Difference between revisions of "2000 AMC 12 Problems/Problem 21"
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~ Nafer | ~ Nafer | ||
| + | === Solution 3 (process of elimination) === | ||
| − | + | Simply testing specific triangles is sufficient. | |
| − | - | + | A triangle with legs of 1 and 2 gives a square of area <math>S=\frac{2}{3}\frac{2}{3}=\frac{4}{9}</math>. The larger sub-triangle has area <math>T_1=\frac{\frac{2}{3}\frac{4}{3}}{2}=\frac{4}{9}</math>, and the smaller triangle has area <math>T_2=\frac{\frac{2}{3}\frac{1}{3}}{2}=\frac{1}{9}</math>. Computing ratios you get <math>\frac{T_1}{S}=1</math> and <math>\frac{T_2}{S}=\frac{1}{4}</math>. Plugging <math>m=1</math> in shows that the only possible answer is <math>\text{(D)}</math> |
| − | + | ~ Snacc | |
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== See also == | == See also == | ||
Revision as of 18:26, 19 October 2020
- The following problem is from both the 2000 AMC 12 #21 and 2000 AMC 10 #19, so both problems redirect to this page.
Contents
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
Solution 1
![[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)
WLOG, 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 
.
Solution 2
![[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label("$a$",(0,1.5),W); label("$c$",(2.5,1),W); label("$A$",(0.5,2.5),W); label("$B$",(3.5,0.75),W); label("$C$",(1,1),W); [/asy]](http://latex.artofproblemsolving.com/1/f/d/1fd5c48ee1cfa3bcf3bd117fe4798e0440cbb6c5.png)
From the diagram from the previous solution, we have 
, 
as the legs and 
as the side length of the square. WLOG, let the area of triangle 
be times the area of square 
.
Since triangle is similar to the large triangle, it has 
, 
and ![\[[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2\]](http://latex.artofproblemsolving.com/5/c/a/5caa9226761852a91d79f81f0de0ea044cd7f2ee.png)
Thus 
Now since triangle 
is similar to the large triangle, it has 
, 
and ![\[[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]\]](http://latex.artofproblemsolving.com/c/6/b/c6b5a58f0fd075d73ce1504285a534f546542588.png)
Thus 
. 
.
~ Nafer
Solution 3 (process of elimination)
Simply testing specific triangles is sufficient.
A triangle with legs of 1 and 2 gives a square of area 
. The larger sub-triangle has area 
, and the smaller triangle has area 
. Computing ratios you get 
and 
. Plugging 
in shows that the only possible answer is
~ Snacc
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
