20. Given a rectangle $A B C D$, let $P$ be a point on the side $C D$. The line $A P$ intersects the line $B C$ at point $T$. Let $M$ be the midpoint of the side $B C$, and it is known that $\widehat{A P M}=2 \widehat{A T C}$. Knowing that the area of triangle $C P T$ is $10 \mathrm{~cm}^{2}$, what is the area of rectangle $A B C D$?
(A) $90 \mathrm{~cm}^{2}$
(B) $120 \mathrm{~cm}^{2}$
(C) $60 \mathrm{~cm}^{2}$
(D) $80 \mathrm{~cm}^{2}$
(E) $160 \mathrm{~cm}^{2}$

## Unione Matematica Italiana  PROGETTO OLIMPIADI DI MATEMATICA

MINISTRY OF EDUCATION,

UNIVERSITY AND RESEARCH

$\mathrm{T} 2$

## The Games of Archimedes - Triennio Competition  23 November 2017

- 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, the other 4 are wrong.
- Each correct answer is worth 5 points, each wrong answer is worth 0 points. For each unanswered or illegible answer, 1 point will be assigned.
- For each of the problems, you must write the letter corresponding to the answer you believe is correct in the grid provided below. No erasures or corrections on the grid are allowed. The use of any type of calculator or communication device is not permitted.

You have 110 minutes to complete the test. Good luck and have fun!

NAME

SURNAME

CLASS

date of birth: email (optional)
![](https://cdn.mathpix.com/cropped/2024_04_17_e21ecd5de9c42efa6d1eg-3.jpg?height=102&width=1308&top_left_y=1218&top_left_x=88)
(20) The correct answer is $(B)$.

We observe that $\widehat{A P M}$ is an exterior angle to triangle $M P T$, so $\widehat{A P M}=\widehat{P M T}+\widehat{P T M}$. Given that $\widehat{A P M}=2 \widehat{P T M}$, it follows that $\widehat{P M T}=\widehat{P T M}$. Triangle $M P T$ is therefore isosceles, and since $C$ is the foot of its altitude, we have $T C=C M=M B$, meaning $T B$ is three times $T C$. By the similarity of triangles $T P C$ and $T A B$, $A B$ will also be three times $C P$. In conclusion, the sides of rectangle $A B C D$ are, respectively, twice and three times the legs of right triangle $T P C$. The area of $A B C D$ is thus equal to $2 \cdot(2 \cdot 3)=12$ times the area of $T P C$.

Problem proposed by Camilla Casamento Tumeo.

N.B. As usual, the proposed problems 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 organizers themselves and in no way with the proposers, whom we wish to thank for their significant contribution, the richness of the ideas offered, the originality, and the quality of the proposals presented.

The organizers of the Archimedes Games, Alessandro D'Andrea, Paolo Francini, Andrea Sambusetti