AIME 2018 I, Problem 4

In $\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution: By the Law of Cosines on $\triangle ABC$, we have: \[\cos(A) = \frac{10^2+10^2-12^2}{2*10*10}  = \frac{7}{25}\] By the Law of Cosines on $\triangle ADE$, then \[\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}\] So, our answer is $250+39=\fbox{289}$.

Answer: 289