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2012 MPFG Problem 10: Difference between revisions

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==Solution 1==
==Solution 1==


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




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


<math>\#1</math> <math>\angle FGD = 90^\circ</math>
[[File:DGF.png|200px|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 <imath>\Delta BEC</imath> and <imath>\Delta ABC</imath>.  


<math>AE = EC = BE = \frac{1}{2} AC</math>
<imath>AE = EC = BE = \frac{1}{2} AC</imath>


<math>GD = \frac{1}{2}BE = \frac{1}{2}EC</math>
<imath>GD = \frac{1}{2}BE = \frac{1}{2}EC</imath>


<math>FD = \frac{1}{2}AC = EC</math>
<imath>FD = \frac{1}{2}AC = EC</imath>


This gives us <math>FD = 2GD</math>, which means <math>\Delta FDG</math> is a <math>30^\circ - 60^\circ - 90^\circ</math> triangle.  
This gives us <imath>FD = 2GD</imath>, which means <imath>\Delta FDG</imath> is a <imath>30^\circ - 60^\circ - 90^\circ</imath> triangle.  


<math>\angle CGD = \angle GDF = 60^\circ</math>. Because <math>DG = GC = \frac{1}{2} EC</math>, <math>\angle C = \frac{180^\circ-60^\circ}{2} = 60^\circ</math>
<imath>\angle CGD = \angle GDF = 60^\circ</imath>. Because <imath>DG = GC = \frac{1}{2} EC</imath>, <imath>\angle C = \frac{180^\circ-60^\circ}{2} = 60^\circ</imath>


<math>\Delta ABC</math> is also a <math>30^\circ - 60^\circ - 90^\circ</math> triangle.
<imath>\Delta ABC</imath> is also a <imath>30^\circ - 60^\circ - 90^\circ</imath> triangle.


<math>\frac{BC}{AG} = \frac{1}{2\cdot\frac{3}{4}} = \frac{2}{3}</math>
<imath>\frac{BC}{AG} = \frac{1}{2\cdot\frac{3}{4}} = \frac{2}{3}</imath>






<math>\#2</math> <math>\angle FDG = 90^\circ</math>
<imath>\#2</imath> <imath>\angle FDG = 90^\circ</imath>


Because <math>\angle DGC = \angle FDG = 90^\circ</math>, <math>\angle C = \frac{180^\circ-90^\circ}{2} = 45^\circ</math>
[[File:FDG.png|250px|center]]


<math>\Delta ABC</math> is a <math>45^\circ - 45^\circ - 90^\circ</math> triangle.
Because <imath>\angle DGC = \angle FDG = 90^\circ</imath>, <imath>\angle C = \frac{180^\circ-90^\circ}{2} = 45^\circ</imath>


<math>\frac{BC}{AG} = \frac{1}{\sqrt{2}\cdot\frac{3}{4}} = \frac{2\sqrt{2}}{3}</math>
<imath>\Delta ABC</imath> is a <imath>45^\circ - 45^\circ - 90^\circ</imath> triangle.


The least possible value of <math>\frac{BC}{AG}</math> is <math>\boxed{\frac{2}{3}}</math>.
<imath>\frac{BC}{AG} = \frac{1}{\sqrt{2}\cdot\frac{3}{4}} = \frac{2\sqrt{2}}{3}</imath>
 
The least possible value of <imath>\frac{BC}{AG}</imath> is <imath>\boxed{\frac{2}{3}}</imath>.
 
~cassphe

Latest revision as of 09:38, 7 November 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 $\boxed{\frac{2}{3}}$.

~cassphe

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