Subfield: Geometry


Question: Point $P$ lies inside triangle $A B C$. Lines $A P, B P, C P$ meet the circumcircle of $A B C$ again at points $K, L, M$, respectively. The tangent to the circumcircle at $C$ meets line $A B$ at $S$. Prove that $S C=S P$ if and only if $M K=M L$.

Solution:
We assume that $C A>C B$, so point $S$ lies on the ray $A B$.

From the similar triangles $\triangle P K M \sim \triangle P C A$ and $\triangle P L M \sim \triangle P C B$ we get $\frac{P M}{K M}=\frac{P A}{C A}$ and $\frac{L M}{P M}=\frac{C B}{P B}$. Multiplying these two equalities, we get

$$
\frac{L M}{K M}=\frac{C B}{C A} \cdot \frac{P A}{P B}
$$

Hence, the relation $M K=M L$ is equivalent to $\frac{C B}{C A}=\frac{P B}{P A}$.

Denote by $E$ the foot of the bisector of angle $B$ in triangle $A B C$. Recall that the locus of points $X$ for which $\frac{X A}{X B}=\frac{C A}{C B}$ is the Apollonius circle $\Omega$ with the center $Q$ on the line $A B$, and this circle passes through $C$ and $E$. Hence, we have $M K=M L$ if and only if $P$ lies on $\Omega$, that is $Q P=Q C$.

<img_3953>

Fig. 1

Now we prove that $S=Q$, thus establishing the problem statement. We have $\angle C E S=$ $\angle C A E+\angle A C E=\angle B C S+\angle E C B=\angle E C S$, so $S C=S E$. Hence, the point $S$ lies on $A B$ as well as on the perpendicular bisector of $C E$ and therefore coincides with $Q$.
As in the previous solution, we assume that $S$ lies on the ray $A B$.

1. Let $P$ be an arbitrary point inside both the circumcircle $\omega$ of the triangle $A B C$ and the angle $A S C$, the points $K, L, M$ defined as in the problem. We claim that $S P=S C$ implies $M K=M L$.

Let $E$ and $F$ be the points of intersection of the line $S P$ with $\omega$, point $E$ lying on the segment $S P$ (see Fig. 2).



<img_3367>

Fig. 2

We have $S P^{2}=S C^{2}=S A \cdot S B$, so $\frac{S P}{S B}=\frac{S A}{S P}$, and hence $\triangle P S A \sim \triangle B S P$. Then $\angle B P S=\angle S A P$. Since $2 \angle B P S=\overparen{B E}+\overparen{L F}$ and $2 \angle S A P=\overparen{B E}+\overparen{E K}$ we have

$$
\overparen{L F}=\overparen{E K} .\tag{1}
$$

On the other hand, from $\angle S P C=\angle S C P$ we have $\overparen{E C}+\overparen{M F}=\overparen{E C}+\overparen{E M}$, or

$$
\overparen{M F}=\overparen{E M}\tag{2}
$$

From (1) and (2) we get $\overparen{M F L}=\overparen{M F}+\overparen{F L}=\overparen{M E}+\overparen{E K}=\overparen{M E K}$ and hence $M K=M L$. The claim is proved.

2. We are left to prove the converse. So, assume that $M K=M L$, and introduce the points $E$ and $F$ as above. We have $S C^{2}=S E \cdot S F$; hence, there exists a point $P^{\prime}$ lying on the segment $E F$ such that $S P^{\prime}=S C$ (see Fig. 3).

<img_3989>

Fig. 3



Assume that $P \neq P^{\prime}$. Let the lines $A P^{\prime}, B P^{\prime}, C P^{\prime}$ meet $\omega$ again at points $K^{\prime}, L^{\prime}, M^{\prime}$ respectively. Now, if $P^{\prime}$ lies on the segment $P F$ then by the first part of the solution we have $\overparen{M^{\prime} F L^{\prime}}=\overparen{M^{\prime} E K^{\prime}}$. On the other hand, we have $\overparen{M F L}>\overparen{M^{\prime} F L^{\prime}}=\overparen{M^{\prime} E K^{\prime}}>\overparen{M E K}$, therefore $\overparen{M F L}>\overparen{M E K}$ which contradicts $M K=M L$.

Similarly, if point $P^{\prime}$ lies on the segment $E P$ then we get $\overparen{M F L}<\overparen{M E K}$ which is impossible. Therefore, the points $P$ and $P^{\prime}$ coincide and hence $S P=S P^{\prime}=S C$.
We present a different proof of the converse direction, that is, $M K=M L \Rightarrow$ $S P=S C$. As in the previous solutions we assume that $C A>C B$, and the line $S P$ meets $\omega$ at $E$ and $F$.

From $M L=M K$ we get $\overparen{M E K}=\overparen{M F L}$. Now we claim that $\overparen{M E}=\overparen{M F}$ and $\overparen{E K}=\overparen{F L}$.

To the contrary, suppose first that $\overparen{M E}>\overparen{M F}$; then $\overparen{E K}=\overparen{M E K}-\overparen{M E}<\overparen{M F L}-\overparen{M F}=$ $\overparen{F L}$. Now, the inequality $\overparen{M E}>\overparen{M F}$ implies $2 \angle S C M=\overparen{E C}+\overparen{M E}>\overparen{E C}+\overparen{M F}=2 \angle S P C$ and hence $S P>S C$. On the other hand, the inequality $\overparen{E K}<\overparen{F L}$ implies $2 \angle S P K=$ $\overparen{E K}+\overparen{A F}<\overparen{F L}+\overparen{A F}=2 \angle A B L$, hence

$$
\angle S P A=180^{\circ}-\angle S P K>180^{\circ}-\angle A B L=\angle S B P .
$$

<img_3845>

Fig. 4

Consider the point $A^{\prime}$ on the ray $S A$ for which $\angle S P A^{\prime}=\angle S B P$; in our case, this point lies on the segment $S A$ (see Fig. 4). Then $\triangle S B P \sim \triangle S P A^{\prime}$ and $S P^{2}=S B \cdot S A^{\prime}<S B \cdot S A=S C^{2}$. Therefore, $S P<S C$ which contradicts $S P>S C$.

Similarly, one can prove that the inequality $\overparen{M E}<\overparen{M F}$ is also impossible. So, we get $\overparen{M E}=\overparen{M F}$ and therefore $2 \angle S C M=\overparen{E C}+\overparen{M E}=\overparen{E C}+\overparen{M F}=2 \angle S P C$, which implies $S C=S P$.