The racer at A is running counter-clockwise around the inside of the outer purple circle, with a “shadow racer” B following behind, related by the three chords that are tangent to the inner orange circle. Does the shadow B ever catch up with A?

Scroll down for a solution to this problem.

## Solution

No, B never catches up.

If B catches up, there is a triangle with inradius r, circumradius R, distance between centres d, so by Euler d² = R(R-r). But by continuity there is a triangle on XY with inradius r, so the incentre is also at distance d, but then the incircle is the same as the orange circle.

## One reply on “The hippodrome shadow”

The answer is: A = B if and only if in the figure of M. Arcus R²-d² = 2Rr.

So A = B all the time or never!

This is the consequence of the theorem proved in the figure below.

See:

https://en.wikipedia.org/wiki/Modern_triangle_geometry#Poristic_triangles

with link [16]