Subfield: Geometry


Question: Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$.

Solution:
Since

$$
\angle A P B+\angle B P C+\angle C P A=2 \pi=(\pi-\angle A C B)+(\pi-\angle B A C)+(\pi-\angle C B A),
$$

at least one of the following inequalities holds:

$$
\angle A P B \geqslant \pi-\angle A C B, \quad \angle B P C \geqslant \pi-\angle B A C, \quad \angle C P A \geqslant \pi-\angle C B A .
$$

Without loss of generality, we assume that $\angle B P C \geqslant \pi-\angle B A C$. We have $\angle B P C>\angle B A C$ because $P$ is inside $\triangle A B C$. So $\angle B P C \geqslant \max (\angle B A C, \pi-\angle B A C)$ and hence

$$
\sin \angle B P C \leqslant \sin \angle B A C .
\tag{*}
$$

Let the rays $A P, B P$, and $C P$ cross the circumcircle $\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\frac{P B_{1}}{B_{1} B_{3}}$ and $\frac{P C_{1}}{C_{1} C_{3}}$ is at least 1 , which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\Omega$.

Because $A, B, C, B_{3}$ lie on a circle, the triangles $C B_{1} B_{3}$ and $B B_{1} A$ are similar, so

$$
\frac{C B_{1}}{B_{1} B_{3}}=\frac{B B_{1}}{B_{1} A}
$$

Applying the sine rule we obtain

$$
\frac{P B_{1}}{B_{1} B_{3}}=\frac{P B_{1}}{C B_{1}} \cdot \frac{C B_{1}}{B_{1} B_{3}}=\frac{P B_{1}}{C B_{1}} \cdot \frac{B B_{1}}{B_{1} A}=\frac{\sin \angle A C P}{\sin \angle B P C} \cdot \frac{\sin \angle B A C}{\sin \angle P B A}
$$

Similarly,

$$
\frac{P C_{1}}{C_{1} C_{3}}=\frac{\sin \angle P B A}{\sin \angle B P C} \cdot \frac{\sin \angle B A C}{\sin \angle A C P}
$$

Multiplying these two equations we get

$$
\frac{P B_{1}}{B_{1} B_{3}} \cdot \frac{P C_{1}}{C_{1} C_{3}}=\frac{\sin ^{2} \angle B A C}{\sin ^{2} \angle B P C} \geqslant 1
$$

using (*), which yields the desired conclusion.
Let the rays $A P, B P$, and $C P$ cross the circumcircle $\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. 
Assume for the sake of contradiction that $A_{2}, B_{2}$, and $C_{2}$ all lie strictly inside circle $A B C$. It follows that $P A_{1}<A_{1} A_{3}, P B_{1}<B_{1} B_{3}$, and $P C_{1}<C_{1} C_{3}$.

Observe that $\triangle P B C_{3} \sim \triangle P C B_{3}$. Let $X$ be the point on side $P B_{3}$ that corresponds to point $C_{1}$ on side $P C_{3}$ under this similarity. In other words, $X$ lies on segment $P B_{3}$ and satisfies $P X: X B_{3}=P C_{1}: C_{1} C_{3}$. It follows that

$$
\angle X C P=\angle P B C_{1}=\angle B_{3} B A=\angle B_{3} C B_{1} .
$$

Hence lines $C X$ and $C B_{1}$ are isogonal conjugates in $\triangle P C B_{3}$.

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Let $Y$ be the foot of the bisector of $\angle B_{3} C P$ in $\triangle P C B_{3}$. Since $P C_{1}<C_{1} C_{3}$, we have $P X<X B_{3}$. Also, we have $P Y<Y B_{3}$ because $P B_{1}<B_{1} B_{3}$ and $Y$ lies between $X$ and $B_{1}$. By the angle bisector theorem in $\triangle P C B_{3}$, we have $P Y: Y B_{3}=P C: C B_{3}$. So $P C<C B_{3}$ and it follows that $\angle P B_{3} C<\angle C P B_{3}$. Now since $\angle P B_{3} C=\angle B B_{3} C=\angle B A C$, we have

$$
\angle B A C<\angle C P B_{3}
$$

Similarly, we have

$$
\angle C B A<\angle A P C_{3} \text { and } \angle A C B<\angle B P A_{3}=\angle B_{3} P A \text {. }
$$

