BL/BC + CM/CA+ AN/AB is constant, // to medians

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BL/BC + CM/CA+ AN/AB is constant, // to medians

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Source: 1976 British FIST p1 - Further International Selection Test

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#1 h Jan 13, 2021, 12:05 PM

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Through a point $P$ in the interior of a fixed triangle $ABC$ lines $PL, PM,PN$ are drawn parallel to the medians through $A,B,C$ respectively to meet $BC, CA, AB$ at $L,M,N$ respectively. Prove that $\frac{BL}{BC} + \frac{CM}{CA}+ \frac{AN}{AB}$ is constant (independent of $P$ ).

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#2 h Oct 10, 2021, 9:36 AM

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$\frac{BL}{BC} + \frac{CM}{CA}+ \frac{AN}{AB}$ is $\cancel{not}$ constant.

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#3 h Oct 10, 2021, 4:29 PM

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this is the official wording, see here

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#4 h Oct 11, 2021, 4:26 PM • 1 Y

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Before formulating my answer in #2, I consulted your blogspot https://imogeometry.blogspot.com/
I found the original text from 1976, which corresponds to the text above.

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#5 h Oct 12, 2021, 1:32 PM • 1 Y

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Given $A(-2a,0),B(2b,0),C(0,2c)$ ; choose $P(k,l)$ .

On $AB$ , point $N(k+\frac{l(b-a)}{2c},0)$ and $s=\frac{AN}{AB}=\frac{l(b-a)+2ck+4ac}{4c(a+b)}$ .

On $BC$ , point $L(\frac{b(ck+(2c-l)(2a+b)}{2c(a+b)},/)$ and $t=\frac{BL}{BC}=\frac{2al+b(2c+l)-ck}{4c(a+b)}$ .

On $CA$ , point $M(\frac{a(ck+(a+2b)(l-2c))}{2c(a+b)},/)$ and $w=\frac{CM}{CA}=\frac{(a+2b)(2c-l)-ck}{4c(a+b)}$ .

$s+t+w=\frac{3}{2}$ .

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