2012 IMO Problems/Problem 5


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 $MK=ML$.


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.

See Also

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