Exercise 223

Let $ABCDE$ be a pentagon inscribed in a circle with $AB=CD=5,BC=DE=12,AE=14$. The sum of the lengths of all diagonals is $\frac mn$. What is $m+n$?

Solution

With $AC=BD=CE=x, AD=y, BE=z$: Ptolemy theorem (45) on $ABCD,ABCE,BCDE$ gives $25+12y=x^2$, $5x+168=xz$, $5z+144=x^2$.
Solving the resulting cubic $x^3=169x+840$ gives $x=15$, and then $z=\frac{81}5,y=\frac{50}3$.
Sum $=3x+y+z=45+\frac{50}3+\frac{81}5=\frac{1168}{15}$, so $m+n=1168+15=\boxed{1183}$.