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Problem 1698

National Olympiad, first round

Geometry Difficulty 6.3 Prove it Ukrainian National Mathematical Olympiad - Fourth Round · Ukraine

Let $ABC$ be an acute-angled triangle and altitudes $AA_1$ and $BB_1$ intersect at $H$ . Consider circles $w_1$ and $w_2$ with centers $H$ and $B$ and with radii $HB_1$ and $BB_1$ respectively. Let $CN$ and $CK$ be the tangent lines from $C$ to circles $w_1$ and $w_2$ respectively $(N \neq B_1, K \neq B_1)$ . Prove that $A_1, N$ and $K$ are collinear.
(Igor Nagel)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $CC_1$ be the altitude (Fig. 42). Since quadrilateral $AB_1HC_1$ is cyclic, we have $\angle A = 180^\circ - \angle C_1HB_1 = \angle B_1HC_1$ . Since $\angle B_1HC_1 = \angle NHC_1$ , we obtain $\angle A = \angle CHN_1$ .

Taking into account $\angle HA_1C = 90^\circ$ , $\angle HNC = 90^\circ$ , $\angle AC_1C = 90^\circ$ , we get that quadrilaterals $HA_1NC$ and $AC_1A_1C$ are cyclic. Hence $180^\circ - \angle C_1A_1C = \angle BAC = \angle CHN = \angle CA_1N$ and therefore points $C_1, A_1$ and $N$ are collinear.

Since $BA_1HC_1$ is cyclic, we have $\angle HC_1A_1 = \angle HBA_1$ . Note that $\angle A_1BK = \angle HBA_1$ . Since $C, C_1, B, K$ are cyclic, it follows that $\angle CBK = \angle KC_1C$ . This means that $C_1, A_1$ and $K$ are collinear. Notice that points $C_1$ and $A_1$ are different. This proves that $C_1, A_1, N$ and $K$ are collinear.

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