India — Team Selection Test

We have that $AB \neq AC$, so $(AEF)$ and $(ABC)$ are not tangent at $A$, thus there is a point $S \neq A$ which is the second intersection of $(AEF)$ and $(ABC)$.

Consider an inversion about the incircle and let the inverse of a point $P$ be denoted by $P'$. Note that $A'$ is the midpoint of $EF$. $S'$ is the foot from $D$ onto $EF$ as $S' \neq A'$ is on the nine point circle of $\triangle DEF$ as well as on $EF$.

Also, since $X$ satisfies $\angle XEI = \angle XDI = 45^\circ$, point $X'$ satisfies $\angle EX'I = \angle DX'I = 45^\circ$. Note that $X$ lies on the same side of $DE$ as $I$, which in turn is the same side of $DE$ as $F$ since $\triangle DEF$ is acute.

Taking $DEF$ to be the reference triangle, the problem becomes the following:

Inverted Problem: In acute $\triangle DEF$, let $X', Y'$ be points such that $DX'E$ and $FY'D$ are isosceles right triangles with $X'$ on the same side of $DE$ as $F$, and $Y'$ on the same side of $DF$ as $E$. Let $A'$ be the midpoint of $EF$ and $S'$ the foot onto $EF$ from $D$. Prove that points $A', S', X', Y'$ are concyclic.

We will prove this by showing that $\angle X'A'Y' = \angle X'S'Y' = 90^\circ$.

Claim 1. $\angle X'S'Y' = 90^\circ$.

Proof. Observe that $\angle X'S'Y' = \angle X'S'D + \angle DS'Y' = \angle X'ED + \angle DFY' = 45^\circ + 45^\circ = 90^\circ$ □

Claim 2. $X'A'Y'$ is a right isosceles triangle with a right angle at $A'$.

Proof. To prove this, we will prove that $\triangle X'A'Y' \sim \triangle X'MD$ where $M$ is the midpoint of $DE$. But observe that by spiral similarity, this is equivalent to showing that $\triangle X'MA' \sim \triangle X'DY'$.

Now,
$$\angle X'MA' = 90^{\circ} - \angle FDE = 45^{\circ} + 45^{\circ} - \angle FDE = \angle FDX' + \angle Y'DE - \angle FDE = \angle X'DY'$$

$$\triangle X'MA' \sim \triangle X'DY' \implies \triangle X'A'Y' \sim \triangle X'MD' \implies \angle X'A'Y = \angle X'MD = 90^{\circ}$$
Thus, we are done!

An alternate proof of the Claim 2 of the first solution:

Claim 2 of the first solution. In acute $\triangle DEF$, let $X', Y'$ be points such that $DX'E$ and $FY'D$ are isosceles right triangles with $X'$ on the same side of $DE$ as $F$, and $Y'$ on the same side of $DF$ as $E$. Let $A'$ be the midpoint of $EF$. Prove that $\triangle X'A'Y'$ is a isosceles right triangle with a right angle at $A'$.

Proof. Consider the composition of rotations
$$\Gamma = \operatorname{Rot}(A', 180^\circ) \circ \operatorname{Rot}(Y', 90^\circ) \circ \operatorname{Rot}(X', 90^\circ)$$
The angles add up to $360^\circ$, so $\Gamma$ is a translation. However, $\Gamma$ keeps $E$ fixed, thus $\Gamma$ is the identity map. Hence, we have
$$\operatorname{Rot}(A', 180^\circ) = \operatorname{Rot}(Y', 90^\circ) \circ \operatorname{Rot}(X', 90^\circ)$$
and then we finish by noting that the center of rotation of the composition of the two rotations $\operatorname{Rot}(Y', 90^\circ)$, $\operatorname{Rot}(X', 90^\circ)$ is given by a point $O$ such that $\angle OX'Y' = 45^\circ$, and $\angle OY'X' = 45^\circ$.

Lemma 1. In triangle $ABC$, let midpoints $M, N, K$ of sides $BC, CA, AB$ respectively and $P$ be the foot of the $A$-altitude on $BC$. Suppose $X, Y$ are points in the plane with $XN = AN$ and $XN \perp AC$ and $YK = AK$ and $YK \perp AB$. Then $X, Y, P, M$ are concyclic.

Proof. Let $Q$ be the point on the $A$-altitude such that $AQ = \frac{1}{2}BC$. Observe that the rotation with centre $X$ of measure $90^\circ$ that sends $C$ to $A$ maps line $BC$ to the line through $A$ perpendicular to $BC$. In particular, since $AQ = CM$, we see that $M$ maps to $Q$. Hence $\angle MXQ = 90^\circ$. Similarly, $\angle MYQ = 90^\circ$ and combined with $\angle MPQ = 90^\circ$, we conclude that $X, Y, P, M$ lie on the circle with diameter $MQ$.

Lemma 2. In triangle $ABC$, the bisector of angle $BAC$ meets the circumcircle of the triangle at $M$. Points $P$ and $Q$ lie on line $AM$ such that $\angle PBA = \angle QBC$. Then $MP \cdot MQ = MB^2$.

Proof. It suffices to show $\angle MBQ = \angle MPB$. This follows as
$$\angle MBQ = \angle MBC + \angle QBC = \angle MAC + \angle PBA = \angle PAB + \angle PBA = \angle MPB$$
and the other configurations can be dealt with by directed angles.

Now back to the original problem. Erect isosceles right-angled triangles $DX'E$ and $DY'F$ in-wards of $\triangle DEF$ with $X', Y'$ the vertices with the right angle. By Lemma 2 applied to $\triangle CDE, \triangle BDF$, it follows that $X, X'$ and $Y, Y'$ are inverses in the incircle. The inverse of $(AEF) \cap (ABC)$ in the incircle is the foot of perpendicular $K$ from $D$ to $EF$ and that of $A$ is the midpoint of $EF$. The conclusion follows from Lemma 1 in $\triangle DEF$. $\square$