integer solutions for $k.$ The only further restriction we must impose on $k$ is that it cannot be divisible by $7,$ so the total for this case is $$ 25-3=\underline{22}. $$ **8:** Note that the number must be divisible by
\begin{align*}
\frac{\text{lcm}(1,2,3,\ldots,10)}{2} &= \frac{2520}{2}
&= 1260.
\end{align*}
If we let the number be $1260,$ then the $4$ -digit condition implies that $$ 1 \leq k \leq 7, $$ which has $7$ integer solutions for $k.$ The only further condition we must impose on $k$ is that it cannot be even, so the total for this case is $$ 7 - 3 = \underline{4}. $$ **9:** Note that the number must be divisible by
\begin{align*}
\frac{\text{lcm}(1, 2, 3, \ldots, 10)}{3} &= \frac{2520}{3}
&= 840.
\end{align*}
If we let the number be $840k,$ then the $4$ -digit condition implies that $$ 2 \leq k \leq 11, $$ which has $10$ integer solutions for $k.$ We must impose the additional restriction on $k$ that it cannot be a multiple of $3,$ so the total for this case is $$ 10 - 3 = \underline{7}. $$ Adding up the totals from these three cases yields $$ 22+4+7=\boxed{33}. $$ $\square$</details> | [
"<details><summary>Solution</summary>Note that $\\operatorname{lcm}(1,2,3,\\cdots,10) = 2^3 \\cdot 3^2 \\cdot 5 \\cdot 7$ . **Case 1:** The divisor that is not included from the set is $7$ .\n\nThen the 4-digit number is a multiple of $\\tfrac{2520}{7} = 360$ . The smallest 4-digit multiple of $360$ is $360 \... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1168,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791441.json"
} |
A shipping company charges $.30l+.40w+.50h$ dollars to process a right rectangular prism-shaped box with dimensions $l,w,h$ in inches. The customers themselves are allowed to label the three dimensions of their box with $l,w,h$ for the purpose of calculating the processing fee. A customer finds that there are two different ways to label the dimensions of their box $B$ to get a fee of $\$ 8.10 $, and two different ways to label $ B $ to get a fee of $ \ $8.70$ . None of the faces of $B$ are squares. Find the surface area of $B$ , in square inches.
*Proposed by Justin Hsieh* | <details><summary>Solution</summary>Let the dimensions of $B$ be $a \times b \times c.$ Then, suppose, without loss of generality, that we have $$ 0.3a+0.4b+0.5c=0.3b+0.4c+0.5a=8.1 $$ This implies that $2a=b+c.$ Note that if we have $2b=a+c$ or $2c=a+b,$ then it would follow that $a=b=c,$ contradiction. Therefore, it follows that $$ 0.3a=0.4c+0.5b=0.3c+0.4b+0.5a=8.7. $$ Now, we have a system of equations: $$ \begin{cases}
0.3b+0.4c+0.5a=8.1
0.3c+0.4b+0.5a=8.7
2a=b+c
\end{cases}
\implies
\begin{cases}
5a+3b+4c=81
5a+4b+3c=87
2a=b+c
\end{cases}. $$ Solving this system of equations yields $$ (a,b,c)=(7,10,4), $$ so it follows that the surface area of $B$ is
\begin{align*}
2(ab+bc+ca) &= 2(7\cdot10+10\cdot4+4\cdot7)
&= 2(70+40+28)
&= 2 \cdot 138
&= \boxed{276}.
\end{align*} $\square$</details> | [
"<details><summary>Systems of Equations Galore</summary>$$ 5l+3w+4h=81 $$ $$ 4l+5w+3h=81 $$ $$ 3l+5w+4h=87 $$ $$ 4l+3w+5h=87 $$ $$ l+h=2w $$ $$ h-l=6 $$ $$ 2w-2l=6 $$ $$ h=l+6 $$ $$ w=l+3 $$ $$ 5l+3(l+3)+4(l+6)=81 $$ $$ l=4 $$ $$ h=10 $$ $$ w=7 $$ $$ 2(4*7+4*10+7*10)=\\boxed{2... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1132,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791443.json"
} |
Alan is assigning values to lattice points on the 3d coordinate plane. First, Alan computes the roots of the cubic $20x^3-22x^2+2x+1$ and finds that they are $\alpha$ , $\beta$ , and $\gamma$ . He finds out that each of these roots satisfy $|\alpha|,|\beta|,|\gamma|\leq 1$ On each point $(x,y,z)$ where $x,y,$ and $z$ are all nonnegative integers, Alan writes down $\alpha^x\beta^y\gamma^z$ . What is the value of the sum of all numbers he writes down?
*Proposed by Alan Abraham* | <blockquote><details><summary>Solution</summary>Let the given polynomial be $f(x).$
We have
\begin{align*}
\sum_{x,y,z \geq 0} \alpha^x \beta^y \gamma^z &= \left(\sum_{x \geq 0} \alpha^x \right)\left(\sum_{y \geq 0} \beta^y \right)\left(\sum_{z \geq 0} \gamma^z\right)
&= \frac{1}{1-\alpha} \cdot \frac{1}{1-\beta} \cdot \frac{1}{1-\gamma}
&= \frac{1}{(1-\alpha)(1-\beta)(1-\gamma)}
&= \frac{1}{f(1)}
&= \frac{1}{1}
&= \boxed{1}.
\end{align*} $\square$</details></blockquote>
Your solution is wrong
You considered $p(x)=(x-\alpha)(x-\beta)(x-\gamma)$ even though the leading coefficient is not 1, it is 20. $p(x)=20(x-\alpha)(x-\beta)(x-\gamma)$ | [
"<details><summary>Solution</summary>Note that $$ \\sum_{x=0}^{\\infty}\\sum_{y=0}^{\\infty}\\sum_{z=0}^{\\infty} \\alpha^x\\beta^y\\gamma^z = (\\alpha^0+\\alpha^1+\\cdots)(\\beta^0+\\beta^1+\\cdots)(\\gamma^0+\\gamma^1+\\cdots). $$ Since $\\alpha^0=\\beta^0=\\gamma^0=1$ and $|\\alpha|,|\\beta|,|\\gamma| \\l... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1108,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791444.json"
} |
Find the smallest positive integer $N$ such that each of the $101$ intervals $$ [N^2, (N+1)^2), [(N+1)^2, (N+2)^2), \cdots, [(N+100)^2, (N+101)^2) $$ contains at least one multiple of $1001.$ *Proposed by Kyle Lee* | <details><summary>Solution</summary>Intuitively, we see that the smallest desired $N$ is the smallest $N$ that guarantees that the desired result holds true minus a little bit, because the difference between consecutive squares getting smaller and smaller makes it harder and harder to find $101$ consecutive intervals, none of which is in between two multiples of $1001.$ (Alternatively, we can note that $N=101, 201, 302, 404,$ and $k-1,$ where $k$ is the answer we will find, all make the interval $[N^2, (N+1)^2)$ not have a multiple of $1001$ in it.)
The $N$ that guarantees the desired result is $N=500$ because for $N \ge 500,$ the interval $[N^2, (N+1)^2)$ has $\ge 1001$ numbers in it. Now, looking at the residues of $500^2, 499^2, 498^2, \ldots$ modulo $1001,$ we see that it goes $751, 753, 757, \ldots,$ with the difference increasing by $2$ each time. The moment this residue passes from just below $1001$ to just above it (i.e. just above $0$ ), the difference between $N^2$ and $(N+1)^2$ will be small enough to make the interval $[N^2, (N+1)^2)$ not have a multiple of $1001.$ Therefore, we want to stop just before then. We have
\begin{align*}
751+2 \cdot \frac{i(i+1)}{2} < 1001 &\implies i(i+1) < 250
&\implies i \le 15,
\end{align*}
so the answer is $500-15=\boxed{485}.$ $\square$</details> | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1152,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791445.json"
} |
For polynomials $P(x) = a_nx^n + \cdots + a_0$ , let $f(P) = a_n\cdots a_0$ be the product of the coefficients of $P$ . The polynomials $P_1,P_2,P_3,Q$ satisfy $P_1(x) = (x-a)(x-b)$ , $P_2(x) = (x-a)(x-c)$ , $P_3(x) = (x-b)(x-c)$ , $Q(x) = (x-a)(x-b)(x-c)$ for some complex numbers $a,b,c$ . Given $f(Q) = 8$ , $f(P_1) + f(P_2) + f(P_3) = 10$ , and $abc>0$ , find the value of $f(P_1)f(P_2)f(P_3)$ .
*Proposed by Justin Hsieh* | <details><summary>Storage</summary>$P_1 = (x - a)(x - b) = x^2 - (a + b)x + ab$ $P_2 = (x - a)(x - c) = x^2 - (c + a)x + ca$ $P_3 = (x - b)(x - c) = x^2 - (b + c)x + bc$ $Q(x) = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc$ $f(Q) = 8 \implies abc(a + b + c)(ab + bc + ca) = 8$ $\implies (a + b + c)(ab + bc + ca) = \frac{8}{abc}---(1)$ $f(P_1) + f(P_2) + f(P_3) = 10 $ $\implies ab(a + b) + bc(b + c) + ca(c + a) = -10$ $\implies (a + b + c)(ab + bc + ca) - 3abc = -10$ $\implies \frac{8}{abc} - 3abc = -10$ [from $(1)$ ] $\implies (abc - 4)(3abc + 2) = 0$ $\implies abc = 4$ $\therefore f(P_1).f(P_2).f(P_3) = -[(abc)^2(a + b)(b + c)(c + a)] = -[(abc)^2{(a + b + c)(ab + bc + ca) - abc}] = -[16(2 - 4)] = \boxed{\textcolor{blue}{32}}$</details> | [
"<details><summary>solution</summary>We have $f(Q) = (a+b+c)(ab + ac + bc)(abc) = 8$ and $f(P_1)+f_(P_2) + f(P_3) = -(a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2) = 10$ , thus $a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 = -10$ .\n\nNow we see that $(a+b+c)(ab + ac + bc) = a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 + 3ab... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791446.json"
} |
Let $z$ be a complex number that satisfies the equation \[\frac{z-4}{z^2-5z+1} + \frac{2z-4}{2z^2-5z+1} + \frac{z-2}{z^2-3z+1} = \frac{3}{z}.\] Over all possible values of $z$ , find the sum of the values of \[\left| \frac{1}{z^2-5z+1} + \frac{1}{2z^2-5z+1} + \frac{1}{z^2-3z+1} \right|.\]
*Proposed by Justin Hsieh* | <details><summary>Solution</summary>First, multiplying both sides of the given equation by $z$ yields $$ \frac{z^2-4z}{z^2-5z+1} + \frac{2z^2-4z}{2z^2-5z+1} + \frac{z^2-2z}{z^2-3z+1}=3. $$ Then, subtracting $3$ from both sides of the equation yields
\begin{align*}
\frac{z^2-4z}{z^2-5z+1} + \frac{2z^2-4z}{2z^2-5z+1} + \frac{z^2-2z}{z^2-3z+1}-3=0 &\implies \frac{z^2-4z}{z^2-5z+1} -1 + \frac{2z^2-4z}{2z^2-5z+1} -1 + \frac{z^2-2z}{z^2-3z+1}-1=0
&\implies \frac{z^2-4z-z^2+5z-1}{z^2-5z+1} + \frac{2z^2-4z-2z^2+5z-1}{2z^2-5z+1} + \frac{z^2-2z-z^2+z-1}{z^2-3z+1} = 0
&\implies \frac{z-1}{z^2-5z+1} + \frac{z-1}{2z^2-5z+1} + \frac{z-1}{z^2-3z+1} = 0
&\implies (z-1) \cdot \left(\frac{1}{z^2-5z+1} + \frac{1}{2z^2-5z+1} + \frac{1}{z^2-3z+1}\right)=0.
\end{align*}
Therefore, it follows that either $z-1=0$ or $\tfrac{1}{z^2-5z+1} + \tfrac{1}{2z^2-5z+1} + \tfrac{1}{z^2-3z+1}=0.$ The latter case yields $|\tfrac{1}{z^2-5z+1} +\tfrac{1}{2z^2-5z+1} +$ $\tfrac{1}{z^2-3z+1}|=0,$ so we can ignore it. In the former case, we have $z=1,$ so it follows that
\begin{align*}
\left|\frac{1}{z^2-5z+1} + \frac{1}{2z^2-5z+1} + \frac{1}{z^2-3z+1}\right| &= \left|\frac{1}{1-5+1}+\frac{1}{2-5+1} + \frac{1}{1-3+1}\right|
&= \left|\frac{1}{-3} + \frac{1}{-2} + \frac{1}{-1}\right|
&= \left|-\frac{1}{3} - \frac{1}{2} - \frac{1}{1}\right|
&= \frac{1}{3} + \frac{1}{2} + \frac{1}{1}
&= \frac{5}{6} + 1
&= \boxed{\frac{11}{6}}.
