Difference between revisions of "2008 AMC 12A Problems/Problem 22"
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=== Solution 2 (without trigonometry) === | === Solution 2 (without trigonometry) === | ||
+ | Draw <math>OD</math> and <math>OC</math> as in the diagram. Draw the altitude from <math>O</math> to <math>DC</math> and call the intersection <math>E</math> | ||
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<asy>unitsize(8mm); | <asy>unitsize(8mm); | ||
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draw((0,0)--(-0.5,3.9686));</asy> | draw((0,0)--(-0.5,3.9686));</asy> | ||
− | + | As proved in the first solution, <math> \angle OCD = 150^\circ</math>. | |
− | + | That makes <math>\Delta OCE</math> a <math>30-60-90</math> triangle, so <math>OE = \frac{x}{2}</math> and <math>CE= \frac{x\sqrt 3}{2}</math> | |
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− | + | Since <math> \Delta OED</math> is a right triangle, | |
+ | <math>\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0</math> | ||
− | + | Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math> | |
==See Also== | ==See Also== |
Revision as of 17:09, 27 January 2011
- The following problem is from both the 2008 AMC 12A #22 and 2004 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
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 |