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

Lets draw an circumcircle around triangle ABC (= circle a), a circle with it's center as A and radius as AC (= circle b), a circle with it's center as B and radius as BC (= circle c). Since the center of a lies on the line BC the three circles above are coaxial to line CD. Let ) Line AX and Line BX collide with a on P (not A) and Q (not B). Also let R be the point where AQ and BP intersects. Then since angle AYB = angle AZB = 90, by ceva's theorem in the opposite way, the point R lies on the line CD.

Since triangles ABC and ACD are similar, AL^2 = AC^2 = AD X AB, so angle ALD = angle ABL In the same way angle BKD = angle BAK So in total because angle ARD = angle ABQ = angle ALD, (A, R, L, D) is concyclic In the same way (B, R, K, D) is concyclic So angle ADR = angle ALR = 90, and in the same way angle BKR = 90 so the line RK and RL are tangent to each c and b. Since R is on the line CD, and the line CD is the concentric line of b and c, the equation RK^2 = RL^2 is true. Which makes the result of RK = RL. Since RM is in the middle and angle ADR = angle BKR = 90, we can say that the triangles RKM and RLM are the same. So KM = LM.

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

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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