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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Soviet Union

Problem:
Given a fixed regular pentagon $ABCDE$ with side $1$ . Let $M$ be an arbitrary point inside or on it. Let the distance from $M$ to the closest vertex be $\mathbf{r}_1$ , to the next closest be $\mathbf{r}_2$ and so on, so that the distances from $M$ to the five vertices satisfy $\mathbf{r}_1 \leq \mathbf{r}_2 \leq \mathbf{r}_3 \leq \mathbf{r}_4 \leq \mathbf{r}_5$ . Find

a) the locus of $M$ which gives $\mathbf{r}_3$ the minimum possible value,

b) the locus of $M$ which gives $\mathbf{r}_3$ the maximum possible value.

Solution

Solution:
Let $X$ be the midpoint of $AB$ and $O$ the center of $ABCDE$ . Suppose $M$ lies inside $AXO$ . Then $ME = \mathbf{r}_3$ . So we maximise $\mathbf{r}_3$ by taking $M$ at $X$ , with distance $1.5590$ , and we minimise $\mathbf{r}_3$ by taking $M$ as the intersection of $AO$ and $EB$ with distance $0.8090$ . $AXO$ is one of $10$ congruent areas, so the required loci are

a) the $5$ midpoints of the diagonals,

b) the $5$ midpoints of the sides.

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