16. Given a right trapezoid $P Q R S$, with right angles at $P$ and $S$, where $\overline{P Q} > \overline{R S}$ and $\overline{P S} = \overline{R S} = 31$. Let $K$ be the point on side $P S$ such that $\overline{P K} = 14$.

Knowing that $\widehat{S K R} = \widehat{P Q R}$, what will be the measure of $K Q$?
(A) 45
(B) 48
(C) 52
(D) 49
(E) 50

## Unione Matematica Italiana  PROGETTO OLIMPIADI DI MATEMATICA  MinisTERo DElL'ISTRUZione,  DELL'UNIVERSITÀ E DELLA RICERCA  I Giochi di Archimede - Gara Biennio

21 November 2019

- The test consists of 16 problems. Each question is followed by 5 answers indicated by the letters (A), (B), (C), (D), (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 month of birth, 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_681a59a021339eeafb7fg-3.jpg?height=495&width=600&top_left_y=1028&top_left_x=86)

ANSWER GRID $\quad$ T2

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

Let $T$ be the foot of the perpendicular from $R$ to $P Q$. From the data given in the problem, it follows that the right triangles $K S R$ and $Q T R$ are congruent. Therefore, $\overline{T Q}=\overline{K S}=\overline{P S}-\overline{P K}=31-14=17$. It follows that $\overline{P Q}=\overline{P T}+\overline{T Q}=31+17=48$.

Using the Pythagorean theorem, we conclude that $\overline{K Q}=\sqrt{\overline{P K}^{2}+\overline{P Q}^{2}}=\sqrt{14^{2}+48^{2}}=50$. Problem proposed by Camilla Casamento Tumeo.

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. Our thanks go to all of them, as well as to 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