2012 IMO Problems/Problem 5

Problem

Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$ , and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK=BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$ . Let $M=\overline{AL}\cap \overline{BK}$ . Prove that $MK=ML$ .

Solution

Let $\Gamma$ , $\Gamma'$ , $\Gamma''$ be the circumcircle of triangle $ABC$ , the circle with its center as $A$ and radius as $AC$ , and the circle with its center as $B$ and radius as $BC$ , Respectively. Since the center of $\Gamma$ lies on $BC$ , the three circles above are coaxial to line $CD$ .

Let Line $AX$ and Line $BX$ collide with $\Gamma$ on $P$ ( $\neq A$ ) and $Q$ ( $\neq B$ ), Respectively. Also let $R = AQ \cap BP$ .

Then, since $\angle AYB = \angle AZB = 90^{\circ}$ , by ceva's theorem, $R$ lies on $CD$ .

Since triangles $ABC$ and $ACD$ are similar, $AL^2 = AC^2 = AD \cdot AB$ , Thus $\angle ALD = \angle ABL$ . In the same way, $\angle BKD = \angle BAK$ .

Therefore, $\angle ARD = \angle ABQ = \angle ALD$ making $(A, R, L, D)$ concyclic. In the same way, $(B, R, K, D)$ is concyclic.

So $\angle ADR = \angle ALR = 90^{\circ}$ , and in the same way $\angle BKR = 90^{\circ}$ . Therefore, the lines $RK$ and $RL$ are tangent to $\Gamma'$ and $\Gamma''$ , respectively.

Since $R$ is on $CD$ , and $CD$ is the concentric line of $\Gamma'$ and $\Gamma''$ , $RK^2 = RL^2$ , Thus $RK = RL$ . Since $RM$ is in the middle and $\angle ADR = \angle BKR = 90^{\circ}$ , we can say triangles $RKM$ and $RLM$ are congruent. Therefore, $KM = LM$ .

Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.

~ Latex edit by Kscv

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2012 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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2012 IMO Problems/Problem 5

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