Question

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 $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Answer: 289

Concepts:law of cosines isosceles triangle

Difficulty rating: 2410

Small Hint:

Let $AD = x,$ so $AE = 10 - x;$ compute $\cos A$ from triangle $ABC$

Big Hint:

The law of cosines in triangle $ADE$ gives $x^2 = x^2 + (10-x)^2$ $- 2x(10-x)\cos A;$ cancel a factor of $10 - x$

Solution:

By the law of cosines in $\triangle ABC,$ $\begin{aligned} \cos A &= \frac{10^2 + 10^2 - 12^2}{2 \cdot 10 \cdot 10} \\ &= \frac{56}{200} = \frac{7}{25}. \end{aligned}$

Let $x = AD = DE = EC,$ so $AE = 10 - x.$ The law of cosines in $\triangle ADE$ gives $\begin{aligned} &x^2 = x^2 + (10 - x)^2 \\ &\quad {}- 2x(10 - x)\cdot\frac{7}{25}, \end{aligned}$ so $(10 - x)^2 = \frac{14}{25}\,x(10 - x).$ Since $x \lt 10,$ we may divide by $10 - x$ to get $10 - x = \frac{14x}{25},$ hence $250 = 39x$ and $x = \frac{250}{39}.$

As $\gcd(250, 39) = 1,$ the answer is $250 + 39 = 289.$

2018 AIME I Problem 4

Answer: 289

Difficulty rating: 2410

Small Hint:

Big Hint:

Solution:

Problem 4 in Other Years