Difference between revisions of "2008 AMC 12A Problems/Problem 22"
(→Solution 2 (without trigonometry)) |
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Line 58: | Line 58: | ||
defaultpen(linewidth(.8)+fontsize(8)); | defaultpen(linewidth(.8)+fontsize(8)); | ||
draw(Circle((0,0),4)); | draw(Circle((0,0),4)); | ||
− | path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; | + | path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); |
draw(mat); | draw(mat); | ||
draw(rotate(60)*mat); | draw(rotate(60)*mat); | ||
Line 65: | Line 65: | ||
draw(rotate(240)*mat); | draw(rotate(240)*mat); | ||
draw(rotate(300)*mat); | draw(rotate(300)*mat); | ||
+ | pair D = rotate(300)*(-3.687,1.5513); | ||
+ | pair C = rotate(300)*(-2.687,1.5513); | ||
+ | pair EE = foot((0.00,0.00),D,C); | ||
+ | draw(D--EE--(0,0)); | ||
label("\(x\)",(-1.55,2.1),E); | label("\(x\)",(-1.55,2.1),E); | ||
label("\(x\)",(0.03,1.5),E); | label("\(x\)",(0.03,1.5),E); | ||
Line 76: | Line 80: | ||
draw((0,0)--(0,3.103)); | draw((0,0)--(0,3.103)); | ||
draw((0,0)--(-2.687,1.5513)); | draw((0,0)--(-2.687,1.5513)); | ||
− | draw((0,0)--(-0.5,3.9686));</asy> | + | draw((0,0)--(-0.5,3.9686)); |
+ | label("\(E\)", EE,SE); | ||
+ | </asy> | ||
As proved in the first solution, <math> \angle OCD = 150^\circ</math>. | As proved in the first solution, <math> \angle OCD = 150^\circ</math>. |
Revision as of 10:58, 23 June 2019
- 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
Solution 1 (trigonometry)
Let one of the mats be , and the center be as shown:
Since there are mats, is equilateral. 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
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, .
Use the Pythagorean theorem on triangle , we get Using the pythagorean theorem to solve, we get must be positive, therefore
~Zeric Hang
See Also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.