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2012 MPFG Problem 10: Difference between revisions

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<math>\#1</math> <math>\angle FGD = 90^\circ</math>


<math>\#1</math> <math>\angle FGD = 90^\circ</math>
[[File:DGF.png|600px|center]]


Connect BE. We discover that DG and FD are consecutively the midlines of <math>\Delta BEC</math> and <math>\Delta ABC</math>.  
Connect BE. We discover that DG and FD are consecutively the midlines of <math>\Delta BEC</math> and <math>\Delta ABC</math>.  
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<math>\#2</math> <math>\angle FDG = 90^\circ</math>
<math>\#2</math> <math>\angle FDG = 90^\circ</math>
[[File:Forgot_line.png|600px|center]]


Because <math>\angle DGC = \angle FDG = 90^\circ</math>, <math>\angle C = \frac{180^\circ-90^\circ}{2} = 45^\circ</math>
Because <math>\angle DGC = \angle FDG = 90^\circ</math>, <math>\angle C = \frac{180^\circ-90^\circ}{2} = 45^\circ</math>

Revision as of 04:47, 26 August 2025

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 $\boxed{\frac{2}{3}}$.

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