2008 AMC 12A Problems/Problem 22
- The following problem is from both the 2008 AMC 12A #22 and 2008 AMC 10A #25, so both problems redirect to this page.
Contents
Problem
A round table has radius . Six rectangular place mats are placed on the table. Each place mat has width and length as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length . Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is ?
Solution 1 (Trigonometry)
Let one of the mats be , and the center be as shown:
Since there are mats, is equilateral (the hexagon with side length is regular). So, . Also, .
By the Law of Cosines: .
Since must be positive, .
Solution 2 (without trigonometry)
Draw and as in the diagram. Draw the altitude from to and call the intersection .
As proved in the first solution, . That makes a triangle, so and
Since is a right triangle,
Solving for gives
Solution 3 (simply Pythagorean Theorem)
By symmetry, is the midpoint of and is an extension of . Thus . Since and , . Since is , (this can also be deduced from Pythagoras on ).
Thus . As previous solutions noted, is equilateral, and thus the desired length is .
Solution 3
Looking at the diagram above, we know that is a diameter of circle due to symmetry. Due to Thales' theorem, triangle is a right triangle with . lies on and because is also a right angle. To find the length of , notice that if we draw a line from to , the midpoint of line , it creates two - - triangles. Therefore, .
Applying the Pythagorean theorem to triangle , we get Using the quadratic formula to solve, we get must be positive, therefore
~Zeric Hang
Solution 4 (coordinate bashing)
We will let be the origin. This way the coordinates of will be . By , the coordinates of will be . The distance any point with coordinates is from the origin is . Therefore, the distance is from the origin is and . We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that
Note: Since and are not labeled in the diagram, refer to solution 1 for the location of points and .
Solution 5
Notice that is the circumference of the circle. Therefore, is the side length of an inscribed hexagon with side length . is a right triangle with . The length of . Using the Pythagorean Theorem, we get
Solving for , we get
Solution 6 (Answer Choices & Estimation)
The smallest distance between the intersection between two adjacent placemats and the circle is a bit less than . Thus, the answer will be a bit more than Going up, we guess that the next one up is the answer, which is about . The next one up after that is which is about and seems too big. So we circle and are happy we just cheesed an problem
See Also
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 |
2008 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.