jallenmorris wrote:
Here is how I approached the problem.
AD (side of square) = 1
BE (side of 45:45:90 triangle) = y
CB (side of equilateral triangle) = z
We're looking for the ratio of the area of BEC to ADB, or \(\frac{BEC}{ADB}\)
If area of Tri. BEC = \(\frac{y^2}{2}\) and Tri.ADB = \(\frac{1(1-y)}{2}\) then we have
\(\frac{y^2}{2} * \frac{2}{1-y} = \frac{y^2}{1-y}\)
if y = \(\frac{1}{2}\), then y^2 = 1/4 and 1-y = 1/2, so we get 1/4 : 1/2 or 1/2....that's not an option.
Attachment:
TriangleInscribed.jpg
We can't take any value for y. There's only one way to put an equilateral triangle inside a square if we fix the square side, so if we chose that side then y is not flexible, and if you try for y=1/3 you'll not find the correct ratio for example.
I have a solution:
Let's take a=1 as the side of the square, and let's name BE=x.
Area of ADB = 1.(1-x)/2=(1-x)/2
Area of BEC=x^2/2
Therefore, ratio = x^2/(1-x)
Let's find x:
AE=sqrt(2)
Side of the Equilateral Triangle = x*sqrt(2) because BEC is isosceles.
On the other hand, if we apply Pythagoras on triangle ADB:
(1-x)^2 = (x*sqrt(2))^2 -1 = 2x^2-1 which leads to a quadratic equation with one positive solution: x=sqrt(3) - 1
Let's then replace x with its value in ratio = x^2/(1-x) --> ratio= 2