2004 AMC 12A Problems/Problem 18
- The following problem is from both the 2004 AMC 12A #18 and 2004 AMC 10A #22, so both problems redirect to this page.
Contents
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 ?
Solutions
Solution 1
Let the point of tangency be . By the Two Tangent Theorem and . Thus . The Pythagorean Theorem on yields
Hence .
Solution 2
Call the point of tangency point and the midpoint of as . by Tangent Theorem. Notice that . Thus, and . Solving . Adding, the answer is .
Solution 3
Clearly, . Thus, the sides of right triangle are in arithmetic progression. Thus it is similar to the triangle and since , .
Solution 4
Let us call the midpoint of side , point . Since the semicircle has radius 1, we can do the Pythagorean theorem on sides . We get . We then know that by Pythagorean theorem. Then by connecting , we get similar triangles and . Solving the ratios, we get , so the answer is .
Solution 5
Using the diagram as drawn in Solution 4, let the total area of square be divided into the triangles , , , and . Let x be the length of AE. Thus, the area of each triangle can be determined as follows:
(the length of CE is calculated with the Pythagorean Theorem, lines GE and CE are perpendicular by definition of tangent)
Adding up the areas and equating to the area of the total square , we get
So, .
~Typo Fix by doulai1
Video Solution
Education, the Study of Everything
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.