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Difference between revisions of "2000 AMC 10 Problems/Problem 19"

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This is choice <math>\boxed{\text{D}}</math>
 
This is choice <math>\boxed{\text{D}}</math>
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 +
(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle <math>k</math> times changes each of the areas <math>k^2</math> times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2000|num-b=18|num-a=20}}
 
{{AMC10 box|year=2000|num-b=18|num-a=20}}

Revision as of 12:25, 11 January 2009

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

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

Solution

[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]

Let the square have area $1$, then it follows that the altitude of one of the triangles is $2m$. The area of the other triangle is $\frac{x}{2}$.

By similar triangles, we have $\frac{x}{1}=\frac{1}{2m}\Rightarrow \frac{x}{2}=\frac{1}{4m}$

This is choice $\boxed{\text{D}}$

(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle $k$ times changes each of the areas $k^2$ times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)

See Also

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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