BTGmoderatorDC wrote: ↑

Wed Nov 04, 2020 5:20 pm

A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%


OA D

Solution:

If the side length of the regular octagon is s, then the area of the octagon is given by 2(1 + √2)(s^2). Partitioning the regular octagon into a bunch of isosceles right triangles, rectangles and a square is one way of obtaining this formula.

Since the regular octagon has 8 sides, 8 equilateral triangles with a side length of s will be drawn and folded over the octagon. Since the area of each equilateral triangle is [(s^2)√3]/4 and since there are 8 such equilateral triangles, the total area of the equilateral triangles is 8 * [(s^2)√3]/4 = 2√3(s^2). Hence, the uncovered area equals 2(1 + √2)(s^2) - 2√3(s^2) = 2(s^2)(1 + √2 - √3). It follows that the ratio we need to approximate is [2(s^2)(1 + √2 - √3)]/[2(1 + √2)(s^2)] = (1 + √2 - √3)/(1 + √2). Since √2 is approximately 1.4 and √3 is approximately 1.7, we get:

(1 + √2 - √3)/(1 + √2)

(1 + 1.4 - 1.7)/(1 + 1.4)

0.7/2.4 ≈ 0.29

Expressed as a percentage, this is 29%, and the closest answer choice is D.

Answer: D