2011 IMO Problems/Problem 6

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$.


Without loss of generality, let $\Gamma$ be the unit circle and let $\ell$ be the line $y=1$.

Denote the coordinates of $A,B,C$ by $(x_a, y_a)$ and similarly for B and C.

We get $x_a^2+y_a^2=1$, ...

The equation for line $AB$ is $\frac{x-x_a}{y-y_a}=\frac{x-x_b}{y-y_b}$

Through a little bash, (0,1) reflects to $\left(\frac{(-x_a+x_b+x_by_a)(y_b-_a)}{(x_a-x_b)^2+(y_a-y_b)^2}, \frac{(x_by_a-x_ay_b)(x_a-x_b)+(y_a-y_b)^2}{(x_a-x_b)^2+(y_a-y_b)^2}\right)$. $(x_a-x_b)^2+(y_a-y_b)^2$ simplifies to $-(2x_ax_b+2y_ay_b)$. The terms for the other 2 are symmetric. The intersection point must reflect to itself, and the equation is $\left(\frac{x_by_a-x_ay_b-x_a+x_b}{y_b-y_a}, 1\right)$.

It is trivial to find the intersections of a,b and their perpendicular bisectors, so this is left to the reader as an exercise.

Regardless, the circumcenter and an intersection of the circles are collinear with (0,0), so it is a tangency.


See Also

2011 IMO (Problems) • Resources
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