As shown in the following figure, a heart is a shape consist of three semicircles with diameters $AB$, $BC$ and $AC$ such that $B$ is midpoint of the segment $AC$. A heart $\omega$ is given. Call a pair $(P, P')$ bisector if $P$ and $P'$ lie on $\omega$ and bisect its perimeter. Let $(P, P')$ and $(Q,Q')$ be bisector pairs. Tangents at points $P, P', Q$, and $Q'$ to $\omega$ construct a convex quadrilateral $XYZT$. If the quadrilateral $XYZT$ is inscribed in a circle, find the angle between lines $PP'$ and $QQ'$. Proposed by Mahdi Etesamifard - Iran
Problem
Source: Iranian Geometry Olympiad 2021 IGO Elementary p3
Tags: geometry, perimeter, angles, semicircle
qwertyboyfromalotoftime
25.01.2022 19:52
parmenides51 wrote: As shown in the following figure, a heart is a shape consist of three semicircles with diameters $AB$, $BC$ and $AC$ such that $B$ is midpoint of the segment $AC$. A heart $\omega$ is given. Call a pair $(P, P')$ bisector if $P$ and $P'$ lie on $\omega$ and bisect its perimeter. Let $(P, P')$ and $(Q,Q')$ be bisector pairs. Tangents at points $P, P', Q$, and $Q'$ to $\omega$ construct a convex quadrilateral $XYZT$. If the quadrilateral $XYZT$ is inscribed in a circle, find the angle between lines $PP'$ and $QQ'$. Proposed by Mahdi Etesamifard - Iran
$\boxed{\text{Solution.}}$ We prove the statement of the problem for both convex and non-convex quadrilaterals.
Lemma 1. If a pair $(P, P')$ is bisector, then the points $P , P'$, and $B$ are collinear.
Proof. Without loss of generality suppose that $P$ lies on the arc $AC$ in such a way that the intersection of $P B$ with the arc $AB$ is $P''.$ It’s clear that the length of the arc $P''A$ is equal to $$\frac{2 \angle P''BA}{180^{\circ}}\times (\text{the perimeter of the semicircle with diameter } AB)$$and length of the arc $PC$ is equal to $$\frac{\angle PBC}{180^{\circ}} \times (\text{the perimeter of the semicircle with diameter } AC)$$We know that the perimeter of semicircle with diameter $AB$ is $\frac{\pi \cdot AB}2$ and the perimeter of semicircle
with diameter $AC$ is $\frac{\pi \cdot AC}2$ thus the length of the arc $P''A$ is equal to $\frac{2 \angle P''BA}{180^{\circ}} \times \frac {\pi \cdot AB}2$ and the length of the arc $PC$ is equal to $\frac {\angle PBC}{180^{\circ}}\times \frac{\pi \cdot AC}2$ which are equal with each other. So $(P, P'')$ is a bisector pair. But there is exactly one point for each $P$ like $P'$ such that the pair $(P, P')$ is bisector, so $P' \equiv P''.$ Hence $PP'$ passes through $B. \square$
Without loss of generality suppose $P$ and $Q$ are on the arc $AC.$ Now we consider these 2 cases:
Case 1. $P$ and $Q$ lie on the arc $AC$ in such a way that $P'$ and $Q'$ both lie on the arc $AB.$ Notice that $ \angle PBQ = 2\angle XPQ = 2\angle XQP$ and $\angle P'BQ' = \angle ZP'Q' = \angle ZQ'P'.$ Therefore $$180^{\circ}=\angle PXQ + \angle P'Q'X = (180^{\circ} - \angle PBQ) + (180^{\circ} - 2\angle P'BQ' = 360^{\circ} - 3\angle PBQ,$$hence $\angle PBQ=60 ^{\circ}$
Case 2. $P$ and $Q$ lie on the arc $AC$ in such a way that $P'$ lies on the arc $AB$ and $Q'$ lies on the arc $BC.$
Let $l$ be the tangent line from $B$ to the heart. Thus $\angle P'Bl = \angle BP'Z$ and $\angle Q'Bl = \angle BQ'Z.$ So $\angle P'ZQ' = 360^{\circ} - 2\angle P'BQ'.$ Now it’s not hard to see that $\angle PBQ = 2\angle XPQ = 2\angle XQP$ , hence $\angle PXQ = 180^{\circ} - \angle PBQ.$ Thus we must have $$180^{\circ} = \angle PXQ + \angle P'ZQ' = (180^{\circ} - \angle PBQ) + (360^{\circ} - 2\angle P'BQ') = 540^{\circ} - 3\angle PBQ,$$so $\angle PBQ=120^{\circ}.$
Hence in both cases the angle between the lines $PP'$ and $QQ'$ is $60^{\circ}.$
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