Problem:
Trapezoid ABCD has AB∥CD,BC=CD=43, and AD⊥BD. Let O be the intersection of the diagonals AC and BD, and let P be the midpoint of BD. Given that OP=11, the length AD can be written in the form mn, where m and n are positive integers and n is not divisible by the square of any prime. What is m+n?
Answer Choices:
A. 65
B. 132
C. 157
D. 194
E. 215
Solution:
Note that △BCD is isosceles with vertex angle C, and CP is a median. Thus CP⊥BD. Also, because AD⊥BD, it follows that CP∥AD.
Let E be the intersection of the lines CP and AB. Then AECD is a parallelogram, so AE=CD=43 and ∠DAE≅∠DCE. But ∠DCE≅∠ECB, because CP is also an angle bisector in △BCD, and ∠DAE≅∠CEB, as corresponding angles, because CE∥AD. By transitivity, ∠ECB≅∠CEB, so △BCE is isosceles with EB=CB=43. Thus AB=AE+EB=43+43=86.
Now △COD∼△AOB, so BODO=ABCD=21. Hence DO=31BD. This implies that
11=OP=DP−DO=21BD−31BD=61BD
so BD=66.
The Pythagorean Theorem in △ABD gives
AD=AB2−BD2=862−662=20⋅152=4⋅5⋅8⋅19=4190
Thus m=4,n=190, and m+n=(D)194.
OR
Note that △AOD and △COP are similar, so AD:CP=DO:11. Also, △CDB is isosceles, so ∠PBC=∠CDB=∠DBA. Therefore △DBA and △PBC are also similar. Thus AD:CP=DB:PB=2:1. It follows that DO=22,DP=33, and BD=66. Because △CDO and △ABO are similar with DO:OB=1:2 it follows that AB=86. The solution concludes as above which will result in (D)194.
