# 2000 AMC 12 Problems/Problem 21

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 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 $\text{\boxed{D}}$.