Difference between revisions of "2012 AIME I Problems/Problem 13"
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so our final answer is <math>3+4+25+9 = \boxed{041.}</math> | so our final answer is <math>3+4+25+9 = \boxed{041.}</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== | ||
+ | |||
+ | 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 <math>O</math> of the circles lies in the interior of triangle <math>ABC</math> 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 <math>\theta</math> be the measure of angle <math>XAC</math> so that angle <math>BAX</math> has measure <math>60-\theta</math>. Let <math>AB=BC=AC=x</math>. The law of cosines on triangles <math>BAX</math> and <math>XAC</math> yields <math>\cos(60-\theta)=\frac{x^2+9}{10x}</math> and <math>\cos\theta=\frac{x^2+16}{10x}</math>. Solving this system will yield the value of <math>x</math>. Since <math>\cos\theta=\frac{x^2+16}{10x}</math> we have that <math>\sin\theta=\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}</math>. Substituting these into the equation <math>\frac{x^2+9}{10x}=\cos(60-\theta)=\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta</math> we obtain <math>\frac{x^2+9}{10x}=\frac{1}{2}\frac{x^2+16}{10x}+\frac{\sqrt{3}}{2}\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}</math>. After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain <math>x^4-50x^2+193=0</math> so that by the quadratic formula <math>x^2=25\pm12\sqrt{3}</math>. Under the hypothesis that <math>O</math> lies in the interior of triangle <math>ABC</math>, <math>x^2</math> must be <math>25+12\sqrt{3}</math>. To see this, note that the other value for <math>x^2</math> is roughly <math>4.2</math> so that <math>x\approx 2.05</math>, but since <math>AX=5</math> and <math>AX\leq x</math> we have a contradiction. We then obtain the area as in Solution 1. | ||
+ | |||
+ | Now, suppose <math>O</math> does not lie in the interior of triangle <math>ABC</math>. We then obtain convex quadrilateral <math>OBAC</math> with diagonals <math>CB</math> and <math>OA</math> intersecting at <math>X</math>. Here <math>AX=AB=AC=x</math>. We may let <math>\alpha</math> denote the measure of angle <math>CAX</math> so that angle <math>XAB</math> measures <math>60-\alpha</math>. Note that the law of cosines on triangles <math>CXA</math> and <math>BXA</math> yield the same equations as in the first case with <math>\theta</math> replaced with <math>\alpha</math>. Thus we obtain again <math>x^2=25\pm12\sqrt{3}</math>. If <math>x^2=25+12\sqrt{3}</math> then <math>x\approx 6.8</math>, but this is impossible since <math>AX\leq 5</math> but the shortest possible distance from <math>A</math> to <math>X</math> is the height of equilateral triangle <math>ABC</math> which is <math>\approx6.8\sqrt{3}\approx5.8</math>; a contradiction. Hence in this case <math>x^2=25-12\sqrt{3}</math>. 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 == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=12|num-a=14}} | {{AIME box|year=2012|n=I|num-b=12|num-a=14}} |
Revision as of 06:12, 1 April 2012
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
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 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 |