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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 15"

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== Problem ==
 
== Problem ==
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Triangle <math>ABC</math> has an inradius of <math>5</math> and a circumradius of <math>16</math>. If <math>2\cos{B} = \cos{A} + \cos{C}</math>, then the area of triangle <math>ABC</math> can be expressed as <math>\frac{a\sqrt{b}}{c}</math>, where <math>a, b,</math> and <math>c</math> are positive integers such that <math>a</math> and <math>c</math> are relatively prime and <math>b</math> is not divisible by the square of any prime. Compute <math>a+b+c</math>.
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== Solution ==
 
== Solution ==
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Using the identity <math>\cos A + \cos B + \cos C = 1+\frac{r}{R}</math>, we have that <math>\cos A + \cos B + \cos C = \frac{21}{16}</math>. From here, combining this with <math>2\cos B = \cos A + \cos C</math>, we have that <math>\cos B = \frac{7}{16}</math> and <math>\sin B = \frac{3\sqrt{23}}{16}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have that <math>b = 6\sqrt{23}</math>. By the Law of Cosines, we have that:
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<cmath>b^2 = a^2 + c^2-2ac\cdot \cos B \implies a^2+c^2-\frac{7ac}{8} = 36 \cdot 23.</cmath>But one more thing: noting that <math>\cos A = \frac{b^2+c^2-a^2}{2cb}</math>. and <math>\cos C = \frac{a^2+b^2-c^2}{2ab}</math>, we know that <math>\frac{36 \cdot 23 + b^2+c^2-a^2}{bc} + \frac{36 \cdot 23+a^2+b^2-c^2}{ab} = \frac{7}{4} \implies</math> <math>\frac{36 \cdot 23 + c^2-a^2}{c} + \frac{36 \cdot 23 + a^2-c^2}{a} = \frac{21\sqrt{23}}{2} \implies</math> <math>\frac{(a+c)(36 \cdot 23 + 2ac-c^2-a^2)}{ac} = \frac{21\sqrt{23}}{2}</math>. Combining this with the fact that <math>a^2+c^2 - \frac{7ac}{8} = 36 \cdot 23</math>, we have that: <math>\frac{(a+c)(-2ac \cdot \frac{7}{16}+2ac)}{ac} = \frac{21\sqrt{23}}{2} \implies</math> <math>a+c = \frac{28 \sqrt{23}}{3}</math>. Therefore, <math>s</math>, our semiperimeter is <math>\frac{23\sqrt{23}}{3}</math>. Our area, <math>r \cdot s</math> is equal to <math>\frac{115\sqrt{23}}{3}</math>, giving us a final answer of <math>\boxed{141}</math>.
  
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~AopsUser101
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== See also ==
 
== See also ==

Latest revision as of 00:44, 3 March 2020

Problem

Triangle $ABC$ has an inradius of $5$ and a circumradius of $16$. If $2\cos{B} = \cos{A} + \cos{C}$, then the area of triangle $ABC$ can be expressed as $\frac{a\sqrt{b}}{c}$, where $a, b,$ and $c$ are positive integers such that $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime. Compute $a+b+c$.

Solution

Using the identity $\cos A + \cos B + \cos C = 1+\frac{r}{R}$, we have that $\cos A + \cos B + \cos C = \frac{21}{16}$. From here, combining this with $2\cos B = \cos A + \cos C$, we have that $\cos B = \frac{7}{16}$ and $\sin B = \frac{3\sqrt{23}}{16}$. Since $\sin B = \frac{b}{2R}$, we have that $b = 6\sqrt{23}$. By the Law of Cosines, we have that: \[b^2 = a^2 + c^2-2ac\cdot \cos B \implies a^2+c^2-\frac{7ac}{8} = 36 \cdot 23.\]But one more thing: noting that $\cos A = \frac{b^2+c^2-a^2}{2cb}$. and $\cos C = \frac{a^2+b^2-c^2}{2ab}$, we know that $\frac{36 \cdot 23 + b^2+c^2-a^2}{bc} + \frac{36 \cdot 23+a^2+b^2-c^2}{ab} = \frac{7}{4} \implies$ $\frac{36 \cdot 23 + c^2-a^2}{c} + \frac{36 \cdot 23 + a^2-c^2}{a} = \frac{21\sqrt{23}}{2} \implies$ $\frac{(a+c)(36 \cdot 23 + 2ac-c^2-a^2)}{ac} = \frac{21\sqrt{23}}{2}$. Combining this with the fact that $a^2+c^2 - \frac{7ac}{8} = 36 \cdot 23$, we have that: $\frac{(a+c)(-2ac \cdot \frac{7}{16}+2ac)}{ac} = \frac{21\sqrt{23}}{2} \implies$ $a+c = \frac{28 \sqrt{23}}{3}$. Therefore, $s$, our semiperimeter is $\frac{23\sqrt{23}}{3}$. Our area, $r \cdot s$ is equal to $\frac{115\sqrt{23}}{3}$, giving us a final answer of $\boxed{141}$.

~AopsUser101

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 14 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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