Swirling In

It is generally true that if $\frac{FB}{AF}=\frac{CD}{DB}=\frac{AE}{EC}=n$ ,then $\frac{MN}{NB}=n-1$ .

To prove this, I will work with a right triangle. The result will hold true for any triangle derived from it by a linear transformation (that is for any triangle) since linear transformation does not affect ratios of lengths on the same straight line.

Assume $AC=AB=n+1$ . Then equations of the internal lines will be:

$y=nx$

$y=1-\frac{1}{n+1}x$

$y=n+1-\frac{n+1}{n}x$

The $x$ -coordinate of $M$ will come from the intersection of the first two and it will be

$x_1=\frac{n+1}{n^2+n+1}$

The $x$ -coordinate of $N$ will come from the intersection of the last two lines and it will be

$x_2=\frac{n^2(n+1)}{n^2+n+1}$

The difference between them will need to be divided by the difference between $x_2$ and $n+1$ .

$\frac{\frac{(n^2-1)(n+1)}{n^2+n+1}}{\frac{(n+1)(n^2+n+1)-n^2(n+1)}{n^2+n+1}}=\frac{(n-1)(n+1)^2}{n^2+2n+1}=n-1$

The ratio has to be the same for the $y$ -coordinates and for the lengths themselves, as they all refer to the same straight line and are non-zero.

For $n=17$ the result is $16=\frac{16}{1}$ , so the answer is $16+1=17$ .