2009 AIME II Problem 3

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Concepts:coordinate geometrysloperectangle

Difficulty rating: 1890

3.

In rectangle ABCD,ABCD, AB=100.AB = 100. Let EE be the midpoint of AD.\overline{AD}. Given that line ACAC and line BEBE are perpendicular, find the greatest integer less than AD.AD.

Solution:

Let AD=h,AD = h, and place A=(0,0),A = (0, 0), B=(100,0),B = (100, 0), C=(100,h),C = (100, h), D=(0,h),D = (0, h), so E=(0,h2).E = \left(0, \frac{h}{2}\right). Line ACAC has slope h100\frac{h}{100} and line BEBE has slope h/2100=h200.\frac{h/2}{-100} = -\frac{h}{200}. Perpendicularity gives h100(h200)=1,\frac{h}{100} \cdot \left(-\frac{h}{200}\right) = -1, so h2=20000h^2 = 20000 and h=1002141.42.h = 100\sqrt{2} \approx 141.42.

The greatest integer less than ADAD is 141.141.

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