Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.9 AIME, harder Prove it Greece

$ABCD$ is a convex cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at the point $E$ . It is given that $AB = 39$ , $AE = 45$ , $AD = 60$ and $BC = 56$ .
Determine the length of $CD$ .

Solution

From the similarity of triangles $BEC$ and $AED$ we $BE = 42$ .

We put $DE = x$ and $CE = y$ . From the similarity of triangles $ABE$ and $DEC$ we have
$\frac{x}{AE} = \frac{y}{BE} = \frac{CD}{AB} \Leftrightarrow \frac{x}{15} = \frac{y}{14} = \frac{CD}{13} = t. \quad (1)$
Moreover, from theorem of Ptolemy we find
$\begin{aligned} BD \cdot AC &= AD \cdot BC + AB \cdot CD \\ \Leftrightarrow (45+14t)(42+15t) = 60 \cdot 56 + 39 \cdot 13t \\ \Leftrightarrow 35t^2 + 126t - 245 = 0 \Leftrightarrow t = \frac{7}{5} \text{ or } t = -5. \end{aligned}$
Since $t$ must be positive, from (1) we have $t = \frac{7}{5}$ and $CD = \frac{91}{5}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.