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Question

FIGURE Triangle ABC has a right angle at C, AC=3 and BC=4. Triangle ABD has a right angle at A and AD=12. Points C and D are on opposite sides of AB. The line through D parallel to AC meets CB extended at E. If DEDB=mn, where m and n are relatively prime positive integers, then m+n=

Accepted Answer

Let F be the point such that DF and CF are parallel to CE and DE, respectively, and let DE = x and BE^2 = 169-x^2. Then, [FDEC] = x(4+169-x^2) = [ABC] + [BED] + [ABD] + [AFD] = 6 + x169-x^22 + 30 + (x-3)(4+169-x^2)2. So, 4x+x169-x^2 = 60 + x169-x^2 - 3169-x^2. Simplifying 3169-x^2 = 60 - 4x, and 1521 - 9x^2 = 16x^2 - 480x + 3600. Therefore 25x^2 - 480x + 2079 = 0, and x = 48155. Checking, x = 635 is the answer, so DEDB = 63513 = 6365. The answer is 128.