Mock AIME 1 Pre 2005 Problems/Problem 15

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

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by Problem 14	Followed by Problem 15
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Mock AIME 1 Pre 2005 Problems/Problem 15

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