Difference between revisions of "2008 AMC 12A Problems/Problem 22"
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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> | 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> | ||
− | Since <math> \Delta | + | Since <math> \Delta ODE</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> | <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> | ||
Revision as of 15:34, 5 October 2020
- 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 quadratic formula to solve, we get
must be positive, therefore
~Zeric Hang
Soultion 4 (coordinate bashing)
We will let be the origin. This way the coordinates of C would be
. By 30-60-90, the coordinates of D would be
. The distance
is from the origin is just
. Therefore, the distance D is from the origin is both 4 and
. We get the equation mentioned in all the previous solution, using the quadratic formula, we get that
Solution 5
Notice that is one-sixth the circumference of the circle. Therefore,
is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius, or
.
is a right angle, therefore
is a right triangle.
is half the length of
, or
. The length of
is
plus the altitude length of one of the equilateral triangles, or
. Using the Pythagorean Theorem, we get
Solving for , we get
, or
.
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.