Difference between revisions of "2000 AMC 12 Problems/Problem 21"
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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 | ||
− | <math>\ | + | <math>\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}</math> |
− | |||
− | |||
− | |||
− | |||
+ | == Solutions == | ||
+ | === Solution 1 === | ||
<center><asy> | <center><asy> | ||
unitsize(36); | unitsize(36); | ||
Line 20: | Line 18: | ||
</asy></center> | </asy></center> | ||
− | WLOG, let a side of the square be <math>1</math>. Simple | + | WLOG, 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 base 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 height 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{\boxed{D}}</math>. |
+ | |||
+ | === Solution 2 (Video Solution) === | ||
− | + | https://youtu.be/HTHveknJFpk | |
+ | https://m.youtube.com/watch?v=TUQsHeJ6RSA&feature=youtu.be | ||
+ | |||
+ | === Solution 3 === | ||
<center><asy> | <center><asy> | ||
unitsize(36); | unitsize(36); | ||
Line 48: | Line 51: | ||
~ Nafer | ~ Nafer | ||
− | == Solution | + | === Solution 4 (process of elimination) === |
− | |||
Simply testing specific triangles is sufficient. | 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{\boxed{D}}</math> | + | A triangle with legs of 1 and 2 gives a square of area <math>S=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}</math>. The larger sub-triangle has area <math>T_1=\frac{\frac{2}{3}\times\frac{4}{3}}{2}=\frac{4}{9}</math>, and the smaller triangle has area <math>T_2=\frac{\frac{2}{3}\times\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{\boxed{D}}</math> |
~ Snacc | ~ Snacc | ||
+ | === Solution 5 === | ||
+ | <center><asy> | ||
+ | unitsize(36); | ||
+ | draw((0,0)--(6,0)--(0,3)--cycle); | ||
+ | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | ||
+ | draw((0,0)--(6,0)--(6,3)--(0,3)--cycle); | ||
+ | draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); | ||
+ | draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); | ||
+ | label("$1$",(1,2),S); | ||
+ | label("$1$",(2,1),W); | ||
+ | label("$2m$",(4,0),S); | ||
+ | label("$x$",(0,2.5),W); | ||
+ | label("$A$", (1.25,1),W); | ||
+ | label("$B$", (4, 2.25),N); | ||
+ | </asy></center> | ||
+ | WLOG, let the length of the square be <math>1</math> (Like Solution 1). Then the length of the larger triangle is <math>2m</math>. Let the length of the smaller triangle be <math>x</math>. | ||
+ | Therefore, since <math>A = B</math> (try to prove that yourself), <math>1 = 2mx</math> or <math>x = 1/2m</math> | ||
+ | The area of the other triangle is <math>1/4m</math>. | ||
+ | From here, the answer is <math>\text{\boxed{D}}</math>. | ||
+ | |||
+ | == Solution 6 == | ||
+ | Because we know that the right triangles are always similar, in a figure | ||
+ | |||
+ | <center><asy> | ||
+ | unitsize(36); | ||
+ | draw((0,0)--(6,0)--(0,3)--cycle); | ||
+ | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | ||
+ | label("$y$",(1,2),S); | ||
+ | label("$y$",(2,1),W); | ||
+ | label("$a$",(4,0),S); | ||
+ | label("$x$",(0,2.5),W); | ||
+ | </asy></center> | ||
+ | , x/y = y/a. Using cross products, we can say that xa= y^2. We are trying to solve for x because m is in terms of a and y; m= a/2y (the ratio of the area of one triangle to the area of the square). So solving for x, x= y^2/a. Then the question is asking us for the answer to x/2y (the ratio of the area of the other triangle to the area of the square). So x/2y is y^2/a2y which is simplified to y/2a- also 1/4m, the answer. | ||
+ | |||
+ | -Smartgrowth | ||
== See also == | == See also == |
Latest revision as of 01:22, 13 May 2024
- 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
Solutions
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 base of the triangle with area
is
. Therefore
where
is the height of the other triangle.
, and the area of that triangle is
.
Solution 2 (Video Solution)
https://m.youtube.com/watch?v=TUQsHeJ6RSA&feature=youtu.be
Solution 3
![[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
Thus
Now since triangle is similar to the large triangle, it has
,
and
Thus .
.
~ Nafer
Solution 4 (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
Solution 5
![[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(6,0)--(6,3)--(0,3)--cycle); draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); label("$A$", (1.25,1),W); label("$B$", (4, 2.25),N); [/asy]](http://latex.artofproblemsolving.com/0/e/8/0e8102e9cd9a9192faacbc07ff9ae3ec30aa634b.png)
WLOG, let the length of the square be (Like Solution 1). Then the length of the larger triangle is
. Let the length of the smaller triangle be
.
Therefore, since
(try to prove that yourself),
or
The area of the other triangle is
.
From here, the answer is
.
Solution 6
Because we know that the right triangles are always similar, in a figure
![[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$y$",(1,2),S); label("$y$",(2,1),W); label("$a$",(4,0),S); label("$x$",(0,2.5),W); [/asy]](http://latex.artofproblemsolving.com/7/d/c/7dc2aedc8013ddc687e997f20ab6da071ca74627.png)
, x/y = y/a. Using cross products, we can say that xa= y^2. We are trying to solve for x because m is in terms of a and y; m= a/2y (the ratio of the area of one triangle to the area of the square). So solving for x, x= y^2/a. Then the question is asking us for the answer to x/2y (the ratio of the area of the other triangle to the area of the square). So x/2y is y^2/a2y which is simplified to y/2a- also 1/4m, the answer.
-Smartgrowth
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.