p1. A number of $n$ students sit around a round table. It is known that there are as many male students as female students. If the number of pairs of $2$ people sitting next to each other is counted, it turns out that the ratio between adjacent pairs of the same sex and adjacent pairs of the opposite sex is $3:2$. Find the smallest possible $n$. p2. Let $a, b$, and $c$ be positive integers so that $$c=a+\frac{b}{a}-\frac{1}{b}.$$Prove that $c$ is the square of an integer. p3. Let $ \Gamma_1$ and $\Gamma_2$ be two different circles with the radius of same length and centers at points $O_1$ and $O_2$, respectively. Circles $\Gamma_1$ and $\Gamma_2$ are tangent at point $P$. The line $\ell$ passing through $O_1$ is tangent to $\Gamma_2$ at point $A$. The line $\ell$ intersects $\Gamma_1$ at point $X$ with $X$ between $A$ and $O_1$. Let $M$ be the midpoint of $AX$ and $Y$ the intersection of $PM$ and $\Gamma_2$ with $Y\ne P$. Prove that $XY$ is parallel to $O_1O_2$. p4. Let $a, b, c$ be positive real numbers with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$a+b+c+\frac{4}{1+(abc)^{2/3}}\ge 5$$ p5. On a chessboard measuring $200 \times 200$ square units are placed red or blue marbles so that each unit square has at most $ 1$ marble. Two marbles are said to be in a row if they are in the same row or column. It is known that for every red marble there are exactly $5$ blue marbles in a row and for every blue marble there are exactly $5$ red marbles in a row. Determine the maximum number of marbles possible on the chessboard.
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Tags: algebra, geometry, combinatorics, number theory, Indonesia Regional MO
04.10.2021 19:10
P1. Since there's same number of male and female n should be even. If there are n people are sitting around a round table then the number of adjacent pairs is n. Also, for any adjacent pairs it is either same sex or opposite sex. Therefor we have n = 3k+2k for some k. Since n is even we must have n = 6k_1+4k_2. If we can show a solution for 10 we are done. Here's a solution for 10 bbggbbgggb.
23.04.2023 13:02
P2 Case1: If $a\mid b \Rightarrow \frac{b}{a}=k, k \in \mathbb{N}$. Because $c$ must be a positive integer, then $\frac{1}{b}$ also must be a positive integer, therefore the only solution that satifsfy this equation is $b=1$ and because $a\mid b \Rightarrow a=1$. Substitute to the equation we got $c=a+\frac{b}{a}-\frac{1}{b}\iff c=1+\frac{1}{1}-\frac{1}{1}=1$ which is a perfect square. Case 2: If $a\nmid b$, let $b=ka+n, 1\leq n < a, k \in \mathbb{N}$. Substitute to the equation we got $c=a+\frac{ka+n}{a}-\frac{1}{ka+n}\iff c=a+k+\frac{n}{a}-\frac{1}{ka+n}$ Because $1\leq n < a$ then $\frac{n}{a}$ is not an integer $\Rightarrow c$ will be an integer if $\frac{n}{a}=\frac{1}{ka+n}\iff n(ka+n)=a$ because $a,n,k$ are positive integer so $ n(ka+n)\neq a$ which is a contradiction. There is no positive integer $n$ that satisfy the equation when $a\nmid b$.
24.04.2023 09:28
24.04.2023 10:44
P3: Let $Q$ be the intersection of $PO_2$ and $O_1O_2$, and let $N$ be the intersection of $XY$ and $O_1O_2$, as shown below: [asy][asy] size(600); import olympiad; pair O1=(0,0), O2=(2,0), P=(1,1.5), A=(4,0), X=(1.5,1.5), M=(1.75,1.25), Y=(2.5,0.75), Q=(2,1.5), N=(1.875,0); draw(circle(O1,1.5)); draw(circle(O2,1.5)); draw(O1--O2); draw(P--Y); draw(A--X); draw(P--M); draw(X--N); draw(P--O2); draw(Y--O2); draw(Q--O2); draw(Q--P); draw(rightanglemark(O2,Q,P,3.5)); label("$O_1$",O1,SW); label("$O_2$",O2,SE); label("$P$",P,N); label("$A$",A,SE); label("$X$",X,NW); label("$M$",M,NW); label("$Y$",Y,NE); label("$Q$",Q,NE); label("$N$",N,S); [/asy][/asy] Since $\ell$ is tangent to $\Gamma_2$ at $A$, we have $\angle APO_2 = \angle AOX$. But $\angle APO_2 = \angle O_1PQ$ since $PO_2 \parallel O_1Q$, so $\angle O_1PQ = \angle AOX$. Therefore, $AX \parallel PQ$. Since $M$ is the midpoint of $AX$ and $O_1$ lies on $AX$, we have $OM \parallel PQ$. Furthermore, $PY$ intersects $O_1O_2$ at $N$ and $O_1N = O_2N$ since $O_1O_2$ is the perpendicular bisector of $PP'$. Therefore, $OMNY$ is a parallelogram. Finally, since $OM \parallel PQ$ and $PQ \parallel XY$ (since $AX \parallel PQ$ and $M$ is the midpoint of $AX$), we have $OM \parallel XY$. Combining this with $OM \parallel NY$, we obtain $XY \parallel NY$, which is parallel to $O_1O_2$. Hence, $XY$ is parallel to $O_1O_2$.
28.05.2023 06:52
P4 Let $\sqrt[3]{abc}=x$, By AM-GM we have $$a+b+c+\frac{4}{1+x^2}\geq 3x+\frac{4}{1+x^2}=\frac{4+3x+3x^3}{1+x^2}\geq \frac{4+2\sqrt{3x\cdot 3x^3}}{1+x^2}=\frac{4+6x^2}{1+x^2}$$Notice that by AM-GM we have $$3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\sqrt[3]{\frac{1}{abc}}\iff x \geq 1$$Then we have $$\frac{4+6x^2}{1+x^2}\geq \frac{5+5x^2}{1+x^2}=5$$