\end{align*} $\square$</details> | [
"I fakesolved this by assuming all the terms were equal, for which the only solution is $z=1$ . Therefore the answer is $\\boxed{\\tfrac{11}{6}}$ ",
"oops I did same as above lol",
"<details><summary>DeToasty3's Solution</summary>Multiply both sides of the equation by $z$ . We get $$ \\frac{z^2-4z}{z^2-5z+... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1120,
"boxed": true,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791447.json"
} |
Grant is standing at the beginning of a hallway with infinitely many lockers, numbered, $1, 2, 3, \ldots$ All of the lockers are initially closed. Initially, he has some set $S = \{1, 2, 3, \ldots\}$ .
Every step, for each element $s$ of $S$ , Grant goes through the hallway and opens each locker divisible by $s$ that is closed, and closes each locker divisible by $s$ that is open. Once he does this for all $s$ , he then replaces $S$ with the set of labels of the currently open lockers, and then closes every door again.
After $2022$ steps, $S$ has $n$ integers that divide ${10}^{2022}$ . Find $n$ .
*Proposed by Oliver Hayman* | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
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"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791448.json"
} | |
Find the probability such that when a polynomial in $\mathbb Z_{2027}[x]$ having degree at most $2026$ is chosen uniformly at random,
$$ x^{2027}-x | P^{k}(x) - x \iff 2021 | k $$ (note that $2027$ is prime).
Here $P^k(x)$ denotes $P$ composed with itself $k$ times.
*Proposed by Grant Yu* | This is a nice "cycles"/chains problem!
Let $p=2027$ . Note $P(x)$ can be any map from $\mathbb{Z}_p$ to (a subset of) $\mathbb{Z}_p$ .
Let $ord(z)$ be the minimal $k$ such that $P^k(z)=z$ . The condition is implying $ord(z)$ exists and $lcm_{z\in \mathbb{Z}_p} ord(z)=43\cdot 47$ .
This means all cycles ( $x\to P(x)\to \cdots P^m(x)=x$ has $m\in \{1,43,47,2021\}$ )
Now we casework. Note if we have a $m$ cycle there are $(m-1)!$ ways to orient it.
Suppose there are $a$ self cycles, $b$ cycles of length 43, $c$ cycles of length 47 and $d$ cycles of length 2021. Then we have $a+43b+47c+2021d=2027$ . Also, either $bc\ge 1$ or $d=1$ .
If $d=1$ then this is easy: $a=6$ so we have $\binom{p}{6} \cdot (p-7)!$ so there are $\frac{p!(p-7)!}{6! (p-6)!} = \frac{p!}{6! (p-6)}$ The number of ways to distribute $b$ cycles of $43$ is $\frac{(43b)!}{b! (43!)^b} (42!)^b = \frac{(43b)!}{b! (43)^b}$ Therefore, the number of functions in this case is the sum of $\frac{p!}{a!(43b)!(47c)!} \frac{(43b)!}{b! 43^b} \frac{(47c)!}{c! (47)^c} = \frac{p!}{a!b!c! 43^b 47^c}$ over all $a+43b+47c=2027$ . | [
"2021 was last year y'all gotta know it's not prime :( i lost so much time"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791450.json"
} |
Let $f(n)$ count the number of values $0\le k\le n^2$ such that $43\nmid\binom{n^2}{k}$ . Find the least positive value of $n$ such that $$ 43^{43}\mid f\left(\frac{43^{n}-1}{42}\right) $$ *Proposed by Adam Bertelli* | [
"this is too hard :("
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791452.json"
} | |
Find the largest $c > 0$ such that for all $n \ge 1$ and $a_1,\dots,a_n, b_1,\dots, b_n > 0$ we have
$$ \sum_{j=1}^n a_j^4 \ge c\sum_{k = 1}^n \frac{\left(\sum_{j=1}^k a_jb_{k+1-j}\right)^4}{\left(\sum_{j=1}^k b_j^2j!\right)^2} $$ *Proposed by Grant Yu* | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791453.json"
} | |
An equilateral $12$ -gon has side length $10$ and interior angle measures that alternate between $90^\circ$ , $90^\circ$ , and $270^\circ$ . Compute the area of this $12$ -gon.
*Proposed by Connor Gordon* | <details><summary>Solution</summary>It is easy to see that the $12$ -gon is in the shape of a cross which is made up of $5$ congruent squares, each of side length $10.$ Hence, the area of the $12$ -gon is
\begin{align*}
5 \cdot 10^2 &= 5 \cdot 100
&= \boxed{500}.
\end{align*} $\square$</details> | [
"<details><summary>Solution</summary>Note you can split the $12$ -gon into five squares of side length $10$ , so the total area is $5(10^2)=\\boxed{500}$ [asy]\ndraw((0,0)--(1,0)--(1,1)--(2,1)--(2,0)--(3,0)--(3,-1)--(2,-1)--(2,-2)--(1,-2)--(1,-1)--(0,-1)--cycle);\n[/asy]</details>",
"This is quite literally th... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1110,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791457.json"
} |
A circle has radius $52$ and center $O$ . Points $A$ is on the circle, and point $P$ on $\overline{OA}$ satisfies $OP = 28$ . Point $Q$ is constructed such that $QA = QP = 15$ , and point $B$ is constructed on the circle so that $Q$ is on $\overline{OB}$ . Find $QB$ .
*Proposed by Justin Hsieh* | <details><summary>Solution</summary>First, note that
\begin{align*}
AP &= AO - PO
&= 52 - 28
&= 24.
\end{align*}
Let the foot of the altitude from $Q$ to $\overline{AO}$ be $X.$ Then, since $QA=QP,$ it follows that
\begin{align*}
AX = XP &= \frac{AP}{2}
&= \frac{24}{2}
&= 12.
\end{align*}
Therefore,
\begin{align*}
XO &= XP + PO
&= 12 + 28
&= 40,
\end{align*}
and the Pythagorean Theorem tells us that
\begin{align*}
XQ &= \sqrt{PQ^2-XP^2}
&= \sqrt{15^2-12^2}
&= \sqrt{225-144}
&= \sqrt{81}
&= 9.
\end{align*}
Applying the Pythagorean Theorem once more yields
\begin{align*}
QO &= \sqrt{QX^2+XO^2}
&= \sqrt{9^2+40^2}
&= \sqrt{81+1600}
&= \sqrt{1681}
&= 41,
\end{align*}
so the answer is
\begin{align*}
QB &= OB - OQ
&= 52 - 41
&= \boxed{11}.
\end{align*} $\square$</details> | [
"<details><summary>Quick Projection</summary>Let the projection of $Q$ onto $AO$ be $M.$ Then $OM=40$ and $MQ=9$ thus $OQ=41$ and $QB=\\boxed{11}$</details>"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1110,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791458.json"
} |
Let $ABC$ be an acute triangle with $\angle ABC=60^{\circ}.$ Suppose points $D$ and $E$ are on lines $AB$ and $CB,$ respectively, such that $CDB$ and $AEB$ are equilateral triangles. Given that the positive difference between the perimeters of $CDB$ and $AEB$ is $60$ and $DE=45,$ what is the value of $AB \cdot BC?$ *Proposed by Kyle Lee* | Neat problem!
<details><summary>Solution</summary>Without loss of generality, suppose that $AB>BC.$ Then, it follows that $D$ is inside the triangle and $E$ is outside the triangle.
Now, since the perimeters of $CDB$ and $AEB$ differ by $60,$ it follows that their side lengths differ by $20,$ i.e. $AB-BC=20.$ Then, note that
\begin{align*}
CE=DA &= AB-BD
&= AB-BC
&= 20,
\end{align*}
Furthermore, we have $AE=AB, CD=BC,$ and $AC=DE,$ so applying Ptolemy's Theorem in quadrilateral $ADCE$ (we can apply it because all isosceles trapezoids are cyclic) yields
\begin{align*}
AB \cdot BC &= AE \cdot CD
&= AC \cdot DE - AD \cdot CE
&= 45 \cdot 45 - 20 \cdot 20
&= 2025-400
&= \boxed{1625}.
\end{align*} $\square$</details> | [
"<details><summary>Slick Solution</summary>WLOG let $D$ be outside triangle $ABC$ and $AB=x.$ Therefore we are looking for $x(x+20).$ We know that $AD=20.$ Law of Cosines on $\\triangle BED$ gives us $$ x^2+(x+20)^2-(x)(x+20)=45^2 $$ $$ x^2+20x=\\boxed{1625} $$</details>"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1124,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791461.json"
} |
Circle $\Gamma$ has diameter $\overline{AB}$ with $AB = 6$ . Point $C$ is constructed on line $AB$ so that $AB = BC$ and $A \neq C$ . Let $D$ be on $\Gamma$ so that $\overleftrightarrow{CD}$ is tangent to $\Gamma$ . Compute the distance from line $\overleftrightarrow{AD}$ to the circumcenter of $\triangle ADC$ .
*Proposed by Justin Hsieh* | <details><summary>Solution</summary>Let $X$ be the center of circle $\Gamma,$ let $Y$ be the midpoint of $\overline{AD},$ and let $Z$ be the center of $(ADC).$ We wish to find $YZ.$ First, note that $AX=BX=DX=3,$ and we also have
\begin{align*}
CX &= BC+BX
&= 6+3
&= 9.
\end{align*}
Therefore, by Pythagoras, we have
\begin{align*}
CD &= \sqrt{CX^2-DX^2}
&= \sqrt{9^2-3^2}
&= \sqrt{81-9}
&= \sqrt{72}
&= 6\sqrt{2}.
\end{align*}
Additionally, note that
\begin{align*}
\cos(\angle{CXD}) &= \frac{DX}{CX}
&= \frac{3}{9}
&= \frac{1}{3},
\end{align*}
so it follows that
\begin{align*}
\cos(\angle{AXD}) &= \cos(180^\circ-\angle{CXD})
&= -\cos(\angle{CXD})
&= -\frac{1}{3}.
\end{align*}
Therefore, by the Law of Cosines, we have
\begin{align*}
AD &= \sqrt{AX^2+DX^2 - 2 \cdot AX \cdot DX \cdot \cos(\angle{AXD})}
&= \sqrt{3^2+3^2-2\cdot3\cdot3\cdot(-\tfrac{1}{3})}
&= \sqrt{9+9+6}
&= \sqrt{24}
&= 2\sqrt{6}.
\end{align*}
This means that $AY=\sqrt{6}.$ Now, since
\begin{align*}
AC &= AB+BC
&= 6+6
&= 12,
\end{align*}
we can use Heron's Formula to get $$ [ACD]=12\sqrt{2}. $$ Therefore, it follows from the triangle circumradius formula that
\begin{align*}
AZ &= \frac{AC \cdot CD \cdot DA}{4 \cdot [ACD]}
&= \frac{12\cdot6\sqrt{2}\cdot2\sqrt{6}}{4\cdot12\sqrt{2}}
&= \frac{288\sqrt{3}}{48\sqrt{2}}
&= \frac{6\sqrt{3}}{\sqrt{2}}
&= 3\sqrt{6}.
\end{align*}
Finally, by Pythagoras, the answer is
\begin{align*}
YZ &= \sqrt{AZ^2-AY^2}
&= \sqrt{(3\sqrt{6})^2-(\sqrt{6})^2}
&= \sqrt{54-6}
&= \sqrt{48}
&= \boxed{4\sqrt{3}}.
\end{align*} $\square$</details> | [
"Let the center of circle $\\Gamma$ be $O$ , it is easy to get $OD=3,OC=9,\\cos\\angle{DOC}=\\frac{1}{3}$ Let the midpoint of $AD$ be $F$ , the intersection of perpendicular bisectors of $AD,AC$ , assuming $J$ , the length of $FJ$ is the desired value\n\nSo since $\\cos\\angle{AOD}=-\\cos\\angle{DOC};\\... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
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"answer_score": 1124,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791463.json"
} |
Let $ABC$ be an equilateral triangle of unit side length and suppose $D$ is a point on segment $\overline{BC}$ such that $DB<DC.$ Let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively. Suppose $X$ and $Y$ are the intersections of lines $AB$ and $ND,$ and lines $AC$ and $MD,$ respectively. Given that $XY=4,$ what is the value of $\frac{DB}{DC}?$ *Proposed by Kyle Lee* | [] | [
"origin:aops",
"2022 CMIMC",
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 CMIMC/2791465.json"
} | |
A triangle $\triangle ABC$ satisfies $AB = 13$ , $BC = 14$ , and $AC = 15$ . Inside $\triangle ABC$ are three points $X$ , $Y$ , and $Z$ such that:
- $Y$ is the centroid of $\triangle ABX$
- $Z$ is the centroid of $\triangle BCY$
- $X$ is the centroid of $\triangle CAZ$
What is the area of $\triangle XYZ$ ?
