Problem 1233

Let $A^{\prime}$ be the reflection of $A$ at the central perpendicular of $B C$. The straight line $A A^{\prime}$ is then parallel to $B C$ and $A^{\prime}$ lies on the circumcircle of $A B C$. Let $P^{\prime}$ be the second point of intersection of $A^{\prime} D$ with the circumcircle of $A B C$. Now applies

$$\angle X Y P^{\prime}=\angle A Y P^{\prime}=\angle A A^{\prime} P^{\prime}=\angle C D P^{\prime}=180^{\circ}-\angle X D P^{\prime}$$

therefore $X Y P^{\prime} D$ is a chordal quadrilateral and $P=P^{\prime}$. Thus $P$ is independent of the choice of $X$.

## 2nd solution

We assume oBdA that $X$ lies inside the distance $B D$ and first consider the case when the points $B, Y, P, C$ lie in this order on the circumcircle of $\triangle A B C$ (see Figure 2 (a)). Let $\alpha=\triangle B D P, \beta=\triangle P B C$ and $\gamma=A C B$. Since $X Y P D$ is a chordal quadrilateral by construction, $\angle P Y X=180^{\circ}-\alpha$, and since $A Y P C$ is also a chordal quadrilateral, $\angle P C A=\alpha$. Using the peripheral angle theorem over the distance $P C$, we show that $\angle P A C=\beta$ applies. We have thus shown that the two triangles $B P D$ and $A P C$ are similar and conclude that $\angle D P B=\angle C P A$. If we subtract the angle $\angle D P A$ from this equation, we get $\angle C P D=\angle A P B$. Using the peripheral angle theorem over the chord $A B$, we see that $\angle A P B=\gamma$, i.e. it is independent of the choice of $X$. The point $P$ is therefore one of the intersections of the circumcircle of $\triangle A B C$ and the local arc with angle $\gamma$ over the distance $D C$. One of these intersection points is $C$ and cannot coincide with $P$ because $X, D$ and $C$ lie on a straight line. We have thus shown that

!

(a)

!

(b)

Figure 2: Constructions for solution 2 of task 8

$P$ for all $X$ coincides with the second of these intersection points.

There is still the case that the points $B, P, Y, C$ are in this order on the circumference of $\triangle A B C$ (see Figure 2(b)). In this case, we find $\triangle P C A=\triangle P Y A=$ $\triangle P D B$, i.e. also $\alpha$. The rest works exactly the same as above.