A-center

Geometry Level 3

The rectangle $ABCD$ has $AB = 4 , BC = 6$ . The square $AEFG$ has side length $\sqrt{13}$ . The latter rotates around the former along the center $A$ . The range of the length $CE$ is $\sqrt{a} \le CE \le \sqrt{b}$ . Evaluate $a+b$ .

The answer is 130.

1 solution

Mark Hennings
Jan 27, 2021

We have $AC = \sqrt{16 + 36} = 2\sqrt{13}$ , and hence $\begin{aligned} CE & \le \; AC + AE = 3\sqrt{13} = \sqrt{117} \\ CE & \ge \; AC - AE = \sqrt{13} \end{aligned}$ Both extremes are possible by rotating $ABCD$ so that $A,C,E$ are collinear. Thus $a = 117$ and $b = 13$ , so that $a+b=\boxed{130}$ .

@Mark Hennings , we really liked your comment, and have converted it into a solution.

Brilliant Mathematics Staff - 4 months, 2 weeks ago

A-center

The answer is 130.

1 solution

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