B3. Let $\mathcal{K}$ be the circumcircle of triangle $ABC$, and $I$ the center of the incircle of triangle $ABC$. The midpoint of the arc $BC$ of circle $\mathcal{K}$ that does not contain point $A$ is denoted by $D$, and the midpoint of the arc $CA$ of circle $\mathcal{K}$ that does not contain point $B$ is denoted by $E$. Let $F$ be the reflection of point $I$ across side $AB$, and suppose that $F$ lies on circle $\mathcal{K}$. What is the measure of angle $\angle DFE$?

## 59th Mathematical Competition for High School Students in Slovenia

Regional Competition, April 1, 2015

## Solutions for 1st Year

In section A, a correct answer will be awarded two points, while a wrong answer will result in the deduction of one point. To avoid a potential negative final score, each contestant is given an initial 3 points.

| 1 | 2 | 3 |
| :--- | :--- | :--- |
| C | C | E |

Explanations:
B3.

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-27.jpg?height=668&width=654&top_left_y=1485&top_left_x=313)

Since points $D$ and $E$ are the midpoints of the respective arcs of the circle $\mathcal{K}$, they lie on the angle bisectors of $\angle BAC$ and $\angle CBA$. Due to reflection, $\angle FAB = \angle BAI = \frac{1}{2} \angle BAC$, so $\angle FED = \angle FAD = \angle BAC$. Similarly, we get $\angle EDF = \angle CBA$. Therefore, $\angle DFE = \angle ACB$. Let's calculate the measure of the angle $\angle ACB$. Points $F, B, C$, and $A$ are concyclic, so $\angle ACB = \pi - \angle BFA$. Since $F$ is the reflection of point $I$, we have

\[
\begin{aligned}
\angle BFA & = \angle AIB = \pi - \frac{1}{2} \angle BAC - \frac{1}{2} \angle CBA = \pi - \frac{1}{2} (\angle BAC + \angle CBA) = \\
& = \pi - \frac{1}{2} (\pi - \angle ACB) = \frac{\pi}{2} + \frac{1}{2} \angle ACB.
\end{aligned}
\]

Thus, $\angle ACB = \pi - \left( \frac{\pi}{2} + \frac{1}{2} \angle ACB \right) = \frac{\pi}{2} - \frac{1}{2} \angle ACB$, from which it follows that $\angle ACB = \frac{\pi}{3}$ and therefore $\angle DFE = \frac{\pi}{3}$.

Clear and labeled diagram .............................................................. Determination that points $D$ and $E$ lie on the angle bisectors ............................. 1 point Justified conclusions that $\angle FED = \angle FAD = \angle BAC$, $\angle EDF = \angle CBA$, and $\angle DFE = \angle ACB$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-28.jpg?height=63&width=1579&top_left_y=588&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-28.jpg?height=63&width=1574&top_left_y=637&top_left_x=278)