Problem:
On square ABCD, points E,F,G, and H lie on sides AB,BC,CD, and DA, respectively, so that EG⊥FH and EG=FH=34. Segments EG and FH intersect at a point P, and the areas of quadrilaterals AEPH,BFPE,CGPF, and DHPG are in the ratio 269:275:405:411. Find the area of square ABCD.
Solution:
Let s be the side length of ABCD, let Q and R be the midpoints of EG and FH, respectively, let S be the foot of the perpendicular from Q to CD, and let T be the foot of the perpendicular from R to AD. The fraction of the area of the square ABCD which is occupied by trapezoid BCGE is
269+275+405+411275+405=21,
so Q is the center of ABCD. Thus R,Q, and S are collinear, and RT=QS=21s. Similarly, the fraction of the area occupied by trapezoid CDHF is 53, so RS=53s and RQ=101s.
Because △QSG≅△RTH, the area of DHPG is the sum of the areas of rectangle DTRS and △RPQ. Rectangle DTRS has area RS⋅RT=53s⋅21s=103s2. If θ=∠QRP, then △RPQ has area 21⋅101ssinθ⋅101scosθ=4001s2sin2θ. Therefore the area of DHPG is s2(103+4001sin2θ). Because the area of trapezoid CDHF is 53s2, the area of CGPF is s2(103−4001sin2θ). Because these areas are in the ratio 411:405=(408+3):(408−3), it follows that
1034001sin2θ=4083
from which sin2θ=1715. Note that θ=∠RHT>∠QAT=45∘, so cos2θ=−1−sin22θ=−178 and sin2θ=21(1−cos2θ)=3425. The area of ABCD is s2=EG2sin2θ=342⋅3425=850.
