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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 5"

(Created page with "==Problem== In triangle <math>ABC,</math> <math>AB=36,BC=40,CA=44.</math> The bisector of angle <math>A</math> meet <math>BC</math> at <math>D</math> and the circumcircle at <ma...")
 
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Let <math>H</math> be the foot of the perpendicular from <math>E</math> to <math>BC</math>. As <math>\Delta BEC</math> is isosceles, it follows that <math>H</math> is the midpoint of <math>BC</math> , and so <math>HC=20</math>. From the angle bisector theorem, <math>\frac{36}{BD}=\frac{44}{CD}</math>. We have <math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>.  Thus,  <math>DH=CD-CH=22-20=2</math>.
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Let <math>H</math> be the foot of the perpendicular from <math>E</math> to <math>BC</math>. As <math>\Delta BEC</math> is isosceles, it follows that <math>H</math> is the midpoint of <math>BC</math>, and so <math>HC=20</math>. From the angle bisector theorem, <math>\frac{36}{BD}=\frac{44}{CD}</math>. We have <math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>.  Thus,  <math>DH=CD-CH=22-20=2</math>.
  
  

Revision as of 20:55, 1 January 2012

Problem

In triangle $ABC,$ $AB=36,BC=40,CA=44.$ The bisector of angle $A$ meet $BC$ at $D$ and the circumcircle at $E$ different from $A$. Calculate the value of $DE^2$

Solution

$\angle BAE \cong \angle BCE$ because they are both subscribed by arc $BE$. $\angle CAE \cong \angle CBE$ because they are both subscribed by arc $CE$. Hence $\angle BCE \cong \angle CBE$, because $\angle BAD \cong CAD$. Then $\Delta BEC$ is isosceles.


Let $H$ be the foot of the perpendicular from $E$ to $BC$. As $\Delta BEC$ is isosceles, it follows that $H$ is the midpoint of $BC$, and so $HC=20$. From the angle bisector theorem, $\frac{36}{BD}=\frac{44}{CD}$. We have $BD+CD=BC=40$. Solving this system of equations yields $BD=18,CD=22$. Thus, $DH=CD-CH=22-20=2$.


$\angle ADB \cong \angle CDE$ because they are vertical angles. It was shown $\angle BAE \cong \angle BCE$, and so $\Delta ADB \sim \Delta CDE$ by $AA$ similarity. Then $\frac{CE}{DE}=\frac{AB}{BD}=\frac{36}{18}$ and so $CE=2DE$.


Then by the Pythagorean Theorem on $\Delta DHE$, $4+HE^2=DE^2$. Also from $\Delta CHE$, $400+HE^2=CE^2=4DE^2$. Subtracting these equations yields $396=3DE^2$, and so $DE^2=\boxed{132}$.

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