*Proposed by Adam Bertelli* | <details><summary>Solution</summary>We proceed with barycentric coordinates. Let $\triangle{ABC}$ be the reference triangle with $$ A=(1,0,0), B=(0,1,0), C=(0,0,1). $$ Also, let $X=(m, n, 1-m-n)$ for some $0 < m,n < 1.$ Then, since $X, Y,$ and $Z$ are cyclical, it follows that $$ Y=(1-m-n, m, n), Z=(n, 1-m-n, m). $$ Then, since $X$ is the centroid of $\triangle{CAZ},$ we get the system of equations: $$ \begin{cases}
\frac{n+1}{3}=m,
\frac{1-m-n}{3}=n,
\end{cases} $$ which results in $$ (m,n)=\left(\frac{5}{13}, \frac{2}{13}\right). $$ Hence, we have $$ X=\left(\frac{5}{13}, \frac{2}{13}, \frac{6}{13}\right), Y = \left(\frac{6}{13}, \frac{5}{13}, \frac{2}{13}\right), Z=\left(\frac{2}{13}, \frac{6}{13}, \frac{5}{13}\right), $$ so it follows that
\begin{align*}
[XYZ] &= [ABC] \cdot \det\begin{pmatrix} \tfrac{5}{13} & \tfrac{2}{13} & \tfrac{6}{13} \tfrac{6}{13} & \tfrac{5}{13} & \tfrac{2}{13} \tfrac{2}{13} & \tfrac{6}{13} & \tfrac{5}{13}\end{pmatrix}
&= 84 \cdot \frac{1}{13}
&= \boxed{\frac{84}{13}}.
\end{align*} $\square$</details> | [
"Define points $B(0,0), C(14,0), A(5,12).$ Then let $X(a,b)$ . Using the fact that the centroid in the coordinate plane is equal to the average of the x-coordinates and y-coordinates, we get $Y(\\frac{a+5}{3}, \\frac{b+12}3)$ and then $Z(\\frac{a+47}{9}, \\frac{b+12}{9})$ . We can solve again for $X$ , getti... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1136,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791468.json"
} |
Let $\Gamma_1, \Gamma_2, \Gamma_3$ be three pairwise externally tangent circles with radii $1,2,3,$ respectively. A circle passes through the centers of $\Gamma_2$ and $\Gamma_3$ and is externally tangent to $\Gamma_1$ at a point $P.$ Suppose $A$ and $B$ are the centers of $\Gamma_2$ and $\Gamma_3,$ respectively. What is the value of $\frac{{PA}^2}{{PB}^2}?$ *Proposed by Kyle Lee* | Solved with **Sleepy_Head**.
<details><summary>Solution</summary>[asy]
import geometry;
import olympiad;
size(5cm,0);
pair C = (0,0);
pair A = (0,3);
pair B = (4,0);
pair O = (3.5,3.5);
line CO = line(C,O);
path w1 = circle(C,1);
path w2 = circle(A,2);
path w3 = circle(B,3);
pair P = intersectionpoints(CO,w1)[0];
draw(A--B--C--cycle);
draw(A--O--B);
draw(C--O);
draw(w1);
draw(w2);
draw(w3);
label(" $A$ ",A,NW);
label(" $B$ ",B,SE);
label(" $C$ ",C,SW);
label(" $O$ ",O,NE);
label(" $P$ ",P-(0,0.1),W);
clip((-2,-2)--(-2,6)--(6,6)--(6,-2)--cycle);
[/asy]
Let the fourth circle be $\Gamma_4,$ and let the centers of $\Gamma_1$ and $\Gamma_4$ be $C$ and $O$ respectively. Furthermore, let the radius of $\Gamma_4$ be $r.$ Then, since $\Gamma_1$ and $\Gamma_4$ are tangent, it follows that $O, P,$ and $C$ are collinear.
Now, applying Stewart's Theorem in $\triangle{ACO}$ on the cevian $\overline{PA}$ yields $$ PA^2 = \frac{8r}{r+1}, $$ and applying Stewart's Theorem in $\triangle{BCO}$ on the cevian $\overline{PB}$ yields $$ PB^2 = \frac{15r}{r+1}. $$ Hence, the answer is
\begin{align*}
\frac{PA^2}{PB^2} &= \frac{\frac{8r}{r+1}}{\frac{15r}{r+1}}
&= \boxed{\frac{8}{15}}.
\end{align*} $\square$</details> | [
"this one was nice, just set random variables and things cancel out very nicely. ",
"Lol two applications of Stewart’s Theorem kills this problem\n\nThe answer is $\\frac{8}{15}$ ",
"Denote the circle that passes through the centers of $\\Gamma_2$ and $\\Gamma_3$ by $\\Omega$ , denote its center by $O$ ,... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1150,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791470.json"
} |
Let $A$ and $B$ be points on circle $\Gamma$ such that $AB=\sqrt{10}.$ Point $C$ is outside $\Gamma$ such that $\triangle ABC$ is equilateral. Let $D$ be a point on $\Gamma$ and suppose the line through $C$ and $D$ intersects $AB$ and $\Gamma$ again at points $E$ and $F \neq D.$ It is given that points $C, D, E, F$ are collinear in that order and that $CD=DE=EF.$ What is the area of $\Gamma?$ *Proposed by Kyle Lee* | Let altitude from $C$ to $AB$ meets $AB$ at $J$ , it is obvious to see that $CJ=\frac{\sqrt{30}}{2}$ , denote the distance $OJ=m,CD=DC=EF=x$ Since similarity between $\triangle{CJE},\triangle{EJO}, JE^2=CJ*OJ$ , we can now express Pythagorean theorem in $Rt\triangle{COE}, 4x^2+\frac{\sqrt{30}m}{2}+m^2=(\frac{\sqrt{30}}{2}+m)^2$ , after simplifying, we have equation $\sqrt{30}m+15=8x^2$ Now we denote the radius of the circle be $R$ , in $RT\triangle{DOE},x^2+m^2+\frac{\sqrt{30}m}{2}=R^2$ ,in $Rt\triangle{AJO},m^2+\frac{5}{2}=R^2$ , connect these two equations together, we can see that $x^2+\frac{\sqrt{30}m}{2}=\frac{5}{2}, 2x^2+\sqrt{30}m=5$ Now we can write $\sqrt{30}m=5-2x^2$ , put it back to the first equation I wrote, it is $20-2x^2=8x^2, x=\sqrt{2}$ Hence, we can get that $m=\frac{\sqrt{30}}{30}$ , as we know, $m^2+\frac{5}{2}=R^2, R^2=\frac{1}{30}+\frac{5}{2}=\frac{38}{15}$ , the desired answer is $\boxed{\frac{38\pi}{15}}$ | [
"<details><summary>kinda bashy</summary>Let the center of the circle be $O$ . Note that $OE$ perpendicularly bisects DF. Now AOBC is a kite, so $CO$ perpendicularly bisects $AB$ too. Let the midpoint of $AB$ be $M$ . Let the radius of the circle be $R$ , and let $CD=DE=EF=x$ . $\\Rightarrow OE^2 = R^... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791472.json"
} |
In triangle $ABC$ , let $I, O, H$ be the incenter, circumcenter and orthocenter, respectively. Suppose that $AI = 11$ and $AO = AH = 13$ . Find $OH$ .
*Proposed by Kevin You* | <blockquote>Nice problem. My solution is the same as **dchenmathcounts**'s but it has a different finish.
<details><summary>Sol</summary>$AH=AO$ means $2R\cos A=R$ so $\angle A=60^\circ$ . It's well-known that when $\angle A=60^\circ$ , $BHIOC$ is cyclic with center $M$ , where $M$ is the midpoint of minor arc $\widehat{BC}$ . In particular the radius of this circle is the circumradius of $(ABC)$ , which is $13$ . Hence, $AM=AI+IM=AI+13=24$ . But $AOMH$ is a rhombus, so the parallelogram law gives $OH^2+AM^2=4\cdot 13^2\implies OH=10$ . The end.</details></blockquote>
Instead the formula should be $2R|\cos A| = AH$ ? So, there are two possible angles of $A$ .
| [
"<details><summary>Good stuff</summary>We use complex coordinates. Let $O=0$ and note $H=a+b+c$ . Then $|O-a|=|H-a|$ or $|a|=|b+c|=13$ . Trig here gives $\\angle A = 60^{\\circ}$ . Now let $D$ be the arc midpoint of $BC$ (note this is equal to $|b+c|$ ) and note $AODH$ is a parallelogram. As $AD=AI+I... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791475.json"
} |
Let $\Gamma_1$ and $\Gamma_2$ be two circles with radii $r_1$ and $r_2,$ respectively, where $r_1>r_2.$ Suppose $\Gamma_1$ and $\Gamma_2$ intersect at two distinct points $A$ and $B.$ A point $C$ is selected on ray $\overrightarrow{AB},$ past $B,$ and the tangents to $\Gamma_1$ and $\Gamma_2$ from $C$ are marked as points $P$ and $Q,$ respectively. Suppose that $\Gamma_2$ passes through the center of $\Gamma_1$ and that points $P, B, Q$ are collinear in that order, with $PB=3$ and $QB=2.$ What is the length of $AB?$ *Proposed by Kyle Lee* | Note that since $C$ lies on the radical axis, we have $CP=CQ$ . Furthermore, due to the tangencies, $$ \angle PAB=\angle BPC,\angle QAB=\angle BQC. $$ Since $\angle BPC=\angle BQC$ as well since $\triangle CPQ$ is isosceles, we have that $AB$ is an angle bisector in $\triangle PAQ$ .
Thus, we remove all the fluff and the problem is essentially asking
<blockquote>In triangle $\triangle APQ$ , $B$ is the foot of the angle bisector from $A$ with $PB=3,PQ=2$ . If the circumcenter of $\triangle PAB$ lies on the circumcircle of $\triangle QAB$ , find $AB$ .</blockquote>
Note that $$ \angle AQP=180-\angle AOB=180-2\angle APQ, $$ which means that $AP=PQ$ . Thus, $AQ=5$ and by angle bisector theorem $AP=\frac{15}{2}$ . Finally, by Stewart's theorem the length of the angle bisector is $$ \frac{3\sqrt{14}}{2}. $$ | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791476.json"
} |
In acute $\triangle ABC,$ let $I$ denote the incenter and suppose that line $AI$ intersects segment $BC$ at a point $D.$ Given that $AI=3, ID=2,$ and $BI^2+CI^2=64,$ compute $BC^2.$ *Proposed by Kyle Lee* | One of my favorite computational geometry problems of all time.
<span style="color:blue">**First Solution, Trigonometry**</span> By the Law of Sines on $\triangle AIB$ , $$ 3 = AI = \frac{AB\sin \frac B2}{\sin\left(\frac A2+\frac B2\right)} = 4R\sin \frac B2 \sin \frac C2 $$ via the double angle formula. Similarly, by the Law of Sines on $\triangle BID$ , $$ 2 = ID = \frac{BI \sin \frac B2}{\sin\left(B + \frac A2\right)} = \frac{4R\sin \frac A2\sin \frac B2 \sin \frac C2}{\sin\left(B+\frac A2\right)}. $$ Combining the two equations, note that $$ \frac{3 \sin \frac A2}{\sin\left(B+\frac A2\right)} = 2 \iff 2\sin\left(B+\frac A2\right) = 3\sin \frac A2. $$ So sum-to-product yields
\begin{align*}
\sin \frac A2 &= 2\left[\sin\left(B+\frac A2\right)-\sin \frac A2\right]
&= 4\cos\left(\frac{A+B}2\right)\sin \frac B2
&= 4 \sin \frac C2 \sin \frac B2. \ (\diamondsuit)
\end{align*}
This is powerful, as the third condition now collapses:
\begin{align*}
BI^2+CI^2&= 64
\iff 16R^2 \sin^2\frac A2\left(\sin^2\frac B2+\sin^2 \frac C2\right)&=64
\iff 256R^2 \sin^2 \frac B2 \sin^2 \frac C2\left(\sin^2\frac B2+\sin^2 \frac C2\right)&=64
\iff \sin^2 \frac B2 + \sin^2 \frac C2 &= \frac 49.
\end{align*}
Going back to $(\diamondsuit)$ , we also have
\begin{align*}
\cos\left(\frac B2+\frac C2\right) &= 4\sin \frac C2 \sin \frac B2
\cos \frac C2 \cos \frac B2 - \sin \frac C2 \sin \frac B2 &= 4 \sin \frac C2 \sin \frac B2
\cos \frac C2 \cos \frac B2 &= 5 \sin \frac C2 \sin \frac B2
\sqrt{\left(1-\sin^2 \frac C2\right)\left(1-\sin^2 \frac B2\right)} &= 5 \sin \frac C2 \sin \frac B2
\sqrt{\frac 59 + \sin^2 \frac B2 \sin^2 \frac C2} &= 5 \sin \frac C2 \sin \frac B2.