Adding these three inequalities yields $\pi<\pi$, and this contradiction concludes the proof.
Choose coordinates such that the circumcentre of $\triangle A B C$ is at the origin and the circumradius is 1 . Then we may think of $A, B$, and $C$ as vectors in $\mathbb{R}^{2}$ such that

$$
|A|^{2}=|B|^{2}=|C|^{2}=1 \text {. }
$$

$P$ may be represented as a convex combination $\alpha A+\beta B+\gamma C$ where $\alpha, \beta, \gamma>0$ and $\alpha+\beta+\gamma=1$. Then

$$
A_{1}=\frac{\beta B+\gamma C}{\beta+\gamma}=\frac{1}{1-\alpha} P-\frac{\alpha}{1-\alpha} A
$$

so

$$
A_{2}=2 A_{1}-P=\frac{1+\alpha}{1-\alpha} P-\frac{2 \alpha}{1-\alpha} A
$$

Hence

$$
\left|A_{2}\right|^{2}=\left(\frac{1+\alpha}{1-\alpha}\right)^{2}|P|^{2}+\left(\frac{2 \alpha}{1-\alpha}\right)^{2}|A|^{2}-\frac{4 \alpha(1+\alpha)}{(1-\alpha)^{2}} A \cdot P \text {. }
$$

Using $|A|^{2}=1$ we obtain

$$
\frac{(1-\alpha)^{2}}{2(1+\alpha)}\left|A_{2}\right|^{2}=\frac{1+\alpha}{2}|P|^{2}+\frac{2 \alpha^{2}}{1+\alpha}-2 \alpha A \cdot P
\tag{1}
$$

Likewise

$$
\frac{(1-\beta)^{2}}{2(1+\beta)}\left|B_{2}\right|^{2}=\frac{1+\beta}{2}|P|^{2}+\frac{2 \beta^{2}}{1+\beta}-2 \beta B \cdot P
\tag{2}
$$

and

$$
\frac{(1-\gamma)^{2}}{2(1+\gamma)}\left|C_{2}\right|^{2}=\frac{1+\gamma}{2}|P|^{2}+\frac{2 \gamma^{2}}{1+\gamma}-2 \gamma C \cdot P
\tag{3}
$$

Summing (1), (2) and (3) we obtain on the LHS the positive linear combination

$$
\operatorname{LHS}=\frac{(1-\alpha)^{2}}{2(1+\alpha)}\left|A_{2}\right|^{2}+\frac{(1-\beta)^{2}}{2(1+\beta)}\left|B_{2}\right|^{2}+\frac{(1-\gamma)^{2}}{2(1+\gamma)}\left|C_{2}\right|^{2}
$$

and on the RHS the quantity

$$
\left(\frac{1+\alpha}{2}+\frac{1+\beta}{2}+\frac{1+\gamma}{2}\right)|P|^{2}+\left(\frac{2 \alpha^{2}}{1+\alpha}+\frac{2 \beta^{2}}{1+\beta}+\frac{2 \gamma^{2}}{1+\gamma}\right)-2(\alpha A \cdot P+\beta B \cdot P+\gamma C \cdot P) .
$$

The first term is $2|P|^{2}$ and the last term is $-2 P \cdot P$, so

$$
\begin{aligned}
\text { RHS } & =\left(\frac{2 \alpha^{2}}{1+\alpha}+\frac{2 \beta^{2}}{1+\beta}+\frac{2 \gamma^{2}}{1+\gamma}\right) \\
& =\frac{3 \alpha-1}{2}+\frac{(1-\alpha)^{2}}{2(1+\alpha)}+\frac{3 \beta-1}{2}+\frac{(1-\beta)^{2}}{2(1+\beta)}+\frac{3 \gamma-1}{2}+\frac{(1-\gamma)^{2}}{2(1+\gamma)} \\
& =\frac{(1-\alpha)^{2}}{2(1+\alpha)}+\frac{(1-\beta)^{2}}{2(1+\beta)}+\frac{(1-\gamma)^{2}}{2(1+\gamma)} .
\end{aligned}
$$

Here we used the fact that

$$
\frac{3 \alpha-1}{2}+\frac{3 \beta-1}{2}+\frac{3 \gamma-1}{2}=0 .
$$

We have shown that a linear combination of $\left|A_{1}\right|^{2},\left|B_{1}\right|^{2}$, and $\left|C_{1}\right|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $\left|A_{1}\right|^{2},\left|B_{1}\right|^{2}$, and $\left|C_{1}\right|^{2}$ must be at least 1 , as required.