Subfield: Geometry


Question: Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle D P A+\angle A Q D=\angle Q I P$.

Solution:
Let $N$ and $M$ be the midpoints of the $\operatorname{arcs} \overparen{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\angle F A E=\angle B A C=\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\varphi$ (dilation in case of $A B=A C$ ) that moves $A F I E$ to $N B M C$. The directed angle in which $\varphi$ changes directions is $\angle(A F, N B)$, same as $\angle(A P, N P)$ and $\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P=E F \cap A P$ and $Q=E F \cap A Q$ are mapped to points $B C \cap N P$ and $B C \cap N Q$, respectively. Denote these points by $P^{\prime}$ and $Q^{\prime}$, respectively.

<img_3581>

Let $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D=L$ then line $B C$ is tangent to circle $P Q D$ ). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have

$$
\frac{B D}{D C}=\frac{B F}{F A} \cdot \frac{A E}{E C}=-\frac{B Z}{Z C}
$$

so the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \cdot Z C=Z D \cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \cdot Z L=Z B \cdot Z C=Z P \cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim.

By $\angle M P P^{\prime}=\angle M Q Q^{\prime}=\angle M L P^{\prime}=\angle M L Q^{\prime}=90^{\circ}$, the quadrilaterals $M L P P^{\prime}$ and $M L Q Q^{\prime}$ are cyclic. Then the problem statement follows by

$$
\begin{aligned}
\angle D P A+\angle A Q D & =360^{\circ}-\angle P A Q-\angle Q D P=360^{\circ}-\angle P N Q-\angle Q L P \\
& =\angle L P N+\angle N Q L=\angle P^{\prime} M L+\angle L M Q^{\prime}=\angle P^{\prime} M Q^{\prime}=\angle P I Q .
\end{aligned}
$$
Define the point $M$ and the same spiral similarity $\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$.

By $\varphi(F)=B$ and $\varphi(E)=C$ the triangles $S B F$ and $S C E$ are similar, so we have

$$
\frac{S B}{S C}=\frac{B F}{C E}=\frac{B D}{C D}
$$

By the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\angle B S C$ and hence passes through $M$.

Let $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\varphi(K)=\varphi(F E \cap S I)=B C \cap S M=D$. By $\varphi(I)=M$, we have $K D \| I M$.

<img_3237>

We claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\angle A M L=\angle A S L=\angle A S I=90^{\circ}$, we have $M L \| P Q$. Hence, $\overparen{P L}=\overparen{M Q}$ and therefore $\angle P S L=\angle M S Q=\angle D S Q$. By $\angle Q P S=\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally,

$$
\frac{S P}{S I}=\frac{S P}{S K} \cdot \frac{S K}{S I}=\frac{S M}{S Q} \cdot \frac{S D}{S M}=\frac{S D}{S Q}
$$

shows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously.

Now the problem statement can be proved by

$$
\angle D P A+\angle A Q D=\angle D P S+\angle S Q D=\angle Q I S+\angle S I P=\angle Q I P
$$
Denote the circumcircle of triangle $A B C$ by $\Gamma$, and let rays $P D$ and $Q D$ meet $\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \perp I P$ and $A V \perp I Q$. Then the problem statement will follow as

$$
\angle D P A+\angle A Q D=\angle V U A+\angle A V U=180^{\circ}-\angle U A V=\angle Q I P .
$$

Let $M$ be the midpoint of arc $\overparen{B U V C}$ and let $N$ be the midpoint of $\operatorname{arc} \overparen{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively.

we observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively.

Combine the inversion with projecting $\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively.

<img_3478>

The combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \perp I F$, $A C \perp I E$ and $A N \perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \perp I P$ and $A V \perp I Q$, and hence completes the solution.