Question

$ABC$ is a triangle with centroid $G$ and side lengths $AB=15,BC=18,AC=25$. $D$ is the mid-point of $BC$. The length of $GD$ can be expressed as $\frac{a\sqrt{d}}{b}$, where, $a$ and $b$ are co-primes positive integers and $d$ is a square free positive integer. Find $a+b+d+1$.

Answer 1

Let $AB=c=15$

$BC=a=18$

$AC=b=25$

$AD=\frac{1}{2}\sqrt{{2b}^2+{2c}^2-a^2}$

$AD=\frac{1}{2}\sqrt{{2\times 25}^2+{2\times 15}^2-{18}^2}$

$AD=\frac{1}{2}\sqrt{1250+450-324}$

$AD=\frac{1}{2}\sqrt{1700-324}$

$AD=\frac{1}{2}\sqrt{1376}$

$AD=2\sqrt{86}$

As $GD=\frac{AD}{3}$ since centroid divides median into $2:1$

$GD=\frac{2}{3}\sqrt{86}$

$a+d+b+1=2+3+86+1=92$