\end{align*}
Solving, $\sin \frac B2 \sin \frac C2 = \frac{\sqrt{30}}{36}$ , and $\sin \frac A2 = \frac{\sqrt{30}}9$ . So $$ \sin A = 2 \cdot \frac{\sqrt{30}}9 \cdot \frac{\sqrt{51}}9 = \frac{2\sqrt{170}}{27}, $$ and $$ R = \frac 3{4 \sin \frac B2 \sin \frac C2} = \frac{9\sqrt{30}}{10}, $$ so $$ BC^2 = \left(\frac{9\sqrt{30}}{10} \cdot \frac{2\sqrt{170}}{27}\cdot 2\right)^2 = \boxed{\frac{272}3}. $$ **<span style="color:blue">Second Solution**</span> Many of the ideas in this solution are very similar to those above: in fact, in some manner this solution is equivalent, if not restated differently.
Stewart's theorem on triangles $ABD$ and $ACD$ yield
\begin{align*}
2 \cdot 5 \cdot 3 + BI^2 \cdot 5 &= 3 \cdot BD^2 + 2 \cdot AB^2
2 \cdot 5 \cdot 3 + CI^2 \cdot 5 &= 3 \cdot CD^2 + 2 \cdot BC^2.
\end{align*}
Summing and writing in standard notation as well as using the third condition, $$ 60+5 \cdot 64 = 2(b^2+c^2)+\frac{3a^2(b^2+c^2)}{(b+c)^2}. \ (\dagger) $$ Furthermore, the Angle Bisector Theorem on triangle $ABD$ and $ABC$ yield $\frac{b+c}a = \frac 32$ , so $(\dagger)$ becomes $$ 380 = 2(b^2+c^2)+\frac 43(b^2+c^2) \iff b^2+c^2=114. $$ Next, Stewart's on $ABC$ (or, for triangle formula geeks, the angle bisector length formula), \begin{align*}
bc-\frac{a^2bc}{(b+c)^2} &= 25
\frac{(b+c)^2(bc-25)}{bc} &= a^2
\frac{\frac 94(bc-25)}{bc} &= 1
bc &= 45.
\end{align*}
Then $$ a^2=\frac 49(b+c)^2 = \frac 49 \cdot 204 = \boxed{\frac{272}3}. $$ | [
"<details><summary>Solution</summary>Let $AB = 3y$ and $AC = 3z.$ Then it follows that $BD = 2y$ and $CD = 2z$ . By Stewart's on $\\triangle{ABD}$ we have\n\\[ IB^2 = \\frac{(2y)^2\\cdot 3 + (3y)^2 \\cdot 2 - 5\\cdot 3 \\cdot 2}{5} = 6y^2 - 6\\]\nSimilarly $IC^2 = 6z^2-6,$ so\n\\[IB^2+IC^2 = 6y^2-6+6z^2 ... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1060,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791478.json"
} |
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ . Rays $\displaystyle \overrightarrow{OB}$ and $\displaystyle \overrightarrow{DC}$ intersect at $E$ , and rays $\displaystyle \overrightarrow{OC}$ and $\displaystyle \overrightarrow{AB}$ intersect at $F$ . Suppose that $AE = EC = CF = 4$ , and the circumcircle of $ODE$ bisects $\overline{BF}$ . Find the area of triangle $ADF$ .
*Proposed by Howard Halim* | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791480.json"
} | |
A particle starts at $(0,0,0)$ in three-dimensional space. Each second, it randomly selects one of the eight lattice points a distance of $\sqrt{3}$ from its current location and moves to that point. What is the probability that, after two seconds, the particle is a distance of $2\sqrt{2}$ from its original location?
*Proposed by Connor Gordon* | <details><summary>Solution</summary>First, note that there are $$ 8^2=64 $$ possible paths that the particle takes in $2$ seconds.
The given condition of the particle traveling to a lattice point a distance of $\sqrt{3}$ from its current location is equivalent to the particle traveling along the vector $$ \langle a_x, a_y, a_z \rangle, $$ from its current location, where $$ a_x,a_y,a_z \in \{-1,1\}. $$ Also, it is not difficult to see that the ending point of the particle after two seconds must be of the form $(\pm 2, \pm 2, 0)$ or permutations. Clearly, for any of these
\begin{align*}
2^2 \cdot 3 &= 4 \cdot 3
&= 12
\end{align*}
permutations for the ending point, there are exactly $2$ two-step paths that end up there (the same two steps but in $2$ possible orders), so the answer is
\begin{align*}
\frac{12\cdot2}{64} &= \frac{24}{64}
&= \boxed{\frac{3}{8}}.
\end{align*} $\square$</details> | [
"<details><summary>Solution</summary>Note that the particle moves by $\\pm1$ unit in each of the $x,y,z$ coordinates each second. To be a distance of $2\\sqrt{2}$ from its original location after two seconds, it must be at $(\\pm2,\\pm2,0)$ or permutations. This requires it to travel the same direction twic... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
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"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791481.json"
} |
Starting with a $5 \times 5$ grid, choose a $4 \times 4$ square in it. Then, choose a $3 \times 3$ square in the $4 \times 4$ square, and a $2 \times 2$ square in the $3 \times 3$ square, and a $1 \times 1$ square in the $2 \times 2$ square. Assuming all squares chosen are made of unit squares inside the grid. In how many ways can the squares be chosen so that the final $1 \times 1$ square is the center of the original $5 \times 5$ grid?
*Proposed by Nancy Kuang* | <details><summary>Solution</summary>Let us assume that the $1 \times 1$ square is in the bottom right corner of the $2 \times 2$ square and, by symmetry, we just need to multiply by $4$ in the end.
Now, if the $2 \times 2$ square is in the bottom right, bottom left, top left, and top right of the $3 \times 3$ square, then there are $1, 2, 4,$ and $2$ possible placements of the $4 \times 4$ square respectively, so the answer is
\begin{align*}
4 \cdot (1+2+4+2) &= 4 \cdot 9
&= \boxed{36}.
\end{align*} $\square$</details> | [
"<details><summary>Casework</summary>It doesn't matter which $4\\times4$ square you choose. \nWLOG let's assume you chose the bottom left $4\\times4$ square. Then you have $4$ cases.\n\n\n- Top right $3\\times3$ square\n- Top left $3\\times3$ square\n- Bottom right $3\\times3$ square\n- Bottom left $3\... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1118,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791483.json"
} |
We say that a set $S$ of $3$ unit squares is \textit{commutable} if $S = \{s_1,s_2,s_3\}$ for some $s_1,s_2,s_3$ where $s_2$ shares a side with each of $s_1,s_3$ . How many ways are there to partition a $3\times 3$ grid of unit squares into $3$ pairwise disjoint commutable sets?
*Proposed by Srinivasan Sathiamurthy* | <details><summary>Solution</summary>Let us define an L as a commutable set in which $s_1$ and $s_3$ share sides with $s_2$ that are *adjacent* and let an I be a commutable set in which $s_1$ and $s_3$ share sides with $s_2$ that are [/i]opposite[/i]. (Note that the letters L and I resemble the shapes of these two types of commutable sets.) It is trivial that any commutable set must be of one of these two types.
We now consider two cases:**Case 1:** The $3 \times 3$ grid is partitioned into $3$ I's. Clearly, the total for this case is $\underline{2}.$ **Case 2:** The $3 \times 3$ grid is partitioned into $2$ L's and $1$ I. There are $4$ ways to choose which side of the $3 \times 3$ grid that the I lies along and $2$ ways to partition the remaining $2 \times 3$ grid into two L's, for a total of $$ 4 \cdot 2 = \underline{8}. $$ Adding up the totals from these two cases yields an answer of $$ 2+8=\boxed{10}. $$ $\square$</details> | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1142,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791485.json"
} |
Dilhan is running around a track for $12$ laps. If halfway through a lap, Dilhan has his phone on him, he has a $\frac{1}{3}$ chance to drop it there. If Dilhan runs past his phone on the ground, he will attempt to pick it up with a $\frac{2}{3}$ chance of success, and won't drop it for the rest of the lap. He starts with his phone at the start of the 5K, what is the chance he still has it when he finished the 5K?
*Proposed by Zack Lee, Daniel Li, Dilhan Salgado* | <details><summary>Solution</summary>If, before running the $12$ th lap, Dilhan has his phone, then there is a $\tfrac{2}{3}$ chance that he will still have it by the end of the $12$ th lap.
On the other hand, if Dilhan *doesn't* have his phone before running the $12$ th lap, then there is still a $\tfrac{2}{3}$ chance that he will have it by the end of that lap.
Hence, the probability that he still has his phone once he has finished the 5K is $$ \boxed{\frac{2}{3}}, $$ regardless. $\square$</details> | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1116,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791486.json"
} |
Daniel, Ethan, and Zack are playing a multi-round game of Tetris. Whoever wins $11$ rounds first is crowned the champion. However Zack is trying to pull off a "reverse-sweep", where (at-least) one of the other two players first hits $10$ wins while Zack is still at $0$ , but Zack still ends up being the first to reach $11$ . How many possible sequences of round wins can lead to Zack pulling off a reverse sweep?
*Proposed by Dilhan Salgado* | <details><summary>Solution</summary>There are $2$ ways to choose which person hits $10$ while Zack is still at $0$ (both other players can do so but we will correct for overcount later), call them Player $A.$ Then, we can view the problem as arranging $21$ $X$ 's and $10$ $Y$ 's if we use an $X$ to represent a win by Zack or Player $A$ (the first $10$ $X$ 's represent a win by Player $A$ and the next $11$ $X$ 's represent a win by Zack) and if we use a $Y$ to represent a win by the third player (if some $Y$ 's come after the last $X,$ then we can scrap those but having $10$ $Y$ 's makes sure that the third player doesn't win before Zack does). Thus, our count is now $$ 2 \cdot \binom{31}{10}. $$ However, the situation in which both other players reach $10$ while Zack is still at $0$ is counted twice, so our final answer is $$ \boxed{2\binom{31}{10}-\binom{20}{10}}. $$ $\square$</details> | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
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"answer_score": 1154,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791487.json"
} |
A sequence of pairwise distinct positive integers is called averaging if each term after the first two is the average of the previous two terms. Let $M$ be the maximum possible number of terms in an averaging sequence in which every term is less than or equal to $2022$ and let $N$ be the number of such distinct sequences (every term less than or equal to $2022$ ) with exactly $M$ terms. What is $M+N?$ (Two sequences $a_1, a_2, \cdots, a_n$ and $b_1, b_2, \cdots, b_n$ are said to be distinct if $a_i \neq b_i$ for some integer $1 \leq i \leq n$ ).
*Proposed by Kyle Lee* | <details><summary>Solution</summary>Note that the length of an averaging sequence with first two terms $a$ and $b$ for $a, b \in \mathbb{Z}_{\ge1}$ is $2+\nu_2|a-b|$ because after the first two terms, the positive difference between consecutive terms of the sequence is cut in half every time, which can happen at most $\nu_2|a-b|$ times.
Since $2^{10}<2022<2^{11},$ it follows that the maximum possible value of $\nu_2|a-b|$ for $a,b \in \{1, 2, \ldots, 2022\}$ is $10,$ which occurs only when $|a-b|=1024,$ so we know that $M=12.$
Clearly, $|a-b|$ cannot be any multiple of $1024$ larger than $1024,$ so to compute $N,$ it suffices to find the number of pairs $(a,b)$ where $a,b \in \{1, 2, \ldots, 2022\}$ and $|a-b|=1024.$ It is not difficult to see that the smaller of the two can be anywhere from $1$ to $998$ and it can be either $a$ or $b,$ so we get $$ N=998\cdot2=1996. $$ Hence, the answer is
\begin{align*}
M+N &= 12 + 1996
&= \boxed{2008}.
\end{align*} $\square$</details> | [] | [
"origin:aops",
"2022 CMIMC",
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"boxed": true,
"end_of_proof": true,
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"path": "Contest Collections/2022 Contests/2022 CMIMC/2791489.json"
} |
For a family gathering, $8$ people order one dish each. The family sits around a circular table. Find the number of ways to place the dishes so that each person’s dish is either to the left, right, or directly in front of them.
