Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 6.1 National olympiad Prove it Ukraine

Let $I$ be the incenter of the triangle $ABC$ , and $P$ be any point on the arc $BAC$ of the circumscribed circle. On the tangent to the circumscribed circle $\omega$ of the triangle $API$ at the point $I$ points $K$ and $L$ were selected so that $BK = KI$ and $CL = LI$ . Prove that the circumcircle of the triangle $PKL$ is tangent to $\omega$ .
(Mykhailo Shtandenko)

Fig. 13

Fig. 14

Solution

Mark the midpoints of the arcs $AB$ , $AC$ , $BC$ that do not contain other points, by the points $W_C$ , $W_B$ , $W_A$ respectively (fig. 13). It is clear that $K$ and $L$ lie on $W_A W_C$ and $W_A W_B$ respectively. Then.

Fig. 13
$\begin{aligned} \angle(KL, PI) &= \angle(AI, AP) = \angle(AW_A, AP) = \angle(W_A W_C, W_C P) = \\ &= \angle(KW_C, W_C P) = \angle(W_A W_B, W_B P) = \angle(LW_B, W_B P), \end{aligned}$

*Archimedes' Lemma.* If a circle is inscribed in a segment of another circle bounded by the chord $BC$ and touches the arc at the point $A_1$ and the chords at the point $A_2$ then the line $A_1A_2$ is a bisector $\angle BA_1C$ (fig. 14).

Fig. 14

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