Difference between revisions of "2008 AMC 12A Problems/Problem 22"
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==Solution 6 (Estimation)== | ==Solution 6 (Estimation)== | ||
− | We will treat the area between two adjacent placemats as an equilateral triangle. Because the inside hexagon is regular, the distance from the center to the closest vertex of an equilateral triangle <math>x.</math> The altitude of an equilateral triangle with side length 1 is <math>\frac{\sqrt{3}}{2},</math> which means <math>x</math> is roughly <math>4-\frac{\sqrt{3}}{2}.</math> Estimating, <math>\frac{\sqrt{7}}{2}</math> is about <math>4</math> (a little less, which is what we'd expect), so we circle <math>\boxed{\text{(C) }\frac{3\sqrt{7}-\sqrt{3}}{2}}</math> and are happy that we just solved an <math>AMC</math> <math>10</math> problem <math>25</math> without having to do any actual thinking. | + | We will treat the area between two adjacent placemats as an equilateral triangle. Because the inside hexagon is regular, the distance from the center to the closest vertex of an equilateral triangle <math>x.</math> The altitude of an equilateral triangle with side length 1 is <math>\frac{\sqrt{3}}{2},</math> which means <math>x</math> is roughly <math>4-\frac{\sqrt{3}}{2}.</math> Estimating, <math>\frac{3\sqrt{7}}{2}</math> (<math>\sqrt{7}</math> is about <math>2.65</math> which can be easily estimated by comparing <math>26^2=676</math> and <math>27^2=729</math>) is about <math>4</math> (a little less, which is what we'd expect), so we circle <math>\boxed{\text{(C) }\frac{3\sqrt{7}-\sqrt{3}}{2}}</math> and are happy that we just solved an <math>AMC</math> <math>10</math> problem <math>25</math> without having to do any actual thinking. |
==See Also== | ==See Also== |
Revision as of 16:49, 16 May 2023
- 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 (Estimation)
We will treat the area between two adjacent placemats as an equilateral triangle. Because the inside hexagon is regular, the distance from the center to the closest vertex of an equilateral triangle The altitude of an equilateral triangle with side length 1 is
which means
is roughly
Estimating,
(
is about
which can be easily estimated by comparing
and
) is about
(a little less, which is what we'd expect), so we circle
and are happy that we just solved an
problem
without having to do any actual thinking.
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.