*Proposed by Nicole Sim* | <details><summary>Solution</summary>We proceed with casework on the number of people who have their own dish. This number clearly must be even (because the remaining people, if any, have to split into pairs of adjacent people who have each other's dishes).**0 people have their own dish:** Then, either the family's dish assignment follows a cyclic pattern around the circle in which case there are $2$ possibilities, or they split up into four pairs of adjacent people who have each other's dishes with also yields $2$ possibilities, for a total of $$ 2+2=\underline{4} $$ for this case.**2 people have their own dish:** If the two people are adjacent, then there are $8$ possibilities, and if they are separated by two people, then there are also $8$ possibilities, for a total of $$ 8+8=\underline{16} $$ for this case.**4 people have their own dish:** We consider the configuration of the pairs formed by the remaining people because this is easier. If the two pairs are next to each other, then there are $8$ possibilities, if the two pairs are separated by one person, there are also $8$ possibilities, and if the two pairs are separated by $2$ people, then there are $4$ possibilities, for a total of $$ 8+8+4=\underline{20}. $$ **6 people have their own dish:** The remaining two people must be adjacent because they have to pair up with each other, so the total for this case is $\underline{8}.$ **8 people have their own dish:** The total for this case is obviously just $\underline{1}.$ Hence, summing the totals of these cases yields
\begin{align*}
4+16+20+8+1=\boxed{49}.
\end{align*} $\square$</details> | [
"oops casework",
"<details><summary>Standard Casework</summary>There are many cases so let's break it down. We first have $2$ easy cases where everyone rotates right or everyone rotates left. \nThen, we need to count the number of *swaps* that occur which is when $2$ adjacent people will swap their dishes. A ... | [
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"boxed": true,
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"n_reply": 3,
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} |
The CMU Kiltie Band is attempting to crash a helicopter via grappling hook. The helicopter starts parallel (angle $0$ degrees) to the ground. Each time the band members pull the hook, they tilt the helicopter forward by either $x$ or $x+1$ degrees, with equal probability, if the helicopter is currently at an angle $x$ degrees with the ground. Causing the helicopter to tilt to $90$ degrees or beyond will crash the helicopter. Find the expected number of times the band must pull the hook in order to crash the helicopter.
*Proposed by Justin Hsieh* | <details><summary>Solution</summary>First, note that the expected number of times the band must pull the hook in order to tilt it to a $1$ degree angle is $2$ because $0$ can go either to itself or to $1,$ each with $\tfrac{1}{2}$ probability.
Then, it is not difficult to see that each of the numbers $32, 33, \ldots, 63$ are equal likely to be the tilt of the helicopter after the hook has been pulled $5$ times from a starting tilt of $1$ degree. For each of $32, 33, \ldots, 44,$ it will take exactly two more hook-pulls to reach a tilt of $90$ degrees or more, and for each of $45, 46, \ldots, 63,$ it will take exactly one more hook-pull to reach a tilt of $90$ degrees or more, so the answer is
\begin{align*}
2 + \frac{13}{32} \cdot 7 + \frac{19}{32} \cdot 6
&= \frac{64}{32} + \frac{91}{32} + \frac{114}{32}
&= \boxed{\frac{269}{32}}.
\end{align*} $\square$</details> | [
"<details><summary>Solution</summary>Let $E_n$ be the expected number of pulls if it is currently $n$ degrees. We have $$ E_n = \\tfrac{1}{2}E_{2n} + \\tfrac{1}{2}E_{2n+1}+1. $$ Now, we have $0=E_{90} = E_{91}=\\cdots$ so $1= E_{45} = E_{46} = \\cdots = E_{89}$ . Now, we can compute $E_{44} = 3/2$ . Con... | [
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} |
At CMIMC headquarters, there is a row of $n$ lightbulbs, each of which is connected to a light switch. Daniel the electrician knows that exactly one of the switches doesn't work, and needs to find out which one. Every second, he can do exactly one of 3 things:
- Flip a switch, changing the lightbulb from off/on or on/off (unless the switch is broken).
- Check if a given lightbulb is on or off.
- Measure the total electricity usage of all the lightbulbs, which tells him exactly how many are currently on.
Initially, all the lightbulbs are off. Daniel was given the very difficult task of finding the broken switch in at most $n$ seconds, but fortunately he showed up to work 10 seconds early today. What is the largest possible value $n$ such that he can complete his task on time?
*Proposed by Adam Bertelli* | [] | [
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Barry has a standard die containing the numbers 1-6 on its faces.
He rolls the die continuously, keeping track of the sum of the numbers he has rolled so far, starting from 0. Let $E_n$ be the expected number of time he needs to until his recorded sum is at least $n$ .
It turns out that there exist positive reals $a, b$ such that $$ \lim_{n \rightarrow \infty} E_n - (an + b) = 0 $$ Find $(a,b)$ .
*Proposed by Dilhan Salgado* | [
"picture this u are a professor in descriptive set theory and giving a talk at your local math cl.ub. You rae talking about unfriendly colorings of graphs and then all of a sudden, you see him there. Barack Obama. You say: holy cow, i never expected barack obama to come to my talk! And then barack obama says given ... | [
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In a class of $12$ students, no two people are the same height. Compute the total number of ways for the students to arrange themselves in a line such that:
- for all $1 < i < 12$ , the person in the $i$ -th position (with the leftmost position being $1$ ) is taller than exactly $i\pmod 3$ of their adjacent neighbors, and
- the students standing at positions which are multiples of $3$ are strictly increasing in height from left to right.
*Proposed by Nancy Kuang* | <details><summary>Solution</summary>First, note that from the first condition, the heights of the people in line have to follow this pattern: $$ \underline{\phantom{n}} < \underline{\phantom{n}} > \underline{\phantom{n}} < \underline{\phantom{n}} < \underline{\phantom{n}} > \underline{\phantom{n}} < \underline{\phantom{n}} < \underline{\phantom{n}} > \underline{\phantom{n}} < \underline{\phantom{n}} < \underline{\phantom{n}} > \underline{\phantom{n}} $$ Now we will provide a method that constructs a line that satisfies the requirements.
- First, pick two students who are not yet in line. Do not pick the two shortest students not yet in line.
- Put the shorter of the two students in the next spot in line, then put in the taller student.
- Take the shortest student still not in line and place them next.
- Repeat this process until all students have been placed in line.
For example, we could pick students $1$ and $3$ to be the line, and then student $2$ will join the line next. (For our purposes, student $i$ is taller than student $j$ if and only if $i > j$ .) Then, we could pick students $10$ and $12$ , and then student $4$ would join. Then, we could pick $6$ and $9$ , and then $5$ . Last, we pick $8$ and $11$ , making $7$ join last. After each iteration, the line looks like this: $$ 1,3,2 \qquad [4,5,6,7,8,9,10,11,12] $$ $$ 1,3,2,10,12,4 \qquad [5,6,7,8,9,11] $$ $$ 1,3,2,10,12,4,6,9,5 \qquad [7,8,11] $$ $$ 1,3,2,10,12,4,6,9,5,8,11,7 \quad\, [\ ] $$ We show that this construction counts each valid line exactly once, and counts each invalid line zero times. First of all, making a different choice at one iteration results in different lines, so no line is counted more than once. This line is also incapable of making invalid lines: both conditions must be satisfied based on the nature of the construction. Finally, any given valid line can be constructed, the reason for which we will motivate by an example:
Suppose the given valid line starts off as $1,2,3$ . This is not allowed because the third person has to be shorter than the second person. This fact allows us to complete the first step. Now suppose the given line starts with students $6$ and $9$ (this is allowed). In a valid line, they must be in the order $6,9$ , in order to follow the pattern, so we can complete the second step. Next, student $1$ (the shortest student not yet in line) cannot go anywhere other than right after $9$ : putting them later would violate the pattern and/or condition two. Therefore the given line must include $1$ as coming next, so we can complete the third step. Finally, all of the remaining students are taller than the third student in line, so the rest of the given line is not affected by the first three students. This allows us to loop back on our method until all students have been placed.
The total number of valid lines is then equal to the total number of constructions. There are $\binom{n}{2} - 1$ choices at each iteration, if there are $n$ people left to place in the line. Thus the total number of lines is $$ \left(\binom{12}{2} - 1 \right) \left(\binom{9}{2} - 1 \right) \left(\binom{6}{2} - 1 \right) \left(\binom{3}{2} - 1 \right) = 65 \times 35 \times 14 \times 2 = \boxed{63700}. $$</details> | [] | [
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} |
Daniel has a (mostly) standard deck of 54 cards, consisting of 4 suits each containing the ranks 1 to 13 as well as 2 jokers. Daniel plays the following game: He shuffles the deck uniformly randomly and then takes all of the cards that end up strictly between the two jokers. He then sums up the ranks of all the cards he has taken and calls that his score.
Let $p$ be the probability that his score is a multiple of 13. There exists relatively prime positive integers $a$ and $b,$ with $b$ as small as possible, such that $|p - a/b| < 10^{-10}.$ What is $a/b?$ *Proposed by Dilhan Salgado, Daniel Li* | <details><summary>solution</summary>The key is to build a generating function with two formal variables: $x$ . whose exponent represents the number of cards strictly between the two jokers, and $y$ . whose exponent represents the sum of the ranks of the cards strictly between the jokers.
To begin, note that if there are $a_1$ $1$ 's, $a_2$ $2$ 's, ..., $a_{13}$ $13$ 's between the two jokers and $S := a_1 + a_2 + \dots + a_{13}$ , then there are $\frac{S!}{a_1! a_2! \dots a_{13}!}$ ways to arrange the cards between the jokers, and $\frac{(53-S)!}{(4-a_1)! (4-a_2)! \dots (4-a_{13})!}$ ways to arrange the rest of the cards, since we may consider all cards between the jokers inclusive as one big card (to make $53-S$ total cards with $4-a_1$ $1$ 's, $4-a_2$ $2$ 's, and so on). Furthermore, there are a total of $\frac{54!}{4!^{13} \cdot 2}$ ways to arrange the cards. Hence if
\[ f(x, y) = \frac{4!^{13} \cdot 2}{54!} \sum_{0 \le a_1, a_2, \dots, a_{13} \le 4} \frac{S!}{a_1! a_2! \dots a_{13}!} \cdot \frac{(53-S)!}{(4-a_1)! (4-a_2)! \dots (4-a_{13})!} x^S y^{a_1 + 2a_2 + \dots +13a_{13}},\]
then the answer is the sum of the coefficients of the terms with exponent of $y$ divisible by $13$ (the exponent of $x$ doesn't matter). We can rewrite this as
\begin{align*}
f(x, y) &= \frac{4!^{13} \cdot 2}{54!} \sum_{0 \le a_1, a_2, \dots, a_{13} \le 4} \frac{53!}{\binom{53}S} \cdot \frac1{4!^{13}} \binom{4}{a_1} \binom{4}{a_2} \dots \binom{4}{a_{13}} x^S y^{a_1 + 2a_2 + \dots +13a_{13}},
&= \frac1{27} \sum_{0 \le a_1, a_2, \dots, a_{13} \le 4} \frac{x^S}{\binom{53}S} \binom{4}{a_1} \left(y^1 \right)^{a_1} \cdot \binom{4}{a_2} \left(y^2 \right)^{a_2} \cdot \dots \cdot \binom{4}{a_{13}} \left(y^{13} \right)^{a_{13}}.
\end{align*}
Thus if
\[g(x, y) := \frac1{27} \left( \sum_{a_1=0}^4 x \binom{4}{a_1} y^1 \right) \left( \sum_{a_2=0}^4 x \binom{4}{a_2} y^2 \right) \dots \left( \sum_{a_{13}=0}^4 x \binom{4}{a_{13}} y^{13} \right) = \frac1{27}(1+xy)^4 (1+xy^2)^4 \dots (1+xy^{13})^4\]
then $g$ has the same terms as $f$ except with each power of $x$ of the form $x^S$ having a coefficient $\binom{53}{S}$ times larger. Now, treating $g$ like a polynomial in $y$ alone, we can compute the sum of the coefficients of the terms with an exponent of $y$ divisible by $13$ to obtain a polynomial in $x$ , then change each term of the form $x^S$ to $\frac1{\binom{53}{S}}$ to obtain our answer. This is pretty easy with the root of unity filter: let $\omega = e^{2\pi i / 13}$ . Then the desired polynomial in $x$ is
\[ h(x) := \frac{g(x, 1) + g(x, \omega) + \dots + g(x, \omega^{12})}{13} \]
For any primitive $13$ th root of unity $\zeta$ , $1, \zeta^2, \dots, \zeta^{12}$ covers all possible roots of unity, so
\begin{align*}
g(x, \zeta) &= \frac1{27} ((1+x\zeta) (1+x\zeta^2) \dots (1+x\zeta^{13}))^4
&= \frac1{27} \left( -x^{13} \left(-\frac1x - \zeta\right) \left(-\frac1x - \zeta^2\right) \dots \left(-\frac1x - \zeta^{13}\right) \right)^4
&= \frac1{27} \left( -x^{13} \left(\left(-\frac1x\right)^{13} - 1\right) \right)^4
&= \frac1{27} (x^{13} + 1)^4.
