Difference between revisions of "2000 AMC 12 Problems/Problem 21"

(Solution 2)
(Added another solution and removed unnecesary comments)
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~ Nafer
 
~ Nafer
  
 +
=== Solution 3 (process of elimination) ===
  
[I think there is a mistake. It says it is a problem 19 for AMC 10 and problem 21 for AMC 12. They are both the same year, 2000. Is this a mistake, or meant to be like this?]
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Simply testing specific triangles is sufficient.
  
-Canadian
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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>
  
 
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~ Snacc
 
 
 
 
Answer to the question asked by Canadian:
 
Many problems overlap in the AMC 10 and AMC 12 for the same year. Yes, it is meant to be like this.
 
  
 
== 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.

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

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

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]

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 height of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the base of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}$.


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]

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

~ 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 $S=\frac{2}{3}\frac{2}{3}=\frac{4}{9}$. 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 $\text{(D)}$

~ Snacc

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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