Subfield: Geometry


Question: Let $A B C$ be a fixed triangle, and let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $B C, C A, A B$, respectively. Let $P$ be a variable point on the circumcircle. Let lines $P A_{1}, P B_{1}, P C_{1}$ meet the circumcircle again at $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Assume that the points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$ are distinct, and lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ form a triangle. Prove that the area of this triangle does not depend on $P$.

Solution:
Let $A_{0}, B_{0}, C_{0}$ be the points of intersection of the lines $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$ (see Figure). We claim that area $\left(A_{0} B_{0} C_{0}\right)=\frac{1}{2} \operatorname{area}(A B C)$, hence it is constant.

Consider the inscribed hexagon $A B C C^{\prime} P A^{\prime}$. By Pascal's theorem, the points of intersection of its opposite sides (or of their extensions) are collinear. These points are $A B \cap C^{\prime} P=C_{1}$, $B C \cap P A^{\prime}=A_{1}, C C^{\prime} \cap A^{\prime} A=B_{0}$. So point $B_{0}$ lies on the midline $A_{1} C_{1}$ of triangle $A B C$. Analogously, points $A_{0}$ and $C_{0}$ lie on lines $B_{1} C_{1}$ and $A_{1} B_{1}$, respectively.

Lines $A C$ and $A_{1} C_{1}$ are parallel, so triangles $B_{0} C_{0} A_{1}$ and $A C_{0} B_{1}$ are similar; hence we have
$$
\frac{B_{0} C_{0}}{A C_{0}}=\frac{A_{1} C_{0}}{B_{1} C_{0}}
$$
Analogously, from $B C \| B_{1} C_{1}$ we obtain
$$
\frac{A_{1} C_{0}}{B_{1} C_{0}}=\frac{B C_{0}}{A_{0} C_{0}}
$$
Combining these equalities, we get
$$
\frac{B_{0} C_{0}}{A C_{0}}=\frac{B C_{0}}{A_{0} C_{0}}
$$
or
$$
A_{0} C_{0} \cdot B_{0} C_{0}=A C_{0} \cdot B C_{0}
$$
Hence we have

<img_3518>
$$
\operatorname{area}\left(A_{0} B_{0} C_{0}\right)=\frac{1}{2} A_{0} C_{0} \cdot B_{0} C_{0} \sin \angle A_{0} C_{0} B_{0}=\frac{1}{2} A C_{0} \cdot B C_{0} \sin \angle A C_{0} B=\operatorname{area}\left(A B C_{0}\right) \text {. }
$$
Since $C_{0}$ lies on the midline, we have $d\left(C_{0}, A B\right)=\frac{1}{2} d(C, A B)$ (we denote by $d(X, Y Z)$ the distance between point $X$ and line $Y Z)$. Then we obtain
$$
\operatorname{area}\left(A_{0} B_{0} C_{0}\right)=\operatorname{area}\left(A B C_{0}\right)=\frac{1}{2} A B \cdot d\left(C_{0}, A B\right)=\frac{1}{4} A B \cdot d(C, A B)=\frac{1}{2} \operatorname{area}(A B C)
$$
. Again, we prove that $\operatorname{area}\left(A_{0} B_{0} C_{0}\right)=\frac{1}{2} \operatorname{area}(A B C)$.

We can assume that $P$ lies on arc $A C$. Mark a point $L$ on side $A C$ such that $\angle C B L=$ $\angle P B A$; then $\angle L B A=\angle C B A-\angle C B L=\angle C B A-\angle P B A=\angle C B P$. Note also that $\angle B A L=\angle B A C=\angle B P C$ and $\angle L C B=\angle A P B$. Hence, triangles $B A L$ and $B P C$ are similar, and so are triangles $L C B$ and $A P B$.

Analogously, mark points $K$ and $M$ respectively on the extensions of sides $C B$ and $A B$ beyond point $B$, such that $\angle K A B=\angle C A P$ and $\angle B C M=\angle P C A$. For analogous reasons, $\angle K A C=\angle B A P$ and $\angle A C M=\angle P C B$. Hence $\triangle A B K \sim \triangle A P C \sim \triangle M B C, \triangle A C K \sim$ $\triangle A P B$, and $\triangle M A C \sim \triangle B P C$. From these similarities, we have $\angle C M B=\angle K A B=\angle C A P$, while we have seen that $\angle C A P=\angle C B P=\angle L B A$. Hence, $A K\|B L\| C M$.



<img_3971>

Let line $C C^{\prime}$ intersect $B L$ at point $X$. Note that $\angle L C X=\angle A C C^{\prime}=\angle A P C^{\prime}=\angle A P C_{1}$, and $P C_{1}$ is a median in triangle $A P B$. Since triangles $A P B$ and $L C B$ are similar, $C X$ is a median in triangle $L C B$, and $X$ is a midpoint of $B L$. For the same reason, $A A^{\prime}$ passes through this midpoint, so $X=B_{0}$. Analogously, $A_{0}$ and $C_{0}$ are the midpoints of $A K$ and $C M$.

Now, from $A A_{0} \| C C_{0}$, we have
$$
\operatorname{area}\left(A_{0} B_{0} C_{0}\right)=\operatorname{area}\left(A C_{0} A_{0}\right)-\operatorname{area}\left(A B_{0} A_{0}\right)=\operatorname{area}\left(A C A_{0}\right)-\operatorname{area}\left(A B_{0} A_{0}\right)=\operatorname{area}\left(A C B_{0}\right)
$$
Finally,
$$
\operatorname{area}\left(A_{0} B_{0} C_{0}\right)=\operatorname{area}\left(A C B_{0}\right)=\frac{1}{2} B_{0} L \cdot A C \sin A L B_{0}=\frac{1}{4} B L \cdot A C \sin A L B=\frac{1}{2} \operatorname{area}(A B C)
$$