Subfield: Geometry


Question: Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circumcircles of the triangles $A M P$ and $B M C$. Prove that as $P$ varies, the point $Q$ lies on a fixed circle.

Solution:
Throughout the solution, we denote by $\sphericalangle(a, b)$ the directed angle between the lines $a$ and $b$.

Let $D$ be the point on the segment $A B$ such that $B D=\lambda \cdot B A$. We will show that either $Q=D$, or $\sphericalangle(D Q, Q B)=\sphericalangle(A B, B C)$; this would mean that the point $Q$ varies over the constant circle through $D$ tangent to $B C$ at $B$, as required.

Denote the circumcircles of the triangles $A M P$ and $B M C$ by $\omega_{A}$ and $\omega_{B}$, respectively. The lines $A P, B C$, and $M Q$ are pairwise radical axes of the circles $\Gamma, \omega_{A}$, and $\omega_{B}$, thus either they are parallel, or they share a common point $X$.

Assume that these lines are parallel (see Figure 1). Then the segments $A P, Q M$, and $B C$ have a common perpendicular bisector; the reflection in this bisector maps the segment $C P$ to $B A$, and maps $M$ to $Q$. Therefore, in this case $Q$ lies on $A B$, and $B Q / A B=C M / C P=$ $B D / A B$; so we have $Q=D$.

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Figure 1

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Figure 2

Now assume that the lines $A P, Q M$, and $B C$ are concurrent at some point $X$ (see Figure 2). Notice that the points $A, B, Q$, and $X$ lie on a common circle $\Omega$ by MiQUEL's theorem applied to the triangle $X P C$. Let us denote by $Y$ the symmetric image of $X$ about the perpendicular bisector of $A B$. Clearly, $Y$ lies on $\Omega$, and the triangles $Y A B$ and $\triangle X B A$ are congruent. Moreover, the triangle $X P C$ is similar to the triangle $X B A$, so it is also similar to the triangle $Y A B$.

Next, the points $D$ and $M$ correspond to each other in similar triangles $Y A B$ and $X P C$, since $B D / B A=C M / C P=\lambda$. Moreover, the triangles $Y A B$ and $X P C$ are equi-oriented, so $\sphericalangle(M X, X P)=\sphericalangle(D Y, Y A)$. On the other hand, since the points $A, Q, X$, and $Y$ lie on $\Omega$, we have $\sphericalangle(Q Y, Y A)=\sphericalangle(M X, X P)$. Therefore, $\sphericalangle(Q Y, Y A)=\sphericalangle(D Y, Y A)$, so the points $Y, D$, and $Q$ are collinear.

Finally, we have $\sphericalangle(D Q, Q B)=\sphericalangle(Y Q, Q B)=\sphericalangle(Y A, A B)=\sphericalangle(A B, B X)=\sphericalangle(A B, B C)$, as desired.
As in the previous solution, we introduce the radical centre $X=A P \cap B C \cap M Q$ of the circles $\omega_{A}, \omega_{B}$, and $\Gamma$. Next, we also notice that the points $A, Q, B$, and $X$ lie on a common circle $\Omega$.

If the point $P$ lies on the $\operatorname{arc} B A C$ of $\Gamma$, then the point $X$ is outside $\Gamma$, thus the point $Q$ belongs to the ray $X M$, and therefore the points $P, A$, and $Q$ lie on the same side of $B C$. Otherwise, if $P$ lies on the $\operatorname{arc} B C$ not containing $A$, then $X$ lies inside $\Gamma$, so $M$ and $Q$ lie on different sides of $B C$; thus again $Q$ and $A$ lie on the same side of $B C$. So, in each case the points $Q$ and $A$ lie on the same side of $B C$.

