Track / Stage 3 / 47 of 260 #47 of 1964

Problem 47

AMC 10/12, early questions

Geometry Difficulty 3.1 Multiple choice

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$ ?

Figure (Asymptote source)

unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6;  path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0));

Pick one

Official solution

Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.

Each of the angles at $X$ is $\frac{180^\circ}9 = 20^\circ$ . From $\triangle XYZ$ , the degree measure of the smaller interior angle of the trapezoid is $\frac{180^\circ - 20^\circ}2 = 80^\circ$ , hence the degree measure of the larger interior angle is $180^\circ - 80^\circ = \boxed{100^\circ \Longrightarrow A}$ .
Proof that all the extended trapezoid legs intersect at the same point: It is sufficient to prove this for any pair of neighboring trapezoids. For two neighboring trapezoids, the situation is symmetric according to their common leg, therefore the extensions of both outside legs intersect the extension of the common leg at the same point, Q.E.D.
Knowing this, we can now easily see that the intersection point must be on the bottom side of our picture, as it lies on the bottom leg of the rightmost trapezoid. And by symmetry the point must be in the center of this side.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.