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| − | ==Problem==
| + | #REDIRECT [[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
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| − | <math>\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}</math>
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| − | ==Solution==
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| − | <asy>
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| − | unitsize(36);
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| − | draw((0,0)--(6,0)--(0,3)--cycle);
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| − | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
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| − | label("$1$",(1,2),S);
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| − | label("$1$",(2,1),W);
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| − | label("$2m$",(4,0),S);
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| − | label("$x$",(0,2.5),W);
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| − | </asy>
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| − | Let the square have area <math>1</math>, then it follows that the altitude of one of the triangles is <math>2m</math>. The area of the other triangle is <math>\frac{x}{2}</math>.
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| − | By similar triangles, we have <math>\frac{x}{1}=\frac{1}{2m}\Rightarrow \frac{x}{2}=\frac{1}{4m}</math>
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| − | 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.)
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=18|num-a=20}}
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