2009 AMC 10B Problems/Problem 24

Problem

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 $x$ ?

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

$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$

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.

$[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted ); label("$X$",(0,0),S); label("$Y$",(R,0),SE); label("$Z$",rotate(20)*(R,0),ENE); draw( arc( (0,0), (1.5,0), rotate(20)*(1.5,0) ) ); label("$20^\circ$", rotate(10)*(1.75,0), E ); [/asy]$

Each of the angles at $X$ has $\frac{180^\circ}9 = 20^\circ$ . From $\triangle XYZ$ , the size of the smaller internal angle of the trapezoid is $\frac{180^\circ - 20^\circ}2 = 80^\circ$ , hence the size of the larger one is $180^\circ - 80^\circ = \boxed{100^\circ}$ .

Proof that all the extended trapezoid legs intersect in 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 in 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.

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2009 AMC 10B Problems/Problem 24

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