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Question

Let ABCD be a parallelogram with area 15. Points P and Q are the projections of A and C, respectively, onto the line BD; and points R and S are the projections of B and D, respectively, onto the line AC. See the figure, which also shows the relative locations of these points. FIGURE Suppose PQ=6 and RS=8, and let d denote the length of BD, the longer diagonal of ABCD. Then d^2 can be written in the form m+n p, where m,n, and p are positive integers and p is not divisible by the square of any prime. What is m+n+p?

Accepted Answer

Let X denote the intersection point of the diagonals AC and BD. Remark that by symmetry X is the midpoint of both PQ and RS, so XP = XQ = 3 and XR = XS = 4. Now note that since APB = ARB = 90^, quadrilateral ARPB is cyclic, and so XR XA = XP XB,which implies XAXB = XPXR = 34. Thus let x> 0 be such that XA = 3x and XB = 4x. Then Pythagorean Theorem on APX yields AP = AX^2 - XP^2 = 3x^2-1, and so[ABCD] = 2[ABD] = AP BD = 3x^2-1 8x = 24xx^2-1=15Solving this for x^2 yields x^2 = 12 + 418, and so(8x)^2 = 64x^2 = 64(12 + 418) = 32 + 841.The requested answer is 32 + 8 + 41 = 81.