Subfield: Geometry


Question: Given an acute triangle $A B C$ with angles $\alpha, \beta$ and $\gamma$ at vertices $A, B$ and $C$, respectively, such that $\beta>\gamma$. Point $I$ is the incenter, and $R$ is the circumradius. Point $D$ is the foot of the altitude from vertex $A$. Point $K$ lies on line $A D$ such that $A K=2 R$, and $D$ separates $A$ and $K$. Finally, lines $D I$ and $K I$ meet sides $A C$ and $B C$ at $E$ and $F$, respectively.

Prove that if $I E=I F$ then $\beta \leq 3 \gamma$.

Solution:
We first prove that
$$
\angle K I D=\frac{\beta-\gamma}{2} \tag{1}
$$
even without the assumption that $I E=I F$. Then we will show that the statement of the problem is a consequence of this fact.

Denote the circumcenter by $O$. On the circumcircle, let $P$ be the point opposite to $A$, and let the angle bisector $A I$ intersect the circle again at $M$. Since $A K=A P=2 R$, triangle $A K P$ is isosceles. It is known that $\angle B A D=\angle C A O$, hence $\angle D A I=\angle B A I-\angle B A D=\angle C A I-$ $\angle C A O=\angle O A I$, and $A M$ is the bisector line in triangle $A K P$. Therefore, points $K$ and $P$ are symmetrical about $A M$, and $\angle A M K=\angle A M P=90^{\circ}$. Thus, $M$ is the midpoint of $K P$, and $A M$ is the perpendicular bisector of $K P$.

<img_3675>

Denote the perpendicular feet of incenter $I$ on lines $B C, A C$, and $A D$ by $A_{1}, B_{1}$, and $T$, respectively. Quadrilateral $D A_{1} I T$ is a rectangle, hence $T D=I A_{1}=I B_{1}$.

Due to the right angles at $T$ and $B_{1}$, quadrilateral $A B_{1} I T$ is cyclic. Hence $\angle B_{1} T I=$ $\angle B_{1} A I=\angle C A M=\angle B A M=\angle B P M$ and $\angle I B_{1} T=\angle I A T=\angle M A K=\angle M A P=$ $\angle M B P$. Therefore, triangles $B_{1} T I$ and $B P M$ are similar and $\frac{I T}{I B_{1}}=\frac{M P}{M B}$.

It is well-known that $M B=M C=M I$. Then right triangles $I T D$ and $K M I$ are also



similar, because $\frac{I T}{T D}=\frac{I T}{I B_{1}}=\frac{M P}{M B}=\frac{K M}{M I}$. Hence, $\angle K I M=\angle I D T=\angle I D A$, and
$$
\angle K I D=\angle M I D-\angle K I M=(\angle I A D+\angle I D A)-\angle I D A=\angle I A D .
$$
Finally, from the right triangle $A D B$ we can compute
$$
\angle K I D=\angle I A D=\angle I A B-\angle D A B=\frac{\alpha}{2}-\left(90^{\circ}-\beta\right)=\frac{\alpha}{2}-\frac{\alpha+\beta+\gamma}{2}+\beta=\frac{\beta-\gamma}{2} .
$$
Now let us turn to the statement and suppose that $I E=I F$. Since $I A_{1}=I B_{1}$, the right triangles $I E B_{1}$ and $I F A_{1}$ are congruent and $\angle I E B_{1}=\angle I F A_{1}$. Since $\beta>\gamma, A_{1}$ lies in the interior of segment $C D$ and $F$ lies in the interior of $A_{1} D$. Hence, $\angle I F C$ is acute. Then two cases are possible depending on the order of points $A, C, B_{1}$ and $E$.
<img_3669>

If point $E$ lies between $C$ and $B_{1}$ then $\angle I F C=\angle I E A$, hence quadrilateral $C E I F$ is cyclic and $\angle F C E=180^{\circ}-\angle E I F=\angle K I D$. By (1), in this case we obtain $\angle F C E=\gamma=\angle K I D=$ $\frac{\beta-\gamma}{2}$ and $\beta=3 \gamma$.

Otherwise, if point $E$ lies between $A$ and $B_{1}$, quadrilateral $C E I F$ is a deltoid such that $\angle I E C=\angle I F C<90^{\circ}$. Then we have $\angle F C E>180^{\circ}-\angle E I F=\angle K I D$. Therefore, $\angle F C E=\gamma>\angle K I D=\frac{\beta-\gamma}{2}$ and $\beta<3 \gamma$.
$$
\angle K I D=\frac{\beta-\gamma}{2} \tag{1}
$$
We give a different proof for (1). Then the solution can be finished in the same way as above.

Define points $M$ and $P$ again; it can be proved in the same way that $A M$ is the perpendicular bisector of $K P$. Let $J$ be the center of the excircle touching side $B C$. It is well-known that points $B, C, I, J$ lie on a circle with center $M$; denote this circle by $\omega_{1}$.

Let $B^{\prime}$ be the reflection of point $B$ about the angle bisector $A M$. By the symmetry, $B^{\prime}$ is the second intersection point of circle $\omega_{1}$ and line $A C$. Triangles $P B A$ and $K B^{\prime} A$ are symmetrical



with respect to line $A M$, therefore $\angle K B^{\prime} A=\angle P B A=90^{\circ}$. By the right angles at $D$ and $B^{\prime}$, points $K, D, B^{\prime}, C$ are concyclic and
$$
A D \cdot A K=A B^{\prime} \cdot A C
$$
From the cyclic quadrilateral $I J C B^{\prime}$ we obtain $A B^{\prime} \cdot A C=A I \cdot A J$ as well, therefore
$$
A D \cdot A K=A B^{\prime} \cdot A C=A I \cdot A J
$$
and points $I, J, K, D$ are also concyclic. Denote circle $I D K J$ by $\omega_{2}$.

<img_3759>

Let $N$ be the point on circle $\omega_{2}$ which is opposite to $K$. Since $\angle N D K=90^{\circ}=\angle C D K$, point $N$ lies on line $B C$. Point $M$, being the center of circle $\omega_{1}$, is the midpoint of segment $I J$, and $K M$ is perpendicular to $I J$. Therefore, line $K M$ is the perpendicular bisector of $I J$ and hence it passes through $N$.

From the cyclic quadrilateral $I D K N$ we obtain
$$
\angle K I D=\angle K N D=90^{\circ}-\angle D K N=90^{\circ}-\angle A K M=\angle M A K=\frac{\beta-\gamma}{2}
$$