A pair of parallel lines divide a parallelogram as shown in the diagram into three regions with the same area. If one of the three regions is an equilateral triangle with perimeter 1, then what is the perimeter of the original parallelogram?

 Feb 19, 2023

Let's label the points in the diagram as follows:

The vertices of the parallelogram are A, B, C, and D, with AB parallel to CD and AD parallel to BC.

The points where the parallel lines intersect the sides of the parallelogram are E, F, G, and H, as shown in the diagram.

The side length of the equilateral triangle is s.

Since the three regions have the same area, the area of each region is 1/3 of the total area of the parallelogram. Let's denote the area of the parallelogram as A.

The area of the equilateral triangle is (s^2√3)/4, and since it is one of the regions, we know that:

(s^2√3)/4 = A/3

Simplifying this equation, we get:

s^2 = 4A/(3√3)

Now, let's look at the parallelogram as a whole. Since AB is parallel to CD, we know that triangle AEF is similar to triangle BGF, and since AD is parallel to BC, we know that triangle ADE is similar to triangle BCF. This means that:


Let's use these ratios to express AE, BF, AD, and BC in terms of s:

AE = s/3 (since the perimeter of the equilateral triangle is 1) EB = 2s/3 BF = 2s/3 FG = s/3 AD = (2s + 2EB)/3 = (2s + 4s/3)/3 = 10s/9 DE = (2s + 2AE)/3 = (2s + 2s/3)/3 = 8s/9 BC = (2s + 2FG)/3 = (2s + s/3)/3 = 7s/9 CF = (2s + 2BF)/3 = (2s + 4s/3)/3 = 10s/9

Now we can find the perimeter of the parallelogram:

Perimeter = 2(AB + AD) = 2(EB + BC) = 2(2s/3 + 7s/9) = 4s/3 + 14s/9 = (22s/9)

Substituting s^2 = 4A/(3√3) from the earlier equation, we get:

Perimeter = (22/9) sqrt(3)


So the perimeter of the original parallelogram is 22/9*sqrt(3).

 Feb 19, 2023

7 Online Users