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Geometry Difficulty 4.3 AIME Find the answer Italy

Problem:

In a square $ABCD$ with side $2$ , a segment $MN$ of length $1$ is constrained to have endpoint $M$ on side $AB$ and endpoint $N$ on side $BC$ . This segment divides the square into a triangle $T$ and a pentagon $P$ . What is the maximum value that the ratio of the area of $T$ to that of $P$ can take?

Pick one

Solution

Solution:

The answer is $\mathbf{(E)}$ . The ratio between the two areas is maximized when the area of $T$ is maximized: in this case, the area of $P$ assumes its minimum value. $T$ is a right triangle whose hypotenuse is $1$ , and it has maximum area when it is isosceles. Indeed, such a triangle can be inscribed in a semicircle of diameter $1$ and attains maximum area when the altitude relative to the hypotenuse is maximized, that is, equals $\frac{1}{2}$ . In this case, the area of $T$ is $\frac{1}{4}$ and that of $P$ is $4-\frac{1}{4}=\frac{15}{4}$ , and thus the ratio is $\frac{1}{15}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from en; metadata (topic, difficulty) added by this project.