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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>\text {(A)}\ \frac{1}{2m+1} \qquad \text {(B)}\ m \qquad \text {(C)}\ 1-m \qquad \text {(D)}\ \frac{1}{4m} \qquad \text {(E)}\ \frac{1}{8m^2}</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>
−
 
 
−
== Solution ==
 
  
 +
== Solutions ==
 
=== Solution 1 ===
 
=== 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 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>.
+
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) ===
  
−
=== Solution 2 ===
+
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 36: Line 39:
 
</asy></center>
 
</asy></center>
  
−
From the diagram above, we have <math>a</math>, <math>b</math> as the legs and <math>c</math> as the side length of the square. WLOG, let the area of triangle A
+
From the diagram from the previous solution, we have <math>a</math>, <math>b</math> as the legs and <math>c</math> as the side length of the square. WLOG, let the area of triangle <math>A</math>
−
be <math>m</math> times the area of square C.
+
be <math>m</math> times the area of square <math>C</math>.
  
−
Since triangle A is similar to the large triangle, it has <math>h = a(\frac{c}{b}) = \frac{ac}{b}</math>, <math>b = c</math> and <cmath>[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2</cmath>
+
Since triangle <math>A</math> is similar to the large triangle, it has <math>h_A = a(\frac{c}{b}) = \frac{ac}{b}</math>, <math>b_A = c</math> and <cmath>[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2</cmath>
 
Thus <math>\frac{a}{2b} = m</math>
 
Thus <math>\frac{a}{2b} = m</math>
  
−
Now since triangle B is similar to the large triangle, it has <math>h = c</math>, <math>b = b\frac{c}{a} = \frac{bc}{a}</math>
+
Now since triangle <math>B</math> is similar to the large triangle, it has <math>h_B = c</math>, <math>b_B = b\frac{c}{a} = \frac{bc}{a}</math> and <cmath>[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]</cmath>
−
=== Solution 2 ===
 
  
 +
Thus <math>n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}</math>. <math>\text{\boxed{D}}</math>.
 +
 +
~ 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 <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
 +
=== Solution 5 ===
 
<center><asy>
 
<center><asy>
 
unitsize(36);
 
unitsize(36);
 
draw((0,0)--(6,0)--(0,3)--cycle);
 
draw((0,0)--(6,0)--(0,3)--cycle);
 
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
 
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
−
label("$b$",(2.5,0),S);
+
draw((0,0)--(6,0)--(6,3)--(0,3)--cycle);
−
label("$a$",(0,1.5),W);
+
draw((0,0)--(6,0)--(6,2)--(0,2)--cycle);
−
label("$c$",(2.5,1),W);
+
draw((0,0)--(2,0)--(2,3)--(0,3)--cycle);
−
label("$A$",(0.5,2.5),W);
+
label("$1$",(1,2),S);
−
label("$B$",(3.5,0.75),W);
+
label("$1$",(2,1),W);
−
label("$C$",(1,1),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>
 
</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>.
  
−
From the diagram above, we have <math>a</math>, <math>b</math> as the legs and <math>c</math> as the side length of the square. WLOG, let the area of triangle A
+
== Solution 6 ==
−
be <math>m</math> times the area of square C.
+
Because we know that the right triangles are always similar, in a figure
  
−
Since triangle A is similar to the large triangle, it has <math>height = a(\frac{c}{b}) = \frac{ac}{b}</math>, <math>base = c</math> and <cmath>[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2</cmath>
+
<center><asy>
−
Thus <math>\frac{a}{2b} = m</math>
+
unitsize(36);
−
 
+
draw((0,0)--(6,0)--(0,3)--cycle);
−
Now since triangle B is similar to the large triangle, it has <math>height = c</math>, <math>base = b\frac{c}{a} = \frac{bc}{a}</math> and <cmath>[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]</cmath>
+
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.
  
−
Thus <math>n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}</math>. <math>(C)</math>
+
-Smartgrowth
  
 
== See also ==
 
== See also ==

Latest revision as of 00:22, 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.

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 $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

$\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}$

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]

WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the height of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.

Solution 2 (Video Solution)

https://youtu.be/HTHveknJFpk

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]

From the diagram from the previous solution, we have $a$, $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$ be $m$ times the area of square $C$.

Since triangle $A$ is similar to the large triangle, it has $h_A = a(\frac{c}{b}) = \frac{ac}{b}$, $b_A = c$ and \[[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2\] Thus $\frac{a}{2b} = m$

Now since triangle $B$ is similar to the large triangle, it has $h_B = c$, $b_B = b\frac{c}{a} = \frac{bc}{a}$ and \[[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]\]

Thus $n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}$. $\text{\boxed{D}}$.

~ 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 $S=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$. The larger sub-triangle has area $T_1=\frac{\frac{2}{3}\times\frac{4}{3}}{2}=\frac{4}{9}$, and the smaller triangle has area $T_2=\frac{\frac{2}{3}\times\frac{1}{3}}{2}=\frac{1}{9}$. Computing ratios you get $\frac{T_1}{S}=1$ and $\frac{T_2}{S}=\frac{1}{4}$. Plugging $m=1$ in shows that the only possible answer is $\text{\boxed{D}}$

~ 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]

WLOG, let the length of the square be $1$ (Like Solution 1). Then the length of the larger triangle is $2m$. Let the length of the smaller triangle be $x$. Therefore, since $A = B$ (try to prove that yourself), $1 = 2mx$ or $x = 1/2m$ The area of the other triangle is $1/4m$. From here, the answer is $\text{\boxed{D}}$.

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]

, 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

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