Subfield: Geometry


Question: Let $A B C$ be a triangle with $\angle B>\angle C$. Let $P$ and $Q$ be two different points on line $A C$ such that $\angle P B A=\angle Q B A=\angle A C B$ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $B Q$ for which $P D=P B$. Let the ray $A D$ intersect the circle $A B C$ at $R \neq A$. Prove that $Q B=Q R$.

Solution:
Denote by $\omega$ the circumcircle of the triangle $A B C$, and let $\angle A C B=\gamma$. Note that the condition $\gamma<\angle C B A$ implies $\gamma<90^{\circ}$. Since $\angle P B A=\gamma$, the line $P B$ is tangent to $\omega$, so $P A \cdot P C=P B^{2}=P D^{2}$. By $\frac{P A}{P D}=\frac{P D}{P C}$ the triangles $P A D$ and $P D C$ are similar, and $\angle A D P=\angle D C P$.

Next, since $\angle A B Q=\angle A C B$, the triangles $A B C$ and $A Q B$ are also similar. Then $\angle A Q B=$ $\angle A B C=\angle A R C$, which means that the points $D, R, C$, and $Q$ are concyclic. Therefore $\angle D R Q=$ $\angle D C Q=\angle A D P$.

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Figure 1

Now from $\angle A R B=\angle A C B=\gamma$ and $\angle P D B=\angle P B D=2 \gamma$ we get

$$
\angle Q B R=\angle A D B-\angle A R B=\angle A D P+\angle P D B-\angle A R B=\angle D R Q+\gamma=\angle Q R B
$$

so the triangle $Q R B$ is isosceles, which yields $Q B=Q R$.
Again, denote by $\omega$ the circumcircle of the triangle $A B C$. Denote $\angle A C B=\gamma$. Since $\angle P B A=\gamma$, the line $P B$ is tangent to $\omega$.

Let $E$ be the second intersection point of $B Q$ with $\omega$. If $V^{\prime}$ is any point on the ray $C E$ beyond $E$, then $\angle B E V^{\prime}=180^{\circ}-\angle B E C=180^{\circ}-\angle B A C=\angle P A B$; together with $\angle A B Q=$ $\angle P B A$ this shows firstly, that the rays $B A$ and $C E$ intersect at some point $V$, and secondly that the triangle $V E B$ is similar to the triangle $P A B$. Thus we have $\angle B V E=\angle B P A$. Next, $\angle A E V=\angle B E V-\gamma=\angle P A B-\angle A B Q=\angle A Q B$; so the triangles $P B Q$ and $V A E$ are also similar.

Let $P H$ be an altitude in the isosceles triangle $P B D$; then $B H=H D$. Let $G$ be the intersection point of $P H$ and $A B$. By the symmetry with respect to $P H$, we have $\angle B D G=\angle D B G=\gamma=$ $\angle B E A$; thus $D G \| A E$ and hence $\frac{B G}{G A}=\frac{B D}{D E}$. Thus the points $G$ and $D$ correspond to each other in the similar triangles $P A B$ and $V E B$, so $\angle D V B=\angle G P B=90^{\circ}-\angle P B Q=90^{\circ}-\angle V A E$. Thus $V D \perp A E$.



Let $T$ be the common point of $V D$ and $A E$, and let $D S$ be an altitude in the triangle $B D R$. The points $S$ and $T$ are the feet of corresponding altitudes in the similar triangles $A D E$ and $B D R$, so $\frac{B S}{S R}=\frac{A T}{T E}$. On the other hand, the points $T$ and $H$ are feet of corresponding altitudes in the similar triangles $V A E$ and $P B Q$, so $\frac{A T}{T E}=\frac{B H}{H Q}$. Thus $\frac{B S}{S R}=\frac{A T}{T E}=\frac{B H}{H Q}$, and the triangles $B H S$ and $B Q R$ are similar.

Finally, $S H$ is a median in the right-angled triangle $S B D$; so $B H=H S$, and hence $B Q=Q R$.

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Figure 2
Denote by $\omega$ and $O$ the circumcircle of the triangle $A B C$ and its center, respectively. From the condition $\angle P B A=\angle B C A$ we know that $B P$ is tangent to $\omega$.

Let $E$ be the second point of intersection of $\omega$ and $B D$. Due to the isosceles triangle $B D P$, the tangent of $\omega$ at $E$ is parallel to $D P$ and consequently it intersects $B P$ at some point $L$. Of course, $P D \| L E$. Let $M$ be the midpoint of $B E$, and let $H$ be the midpoint of $B R$. Notice that $\angle A E B=\angle A C B=\angle A B Q=\angle A B E$, so $A$ lies on the perpendicular bisector of $B E$; thus the points $L, A, M$, and $O$ are collinear. Let $\omega_{1}$ be the circle with diameter $B O$. Let $Q^{\prime}=H O \cap B E$; since $H O$ is the perpendicular bisector of $B R$, the statement of the problem is equivalent to $Q^{\prime}=Q$.

Consider the following sequence of projections (see Fig. 3).

1. Project the line $B E$ to the line $L B$ through the center $A$. (This maps $Q$ to $P$.)
2. Project the line $L B$ to $B E$ in parallel direction with $L E .(P \mapsto D$.
3. Project the line $B E$ to the circle $\omega$ through its point $A$. $(D \mapsto R$.)
4. Scale $\omega$ by the ratio $\frac{1}{2}$ from the point $B$ to the circle $\omega_{1} .(R \mapsto H$.
5. Project $\omega_{1}$ to the line $B E$ through its point $O$. $\left(H \mapsto Q^{\prime}\right.$.)

We prove that the composition of these transforms, which maps the line $B E$ to itself, is the identity. To achieve this, it suffices to show three fixed points. An obvious fixed point is $B$ which is fixed by all the transformations above. Another fixed point is $M$, its path being $M \mapsto L \mapsto$ $E \mapsto E \mapsto M \mapsto M$.



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Figure 3

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Figure 4

In order to show a third fixed point, draw a line parallel with $L E$ through $A$; let that line intersect $B E, L B$ and $\omega$ at $X, Y$ and $Z \neq A$, respectively (see Fig. 4). We show that $X$ is a fixed point. The images of $X$ at the first three transformations are $X \mapsto Y \mapsto X \mapsto Z$. From $\angle X B Z=\angle E A Z=\angle A E L=\angle L B A=\angle B Z X$ we can see that the triangle $X B Z$ is isosceles. Let $U$ be the midpoint of $B Z$; then the last two transformations do $Z \mapsto U \mapsto X$, and the point $X$ is fixed.