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2012 MPFG Problem 10

Problem

Let $\Delta ABC$ be a triangle with a right angle $\angle ABC$ . Let $D$ be the midpoint of $BC$ , let $E$ be the midpoint of $AC$ , and let $F$ be the midpoint of $AB$ . Let $G$ be the midpoint of $EC$ . One of the angles of $\Delta DFG$ is a right angle. What is the least possible value of $\frac{BC}{AG}$ ? Express your answer as a fraction in simplest form.

Solution 1

The question doesn't give us which angle of $\Delta DFG$ is the right angle, so we would have to discuss different cases. Obviously $\angle DFG$ can't be the right angle.

$\#1$ $\angle FGD = 90^\circ$

Connect BE. We discover that DG and FD are consecutively the midlines of $\Delta BEC$ and $\Delta ABC$ .

$AE = EC = BE = \frac{1}{2} AC$

$GD = \frac{1}{2}BE = \frac{1}{2}EC$

$FD = \frac{1}{2}AC = EC$

This gives us $FD = 2GD$ , which means $\Delta FDG$ is a $30^\circ - 60^\circ - 90^\circ$ triangle.

$\angle CGD = \angle GDF = 60^\circ$ . Because $DG = GC = \frac{1}{2} EC$ , $\angle C = \frac{180^\circ-60^\circ}{2} = 60^\circ$

$\Delta ABC$ is also a $30^\circ - 60^\circ - 90^\circ$ triangle.

$\frac{BC}{AG} = \frac{1}{2\cdot\frac{3}{4}} = \frac{2}{3}$

$\#2$ $\angle FDG = 90^\circ$

Because $\angle DGC = \angle FDG = 90^\circ$ , $\angle C = \frac{180^\circ-90^\circ}{2} = 45^\circ$

$\Delta ABC$ is a $45^\circ - 45^\circ - 90^\circ$ triangle.

$\frac{BC}{AG} = \frac{1}{\sqrt{2}\cdot\frac{3}{4}} = \frac{2\sqrt{2}}{3}$

The least possible value of $\frac{BC}{AG}$ is $\boxed{\frac{2}{3}}$ .

~cassphe

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