Equiangular hexagon 𝐴𝐵𝐶𝐷𝐸𝐹 has side lengths 𝐴𝐵 = 𝐶𝐷 = 𝐸𝐹 =
1 and
. 𝐵𝐶 = 𝐷𝐸 = 𝐹𝐴 = 𝑟. The area of 𝐴𝐶𝐸 is 70% of the
area of the hexagon. What is the sum of all possible values of 𝑟?
First, draw a picture.
𝑋
1
𝐴
𝐵
𝑟
𝑟
𝐶
𝐹
1
𝑍
1
𝑟
𝐸
𝐷
𝑌
Extending the sides of the equiangular hexagon (Each angle in the hexagon is 120°) form
a equilateral triangle of side length 𝑟 + 2. We can express the area of the hexagon as the area of the
big triangle minus the three equilateral triangles with side length 1. This is:
𝑟+2
4
2
3
−3
12 3
4
(𝑟 2 + 4𝑟 + 1) 3
=
4
Now we want to find the area of ∆𝐴𝐶𝐸. Draw a perpendicular from 𝐶 to 𝐷𝑌. Let’s call the
foot of the perpendicular 𝑃. We know 𝐶𝑃 =
with side length 1.
3
,
2
1
and 𝐷𝑃 = 2, because ∆𝐶𝑃𝑌 is an equilateral triangle
𝐶
𝑟
𝐸
𝐷
3
,
2
∆𝐸𝑃𝐶 is a right triangle with legs 𝐸𝑃 =, 𝐶𝑃 =
𝑃
𝑌
and hypotenuse 𝐸𝐶, which we want to find.
Using the Pythagorean theorem,
1
𝐸𝐶 = 𝑟 +
2
2
2
+
3
2
2
= 𝑟2 + 𝑟 + 1
By SAS, Triangles 𝐸𝐷𝐶, 𝐴𝐵𝐶, and 𝐴𝐹𝐸 are congruent, so ∆𝐴𝐶𝐸 is equilateral.
The area of ∆𝐴𝐶𝐸 is
𝐸𝐶 2 3
=
4
We know that ∆𝐴𝐶𝐸, is 70% the area of the hexagon, so we can plug these values in.
(𝑟 2 + 𝑟 + 1) 3
7 (𝑟 2 + 4𝑟 + 1) 3
=
4
10
4
Divide both sides by
3
.
4
𝑟2 + 𝑟 + 1 =
7 2
𝑟 + 4𝑟 + 1
10
Multiply both sides by 10.
10𝑟 2 + 10𝑟 + 10 = 7 𝑟 2 + 4𝑟 + 1
3𝑟 2 − 18𝑟 + 3 = 0
𝑟 2 − 6𝑟 + 1 = 0
So the sum of all possible values of 𝑟 is
−(−6)
1
= 𝑬(𝟔)