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Figure 3

Now we prove that the ratio

$$
\frac{Q B}{\sin \angle Q B C}=\frac{Q B}{Q X} \cdot \frac{Q X}{\sin \angle Q B X}
$$

is constant. Since the points $A, Q, B$, and $X$ are concyclic, we have

$$
\frac{Q X}{\sin \angle Q B X}=\frac{A X}{\sin \angle A B C}
$$

Next, since the points $B, Q, M$, and $C$ are concyclic, the triangles $X B Q$ and $X M C$ are similar, so

$$
\frac{Q B}{Q X}=\frac{C M}{C X}=\lambda \cdot \frac{C P}{C X}
$$

Analogously, the triangles $X C P$ and $X A B$ are also similar, so

$$
\frac{C P}{C X}=\frac{A B}{A X}
$$

Therefore, we obtain

$$
\frac{Q B}{\sin \angle Q B C}=\lambda \cdot \frac{A B}{A X} \cdot \frac{A X}{\sin \angle A B C}=\lambda \cdot \frac{A B}{\sin \angle A B C}
$$

so this ratio is indeed constant. Thus the circle passing through $Q$ and tangent to $B C$ at $B$ is also constant, and $Q$ varies over this fixed circle.
Let us perform an inversion centred at $C$. Denote by $X^{\prime}$ the image of a point $X$ under this inversion.

The circle $\Gamma$ maps to the line $\Gamma^{\prime}$ passing through the constant points $A^{\prime}$ and $B^{\prime}$, and containing the variable point $P^{\prime}$. By the problem condition, the point $M$ varies over the circle $\gamma$ which is the homothetic image of $\Gamma$ with centre $C$ and coefficient $\lambda$. Thus $M^{\prime}$ varies over the constant line $\gamma^{\prime} \| A^{\prime} B^{\prime}$ which is the homothetic image of $A^{\prime} B^{\prime}$ with centre $C$ and coefficient $1 / \lambda$, and $M=\gamma^{\prime} \cap C P^{\prime}$. Next, the circumcircles $\omega_{A}$ and $\omega_{B}$ of the triangles $A M P$ and $B M C$ map to the circumcircle $\omega_{A}^{\prime}$ of the triangle $A^{\prime} M^{\prime} P^{\prime}$ and to the line $B^{\prime} M^{\prime}$, respectively; the point $Q$ thus maps to the second point of intersection of $B^{\prime} M^{\prime}$ with $\omega_{A}^{\prime}$ (see Figure 4).

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Figure 4

Let $J$ be the (constant) common point of the lines $\gamma^{\prime}$ and $C A^{\prime}$, and let $\ell$ be the (constant) line through $J$ parallel to $C B^{\prime}$. Let $V$ be the common point of the lines $\ell$ and $B^{\prime} M^{\prime}$. Applying PAppus' theorem to the triples $\left(C, J, A^{\prime}\right)$ and $\left(V, B^{\prime}, M^{\prime}\right)$ we get that the points $C B^{\prime} \cap J V$, $J M^{\prime} \cap A^{\prime} B^{\prime}$, and $C M^{\prime} \cap A^{\prime} V$ are collinear. The first two of these points are ideal, hence so is the third, which means that $C M^{\prime} \| A^{\prime} V$.

Now we have $\sphericalangle\left(Q^{\prime} A^{\prime}, A^{\prime} P^{\prime}\right)=\sphericalangle\left(Q^{\prime} M^{\prime}, M^{\prime} P^{\prime}\right)=\angle\left(V M^{\prime}, A^{\prime} V\right)$, which means that the triangles $B^{\prime} Q^{\prime} A^{\prime}$ and $B^{\prime} A^{\prime} V$ are similar, and $\left(B^{\prime} A^{\prime}\right)^{2}=B^{\prime} Q^{\prime} \cdot B^{\prime} V$. Thus $Q^{\prime}$ is the image of $V$ under the second (fixed) inversion with centre $B^{\prime}$ and radius $B^{\prime} A^{\prime}$. Since $V$ varies over the constant line $\ell, Q^{\prime}$ varies over some constant circle $\Theta$. Thus, applying the first inversion back we get that $Q$ also varies over some fixed circle.

One should notice that this last circle is not a line; otherwise $\Theta$ would contain $C$, and thus $\ell$ would contain the image of $C$ under the second inversion. This is impossible, since $C B^{\prime} \| \ell$.