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

(→‎Solution)
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==Solution==
 
==Solution==
 +
 +
===Solution 1===
 +
 +
<asy>
 +
unitsize(0.15cm);
 +
draw((28,0)--(0,24sqrt(2))--(-12,0)--cycle);
 +
draw((0,24sqrt(2))--(8,-8sqrt(2)));
 +
draw(circumcircle((28,0),(0,24sqrt(2)),(-12,0)));
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draw((8,-8sqrt(2))--(-12,0));
 +
draw((8,-8sqrt(2))--(28,0));
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draw((8,-8sqrt(2))--(8,0));
 +
label("$A$",(0,24sqrt(2)),NNW);
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label("$B$",(-12,0),WSW);
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label("$C$",(28,0),ESE);
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label("$D$",(6,0),NW);
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label("$H$",(8,0),NNE);
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label("$E$",(8,-8sqrt(2)),S);
 +
</asy>
  
 
<math>\angle BAE \cong \angle BCE</math> because they are both subscribed by arc <math>BE</math>. <math>\angle CAE \cong \angle CBE</math> because they are both subscribed by arc <math>CE</math>. Hence <math>\angle BCE \cong \angle CBE</math>, because <math>\angle BAD \cong CAD</math>.  Then <math>\Delta BEC</math> is isosceles.  
 
<math>\angle BAE \cong \angle BCE</math> because they are both subscribed by arc <math>BE</math>. <math>\angle CAE \cong \angle CBE</math> because they are both subscribed by arc <math>CE</math>. Hence <math>\angle BCE \cong \angle CBE</math>, because <math>\angle BAD \cong CAD</math>.  Then <math>\Delta BEC</math> is isosceles.  
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Then by the Pythagorean Theorem on <math>\Delta DHE</math>, <math>4+HE^2=DE^2</math>. Also from <math>\Delta CHE</math>, <math>400+HE^2=CE^2=4DE^2</math>. Subtracting these equations yields <math>396=3DE^2</math>, and so <math>DE^2=\boxed{132}</math>.
 
Then by the Pythagorean Theorem on <math>\Delta DHE</math>, <math>4+HE^2=DE^2</math>. Also from <math>\Delta CHE</math>, <math>400+HE^2=CE^2=4DE^2</math>. Subtracting these equations yields <math>396=3DE^2</math>, and so <math>DE^2=\boxed{132}</math>.
 +
 +
===Solution 2===
 +
<asy>
 +
unitsize(0.15cm);
 +
draw((28,0)--(0,24sqrt(2))--(-12,0)--cycle);
 +
draw((0,24sqrt(2))--(8,-8sqrt(2)));
 +
draw(circumcircle((28,0),(0,24sqrt(2)),(-12,0)));
 +
label("$A$",(0,24sqrt(2)),NNW);
 +
label("$B$",(-12,0),WSW);
 +
label("$C$",(28,0),ESE);
 +
label("$D$",(6,0),NW);
 +
label("$E$",(8,-8sqrt(2)),S);
 +
</asy>
 +
 +
Let <math> BD=x </math>, so that <math> DC=40-x </math>. From the Angle Bisector Theorem, <math> \frac{x}{36}=\frac{40-x}{44} </math>. Cross-multilplying and solving for <math> x </math>, we find that <math> x=18 </math>. Thus, <math> BD=18 </math> and <math> DC=22 </math>.
 +
 +
 +
Now, from Stewart's Theorem, <math> AB^2\cdot CD+AC^2\cdot BD=AD^2\cdot BC+BC\cdot BD\cdot CD </math>. Plugging in values and letting <math> AD=y </math>, we find that <math> 36^2\cdot22+44^2\cdot18-18\cdot22\cdot40=40y^2 </math>.
 +
 +
 +
Dividing both sides by <math> 40 </math> gives <math> \frac{18^2\cdot22+22^2\cdot18}{10}-18\cdot22=y^2 </math>.Factoring a <math> 18\cdot22 </math> out of the numerator of the fraction and continuing to simplify, we find that <math> y^2=18\cdot22\cdot3 </math>.
 +
 +
 +
Now, from Power of a Point on <math> D </math>, we have <math> BD^2\cdot DC^2=AD^2\cdot DE^2 </math>. Now, let <math> DE=z </math>, so we have <math> (18\cdot22)^2=(18\cdot22\cdot3)z^2 </math>. From here, we can find that <math> z^2=\boxed{132} </math>.

Revision as of 22:09, 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

Solution 1

[asy] unitsize(0.15cm); draw((28,0)--(0,24sqrt(2))--(-12,0)--cycle); draw((0,24sqrt(2))--(8,-8sqrt(2))); draw(circumcircle((28,0),(0,24sqrt(2)),(-12,0))); draw((8,-8sqrt(2))--(-12,0)); draw((8,-8sqrt(2))--(28,0)); draw((8,-8sqrt(2))--(8,0)); label("$A$",(0,24sqrt(2)),NNW); label("$B$",(-12,0),WSW); label("$C$",(28,0),ESE); label("$D$",(6,0),NW); label("$H$",(8,0),NNE); label("$E$",(8,-8sqrt(2)),S); [/asy]

$\angle BAE \cong \angle BCE$ 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 $DE^2=\boxed{132}$.

Solution 2

[asy] unitsize(0.15cm); draw((28,0)--(0,24sqrt(2))--(-12,0)--cycle); draw((0,24sqrt(2))--(8,-8sqrt(2))); draw(circumcircle((28,0),(0,24sqrt(2)),(-12,0))); label("$A$",(0,24sqrt(2)),NNW); label("$B$",(-12,0),WSW); label("$C$",(28,0),ESE); label("$D$",(6,0),NW); label("$E$",(8,-8sqrt(2)),S); [/asy]

Let $BD=x$, so that $DC=40-x$. From the Angle Bisector Theorem, $\frac{x}{36}=\frac{40-x}{44}$. Cross-multilplying and solving for $x$, we find that $x=18$. Thus, $BD=18$ and $DC=22$.


Now, from Stewart's Theorem, $AB^2\cdot CD+AC^2\cdot BD=AD^2\cdot BC+BC\cdot BD\cdot CD$. Plugging in values and letting $AD=y$, we find that $36^2\cdot22+44^2\cdot18-18\cdot22\cdot40=40y^2$.


Dividing both sides by $40$ gives $\frac{18^2\cdot22+22^2\cdot18}{10}-18\cdot22=y^2$.Factoring a $18\cdot22$ out of the numerator of the fraction and continuing to simplify, we find that $y^2=18\cdot22\cdot3$.


Now, from Power of a Point on $D$, we have $BD^2\cdot DC^2=AD^2\cdot DE^2$. Now, let $DE=z$, so we have $(18\cdot22)^2=(18\cdot22\cdot3)z^2$. From here, we can find that $z^2=\boxed{132}$.

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