Difference between revisions of "2000 AMC 12 Problems/Problem 21"
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<math>\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}</math> | <math>\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}</math> | ||
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− | + | === Solution 1 === | |
− | == Solution 1 == | ||
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WLOG, let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A = \frac{1}{2}bh = \frac{h}{2}</math>, the base of the triangle with area <math>m</math> is <math>2m</math>. Therefore <math>\frac{2m}{1} = \frac{1}{x}</math> where <math>x</math> is the height of the other triangle. <math>x = \frac{1}{2m}</math>, and the area of that triangle is <math>\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}</math>. | WLOG, let a side of the square be <math>1</math>. Simple angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A = \frac{1}{2}bh = \frac{h}{2}</math>, the base of the triangle with area <math>m</math> is <math>2m</math>. Therefore <math>\frac{2m}{1} = \frac{1}{x}</math> where <math>x</math> is the height of the other triangle. <math>x = \frac{1}{2m}</math>, and the area of that triangle is <math>\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}</math>. | ||
− | == Solution 2 == | + | === Solution 2 === |
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~ Nafer | ~ Nafer | ||
− | == Solution 3 (process of elimination) == | + | === Solution 3 (process of elimination) === |
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Simply testing specific triangles is sufficient. | Simply testing specific triangles is sufficient. | ||
Latest revision as of 13:35, 19 January 2021
- The following problem is from both the 2000 AMC 12 #21 and 2000 AMC 10 #19, so both problems redirect to this page.
Contents
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 times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
Solutions
Solution 1
WLOG, let a side of the square be . Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since , the base of the triangle with area is . Therefore where is the height of the other triangle. , and the area of that triangle is .
Solution 2
From the diagram from the previous solution, we have , as the legs and as the side length of the square. WLOG, let the area of triangle be times the area of square .
Since triangle is similar to the large triangle, it has , and Thus
Now since triangle is similar to the large triangle, it has , and
Thus . .
~ 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 . The larger sub-triangle has area , and the smaller triangle has area . Computing ratios you get and . Plugging in shows that the only possible answer is
~ Snacc
See also
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.