Library / /2 of 68

, 2017

Geometry Difficulty 3.6 AMC 10/12 Find the answer United States

Problem:

Let $A$ , $B$ , $C$ , $D$ be four points on a circle in that order. Also, $AB = 3$ , $BC = 5$ , $CD = 6$ , and $DA = 4$ . Let diagonals $AC$ and $BD$ intersect at $P$ . Compute $\frac{AP}{CP}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Note that $\triangle APB \sim \triangle DPC$ so $\frac{AP}{AB} = \frac{DP}{CD}$ . Similarly, $\triangle BPC \sim \triangle APD$ so $\frac{CP}{BC} = \frac{DP}{DA}$ . Dividing these two equations yields
$\frac{AP}{CP} = \frac{AB \cdot DA}{BC \cdot CD} = \frac{3 \cdot 4}{5 \cdot 6} = \frac{12}{30} = \frac{2}{5}$

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 reproduced verbatim; metadata (topic, difficulty) added by this project.