<s>[INST] In $\triangle ABC$ with integer sides, $BC$ is the average of the other sides, and $\cos C=AB/AC$. Find the smallest possible area. 
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 $\triangle ABC$.

I set an altitude, BD, then tried doing $CD=\frac{AD \cdot BC}{AC}$, then solving. But this does not work. What should I do?
 [/INST] Let $BC=n$, $BA=n-k$ and $AC=n+k$. Then
$$\cos C = \frac{n-k}{n+k}= \frac{ (n+k)^2+n^2 -(n-k)^2}{2n(n+k)}$$
which yields $n=6k$. Thus, the smallest triangle has the integer side lengths 5, 6 and 7, and the corresponding area is $6\sqrt6$.
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