Problem:
In rectangle ABCD,AB=100. Let E be the midpoint of AD. Given that line AC and line BE are perpendicular, find the greatest integer less than AD.
Solution:
Because ∠EBA and ∠ACB are both complementary to ∠EBC, the angles EBA and ACB are equal, and right triangles BAE and CBA are similar. Thus
ABAE=BCAB=2AEAB.
Hence 2⋅AE2=AB2 and AD=2⋅AE=2⋅2100=1002. Because 141<1002<142, the requested answer is 141.