No Problemmo #5

Geometry Level 5

A triangle $ABC$ is drawn such that $D$ divides $BC$ in the ratio $3:2$ internally and the point $E$ divides $AB$ in the ratio $3:2$ externally. Now $AD$ and $CE$ intersect at $F$ and $Q$ be the midpoint of $AC$ . Then find the ratio in which $FQ$ divides $BC$ .

Let the answer be in the form $\dfrac{m}{n}$ where $m$ and $n$ are co-prime positive integers. Input your answer as $m+n$ .

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The answer is 19.

1 solution

Jitarani Nayak
Nov 11, 2017

I will do the repeated application of Menelaus theorem which states that if a transversal cuts the sides BC, CA ,AB(suitabily extended) of ∆ABC in points D, E, F, respectively, then $\frac{BD}{DC}\cdot \frac{CE}{EA}\cdot \frac{AF}{FB}=-1$
For the time being we will take it 1. Applying it on ∆EBC, with ADF as transversal, we get $\frac{CF}{FE}=\frac{2}{9}$
Similarly applying on ∆AEF and ∆ADC successively we get $\frac{AD}{DF}=\frac{11}{4}$ and $\frac{DK}{KC}=\frac{4}{15}$ where K is the point of intersection of FQ with BC. Now you can easily proceed to get the answer as $\boxed{\frac{BK}{KC}=\frac{13}{6}}$

No Problemmo #5

The answer is 19.

1 solution

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