Triangle Area Calculation 2
In triangle $ ABC, $ the length of side $ \overline{BC} $ is equal to the average of the other two sides. Also,
\[\cos C = \frac{AB}{AC}.\]Given that all the side lengths are integers, find the smallest possible area of triangle $ ABC $.
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Solution
We are told that $ a = \frac{b + c}{2} $. Also, $ \cos C = \frac{c}{b}, $ and by the Law of Cosines,
\[\cos C = \frac{a^2 + b^2 - c^2}{2ab}.\]Then $ \frac{a^2 + b^2 - c^2}{2ab} = \frac{c}{b}, $ so
\[a^2 + b^2 - c^2 = 2ac.\]From the equation $ a = \frac{b + c}{2}, $ $ b = 2a - c $. Substituting, we get
\[a^2 + (2a - c)^2 - c^2 = 2ac.\]This simplifies to $ 5a^2 - 6ac = 0, $ which factors as $ a(5a - 6c) = 0 $. Then $ c = \frac{5}{6} a $ and
\[b = 2a - c = 2a - \frac{5}{6} a = \frac{7}{6} a.\]Since we want the smallest possible area of triangle $ ABC, $ and all the side lengths are integers, we take $ a = 6 $. Then $ c = 5 $ and $ b = 7 $. By Heron's formula, the area of the triangle is $ \sqrt{9(9 - 6)(9 - 7)(9 - 5)} = \boxed{6 \sqrt{6}} $.
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