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 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 $T_1=\frac{\frac{2}{3}\frac{4}{3}}{2}=\frac{4}{9}$ , and the smaller triangle has area $T_2=\frac{\frac{2}{3}\frac{1}{3}}{2}=\frac{1}{9}$ . 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

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.

Page

Toolbox

Search