20. Given a triangle $D E F$, obtuse at $E$, let $O$ be the center of the circumcircle of the triangle. Let $P$ be the point where the angle bisector from $E$ intersects the side $D F$. It is known that the convex quadrilateral $F E P O$ is cyclic. Given that $\widehat{E D F}=52^{\circ}$, what is the measure of the angle $\widehat{D F E}$?
(A) $24^{\circ}$
(B) $22^{\circ}$
(C) $26^{\circ}$
(D) $30^{\circ}$
(E) $28^{\circ}$

## Unione Matematica Italiana  PROGETTO OLIMPIADI DI MATEMATICA

Ministry OF EDUCATION,

UNIVERSITY AND RESEARCH

## The Games of Archimedes - Triennio Competition

November 21, 2019

- The test consists of 20 problems. Each question is followed by 5 answers indicated by the letters $\mathbf{( A )},(\mathbf{B}),(\mathbf{C}),(\mathbf{D}),(\mathbf{E})$. Only one of these answers is correct.
- Each correct answer is worth 5 points, each incorrect answer is worth 0 points. For each unanswered or illegible answer, 1 point will be assigned.
- For each of the problems, write IN BLOCK LETTERS the letter corresponding to the answer you believe is correct in the grid below. No erasures or corrections are allowed.
- COMPLETELY BLACKEN your birth month, your gender, and your class. Write the other required information IN BLOCK LETTERS next to the arrows, with the utmost care and precision.

The use of calculators or communication devices is not allowed. You have 110 minutes to complete the test. Good luck!

![](https://cdn.mathpix.com/cropped/2024_04_17_594e72200cb405ca2f14g-3.jpg?height=498&width=613&top_left_y=1049&top_left_x=75)

ANSWER GRID T2

| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 5 | 6 | 7 | 8 |
| 9 | 10 | 11 | 12 |
| 13 | 14 | 15 | 16 |
| 17 | 18 | 19 | 20 |
(20) The correct answer is $(A)$.

Let $\widehat{E D F}=\alpha$ and $\widehat{D E P}=\widehat{P E F}=x$.

By the exterior angle theorem (applied to triangle $D E P$), we have $\widehat{E P F}=\widehat{E D P}+\widehat{D E P}=\alpha+x$. Furthermore, considering the circumcircle of quadrilateral $F E P O$, we have that $\widehat{E P F}=\widehat{E O F}$, as they are angles subtending the same arc $E F$. By the central angle theorem (applied to the circumcircle of $D E F$), we have $2 \widehat{E D F}=\widehat{E O F}$, which means $2 \alpha=\alpha+x$, or $x=\alpha$.

Therefore, $\widehat{D E F}=2 \alpha$ and, considering triangle $D E F$, we conclude that the angle $\widehat{D F E}$ is the supplement of $3 \alpha$. Given that $\alpha=52^{\circ}$, we find that $\widehat{D F E}=180^{\circ}-3 \cdot 52^{\circ}=24^{\circ}$. Problem proposed by Camilla Casamento Tumeo.

![](https://cdn.mathpix.com/cropped/2024_04_17_c43da7de8bb5572fa51ag-5.jpg?height=585&width=506&top_left_y=255&top_left_x=1422)

As every year, the problems proposed for the Archimedes Games have been, in addition to being checked and selected, in many cases extensively modified or reformulated by the competition organizers. Therefore, in indicating the names of the proposers, it is understood that the responsibility for the selection of the questions, as well as for any errors, imprecisions, or questionable formulations, lies exclusively with the competition organizers and not at all with the proposers. We thank all of them, as well as those who contributed with questions that were not included in the competition text, for the variety, originality, and quality of the proposals presented.

The organizers of the Archimedes Games: Paolo Francini, Andrea Sambusetti