2009 AIME II Problems/Problem 3
Revision as of 14:54, 11 April 2009 by Thedabs (talk | contribs) (New page: Image:AIMEII3.JPG From the problem, AB=100 and triangle AFB is a right triangle. As ABCD is a rectangle, triangles ADB, and ABE are also right triangles. By AA, AFB~ADB, and by the alt...)
File:AIMEII3.JPG From the problem, AB=100 and triangle AFB is a right triangle. As ABCD is a rectangle, triangles ADB, and ABE are also right triangles. By AA, AFB~ADB, and by the altitude on hypotenuse theorems, AFB~ABE, so ABE~ADB. This gives AE/AB=AB/BC. AE=AD/2 and BD=AD, so AD/(2AB)=AB/AD, or (AD)^2=2(AB)^2, so AD=AB*sqrt(2), or 100*sqrt(2), so the answer is 141.
