Problem:
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.
Suppose PQ=6 and RS=8, and let d denote the
length of BD, the longer diagonal of ABCD. Then d2 can be written in the form m+np, where m,n, and p are positive integers and p is not divisible by the square of any prime. What is m+n+p ?
Answer Choices:
A. 81
B. 89
C. 97
D. 105
E. 113 Solution:
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 XBXA=XRXP=43.
Thus let x>0 be such that XA=3x and XB=4x. Then Pythagorean Theorem on △APX yields AP=AX2−XP2=3x2−1, and so
[ABCD]=2[ABD]=AP⋅BD=3x2−1⋅8x=24xx2−1=15
Solving this for x2 yields x2=21+841, and so
(8x)2=64x2=64(21+841)=32+841
The requested answer is 32+8+41=(A)81 .
OR
Let X denote the intersection point of the diagonals AC and BD, and let θ=∠COB. Then, by the given conditions, XR=4,XQ=3,[XCB]=415. So,
XC=cosθ3 XBcosθ=4
21XBXCsinθ=415
Combining the above 3 equations, we get
cos2θsinθ=85
Since we want to find d2=4XB2=cos2θ64, we let x=cos2θ1. Then
cos4θsin2θ=cos4θ1−cos2θ=x2−x=6425
Solving this, we get x=84+41, so d^{2}=64 x=32+8 \sqrt{41} \rightarrow 32+8+41= \boxed
