2011 Japan Mathematical Olympiad Preliminary Problem 6

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2011 Japan Mathematical Olympiad Preliminary Problem 6

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Jan 10, 2011, 2:54 PM • 2 Y

Y by Adventure10, Mango247

Given a right-angled triangle with $\angle{ABC}=90^\circ$ . Three points $P,\ Q,\ R$ are on the sides $BC,\ CA,\ AB$ respectively such that $AQ:QC=2:1,\ AR=AQ,\ QP=QR,\ \angle{PQR}=90^\circ$ . If $CP=1$ , then find the length of $AR$ ?

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#2 h Apr 28, 2025, 12:37 PM

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Let right triangle $\Delta ABC$ have lengths $BC=x,BA=y,AC=\sqrt{x^2+y^2}$ and choose points $P,Q,R$ above such that:

$BP=x-1, PC=1, CQ = \frac{\sqrt{x^2+y^2}}{3}, QA = AR = \frac{2\sqrt{x^2+y^2}}{3}, RB = y - \frac{2\sqrt{x^2+y^2}}{3}, QP=QR=z, PR=\sqrt{2}z$

and the following set of Law of Cosines equations hold for triangles $\Delta CPQ, \Delta AQR, \Delta BPR$ :

$z^2 = 1 + \frac{x^2+y^2}{9} - 2(1)\left(\frac{\sqrt{x^2+y^2}}{3}\right)\left(\frac{x}{\sqrt{x^2+y^2}}\right) = \frac{x^2+y^2}{9}+1-\frac{2x}{3}$ (i);

$z^2 = 2\left(\frac{2\sqrt{x^2+y^2}}{3}\right)^2 - 2\left(\frac{2\sqrt{x^2+y^2}}{3}\right)^2\left(\frac{y}{\sqrt{x^2+y^2}}\right) =\frac{8}{9}(x^2+y^2)-\frac{8}{9}y\sqrt{x^2+y^2}$ (ii);

$2z^2 = (x-1)^2 + \left(y - \frac{2\sqrt{x^2+y^2}}{3}\right)^2$ (iii)

which yields the single positive real triplet $(x,y,z) = (5.68328, 11.3666, 3.893)$ (via Wolfram Alpha). Hence:

$AR = \frac{2}{3}\sqrt{5.68328^2+11.3666^2} = \textcolor{red}{8.47216}$ .

This post has been edited 2 times. Last edited by Mathzeus1024, Apr 28, 2025, 12:41 PM

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