A quick problem!

Geometry Level 3

In rectangle $ABCD$ , vertex $B$ is on line $L_{1}$ and vertex $D$ is on line $L_{2}$ and vertex $A$ is a distance $m$ from $L_{1}$ and a distance $n$ from $L_{2}$ .

Let $A_{min}$ be the minimum area of rectangle $ABCD$ . Find $\dfrac{A_{min}}{mn}$ .

The answer is 2.

1 solution

Rocco Dalto
Mar 18, 2021

$m = a\cos(\dfrac{\pi}{2} - \theta) = a\sin(\theta) \implies a = \dfrac{m}{\sin(\theta)}$ and $n = b\cos(\theta) \implies b = \dfrac{n}{\cos(\theta)}$

$\implies A_{ABCD} = 2mn\csc(2\theta) \geq 2mn \implies \dfrac{A_{min}}{mn} = \boxed{2}$

Note: Min occurs at $\theta = \dfrac{\pi}{4}$ .

A quick problem!

The answer is 2.

1 solution

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