Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, | + | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC</math> is similar to <math>\triangle BCD</math>, and <math>AB > BC</math>. There exists a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC</math> is similar to <math>\triangle CEB</math> and the area of Triangle <math>AED</math> is <math>17</math> times the area of Triangle <math>CEB</math>. Find <math>AB/BC</math>. |
<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math> | <math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math> | ||
Revision as of 21:36, 16 February 2017
Problem
Quadrilateral has right angles at and , is similar to , and . There exists a point in the interior of such that is similar to and the area of Triangle is times the area of Triangle . Find .
Solution
Let , , and . Note that . By the Pythagorean Theorem, . Since ~ ~ , the ratios of side lengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Note that ~ ~ . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields: Therefore, the answer is
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
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