In ABC, E, F, G are points on AB, BC, CA respectively such that $\frac{AE}{EB} = \frac{BF}{FC} = \frac{CG}{GA} = \frac{1}{3}$. K, L, M are the intersection points of the lines AF and CE, BG and AF, CE and BG, respectively. Suppose the area of ABC is 1; find the area of KLM.
I know there is a general formula when dividing in m:n ie. $\frac{(m-n)^2}{m^2+mn+n^2}$
P.S. please no Bary/ Complex
To verify the result, I put $m=n$ and it worked. Also, I tried the area lemma, Ceva's theorem but to no use. I tried this problem for a long time and then realised that Menelaus worked but as this excercise was in the area and ratio section, that felt like 'cheating' as I want to know the author's intended method.
Thanks!
Let [.] denote areas. Evaluate the area ratio
$$\frac{[CBM]}{[EBM]}= \frac{[CBG]}{[EBG]} = \frac{\frac 14 [ABC]}{\frac 34 [ABG]}= \frac{\frac 14 [ABC]}{\frac 34 \frac 34[ABC]} =\frac 49 $$
which leads to the area $[CBM]$, and likewise $[ACK]$, $[BAL]$
$$[CBM] = \frac{4}{4+9} [CEB] = \frac{4}{13}\frac 34[ABC] =\frac{3}{13} = [ACK] = [BAL] $$ Then $$[KLM] = [ABC] -[CBM]- [ACK]-[BAL] =1-3\cdot\frac{3}{13}=\frac 4{13}$$
The derivation applies to general ratio $m:n$ as well $$\frac{[CBM]}{[EBM]}=\frac{m(m+n)}{n^2}\implies \frac{[CBM]}{[CBE]}=\frac{m(m+n)}{m(m+n)+n^2} $$ $$\implies [CBM]= \frac{m(m+n)}{m(m+n)+n^2}\cdot\frac{n}{m+n} =\frac{m n}{m^2+mn +n^2}$$ $$[KLM] = 1-3[CBM]= \frac{(m-n)^2}{m^2+mn +n^2} $$