Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it New Zealand

Problem:

In triangle $ABC$ , points $D$ and $E$ lie on the interior of segments $AB$ and $AC$ , respectively, such that $AD = 1$ , $DB = 2$ , $BC = 4$ , $CE = 2$ and $EA = 3$ . Let $DE$ intersect $BC$ at $F$ . Determine the length of $CF$ .

Solution

Solution:

First notice that the sidelengths of $\triangle ABC$ are $3$ , $4$ and $5$ . By Pythagoras this implies that triangle $ABC$ is right-angled at $B$ . Now we can put the diagram on coordinate axes such that $B = (0,0)$ , $A = (0,3)$ and $C = (4,0)$ . Furthermore we get $D = (0,2)$ and since $E$ divides $CA$ into the ratio $2:3$ we get $E = (2.4,1.2)$ , as shown in the diagram.

Now we can calculate the slope of the line $DE$ to be $\frac{-0.8}{2.4} = -\frac{1}{3}$ . This means that the equation of line $DE$ is given by $y = -\frac{x}{3} + 2$ . Therefore the $x$ -intercept of this line is the solution to $0 = -\frac{x}{3} + 2$ . The solution is when $x = 6$ , and thus $F = (6,0)$ . Hence $CF = 2$ .

$\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.