\end{align*}
But $13$ is prime, so all such $\omega^k$ with $0<k<13$ are primitive, hence
\[ h(x) = \frac{g(x, 1) + 12 g(x, \omega)}{13} = \frac1{13} \left(\frac{(1+x)^{52}}{27} + 12 \cdot \frac{(x^{13} + 1)^4}{27} \right).\]
Now it suffices to change each $x^S$ to $\frac1{\binom{53}{S}}$ and sum the resulting expression.
For the first term, each monomial is of the form $\frac1{13 \cdot 27} \binom{52}{S} x^S$ , hence each monomial contributes
\[ \frac1{13 \cdot 27} \cdot \frac{\binom{52}{S}}{\binom{53}{S}} = \frac1{13 \cdot 27} \cdot \frac{52! \cdot S! \cdot (53 - S)!}{53! \cdot S! \cdot (52 - S)!} = \frac{53 - S}{13 \cdot 27 \cdot 53}.\]
Summing from $S = 0$ to $52$ , this contributes a total of $\frac1{13 \cdot 27 \cdot 53} \cdot \frac{53 \cdot 54}{2} = \frac1{13}$ .
Now for the second term, we can expand normally to find that the expression is
\[ \frac{12}{13 \cdot 27} \left( 1 + 4x^{13} + 6x^{26} + 4x^{39} + x^{52} \right).\]
Ignoring the constant at the front, the first and the last monomial give $1$ and $\frac1{53}$ , respectively, while the middle three contribute less than $\frac{9}{\binom{53}{12}} \ll 10^{-9}$ (not enough), hence the answer is
\[ \frac1{13} + \frac{12}{13 \cdot 27} \left(1 + \frac1{53} \right) = \frac1{13} + \frac{12}{13 \cdot 27} \cdot \frac{54}{53} = \frac{53 + 24}{689} = \frac{77}{689}. \]</details> | [
"Solved this as a 1st grader on the CMIMC.\n\n<details><summary>Combinatorial Approach</summary>Let the number of cards between the jokers be $0\\leq d\\leq 52$ . Call a configuration good if the score is a multiple of 13.\n\nFor $d \\neq 0,13,26,39,52$ , we will use the cycle approach. Consider one arrangement o... | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 CMIMC/2791498.json"
} |
Find all integer values of $x$ for which the value of the expression
\[x^2+6x+33\]
is a perfect square. | $(x+3)^2+24 =m^2$ $(m-x-3)(m+x+3) = 24$ $x = (-8,-4,-2,2)$ | [
"StarLex1, the question asks for all <u>integer</u> solutions. You missed the negative solutions. (Of course your method also yields those as well.)",
"<blockquote>Find all integer values of $x$ for which the value of the expression \n\\[x^2+6x+33\\] \nis a perfect square.</blockquote>\nToo easy for HSO\nShould... | [
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} |
Let $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$ . Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area.
If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$ , determine the area of the square. | Area of paralelogram $BEDZ$ is equal to $DE*1=DE$ . On the other hand same area is equal to $EB*AB$ . Area of whole square is $AB^2$ and Is also equal to sum of area of triangles $EAD, ZCB$ and parallelogram $BEDZ$ so we have $AB^2=3AB*EB$ . Since $DE=EB*AB$ we have $DE=\frac{AB^2}{3}$ . From triangle $EAD$ we have that $DE=\frac{AB\sqrt{13}}{3}$ , so from last two equalities we get $AB=\sqrt{13}$ and so the area of square is $13$ . | [] | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785511.json"
} |
If $x,y$ are real numbers with $x+y\geqslant 0$ , determine the minimum value of the expression
\[K=x^5+y^5-x^4y-xy^4+x^2+4x+7\]
For which values of $x,y$ does $K$ take its minimum value?
| $K=(x+y)((x-y)^2(x^2+y^2))+(x+2)^2+3\ge 0+0+3=3$ | [
"[url]https://artofproblemsolving.com/community/c6h2776804p24363062"
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"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785512.json"
} |
Consider the digits $1, 2, 3, 4, 5, 6, 7$ .
(a) Determine the number of seven-digit numbers with distinct digits that can be constructed using the digits above.
(b) If we place all of these seven-digit numbers in increasing order, find the seven-digit number which appears in the $2022^{\text{th}}$ position.
| [
"a) 7!\nb) 1_ _ _ _ _ _ = 720\n 2 _ _ _ _ _ _ = 720 = 1440\n 3 1 _ _ _ _ _ = 120 \n 3 2 _ _ _ _ _ = 120\n 3 4 _ _ _ _ _ = 120\n 3 5 _ _ _ _ _ = 120 = 480\n 3 6 1 _ _ _ _ = 24\n 3 6 2 _ _ _ _ = 24\n 3 6 4 _ _ _ _ = 24\n 3 6 5 _ _ _ _ = 24\n 3 6 7 1 _ _ _ = 6\n3671542\n\n \n\... | [
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} | |
Determine all real numbers $x\in\mathbb{R}$ for which
\[
\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor=x-2022.
\]
The notation $\lfloor z \rfloor$ , for $z\in\mathbb{R}$ , denotes the largest integer which is less than or equal to $z$ . For example:
\[\lfloor 3.98 \rfloor =3 \quad \text{and} \quad \lfloor 0.14 \rfloor =0.\] | <blockquote>Determine all real numbers $x\in\mathbb{R}$ for which
\[
\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor=x-2022
\]
</blockquote> $LHS\in\mathbb Z$ and so $RHS\in\mathbb Z$ and so $x\in\mathbb Z$ If $x=6n$ , equation is $3n+2n=6n-2022$ and so $n=2022$ and $x=12132$ If $x=6n+1$ , equation is $3n+2n=6n+1-2022$ and so $n=2021$ and $x=12127$ If $x=6n+2$ , equation is $3n+1+2n=6n+2-2022$ and so $n=2021$ and $x=12128$ If $x=6n+3$ , equation is $3n+1+2n+1=6n+3-2022$ and so $n=2021$ and $x=12129$ If $x=6n+4$ , equation is $3n+2+2n+1=6n+4-2022$ and so $n=2021$ and $x=12130$ If $x=6n+5$ , equation is $3n+2+2n+1=6n+5-2022$ and so $n=2020$ and $x=12125$ Anbd so $\boxed{x\in\{12125,12127,12128,12129,12130,12132\}}$ | [] | [
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"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785525.json"
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Determine all pairs of prime numbers $(p, q)$ which satisfy the equation
\[
p^3+q^3+1=p^2q^2
\]
| We claim that $\boxed {(p,q)=(3,2)}$ is the only solution.
WLOG $p \ge q$ ,
If $p=q$ $2p^3+1=p^4 \Rightarrow p^3(p-2)=1$ which is not possible.
Hence, $p >q$ Now, $q^3+1 \equiv 0 \pmod{p} \Rightarrow q^6 \equiv 1 \pmod{p}$ So, $\operatorname{ord}_p(q)=1,2,3,6$ **<span style="color:#f00">Case 1:-</span>** $\operatorname{ord}_p(q)=1$ We have $p \mid {q-1}$ and therefore $q=pk+1$ for some k $$ p^3+(pk+1)^3+1=p^2q^2 \Rightarrow 2 \equiv 0 \pmod{p} $$ which implies $p=2$ but that's not possible as $p>q$ **<span style="color:#f00">Case 2:-</span>**If $\operatorname{ord}_p(q)=3$ $$ p \mid {q^3-1} \Rightarrow p \mid {q^3+1 -2} \Rightarrow p \mid 2 $$ which is again not possible.**<span style="color:#f00">Case 3:-</span>**If $\operatorname{ord}_p(q)=2$ $$ p \mid {q^2-1} \Rightarrow p \mid (q+1)(q-1) $$ $p$ can't divide both because it will again give $p=2$ So $p \mid q+1$ or $p \mid q-1$ If $p \mid q+1$ then $p \le q+1 \le p \Rightarrow p=q+1$ which yields $(p,q)=(3,2)$ as solution.
If $p \mid q-1 \Rightarrow p \le q-1 <q $ which is a contradiction.**<span style="color:#f00">Case 4:- </span>** $\operatorname{ord}_p(q)=6$ $$ p \mid q^6-1 \Rightarrow p \mid (q^3+1)(q^3-1) \Rightarrow p \nmid q^3-1 $$ and the rest of the solution follows #2 | [
"Without loss of generality assume that $p \\geq q$ . If $p=q$ then $2p^3+1=p^4$ , but $1=p^3(p-2) \\geq p^3>1,$ absurd.\n\nNow let $p>q$ . If $q=3$ , then $p^3+28=9p^2$ , i.e. $(p-2)(p^2-7p-14)=0$ , so $p=2$ , absurd. If now $q \\neq 3$ , then taking $\\pmod p^2$ we obtain $p^2 \\mid (q^3+1)=(q+1)(q^2-... | [
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Let $ABC$ be an acute-angled triangle, and let $D, E$ and $K$ be the midpoints of its sides $AB, AC$ and $BC$ respectively. Let $O$ be the circumcentre of triangle $ABC$ , and let $M$ be the foot of the perpendicular from $A$ on the line $BC$ . From the midpoint $P$ of $OM$ we draw a line parallel to $AM$ , which meets the lines $DE$ and $OA$ at the points $T$ and $Z$ respectively. Prove that:
(a) the triangle $DZE$ is isosceles
(b) the area of the triangle $DZE$ is given by the formula
\[E_{DZE}=\frac{BC\cdot OK}{8}\]
| Statement is much scarier than the problem itself.
a) Just note that $DMEK$ is isosceles trapezoid, since $D,M,E,K$ are concyclic and $\overline{MK}\parallel \overline{DE}$ . Thus, perpendicular bisectors of $\overline{DE},\overline{MK}$ coincide. Therefore, $Z$ is the center of $(ADEO)$ as $D$ lies on $\overline{AO}$ and perpendicular bisector of $\overline{DE}$ . We conclude that $\triangle DZE$ is isosceles.
b) By homothety, $\triangle DZE\sim\triangle BOC$ with the ratio of similarity $2$ . Thus, $$ E_{DZE}=\frac{DE\cdot TZ}{2}=\frac{\tfrac{BC}{2}\cdot \tfrac{OK}{2}}{2}=\frac{BC\cdot OK}{2}. $$ | [
"Since $ZP \\parallel AM$ and $OP=PM$ $\\implies AZ=ZO$ . And $AD=DB$ , $AE=EC$ $\\to DZ\\parallel OB$ and $ZE\\parallel OC$ . From Thales, we have $2DZ=OB , 2ZE=OC$ $\\to DZ=ZE$ .\n And $$ E_{DZE}=\\frac{DE\\cdot TZ}{2}=\\frac{\\tfrac{BC}{2}\\cdot \\tfrac{OK}{2}}{2}=\\frac{BC\\cdot OK}{2}. $$ and... | [
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Let $A$ be a subset of $\{1, 2, 3, \ldots, 50\}$ with the property: for every $x,y\in A$ with $x\neq y$ , it holds that
\[\left| \frac{1}{x}- \frac{1}{y}\right|>\frac{1}{1000}.\]
Determine the largest possible number of elements that the set $A$ can have. | We begin with the following Claim:**Claim:** If $x,x+1 \in A$ , then $x \leq 31$ . Furthermore, if $x,x+2 \in A$ , then $x \leq 43$ .
*Proof:* Easy to see from the given hypothesis $\blacksquare$ Therefore, by partitioning the elements of the set $X=\{1, 2, 3, \ldots, 50\}$ in 10 groups as follows,**Group 1:** $X_1=\{1,2,\ldots, 31 \}$ ,**Group 2:** $X_2=\{32,33 \}$ ,**Group 3:** $X_3=\{34,35 \}$ , $\ldots$ **Group7:** $X_7=\{42,43 \}$ ,**Group 8:** $X_8=\{44,45,46 \}$ ,**Group 9:** $X_9=\{47,48,49 \}$ και**Group 10:** $X_{10}=\{50\}$ ,
we note that $A$ can contain at most one element from each $X_i$ , with $2 \leq i \leq 7$ , and at most one element from each $X_i$ , with $8 \leq i \leq 9$ . In total, we have $|A| \leq 31+(1+1+\ldots+1)+(1+1)+1=31+6+2+1=40$ .
An example showing that $|A|=40$ is feasible, is the set $\{1,2,\ldots,31,32,34,36,38,40,42,44,47,50 \}$ . To check that this set works, note that for all $a,b,c \in A$ with $a<b<c$ , we have $\dfrac{1}{a}-\dfrac{1}{c}=(\dfrac{1}{a}-\dfrac{1}{b})+(\dfrac{1}{b}-\dfrac{1}{c})>\dfrac{1}{a}-\dfrac{1}{b},$ so it suffices to check the hypothesis for each two consecutive (when written in increasing order) elements. This is immediate from the Claim. | [] | [
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Prove that for every natural number $k$ , at least one of the integers
\[ 2k-1, \quad 5k-1 \quad \text{and} \quad 13k-1\]
is not a perfect square.
| <blockquote>With process of elimination, if $k$ is $1$ , then $2k-1$ would be a perfect square.
