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Ukrainian National Mathematical Olympiad

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Ukraine

Let $ABC$ be an acute angled triangle, $AA_1$ and $CC_1$ be its bisector, $I$ be the incenter of $ABC$ , $M$ and $N$ be the midpoints of $AI$ and $CI$ , respectively. Inside the triangles $AC_1I$ and $A_1CI$ we choose points $K$ and $L$ , such that

$\angle AKI = \angle CLI = \angle AIC$ , $\angle AKM = \angle ICA$ , $\angle CLN = \angle IAC$ . Prove that radiuses of the circumcircles of the triangles $KIL$ and $ABC$ are equal.

Solution

(Anton Trigub)

Fig. 46
Suppose that lines $AI$ and $CI$ meet the circumcircle of $\triangle ABC$ (second time) in the points $W_A$ and $W_C$ , respectively (Fig. 46). Since $\angle AKI = \angle AIC$ , then circumcircle of the triangle $AKI$ tangent to the line $W_C I$ . Consider the triangle $AW_C I$ . According to well-known fact this triangle is isosceles. Then the circumcircle of the triangle $\triangle AKI$ tangents to $W_C A$ as well. Thus a symmedian of the $\triangle AKI$ belongs to the line $W_C K$ . Then $\angle W_C KI = 180^\circ - \angle ICA = 180^\circ - \angle W_C W_A I$ . It means that $K$ belongs to the circumcircle of the $\triangle W_C W_A I$ . In the same way $L$ belongs to this circumcircle as well. Let us note that $\triangle W_C B W_A = \triangle W_C I W_A$ (they have equal sides), and thus the radius of the circumcircle of $\triangle KIL$ equal to the radius of the circumcircle of $\triangle ABC$ .

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