Difference between revisions of "2008 AMC 12A Problems/Problem 22"
CakeIsEaten (talk | contribs) (→Solution 2 (without trigonometry)) |
CakeIsEaten (talk | contribs) (→Solution 2 (without trigonometry)) |
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Then, drop a perpendicular down between the radiis. | Then, drop a perpendicular down between the radiis. | ||
Now we write the length of perpendiculars in terms of the segmented radii and the full radii. | Now we write the length of perpendiculars in terms of the segmented radii and the full radii. | ||
+ | |||
+ | <asy>unitsize(8mm); | ||
+ | defaultpen(linewidth(.8)+fontsize(8)); | ||
+ | draw(Circle((0,0),4)); | ||
+ | path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; | ||
+ | draw(mat); | ||
+ | draw(rotate(60)*mat); | ||
+ | draw(rotate(120)*mat); | ||
+ | draw(rotate(180)*mat); | ||
+ | draw(rotate(240)*mat); | ||
+ | draw(rotate(300)*mat); | ||
+ | label("\(x\)",(-1.55,2.1),E); | ||
+ | label("\(x\)",(0.03,1.5),E); | ||
+ | label("\(A\)",(-3.6,2.5513),E); | ||
+ | label("\(B\)",(-3.15,1.35),E); | ||
+ | label("\(C\)",(0.05,3.20),E); | ||
+ | label("\(D\)",(-0.75,4.15),E); | ||
+ | label("\(O\)",(0.00,-0.10),E); | ||
+ | label("\(1\)",(-0.1,3.8),S); | ||
+ | label("\(4\)",(-0.4,2.2),S); | ||
+ | draw((0,0)--(0,3.103)); | ||
+ | draw((0,0)--(-2.687,1.5513)); | ||
+ | draw((0,0)--(-0.5,3.9686));</asy> | ||
Since i can't write latx stuff, lets skip to the end. | Since i can't write latx stuff, lets skip to the end. |
Revision as of 04:27, 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)
Since that link doesn't really lead to anything now, here it is:
Draw the center of the circle. Then connect the radius to a far corner of the box and a partial radius to the closer corner of the box. Then, drop a perpendicular down between the radiis. Now we write the length of perpendiculars in terms of the segmented radii and the full radii.
Since i can't write latx stuff, lets skip to the end.
we have x^2 + xsqrt3 - 15 = 0. Utilize quadratic equation. We now have x = (-3 + sqrt63)/2 take out the 9 from the sqrt and we have:
(3sqrt7 - 3)/2, answer choice C.
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 |