If $k$ is $13$ , then $5k-1$ can't be a perfect square.
Looking closely at the wording *at least one* of the integers can't be a perfect square for any natural number $k$ .
So, our answer is $13k-1$ </blockquote>
What this a proof your answer doesn't even make sense
<details><summary>sol</summary>We proceed with a proof by contradiction. Assume that all three are squares. We see that $2k-1$ implies that the square is odd. Since the quadratic residues modulo $4$ are $0, 1$ , if the value is a square it must be $1 \pmod 4$ . So we can write for a nonnegative integer $n$ , $k=4n+1$ . Substituting for $5k-1$ and $13k-1$ , we find that $20n+4$ and $52n+12$ are both squares. Dividing both expressions by $4$ , which preserves the property that this value is a square, we get $5n+1$ and $13n+3$ . But these two values differ by $8n+2$ , which is $2 \pmod 4$ , which means that one of the two must not be a square, contradiction.</details> | [
"If $k$ is even, then $2k-1$ cannot be a perfect square. If $k\\equiv 3\\pmod{4}$ , $k\\equiv 5\\pmod{16}$ , or $k\\equiv 9\\pmod{16}$ , then $5k-1$ cannot be a perfect square. If $k\\equiv 1\\pmod{16}$ or $k\\equiv 13\\pmod{16}$ , then $13k-1$ cannot be a perfect square. As we have exhausted all resi... | [
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In a triangle $ABC$ with $\widehat{A}=80^{\circ}$ and $\widehat{B}=60^{\circ}$ , the internal angle bisector of $\widehat{C}$ meets the side $AB$ at the point $D$ . The parallel from $D$ to the side $AC$ , meets the side $BC$ at the point $E$ .
Find the measure of the angle $\angle EAB$ . | [
"https://artofproblemsolving.com/community/c6h2828368p25017716 Romania JBMO tst"
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If $a,b,c$ are positive real numbers with $abc=1$ , prove that
(a) \[2\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) \geqslant \frac{9}{ab+bc+ca}\]
(b)\[2\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) \geqslant \frac{9}{a^2 b+b^2 c+c^2 a}\] | (a) Using $abc=1$ , and AM-HM inequality,
\[
\sum \frac{ab}{a+b} = \sum \frac{1}{c(a+b)} \ge \frac{9}{2(ab+bc+ca)}.
\]
(b) It suffices to show $a^2b + b^2c + c^2 a\ge ab+bc+ca$ . To that end, let $a=x/y$ , $b=y/z$ and $c=z/x$ for some $x,y,z>0$ . After some algebra, it boils down verifying $x^3+y^3+z^3 \ge x^2y + y^2z + z^2x$ . But this is immediate by AM-GM by noticing $(x^3+x^3+y^3)/3 \ge x^2y$ and applying analogous inequalities. | [
"As in Bulgaria JBMO 2022 TST. In a), write $ab = 1/c$ etc. and finish by Titu's lemma. In b) we wish to show $a^2b + b^2c + c^2a \\geq ab + bc + ca$ which follows by the AM-GMs $a^2b + c^2a + c^2a \\geq 3\\sqrt[3]{a^4bc^4} = 3ac$ ."
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The numbers $1, 2, 3, \ldots , 10$ are written on the blackboard. In each step, Andrew chooses two numbers $a, b$ which are written on the blackboard such that $a\geqslant 2b$ , he erases them, and in their place writes the number $a-2b$ .
Find all numbers $n$ , such that after a sequence of steps as above, at the end only the number $n$ will remain on the blackboard. | Can the remaining number be present multiple times? If not, then note that the sum of numbers is preserved mod $3$ and so the remaining
number can be at most $1$ , $4$ , $7$ , $10$ (and there are examples for all of these, I guess). | [] | [
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Find all pairs of integers $(m, n)$ which satisfy the equation
\[(2n^2+5m-5n-mn)^2=m^3n\] | My try $(2n^2+5m-5n-mn)^2=m^3n\Longrightarrow$ either $m,n\le0$ , or $m,n\ge0$ 1) Case with $m=n$ Replacing $m$ with $n\Longrightarrow n^4=n^4\Rightarrow$ infinite solutions $(n,m)=(t,t), (-t,-t)$ 2) Case with $m\neq n$ The LHS is a perfect square, $m^3n$ should also be a square. This occurs if $mn$ is a square. Since $m\neq n$ ,
assuming, in the simplest case that $m=nk^2\Rightarrow mn=n^2k^2$ Subst. $m=nk^2$ in the title equation $\Longrightarrow (2n^2+5nk^2-5n-n^2k^2)^2=n^4k^6 \Longrightarrow 2n^2+5nk^2-5n-n^2k^2=\pm n^2k^3$ 2.1) Case with " $+n^2k^3$ " in the RHS $\Longrightarrow n^2(k^3+k^2-2)=5n(k^2-1)\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}$ $\Longrightarrow \frac{\mid 5k^2-5\mid}{\mid k^3+k^2-2\mid}\ge 1\Longrightarrow -5\le k\le 3$ $k=-5\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-\frac{60}{51}$ $k=-4\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-\frac{75}{50}$ $k=-3\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-2, m=nk^2=(-2)(-3)^2=-18$ $k=-2\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-\frac{15}{6}$ $k=-1\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=0$ $k=0\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=\frac{5}{2}$ $k=1\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=0$ $k=2\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=\frac{15}{10}$ $k=3\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=\frac{40}{34}$ In conclusion, one solution $(n,m)=(-2,-18)$ 2.2) Case with " $-n^2k^3$ " in the RHS $\Longrightarrow n^2(k^2-k^3-2)=5n(k^2-1)\Longrightarrow n=\frac{5k^2-5}{k^2-k^3-2}$ $\Longrightarrow \frac{\mid 5k^2-5\mid}{\mid k^2-k^3-2\mid}\ge 1\Longrightarrow -5\le k\le 5$ Same solution as above $(n,m)=(-2,-18)$ for $k= -3, k=3$ @ below: fixed it | [
"<blockquote>(WLOG $m>n),\\Longrightarrow m=nk^2\\Rightarrow mn=n^2k^2$ Subst. $m=nk^2$ in the title equation</blockquote>\n1) you can not say \"WLOG $m>n$ \" since there is no symetrie\n2) $mn$ perfect square does not imply $m=nk^2$ (look for example at $(m,n)=(18,8)$ )\n\n",
" $(2n^2+5m-5n-mn)^2=m^3n$... | [
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Determine for how many positive integers $n\in\{1, 2, \ldots, 2022\}$ it holds that $402$ divides at least one of
\[n^2-1, n^3-1, n^4-1\] | I claim the answer is $31$ Note that $402 = 2\cdot 3\cdot 67$ . From here, we immediately obtain that $n\equiv 1\pmod{2}$ . Next, assume $402\mid n^2-1$ . Then, $n\in\{1,2\}\pmod{3}$ and $n\in\{\pm 1\}\pmod{67}$ . There are four such residue classes in modulo $402$ . Likewise, if $402\mid n^4-1=(n^2-1)(n^2+1)$ , then we must have (a) $n\in\{1,2\}\pmod{3}$ ; and (b) $67\mid n^2-1$ . Indeed, the last conclusion is due to the fact that $67\nmid n^2+1$ since $67$ is a prime of form $4k+3$ . This is the same residue classes as above. Finally, if $402\mid n^3-1=(n-1)(n^2+n+1)$ , then $n\equiv 1\pmod{3}$ and if $n\not\equiv 1\pmod{67}$ , we must have $67\mid n^2+n+1\implies 67\mid (2n+1)^2+3$ . Thus, $2n+1\equiv \pm 8\pmod{67}\iff n\in\{29,37\}\pmod{67}$ . This brings two extra residue classes modulo $402$ , thus a total of $6$ .
Now that $\lfloor 2022/402\rfloor=5$ , we obtain that among $\{1,2,\dots,2010\}$ , there are $30$ such numbers. Our final focus is on $\{2011,\dots,2022\}$ . From here, $2011$ is the only such number satisfying the modulo $3\cdot 67$ condition. | [
"@above you can't have $n \\equiv 2 \\pmod{3}$ and $n \\equiv 29, 37 \\pmod{67}$ , so there are $6$ residue classes.",
"No, you certainly can: if $n\\equiv 2\\pmod{3}$ , then $3\\mid n^2-1$ . Likewise, if $n\\in\\{29,37\\}$ then $67\\mid n^2+n+1\\implies 67\\mid n^3-1$ .**Edit.** I was wrong, thx to pos... | [
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Let $ABC$ be an obtuse-angled triangle with $ \angle ABC>90^{\circ}$ , and let $(c)$ be its circumcircle. The internal angle bisector of $\angle BAC$ meets again the circle $(c)$ at the point $E$ , and the line $BC$ at the point $D$ . The circle of diameter $DE$ meets the circle $(c)$ at the point $H$ .
If the line $HE$ meets the line $BC$ at the point $K$ , prove that:
(a) the points $K, H, D$ and $A$ are concyclic
(b) the line $AH$ passes through the point of intersection of the tangents to the circle $(c)$ at the points $B$ and $C$ . | Another solution**a)** Let $\angle DAC = a , \angle DCA= b$ then $\angle ADB = a+b = \angle ADK$ also $\angle DCE=a \implies
\angle ACE=a+b \implies \angle AHK=a+b$ Hence $\angle AHK= \angle ADK \implies ADHK$ cyclic.**b)**Let $M$ be midpoint $BC$ .Obviously $EB=EC \implies EM \bot BC \implies M$ lies on $(DHE)$ We have $\angle AHD=90-a-b \implies \angle AKD=90-a-b =\angle AKM(1)$
On the other hand $\angle EDM=a+b \implies \angle DEM=90-a-b=\angle AEM(2)$ By $1$ and $2$ $AKEM$ cyclic
Let $\angle BAH =a-c \implies \angle HAD=c \implies \angle HKD=c \implies \angle EKM=c \implies \angle EAM=c
\implies \angle MAC=a-c$ Thus $AH$ is $A-$ **symmedian** $\implies$ $AH$ passes through the point of intersection of the tangents to the circle $(c)$ at the points $B$ and $C$ .
so we are done :-D | [
"[https://mathematica.gr/forum/viewtopic.php?f=58&t=71185#p345811](https://mathematica.gr/forum/viewtopic.php?f=58&t=71185#p345811)\n",
"For (a), just $\\angle KHA = \\angle ACE = \\angle ACB + \\angle EAC = \\angle KDA$ . Let $O$ be the circumcenter of this circle.\n\nThen $O$ is the midpoint of $KD$ (and... | [
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Let
\[M=\{1, 2, 3, \ldots, 2022\}\]
Determine the least positive integer $k$ , such that for every $k$ subsets of $M$ with the cardinality of each subset equal to $3$ , there are two of these subsets with exactly one common element. | Say that a set of triples, $V$ , is "good" iff each pair of triples $u,v\in V$ is disjoint or has two elements in common. Construct a graph $\mathcal{G}$ on good $V$ by drawing an edge $u\sim v$ between each $u,v\in V$ with $|u\cap v|=2$ . The relation $\sim$ is obviously reflexive and, in a good graph, it's transitive as well.
What do components of this graph look like? NB: we'll illustrate paths in what follows, not all other edges in the component.
- Case $1$ : isolated vertices. This uses up $3$ elements of $M$ for the reward of only a single vertex (triple). Say it has "efficiency" $\eta_3 = \tfrac{1}{3}$ , i.e. #triples divided by # $M$ -elements.
- Case $2$ : isolated component $u\sim v$ . This has efficiency $\tfrac{2}{4}=\tfrac{1}{2}$ . We'll never use this case.
- Case $3$ : $\{a,b,c\}\sim\{a,b,d\}\sim\{a,c,d\}\sim\{b,c,d\}$ . Note that after $\{a,c,d\}$ , no further triple involving a new element together with any of $(a,b,c,d)$ is allowed, since any of them would clash (have intersection size 1) with an existing triple:
\begin{tabular}{c|c}
new &existing \hline
a,b,x & a,c,d
a,c,x & a,b,d
a,d,x & a,b,c
b,c,x & a,b,d
b,d,x & a,b,c
c,d,x & a,b,c
\end{tabular}where $x$ is new. However, we can add $\{b,c,d\}$ without a clash, giving $\eta_4=1$ .
- Case $4$ : $\{a,b,c\}\sim\{a,b,d\}\sim\{a,b,e\}\sim\ldots$ . All subsequent terms must be of the form $\{a,b,\star\}$ in order to avoid a return to case $3$ (which would immediately lead to a contradiction). Supposing this path consists of $k-2\ge 3$ triples, its efficiency is $\eta_k=\tfrac{k-2}{k}$ .
