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Problem 1803

National Olympiad, first round

Geometry Difficulty 6.7 Prove it Mongolian Mathematical Olympiad 46 · Mongolia

Let $ABCD$ be a tangential quadrilateral. Let $\omega$ be externally inscribed circle in $ABCD$ , tangent to $AD$ , $BC$ and $AB$ . Denote by $X_{AB}$ the point of tangency of $\omega$ and the circle that passes through $A$ and $B$ and internally tangent to $\omega$ . Let us define $X_{BC}$ , $X_{CD}$ and $X_{DA}$ , analogously. Prove that the bisectors of the angles $\angle AX_{AB}B$ , $\angle BX_{BC}C$ , $\angle CX_{CD}D$ and $\angle DX_{DA}A$ are concurrent.

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 solutions — 2

Solution 1

Let $I$ be the center of the incircle of $ABCD$ , and $AB$ touches the incircle at $Q$ and $\omega$ at $P$ . It will be sufficient to show that $X_{AB}$ , $P$ , $I$ are collinear and $\angle BX_{AB}P = \angle AX_{AB}P$ (all passes through $I$ ). Let $(IP) \cap \omega = X$ , $PP'$ and $QQ'$ are diameters and $(PQ') \cap \omega = Y$ , $(XP') \cap (AB) = R$ .

$Q'I = IQ$ , $QQ' \parallel PP'$ follows that $(PP'YX)$ is harmonic division.

Therefore, the tangent line to $\omega$ at $Y$ passes through $R$ . The tangent line at $Q'$ is parallel to $AB$ . From this, $BC$ , $AD$ , $PQ'$ are concurrent. If $\omega$ touches $BC$ and $AD$ at $B'$ and $A'$ respectively, we have $(A'YB'P)$ harmonic division and $R \in (B'A')$ . According to Nagel point, $BA'$ , $AB'$ , $PQ'$ lines are concurrent. This implies that $(RBPA)$ is a harmonic division. Considering $\angle PXP' = 90^\circ$ , $\angle BXP = \angle AXP$ holds. If $(XB) \cap \omega = B''$ , $(XA) \cap \omega = A''$ , then $A''B'' \parallel AB$ . Hence $\omega(XBA)$ is tangent to $\omega$ , as needed.

Solution 2

Let $I$ be center of incircle of $ABCD$ , and $AB$ touches incircle at $Q$ and $\omega$ at $P$ . It will be sufficient to show that $X_{AB}$ , $P$ , $I$ are collinear and $\angle BX_{AB}P = \angle AX_{AB}P$ (all passes through $I$ ). Let $(IP) \cap \omega = X$ , $PP'$ and $QQ'$ are diameters and $(PQ') \cap \omega = Y$ , $(XP') \cap (AB) = R$ .

$Q'I = IQ$ , $QQ' \parallel PP'$ follows that $(PP'YX)$ is harmonic division. Therefore, tangent line to $\omega$ at $Y$ passes through $R$ . Tangent line at $Q'$ is parallel to $AB$ . From this $BC$ , $AD$ , $PQ'$ are concurrent. If $\omega$ touches $BC$ and $AD$ at $B'$ and $A'$ respectively, we have $(A'YB'P)$ harmonic division and $R \in (B'A')$ . According to Nagel point $BA'$ , $AB'$ , $PQ'$ lines are concurrent. This implies that $(RBPA)$ harmonic division. Considering $\angle PXP' = 90^\circ$ , $\angle BXP = \angle AXP$ holds. If $(XB) \cap \omega = B''$ , $(XA) \cap \omega = A''$ , then $A''B'' \parallel AB$ . Hence $\omega(XBA)$ is tangent to $\omega$ , as needed.

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