Problem:
In square ABCD, points E and H lie on AB and DA, respectively, so that AE=AH. Points F and G lie on BC and CD, respectively, and points I and J lie on EH so that FI⊥EH and GJ⊥EH. See the figure below. Triangle AEH, quadrilateral BFIE, quadrilateral DHJG, and pentagon FCGJI each has area 1 . What is FI2 ?
Answer Choices:
A. 37​
B. 8−42​
C. 1+2​
D. 47​2​
E. 22​ Solution:
Since the total area is 4 , the side length of square ABCD is 2 . We see that since triangle HAE is a right isosceles triangle with area 1 , we can determine sides HA and AE both to be 2​. Now, consider extending FB and IE until they intersect. Let the point of intersection be K. We note that EBK is also a right isosceles triangle with side 2−2​ and find its area to be 3−22​. Now, we notice that FIK is also a right isosceles triangle (because ∠EKB=45∘ ) and find it's area to be 21​FI2. This is also equal to 1+3−22​ or 4−22​. Since we are looking for FI2, we want two times this. That gives (B)8−42​​.
OR
Draw the auxiliary line AC. Denote by M the point it intersects with HE, and by N the point it intersects with GF. Last, denote by x the segment FN, and by y the segment FI. We will find two equations for x and y, and then solve for y2.
Since the overall area of ABCD is 4⟹AB=2, and AC=22​. In addition, the area of △AME=21​⟹AM=1.
The two equations for x and y are then:
Length of AC:1+y+x=22​⟹x=(22​−1)−y
Area of CMIF: 21​x2+xy=21​⟹x(x+2y)=1.
Substituting the first into the second, yields [(22​−1)−y]⋅[(22​−1)+y]=1
