guerrero wrote:An equilateral triangle ABC is inscribed in square ADEF, forming three right triangles: ADB, ACF and BEC. What is the ratio of the area of triangle BEC to that of triangle ADB?
A. 4/3
B. sqrt(3)
C. 2
D. 5/2
E. sqrt(5)
Dear
guerrero,
I'm happy to help.
This is a really tricky problem. I solved it a way that I consider somewhat "brute force" --- I kept thinking, "There must be a more elegant solution", but it didn't occur to me. See the attached diagram.
I said --- Let the square have a side of 1. Let CE = BC = x. Then DB = (1 - x).
We know BC = AB, because the triangle is equilateral. This means
x^ 2 + x^2 = (BC)^2 = (AB)^2 = 1^2 + (1-x)^2
2(x^2) = 1 + (1 - 2x + x^2)
x^2 = 2 - 2x
x^2 + 2x = 2
I'm going to solve the quadratic by completing-the-square:
x^2 + 2x + 1 = 3
(x + 1)^2 = 3
x + 1 = sqrt(3) --- note: we are only interested in lengths = positive solution
x = sqrt(3) - 1
So CE = BE = sqrt(3) - 1, and BD = (1 - x) = 2 - sqrt(3). We want the ratio of areas, (0.5*bh) --- I am going to ignore the 0.5 part, since that would cancel when we take the ratio.
For triangle ADB, bh = (AD)(DB) = 1*(2 - sqrt(3)) = 2 - sqrt(3)
For triangle BCE, bh = (BE)(CE) = (sqrt(3) - 1)^2 = 3 - 2*sqrt(3) + 1
= 4 - 2*sqrt(3) = 2*[2 - sqrt(3)]
So, the area of BCE is exactly twice the area of ADB. Answer = [spoiler]
(C)[/spoiler].
This was a solution that involved tremendous amount of algebra & arithmetic. I feel as though there should be a purely geometric solution which would be the very epitome of elegance, but I was not able to figure out what that might be.
I hope this helps.
Mike
