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Geometry Difficulty 5.4 AIME, harder Find the answer Italy

$AB$ and $CD$ are two segments, both of length $4$ , having the midpoint $M$ in common and such that $B \widehat{M} D = 60^\circ$ . We denote by $X$ the set of all and only the points that are at distance at most $1$ from at least one of the two segments. What is the measure of the surface of $X$ ?

Pick one

Solution

The answer is (D). Let $X_1$ (respectively $X_2$ ) be the locus of points at distance $\leqslant 1$ from segment $AB$ alone (respectively $CD$ ). Clearly $X_1$ and $X_2$ are congruent and $X$ is the union of the two. Then $\operatorname{Area}(X) = 2\operatorname{Area}(X_1) - \operatorname{Area}(X_1 \cap X_2)$ .

$X_1$ is formed by a rectangle with base $4$ and length $2$ , joined to two semicircles at the ends with diameter on the shorter sides. The area of $X_1$ is therefore $8 + \pi$ .

The intersection $X_1 \cap X_2$ is a parallelogram with both heights of length $2$ and one angle of $60^\circ$ , so it is the union of two equilateral triangles of side $\frac{4\sqrt{3}}{3}$ . Its area is $\frac{8\sqrt{3}}{3}$ .

In conclusion, $\operatorname{Area}(X) = 16 - \frac{8\sqrt{3}}{3} + 2\pi$ .

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