Difference between revisions of "2009 AMC 10B Problems/Problem 24"
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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. | 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. | ||
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| + | '''Alternate Solution''' | ||
| + | The trapezoids and the line form a decagon. The angle sum of the decagon is 144<math>. The two angles formed by the line are 180-x. There are 8 angles left. Each of those angles sums to 360. The angle needed is 360-2x. Putting it all together; </math>8(360-2x) +2(180-x) =1440$. This can be easily solved, especially if a 2 is factored out from (360-2x). The equation does indeed give x=100. '''A''' | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2009|ab=B|num-b=23|num-a=25}} | ||
Revision as of 18:58, 11 January 2012
Problem
The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with 
trapezoids, let 
be the angle measure in degrees of the larger interior angle of the trapezoid. What is ?
![[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]](http://latex.artofproblemsolving.com/d/e/c/dec895c3ce8d2cb187b83306a03e47ef6abc8f75.png)
Solution
Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.
![[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted ); label("$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]](http://latex.artofproblemsolving.com/f/d/8/fd8cd92c472bdeba95f38b0309708383757b3551.png)
Each of the angles at 
has 
. From 
, the size of the smaller internal angle of the trapezoid is 
, hence the size of the larger one is 
.
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.
Alternate Solution
The trapezoids and the line form a decagon. The angle sum of the decagon is 144
8(360-2x) +2(180-x) =1440$. This can be easily solved, especially if a 2 is factored out from (360-2x). The equation does indeed give x=100. A
See Also
| 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 | ||
| All AMC 10 Problems and Solutions | ||
