Subfield: Geometry


Question: Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$.

Solution:
We may assume that $C A>C B$. Observe that $H^{\prime}$ and $I^{\prime}$ lie inside the segments $C F$ and $C G$, respectively. Therefore, $M$ lies outside $\triangle A B C$ (see Figure 1).

Due to the powers of points $A$ and $B$ with respect to the circle $\Gamma$, we have

$$
C H^{\prime} \cdot C A=A H \cdot A C=A D^{2}=B D^{2}=B I \cdot B C=C I^{\prime} \cdot C B .
$$

Therefore, $C H^{\prime} \cdot C F=C I^{\prime} \cdot C G$. Hence, the quadrilateral $H^{\prime} I^{\prime} G F$ is cyclic, and so $\angle I^{\prime} H^{\prime} C=$ $\angle C G F$.

Let $D F$ and $D G$ meet $\Gamma$ again at $R$ and $S$, respectively. We claim that the points $R$ and $S$ lie on the line $H^{\prime} I^{\prime}$.

Observe that $F H^{\prime} \cdot F A=F H \cdot F C=F R \cdot F D$. Thus, the quadrilateral $A D H^{\prime} R$ is cyclic, and hence $\angle R H^{\prime} F=\angle F D A=\angle C G F=\angle I^{\prime} H^{\prime} C$. Therefore, the points $R, H^{\prime}$, and $I^{\prime}$ are collinear. Similarly, the points $S, H^{\prime}$, and $I^{\prime}$ are also collinear, and so all the points $R, H^{\prime}, Q, I^{\prime}, S$, and $M$ are all collinear.

<img_3960>

Figure 1

<img_3866>

Figure 2

Then, $\angle R S D=\angle R D A=\angle D F G$. Hence, the quadrilateral $R S G F$ is cyclic (see Figure 2). Therefore, $M H^{\prime} \cdot M I^{\prime}=M F \cdot M G=M R \cdot M S=M P \cdot M C$. Thus, the quadrilateral $C P I^{\prime} H^{\prime}$ is also cyclic. Let $\omega$ be its circumcircle.

Notice that $\angle H^{\prime} C Q=\angle S D C=\angle S R C$ and $\angle Q C I^{\prime}=\angle C D R=\angle C S R$. Hence, $\triangle C H^{\prime} Q \sim \triangle R C Q$ and $\triangle C I^{\prime} Q \sim \triangle S C Q$, and therefore $Q H^{\prime} \cdot Q R=Q C^{2}=Q I^{\prime} \cdot Q S$.

We apply the inversion with center $Q$ and radius $Q C$. Observe that the points $R, C$, and $S$ are mapped to $H^{\prime}, C$, and $I^{\prime}$, respectively. Therefore, the circumcircle $\Gamma$ of $\triangle R C S$ is mapped to the circumcircle $\omega$ of $\triangle H^{\prime} C I^{\prime}$. Since $P$ and $C$ belong to both circles and the point $C$ is preserved by the inversion, we have that $P$ is also mapped to itself. We then get $Q P^{2}=Q C^{2}$. Hence, $Q P=Q C$.
Let $X=H I \cap A B$, and let the tangent to $\Gamma$ at $C$ meet $A B$ at $Y$. Let $X C$ meet $\Gamma$ again at $X^{\prime}$ (see Figure 3). Projecting from $C, X$, and $C$ again, we have $(X, A ; D, B)=$ $\left(X^{\prime}, H ; D, I\right)=(C, I ; D, H)=(Y, B ; D, A)$. Since $A$ and $B$ are symmetric about $D$, it follows that $X$ and $Y$ are also symmetric about $D$.

Now, Menelaus' theorem applied to $\triangle A B C$ with the line $H I X$ yields

$$
1=\frac{C H}{H A} \cdot \frac{B I}{I C} \cdot \frac{A X}{X B}=\frac{A H^{\prime}}{H^{\prime} C} \cdot \frac{C I^{\prime}}{I^{\prime} B} \cdot \frac{B Y}{Y A}
$$

By the converse of Menelaus' theorem applied to $\triangle A B C$ with points $H^{\prime}, I^{\prime}, Y$, we get that the points $H^{\prime}, I^{\prime}, Y$ are collinear.

<img_3164>

Figure 3

Let $T$ be the midpoint of $C D$, and let $O$ be the center of $\Gamma$. Let $C M$ meet $T Y$ at $N$. To avoid confusion, we clean some superfluous details out of the picture (see Figure 4).

Let $V=M T \cap C Y$. Since $M T \| Y D$ and $D T=T C$, we get $C V=V Y$. Then CEvA's theorem applied to $\triangle C T Y$ with the point $M$ yields

$$
1=\frac{T Q}{Q C} \cdot \frac{C V}{V Y} \cdot \frac{Y N}{N T}=\frac{T Q}{Q C} \cdot \frac{Y N}{N T}
$$

Therefore, $\frac{T Q}{Q C}=\frac{T N}{N Y}$. So, $N Q \| C Y$, and thus $N Q \perp O C$.

Note that the points $O, N, T$, and $Y$ are collinear. Therefore, $C Q \perp O N$. So, $Q$ is the orthocenter of $\triangle O C N$, and hence $O Q \perp C P$. Thus, $Q$ lies on the perpendicular bisector of $C P$, and therefore $C Q=Q P$, as required.

<img_3249>

Figure 4