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, $y^2 = \frac{(xy)^2}{x^2} = \frac{(4*99)^2}{(12*99)} = \boxed{132}$