Skip to main content

2021 Fall AMC 10B Problem 18

Problem 18 of 25IntermediateGeometry

Three identical square sheets of paper each with side length 66 are stacked on top of each other. The middle sheet is rotated clockwise 30∘30^\circ about its center and the top sheet is rotated clockwise 60∘60^\circ about its center, resulting in the 2424-sided polygon shown in the figure below. The area of this polygon can be expressed in the form a−bc,a-b\sqrt{c}, where a,a, b,b, and cc are positive integers, and cc is not divisible by the square of any prime. What is a+b+c?a+b+c?

Answer choices

Show solution

Solution

The boundary can be split into 2424 congruent triangles. Each has angles 15∘,15^\circ, 45∘,45^\circ, and 120∘.120^\circ. For one such triangle, draw the altitude from the center-side direction. The altitude is 3,3, half the side length of a square. The adjacent right triangle has a 30∘30^\circ angle, so the part cut off from a length 33 base is 3tan⁡30∘=3.3\tan30^\circ=\sqrt3. Thus each small triangle has base 3−33-\sqrt3 and height 3,3, giving area 3(3−3)2=9−332.\frac{3(3-\sqrt3)}{2}=\frac{9-3\sqrt3}{2}. The total area is 24⋅9−332=108−363.24\cdot\frac{9-3\sqrt3}{2}=108-36\sqrt3. Hence a+b+c=108+36+3=147.a+b+c=108+36+3=147. Thus, the answer is E .

More practice

Concepts: area decomposition · transformation · trigonometry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.