# 2005 Indonesia MO Problems/Problem 4

## Problem

Let $M$ be a point in triangle $ABC$ such that $\angle AMC=90^{\circ}$, $\angle AMB=150^{\circ}$, $\angle BMC=120^{\circ}$. The centers of circumcircles of triangles $AMC,AMB,BMC$ are $P,Q,R$, respectively. Prove that the area of $\triangle PQR$ is greater than the area of $\triangle ABC$.

## Solution

From the above constraints, we can let $M$ be $(0,0)$, $C$ be $(0,2c)$, $A$ be $(-2a,0)$, and $B$ be $(2b\sqrt{3},-2b)$, where $a,b,c$ are positive. Note that the centers of circumcircles are on the perpendicular bisectors of $AM$, $BM$, or $CM$.

$[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-24.2,xmax=24.2,ymin=-24.2,ymax=24.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=4,gy=4; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((-24,0)--(24,0),Arrows); draw((0,24)--(0,-24),Arrows); pair M=(0,0),A=(-16,0),C=(0,10),B=(6.928,-4),P=(-8,5),R=(7.506,5),Q=(-8,-21.856); dot(M); label("(0,0)",M,NW); draw(A--B--C--A); dot(A); label("(0,2c)",C,N); dot(B); label("(2b\sqrt{3},-2b)",B,SE); dot(C); label("(-2a,0)",A,SW); draw(A--M,dotted); draw(B--M,dotted); draw(C--M,dotted); dot(P); label("P",P,NW); dot(Q); label("Q",Q,S); dot(R); label("R",R,NE); draw(P--Q--R--P,dotted); [/asy]$

Because $AMC$ is a right triangle, the circumcenter of $\triangle AMC$ is the midpoint of $AC$, so the coordinates of $P$ are $(-a,c)$. Additionally, the midpoint of $MB$ is $(b\sqrt{3},-b)$ and the slope of line $MB$ on the coordinate grid is $-\tfrac{\sqrt{3}}{3}$. Therefore, the equation of the perpendicular bisector of $MB$ is $y + b = \sqrt{3} (x - b\sqrt{3})$.

Note that $R$ is on the perpendicular bisector of $AM$, so the y-coordinate of $R$ is $c$. Therefore, the x-coordinate of $R$ is $\tfrac{\sqrt{3}}{3}c + \tfrac{4\sqrt{3}}{3}b$. Thus, the length of $PR$ is $\tfrac{\sqrt{3}}{3}c + \tfrac{4\sqrt{3}}{3}b + a$.

Now we'll calculate the area of $\triangle ABC$ and $\triangle PQR$. By using a triangle area formula, \begin{align*} [ABC] &= \frac12 \cdot 2a \cdot 2c + \frac12 \cdot 2c \cdot 4b \cdot \frac12 + \frac12 \cdot 2a \cdot 4b \cdot \frac{\sqrt{3}}{2} \\ &= 2ac + 2bc + 2ab\sqrt{3}. \end{align*} Since $\triangle PQR$ is a 30-60-90 triangle, $PQ = c + 4b + a\sqrt{3}$. Therefore, \begin{align*} [PQR] &= \frac12 \cdot (\frac{\sqrt{3}}{3}c + \frac{4\sqrt{3}}{3}b + a) \cdot (c + 4b + a\sqrt{3}) \\ &= \frac{\sqrt{3}}{2} \cdot (\frac{\sqrt{3}}{3}c + \frac{4\sqrt{3}}{3}b + a)^2 \\ &= \frac{\sqrt{3}}{2} \cdot (\frac13 c^2 + \frac{16}{3} b^2 + a^2 + \frac83 bc + \frac{2\sqrt{3}}{3}ac + \frac{8\sqrt{3}}{3}ab) \\ &= \frac{\sqrt{3}}{6}c^2 + \frac{8\sqrt{3}}{3}b^2 + \frac{\sqrt{3}}{2}a^2 + \frac{4\sqrt{3}}{3}bc + ac + 4ab. \end{align*} Now what's left is to show that $[PQR] > [ABC]$. Since $a,b,c > 0$, $\tfrac{4\sqrt{3}}{3}bc + 4ab > 2bc + 2ab\sqrt{3}$. Additionally, by using the AM-GM Inequality, \begin{align*} \frac{\frac{\sqrt{3}}{6} c^2 + \frac{\sqrt{3}}{2} a^2}{2} &\ge \sqrt{\frac14 \cdot (ac)^2} \\ \frac{\sqrt{3}}{6} c^2 + \frac{\sqrt{3}}{2} a^2 &\ge ac \\ ac + \frac{\sqrt{3}}{6} c^2 + \frac{\sqrt{3}}{2} a^2 &\ge 2ac. \end{align*} Therefore, $[PQR] > ac + \frac{\sqrt{3}}{6} c^2 + \frac{\sqrt{3}}{2} a^2 + \frac{4\sqrt{3}}{3}bc + 4ab > [ABC]$.