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.
Call the point of tangency point and the midpoint of as . by Tangent Theorem. Notice that . Thus, and . Solving . Adding, the answer is .
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 .
Using the diagram as drawn in Solution 5, 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 (2*2=4), we get
Solving for x:
Solving for length of CE with the value we have for x: