Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.2 AIME Find the answer Italy

Problem:

A convex quadrilateral $ABCD$ is such that $AB = 14$ , $BC = 24$ , $CD = 26$ and $DA = 16$ . Its area is $360$ . Find the area of the region of the plane inside the quadrilateral consisting of the points at distance at most $3$ from its perimeter.

Pick one

Solution

Solution:

The answer is $(\mathbf{E})$ . The solution of the problem is based on two key considerations. First of all, noting that $AB + CD = BC + DA = 40$ , we deduce that the quadrilateral $ABCD$ is circumscribed about a circle. In particular, denoting by $r$ the radius of this circle, we have $r = \frac{360}{40} = 9$ .

Now let $I$ be the center of the circle. Applying a homothety centered at $I$ with ratio $\frac{2}{3}$ , the quadrilateral is sent to $A'B'C'D'$ , which is circumscribed about a circle with center $I$ and radius $6$ . We then note that the sides of $A'B'C'D'$ are at distance exactly $3$ from those of $ABCD$ , so a point interior to $ABCD$ is at distance at most $3$ from the perimeter if and only if it does not lie inside $A'B'C'D'$ .

To conclude, $A'B'C'D'$ has area $\left(\frac{2}{3}\right)^2 \cdot 360 = 160$ , so the answer is $360 - 160 = 200$ .

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