Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 MPFG Problem 10"

(Created page with "==Problem== Let <math>\Delta ABC</math> be a triangle with a right angle <math>\angle ABC</math>. Let <math>D</math> be the midpoint of <math>BC</math>, let <math>E</math> be...")
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
Let <math>\Delta ABC</math> be a triangle with a right angle <math>\angle ABC</math>. Let <math>D</math> be the midpoint of <math>BC</math>, let <math>E</math> be the midpoint of <math>AC</math>, and let <math>F</math> be the midpoint of <math>AB</math>. Let <math>G</math> be the midpoint of <math>EC</math>. One of the angles of <math>\Delta DFG</math> is a right angle. What is the least possible value of <math>\frac{BC}{AG}</math> ? Express your answer as a fraction in simplest form.
 
Let <math>\Delta ABC</math> be a triangle with a right angle <math>\angle ABC</math>. Let <math>D</math> be the midpoint of <math>BC</math>, let <math>E</math> be the midpoint of <math>AC</math>, and let <math>F</math> be the midpoint of <math>AB</math>. Let <math>G</math> be the midpoint of <math>EC</math>. One of the angles of <math>\Delta DFG</math> is a right angle. What is the least possible value of <math>\frac{BC}{AG}</math> ? Express your answer as a fraction in simplest form.
 +
 +
==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.
 +
 +
 +
 +
<math>\#1</math> <math>\angle FGD = 90^\circ</math>
 +
 +
Connect BE. We discover that DG and FD are consecutively the midlines of <math>\Delta BEC</math> and <math>\Delta ABC</math>.
 +
 +
<math>AE = EC = BE = \frac{1}{2} AC</math>
 +
 +
<math>GD = \frac{1}{2}BE = \frac{1}{2}EC</math>
 +
 +
<math>FD = \frac{1}{2}AC = EC</math>
 +
 +
This gives us <math>FD = 2GD</math>, which means <math>\Delta FDG</math> is a <math>30^\circ - 60^\circ - 90^\circ</math> 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>
 +
 +
<math>\Delta ABC</math> is also a <math>30^\circ - 60^\circ - 90^\circ</math> triangle.
 +
 +
<math>\frac{BC}{AG} = \frac{1}{2\cdot\frac{3}{4}} = \frac{2}{3}</math>
 +
 +
 +
 +
<math>\#2</math> <math>\angle FDG = 90^\circ</math>
 +
 +
Because <math>\angle DGC = \angle FDG = 90^\circ</math>, <math>\angle C = \frac{180^\circ-90^\circ}{2} = 45^\circ</math>
 +
 +
<math>\Delta ABC</math> is a <math>45^\circ - 45^\circ - 90^\circ</math> triangle.
 +
 +
<math>\frac{BC}{AG} = \frac{1}{\sqrt{2}\cdot\frac{3}{4}} = \frac{2\sqrt{2}}{3}</math>
 +
 +
The least possible value of <math>\frac{BC}{AG}</math> is <math>\boxed{\frac{2}{3}}</math>.

Revision as of 05:44, 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 $\boxed{\frac{2}{3}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_MPFG_Problem_10&oldid=258926"