The most efficient way to group triples into components is therefore case 3, with $\eta_4=1$ , followed by case 4, with case 1 useless unless we find ourselves with three elements of $M$ not yet assigned.
To finish: if there are $q$ "case 3" components, then there are $2022-4q=2$ (unusable) or $6,10,14,\ldots$ elements of $M$ remaining to be packed into a "case 4" component, making for $4q+(2022-4q)\eta_{2022-4q}=\boxed{2020}$ triples. | [
"Hello @above,\nThe subsets {1,2,3},{1,2,4},…,{1,2,2022} shows that $k \\geq 2021$ .\nAlso could you please explain what you mean by efficiency?\nBest regards ",
"Call the \"repeated\" number set of a number $n \\in M$ the set of duplicates of n in the k elements (Causing n to be counted more than once).\nFor ... | [
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Find all pairs of real numbers $(x,y)$ for which
\[
\begin{aligned}
x^2+y^2+xy&=133
x+y+\sqrt{xy}&=19
\end{aligned}
\] | Squaring and simplifying, we get $(x+y+ \sqrt{xy})^2=133+38 \sqrt{xy}$ . So we obtain $\sqrt{xy}=6 \Rightarrow xy=36$ . Thus, $x+y=19- \sqrt{xy}=13$ . So $x,y$ are roots of $u^2-13u+36=(u-4)(u=9)=0$ by Vieta. So $\boxed{(x,y)=(4,9),(9,4)}$ . | [
" $x^2+y^2+xy=(x+y+\\sqrt{xy})(x+y-\\sqrt{xy})\\implies \\frac{133}{19}=7=x+y-\\sqrt{xy}$ $(x+y+\\sqrt{xy})+(x+y-\\sqrt{xy})=19+7\\implies x+y=13\\implies \\sqrt{xy}=6\\implies xy=36$ By Vieta’s formula, $x,y$ are roots of $$ a^2-13a+36=0\\implies (a-4)(a-9)=0\\implies a=4,9\\implies \\boxed{(x,y)=(4,9),(9,4)}... | [
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Let $n, m$ be positive integers such that
\[n(4n+1)=m(5m+1)\]
(a) Show that the difference $n-m$ is a perfect square of a positive integer.
(b) Find a pair of positive integers $(n, m)$ which satisfies the above relation.
Additional part (not asked in the TST): Find all such pairs $(n,m)$ . | a) Clearly, we have $n>m$ . We get $4n^2+n=5m^2+m$ .
Factorising, we obtain $(n-m)(4n+4m+1)=m^2$ and $(n-m)(5m+5n+1)=n^2$ .
Letting $gcd(m,n)=d \Rightarrow gcd(m^2,n^2)=d^2$ we obtain $\gcd(m^2,n^2)=(n-m)*\gcd(4n+4m+1,5m+5n+1)=(n-m)*\gcd(4m+4n+1,m+n)=(n-m)=d^2$ . Hence proved.
b) Let $n=dx, m=dy$ . Thus $n-m=5m^2-4n^2=d^2 \Rightarrow (2x)^2-5y^2=-1$ , or more conveniently $u^2-5y^2=-1$ .
So trivial (does not satisfy given conditions) solution is $(u_1,y_1)=(2,1)$ which gives $u_n+y_n\sqrt{5}=(2+\sqrt{5})^{2n-1}$ , we get another solution of $(u_2,y_2)=(38,17) \Rightarrow \boxed{(m,n)=(38,34)}$ which indeed works.
All the solutions are of the form $(x,y)=(\frac{u_n}{2},2y_n)$ where $u_n,y_n$ are solutions to the Pell's equation of $u^2-5y^2=-1$ ( $u$ is even). The next solution will be $(x,y)=(341,305)$ .
| [
" $(a)\\quad n(4n+1)=m(5m+1)\\implies (n-m)(4m+4n+1)=m^2, (n-m)(5m+5n+1)=n^2$ $\\gcd(m^2,n^2)=(n-m)\\gcd(4m+4n+1,5m+5n+1)=n-m\\implies \\gcd(m,n)^2=n-m$ Thus, $n-m$ is perfect square.",
"(b) $(n,m)=(38,34)$ ",
"Clearly $m>n$ . Let $d={\\rm gcd}(m,n)$ with $m=dm_1$ and $n=dn_1$ . Then,\n\\[\n4dn_1^2 + ... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 1134,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787233.json"
} |
Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$ and let $(c)$ be its circumcircle with center $O$ . Let $M$ be the midpoint of $BC$ . The line $AM$ meets the circle $(c)$ again at the point $D$ . The circumcircle $(c_1)$ of triangle $\triangle MDC$ intersects the line $AC$ at the points $C$ and $I$ , and the circumcircle $(c_2)$ of $\triangle AMI$ intersects the line $AB$ at the points $A$ and $Z$ .
If $N$ is the foot of the perpendicular from $B$ on $AC$ , and $P$ is the second point of intersection of $ZN$ with $(c_2)$ , prove that the quadrilateral with vertices the points $N, P, I$ and $M$ is a parallelogram. | <blockquote>Could you tell how to verify that BZ is perpendicular to AC，thanks so much</blockquote>
Since $\angle CZM=\angle AIM=\angle BDM=\angle BCA$ we have $ZM=MC$ but $BN \perp NC, BM=MC$ , so we have $MN=BM=MC=MZ$ $\therefore B,N,Z,C$ concyclic.
btw: There is something different between the posts by @Demetres and @Y.J, I hope you can overcome it. | [
"Easy to verify $BZ\\bot AC$ , which implies $N,Z,C,B$ in a circle.",
"\nCould you tell how to verify that BZ is perpendicular to AC，thanks so much\n",
"No problem ,I can learn it ,thanks"
] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787241.json"
} |
Let $m, n$ be positive integers with $m<n$ and consider an $n\times n$ board from which its upper left $ m\times m$ part has been removed. An example of such board for $n=5$ and $m=2$ is shown below.
Determine for which pairs $(m, n)$ this board can be tiled with $3\times 1$ tiles. Each tile can be positioned either horizontally or vertically so that it covers exactly three squares of the board. The tiles should not overlap and should not cover squares outside of the board.
**Attachments:**
 | <details><summary>Hint: Necessary condition:</summary>$3 \mid n^2-m^2$ then $3 \mid (n-m)(n+m)$</details>
<details><summary>Hint: Extension</summary>If $(n,m)$ is possible so is $(n+3,m)$ and so is $(n+3,m+3)$</details> | [] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787243.json"
} |
Charles has some marbles. Their colors are either red, green, or blue. The total number of red and green marbles is $38\%$ more than that of blue marbles. The total number of green and blue marbles is $150\%$ more than that of red marbles. If the total number of blue and red marbles is more than that of green marbles by $n\%$ , find $n$ .
*Proposed by **stayhomedomath*** | <details><summary>Solution</summary>Let $r$ denote red marbles, $g$ denote green marbles, and $b$ denote blue marbles. So, $$ r+g=\frac{69}{50}b $$ $$ g+b=\frac{5}{2}r $$ Subtracting, we get $$ r-b=\frac{69}{50}b-\frac{5}{2}r \implies r=\frac{17}{25}b $$ Plugging into the first equation, $$ \frac{34}{50}b+g=\frac{69}{50}b \implies g=\frac{7}{10}b \implies b=\frac{10}{7}g $$ Plugging into the second equation, $$ g+\frac{25}{17}r=\frac{5}{2}r\implies g=\frac{85-50}{34}r=\frac{35}{34}r \implies r=\frac{34}{35}g $$ Notice we want to find $$ b+r=\frac{100+n}{100}g \implies \frac{100+n}{100}g=\frac{50+34}{35}g=\frac{84}{35}g=\frac{12}{5}g=\frac{240}{100}g $$ Thus, $$ n=240-100=\boxed{140} $$</details> | [
"First! $\\;$ ",
"lol\n\nthis problem is rlly good for a #1 btw",
"I agree! Good job, **stayhomedomath**!",
"<blockquote>I agree! Good job, **stayhomedomath**!</blockquote>\n\nwhy did you bold the name like five times",
"i don't know **pog**",
"<blockquote>i don't know **pog**</blockquote>\n\nVery cool *... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1034,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 DIME/2673235.json"
} |
An up-right path from lattice points $P$ and $Q$ on the $xy$ -plane is a path in which every move is either one unit right or one unit up. The probability that a randomly chosen up-right path from $(0,0)$ to $(10,3)$ does not intersect the graph of $y=x^2+0.5$ can be written as $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
*Proposed by **HrishiP*** | Although the probability that some path is chosen is the same, the number of ways to get to that path is different, thus altering the final probability.
Trivially there are $\binom{13}{3}$ ways to get to $(10,3)$ . Now, the final “correct” grid is an $8 \times 4$ grid connected by $3$ points on the lower left corner: $(0,0)$ , $(1,0)$ , and $(1,1)$ . There are $2$ points that matter: all possible ways are achieved with points $(3,0)$ , and $(2,1)$ , the latter of which has $2$ ways to get to. The final expression is: $$ \frac{2 \binom{10}{2} + \binom{10}{3}}{\binom{13}{3}} = \frac{105}{143} \implies 105+143=\boxed{248} $$ | [
"i'm getting 210/286 = 105/143 => 248 from block walking and also its 2*10c2+10c3=210??",
"…I think the author messed up. Yeah the answer should be 248 I'm pretty sure. I will correct it when I get back home.",
"k thanks :)",
"Actually, each path is not equally likely. Try testing on a 3x3 grid :)\n\nThe answ... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1026,
"boxed": false,
"end_of_proof": false,
"n_reply": 30,
"path": "Contest Collections/2022 Contests/2022 DIME/2673236.json"
} |
Given a regular hexagon $ABCDEF$ , let point $P$ be the intersection of lines $BC$ and $DE$ , and let point $Q$ be the intersection of lines $AP$ and $CD$ . If the area of $\triangle QEP$ is equal to $72$ , find the area of regular hexagon $ABCDEF$ .
*Proposed by **DeToasty3*** | Suppose $s$ is the side length of the hexagon. Assign coordinates to $P$ as the origin: we get two equations: $$ y=x \sqrt{3} + s \sqrt{3} $$ $$ y=-\sqrt{3}{2}s $$ Solving yields the coordinates of $Q$ is $(\tfrac{-2}{3} s, \tfrac{s \sqrt{3}}{3})$ , so the area of $\triangle QEP$ is $\frac{\sqrt{3}}{3} s^2 = 72 \implies s^2 = \frac{72 \cdot 3}{\sqrt{3}}$ The final expression is $$ \frac{3 \sqrt{3}}{2} s^2 = 36 \cdot 3 \cdot 3 = \boxed{324} $$ | [
"how about u catch this ratio",
"coordbash gives 72*9/2=324",
"<blockquote>coordbash gives 72*9/2=324</blockquote>\n\nsimilar triangles + triangle median $\\Longrightarrow 72\\cdot\\frac{1}{2}\\cdot\\frac{3}{2}\\cdot6=324$ ",
":|\n\ni thought that $72\\cdot\\frac{1}{2}\\cdot\\frac{3}{2}\\cdot6 = 432$ ",
... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1024,
"boxed": true,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 DIME/2673237.json"
} |
The four-digit base ten number $\underline{a}\;\underline{b}\;\underline{c}\;\underline{d}$ has all nonzero digits and is a multiple of $99$ . Additionally, the two-digit base ten number $\underline{a}\;\underline{b}$ is a divisor of $150$ , and the two-digit base ten number $\underline{c}\;\underline{d}$ is a divisor of $168$ . Find the remainder when the sum of all possible values of the number $\underline{a}\;\underline{b}\;\underline{c}\;\underline{d}$ is divided by $1000$ .
*Proposed by **treemath*** | I'm gonna use $ab$ and $cd$ cus im too lazy. We can represent $abcd \equiv 100ab + cd \equiv ab + cd \pmod{99}$ . Then we have that the two digit factors of $150$ are: $\{10, 15, 30, 50, 75\}$ and the factors of $168$ are: $\{12, 14, 24, 28, 42, 56, 84\}$ We want these two added up to be divisible by $99$ , and notice the max $84+75$ is less than $99(2)$ meaning that an ordered pair of $(ab, cd)$ must have $ab + cd = 99$ . We can easily check and confirm that $(15, 84)$ and $(75, 24)$ work giving a sum total residue of $1000$ as $\boxed{108}$ . | [
"7524, 1584",
"just bash to get that the two numbers are 7524, 1584",
"wait i just realized the title is misleading this problem isn't about bases it's about digits\n\nok changed :P",
"Bash for two-digit divisors of $168$ and $150$ , bash the condition for multiple of $11$ , and then the multiple of $9$ ... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 DIME/2673238.json"
} |
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