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

$\angle BAE \cong \angle BCE$ 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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Mock Geometry AIME 2011 Problems/Problem 5

Problem

Solution