Based on a Putnam problem.
ABC is a triangle. A', B', C' are points on BC, AC and AB respectively such that AA' bisects BB', BB' bisects CC' and CC' bisects AA'.
Show that A', B', C' divide the sides BC, CA, AB in the same ratio.
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The goal of this post is Solution 2).
Solution 1)
Do an affine transformation to make the triangle a right isosceles triangle and use co-ordinate geometry.
Or make it an equilateral triangle and use some symmetry/continuity arguments (haven't tried it mysef).
Solution 2)
A physics based solution!
(Might help to draw a figure): Suppose BA'/A'C = r and CB'/B'A = s.
Put a mass of 1 at B and a mass of r at C. Center of mass of B and C is at A'.
Put a mass of rs at A. Center of mass of A and C is B'.
Thus the center of mass of A,B,C is the intersection point of AA' and BB', which is the midpoint of BB'.
Thus we must have that $r + rs = 1 \implies r = \frac{1}{1+s}$
SImilarly if AC'/C'B = t, we get
$s + st = 1$ and $t + rt = 1$
Now $f(x) = \frac{1}{1+x}$ is decreasing. Thus if $r \leq s \leq t$, $f(r) \geq f(s) \geq f(t)$ and thus $r \geq s$ and so $r = s$.
This implies $r = s = t$.
Turns out $r = s = t = \frac{1}{\varphi}$ where $\varphi$ is the golden ratio.
The putnam problem was to find the area of the small triangle formed by the midpoints (given area of ABC). This should now be easy once we know the ratio, as we can affine transform to any "nice" triangle to make the compute easier.
