Difference between revisions of "2012 AIME I Problems/Problem 13"
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==Problem 13== | ==Problem 13== | ||
Three concentric circles have radii <math>3,</math> <math>4,</math> and <math>5.</math> An equilateral triangle with one vertex on each circle has side length <math>s.</math> The largest possible area of the triangle can be written as <math>a + \tfrac{b}{c} \sqrt{d},</math> where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive integers, <math>b</math> and <math>c</math> are relatively prime, and <math>d</math> is not divisible by the square of any prime. Find <math>a+b+c+d.</math> | Three concentric circles have radii <math>3,</math> <math>4,</math> and <math>5.</math> An equilateral triangle with one vertex on each circle has side length <math>s.</math> The largest possible area of the triangle can be written as <math>a + \tfrac{b}{c} \sqrt{d},</math> where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive integers, <math>b</math> and <math>c</math> are relatively prime, and <math>d</math> is not divisible by the square of any prime. Find <math>a+b+c+d.</math> | ||
+ | |||
+ | == Solution 1== | ||
+ | Reinterpret the problem in the following manner. Equilateral triangle <math>ABC</math> has a point <math>X</math> on the interior such that <math>AX = 5,</math> <math>BX = 4,</math> and <math>CX = 3.</math> A <math>60^o</math> counter-clockwise rotation about vertex <math>A</math> maps <math>X</math> to <math>X'</math> and <math>C</math> to <math>C'.</math> Note that angle <math>XAX'</math> is <math>60</math> and <math>XA = X'A = 5</math> which tells us that triangle <math>XAX'</math> is equilateral and that <math>XX' = 5.</math> We now notice that <math>XC = 3</math> and <math>X'C = 4</math> which tells us that angle <math>XCX'</math> is <math>90</math> because there is a <math>3</math>-<math>4</math>-<math>5</math> Pythagorean triple. Now note that <math>\angle ABC + \angle ACB = 120</math> and <math>\angle XCA + \angle XBA = 90,</math> so <math>\angle XCB+\angle XBC = 30</math> and <math>\angle BXC = 150.</math> Applying the law of cosines on triangle <math>BXC</math> yields | ||
+ | |||
+ | <cmath>BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos(150) = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}</cmath> | ||
+ | |||
+ | and thus the area of <math>ABC</math> equals <cmath>BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.</cmath> | ||
+ | |||
+ | so our final answer is <math>3+4+25+9 = \boxed{041.}</math> | ||
==Solution 2== | ==Solution 2== |
Revision as of 02:11, 24 December 2014
Contents
[hide]Problem 13
Three concentric circles have radii
and
An equilateral triangle with one vertex on each circle has side length
The largest possible area of the triangle can be written as
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
Solution 1
Reinterpret the problem in the following manner. Equilateral triangle has a point
on the interior such that
and
A
counter-clockwise rotation about vertex
maps
to
and
to
Note that angle
is
and
which tells us that triangle
is equilateral and that
We now notice that
and
which tells us that angle
is
because there is a
-
-
Pythagorean triple. Now note that
and
so
and
Applying the law of cosines on triangle
yields
and thus the area of equals
so our final answer is
Solution 2
We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center of the circles lies in the interior of triangle
or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let
be the measure of angle
so that angle
has measure
. Let
. The law of cosines on triangles
and
yields
and
. Solving this system will yield the value of
. Since
we have that
. Substituting these into the equation
we obtain
. After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain
so that by the quadratic formula
. Under the hypothesis that
lies in the interior of triangle
,
must be
. To see this, note that the other value for
is roughly
so that
, but since
and
we have a contradiction. We then obtain the area as in Solution 1.
Now, suppose does not lie in the interior of triangle
. We then obtain convex quadrilateral
with diagonals
and
intersecting at
. Here
. We may let
denote the measure of angle
so that angle
measures
. Note that the law of cosines on triangles
and
yield the same equations as in the first case with
replaced with
. Thus we obtain again
. If
then
, but this is impossible since
but the shortest possible distance from
to
is the height of equilateral triangle
which is
; a contradiction. Hence in this case
. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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