Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>. | The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>. | ||
− | Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar the 2 pole problem(http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, <math>MM_2</math> must be <math>\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}</math>, by the harmonic mean. Thus, <math>MM_1</math> must be <math>\frac{36}{5}</math>.) | + | Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem(http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, <math>MM_2</math> must be <math>\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}</math>, by the harmonic mean. Thus, <math>MM_1</math> must be <math>\frac{36}{5}</math>.) |
Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 = \boxed{\frac{88}5}</math>. | Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 = \boxed{\frac{88}5}</math>. |
Revision as of 17:55, 19 September 2015
Problem
Consider the -sided polygon , as shown. Each of its sides has length , and each two consecutive sides form a right angle. Suppose that and meet at . What is the area of quadrilateral ?
Solution
Solution 1
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles and .
We know that triangles and are similar because . Also, since , the ratio of the distance from to to the distance from to is also . Solving with the fact that the distance from to is 4, we see that the distance from to is .
The area of is simply , the area of is , and the area of rectangle is .
Taking the area of rectangle and subtracting the combined area of and yields .
Solution 2
Extend and and call their intersection .
The triangles and are clearly similar with ratio , hence and thus . The area of the triangle is .
The triangles and are similar as well, and we now know that the ratio of their dimensions is .
Draw altitudes from onto and , let their feet be and . We get that . Hence . (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem(http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, must be , by the harmonic mean. Thus, must be .)
Then the area of is , and the area of can be obtained by subtracting the area of , which is . Hence the answer is .
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.