Mock Geometry AIME 2011 Problems/Problem 5

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 $z^2=\boxed{132}$.

Solution 3

This is the author's solution. Refer to the above diagrams.

Notice that \[\angle AEC=\angle ABD\] \[\angle EAC =\angle BAD\] So by AA similarity, $\triangle AEC \sim \triangle ABD.$ It follows that $\frac{AE}{AC} = \frac{AB} {AD},$ or $AE*AD= AB*AC.$ Now let $AD=x, DE=y.$ By the previous result, we have \[x(x+y)=x^2+xy=36*44=16*99\] But by the angle bisector theorem, $BD=\frac{36*40}{36+44} = 18, CD = \frac{44*40}{36+44}=22.$ Then, by power of a point on $D,$ we have \[xy=18*22=4*99\] Subtracting the last two equations, we find \[x^2=12*99\] Thus, \[y^2 = \frac{(xy)^2}{x^2} = \frac{(4*99)^2}{(12*99)} = \boxed{132}\]

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