Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 IMO Problems/Problem 6"

m (Removed Spam)
(Tag: Replaced)
(→‎Solution)
Line 10: Line 10:
 
The equation for line <math>AB</math> is <math>\frac{x-x_a}{y-y_a}=\frac{x-x_b}{y-y_b}</math>
 
The equation for line <math>AB</math> is <math>\frac{x-x_a}{y-y_a}=\frac{x-x_b}{y-y_b}</math>
  
−
Through a little bash, (0,1) reflects to <math>(\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}</math>. <math>(x_a-x_b)^2+(y_a-y_b)^2</math> simplifies to <math>-(2x_ax_b+2y_ay_b)</math>. The terms for the other 2 are symmetric.
+
Through a little bash, (0,1) reflects to <math>\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)</math>. <math>(x_a-x_b)^2+(y_a-y_b)^2</math> simplifies to <math>-(2x_ax_b+2y_ay_b)</math>. The terms for the other 2 are symmetric.
 
The intersection point must reflect to itself, and the equation is <math>(\frac{x_by_a-x_ay_b-x_a+x_b}{y_b-y_a}, 1)</math>.
 
The intersection point must reflect to itself, and the equation is <math>(\frac{x_by_a-x_ay_b-x_a+x_b}{y_b-y_a}, 1)</math>.
  

Revision as of 10:05, 31 December 2022

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 $\Gamma$.

Solution

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 $(\frac{x_by_a-x_ay_b-x_a+x_b}{y_b-y_a}, 1)$.

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.

-Trex4days

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_IMO_Problems/Problem_6&oldid=185332"