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:
Answer (A): Let X denote the intersection point of the diagonals AC and BD. Note that by symmetry X is the midpoint of both PQ and RS, so XP=XQ=3 and XR=XS=4. Because ∠APB=∠ARB=90∘, quadrilateral ABRP is cyclic, and the power of the point X is XR⋅XA=XP⋅XB. Therefore
XBXA=XRXP=43.
Thus there is an x>0 such that XA=3x and XB=4x. Applying the Pythagorean Theorem to △APX yields AP=AX2−XP2=3x2−1, so
[ABCD]=2[ABD]=AP⋅BD=3x2−1⋅8x=24xx2−1=15.
Solving this for x2 yields x2=21+841, so
d2=(8x)2=64x2=64(21+841)=32+841.
The requested sum is 32+8+41=81.
OR
Let X denote the intersection of AC and BD, and s=sin∠AXB. Note that ∠AXB is the external angle of right triangles △APX and △BRX, implying that △APX∼△BRX with ratio 3:4. Then there is a real number z such that AC=3z and BD=4z. The area of △AXB is
41⋅15=21⋅AX⋅BX⋅s=23⋅sz2.
Hence
s=2z25
Also, the Pythagorean Theorem applied to △BRX gives BR=4z2−16, so
s=BXBR=2z4z2−16
Equating the two expressions for s gives
2z25=2z4z2−16
which simplifies to 4z4−16z2−25=0. The solution BD2=16z2=32+841 then follows.