Difference between revisions of "2004 AMC 12A Problems/Problem 18"
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Clearly, <math>EA = EF = BG</math>. Thus, the sides of [[right triangle]] <math>CDE</math> are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle <math>3 - 4 - 5</math> and since <math>DC = 2</math>, <math>CE = 5/2</math>. | Clearly, <math>EA = EF = BG</math>. Thus, the sides of [[right triangle]] <math>CDE</math> are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle <math>3 - 4 - 5</math> and since <math>DC = 2</math>, <math>CE = 5/2</math>. | ||
− | + | == Solution 3 == | |
[[Image:2004_AMC12A-18.png]] | [[Image:2004_AMC12A-18.png]] | ||
TBE | TBE |
Revision as of 14:10, 25 July 2015
- The following problem is from both the 2004 AMC 12A #18 and 2004 AMC 10A #22, so both problems redirect to this page.
Problem
Square has side length . A semicircle with diameter is constructed inside the square, and the tangent to the semicircle from intersects side at . What is the length of ?
Solution 1
Let the point of tangency be . By the Two Tangent Theorem and . Thus . The Pythagorean Theorem on yields
Hence .
Solution 2
Clearly, . Thus, the sides of right triangle are in arithmetic progression. Thus it is similar to the triangle and since , .
Solution 3
See also
2004 AMC 12A (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 12 Problems and Solutions |
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.