Difference between revisions of "2000 AMC 12 Problems/Problem 21"
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| + | {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #21]] and [[2000 AMC 10 Problems|2000 AMC 10 #19]]}} | ||
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== Problem == | == Problem == | ||
Through a point on the [[hypotenuse]] of a [[right triangle]], lines are drawn [[parallel]] to the legs of the triangle so that the triangle is divided into a [[square]] and two smaller right triangles. The area of one of the two small right triangles is <math>m</math> times the area of the square. The [[ratio]] of the area of the other small right triangle to the area of the square is | Through a point on the [[hypotenuse]] of a [[right triangle]], lines are drawn [[parallel]] to the legs of the triangle so that the triangle is divided into a [[square]] and two smaller right triangles. The area of one of the two small right triangles is <math>m</math> times the area of the square. The [[ratio]] of the area of the other small right triangle to the area of the square is | ||
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== Solution == | == Solution == | ||
| − | + | ||
| + | <center><asy> | ||
| + | unitsize(36); | ||
| + | draw((0,0)--(6,0)--(0,3)--cycle); | ||
| + | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | ||
| + | label("$1$",(1,2),S); | ||
| + | label("$1$",(2,1),W); | ||
| + | label("$2m$",(4,0),S); | ||
| + | label("$x$",(0,2.5),W); | ||
| + | </asy></center> | ||
[[Without loss of generality]] let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A = \frac{1}{2}bh = \frac{h}{2}</math>, the height of the triangle with area <math>m</math> is <math>2m</math>. Therefore <math>\frac{2m}{1} = \frac{1}{x}</math> where <math>x</math> is the base of the other triangle. <math>x = \frac{1}{2m}</math>, and the area of that triangle is <math>\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}</math>. | [[Without loss of generality]] let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A = \frac{1}{2}bh = \frac{h}{2}</math>, the height of the triangle with area <math>m</math> is <math>2m</math>. Therefore <math>\frac{2m}{1} = \frac{1}{x}</math> where <math>x</math> is the base of the other triangle. <math>x = \frac{1}{2m}</math>, and the area of that triangle is <math>\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}</math>. | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2000|num-b=20|num-a=22}} | {{AMC12 box|year=2000|num-b=20|num-a=22}} | ||
| + | {{AMC10 box|year=2000|num-b=18|num-a=20}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 23:57, 26 November 2011
- The following problem is from both the 2000 AMC 12 #21 and 2000 AMC 10 #19, so both problems redirect to this page.
Problem
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is 
times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
Solution
![[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); [/asy]](http://latex.artofproblemsolving.com/a/f/e/afe29f5d468b8c91dd67e38cbbc66bd25501eaf9.png)
Without loss of generality let a side of the square be 
. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since 
, the height of the triangle with area is 
. Therefore 
where 
is the base of the other triangle. 
, and the area of that triangle is 
.
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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 12 Problems and Solutions | |
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
