# Challenge Problem 11: What Is The Area of The Inscribed Triangle In A Rectangle?

In square ABCD, what is the area of triangle BEF?

The problem can also be generalized. In rectangle ABCD, what is the area of triangle BEF in terms of areas x, y, and z?

Let’s solve the general case of the rectangle.

Suppose AB = a and AF = c. Since the area of triangle ABF is y, we have:

ac/2 = y
c = 2y/a

Suppose BC = b. Then AD = b, and FD = b – 2y/a.

Suppose CE = d. Since the area of triangle BCE is x, we have:

bd/2 = x
d = 2x/b

Since AB = a = CD, we have ED = a – 2x/b.

The area of triangle DEF is z, so we have:

0.5(a – 2x/b)(b – 2y/a) = z
ab – 2x – 2y + 4xy/(ab) = 2z
ab – 2x – 2y – 2z + 4xy/(ab) = 0
ab – 2(x + y + z) + 4xy/(ab) = 0

Now multiply both sides by ab to get a quadratic equation in the variable ab.

(ab)^2 – (ab)2(x + y + z) + 4xy = 0

We can solve this using the quadratic formula to get:

ab = (2(x + y + z) ± √[4(x + y + z)^2 – 4(4xy)])/2
ab = (2(x + y + z) ± 2 √((x + y + z)^2 – 4xy))/2
ab = (x + y + z) ± √((x + y + z)^2 – 4xy)

Now recall that ab is the area of the entire rectangle, so it must be larger than x + y + z. Hence we reject the negative radical and retain the positive radical. Thus we have:

ab = (x + y + z) + √((x + y + z)^2 – 4xy)

Furthermore, ab is the sum of the areas of 4 triangles–the three known areas plus the unknown area w.

ab = (x + y + z) + w

If we equate the two formulas equal to ab, we can conclude:

w = √((x + y + z)^2 – 4xy)

We have a simple formula for any such problem! In the square we had known areas of 3, 4, and 5, so we can solve for w as follows:

w = √((3 + 4 + 5)2 – 4(3)(4))
w = √(96)
w = 4√(6) ≈ 9.80

But it’s amazing we have a general formula for any rectangle!

w = √((x + y + z)^2 – 4xy)

Sources

Square problem adapted from National Museum of Mathematics tweet

Generalization to rectangle from Mustafa Kemal Ince tweet

https://mindyourdecisions.com/blog/2020/01/20/inscribed-triangle-in-a-rectangle/

Yêu Tiếng Anh 