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Difference between revisions of "2016 OIM Problems/Problem 5"

(Created page with "== Problem == The circles <math>\Gamma_1</math> and <math>\Gamma_2</math> intersect at two different points <math>A</math> and <math>K</math>. The tangent common to <math>\Gam...")
 
 
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== Problem ==
 
== Problem ==
 
The circles <math>\Gamma_1</math> and <math>\Gamma_2</math> intersect at two different points <math>A</math> and <math>K</math>. The tangent
 
The circles <math>\Gamma_1</math> and <math>\Gamma_2</math> intersect at two different points <math>A</math> and <math>K</math>. The tangent
−
common to <math>\Gamma_1</math> and <math>\Gamma_2</math> closest to 4K<math> touches </math>\Gamma_1<math> at </math>B<math> and </math>\Gamma_2<math> at </math>C<math>.  
+
common to <math>\Gamma_1</math> and <math>\Gamma_2</math> closest to <math>K</math> touches <math>\Gamma_1</math> at <math>B</math> and <math>\Gamma_2</math> at <math>C</math>. Let <math>P</math> be the foot of the perpendicular from <math>B</math> on <math>AC</math>, and <math>Q</math> the foot of the perpendicular from <math>C</math> on <math>AB</math>. If <math>E</math> and <math>F</math> are the symmetrical points of <math>K</math> with respect to the lines <math>PQ</math> and <math>BC</math>, prove that the points <math>A, E</math> and <math>F</math> are collinear.
−
Let </math>P<math> be the foot of the perpendicular from </math>B<math> on </math>AC<math>, and </math>Q<math> the foot of the perpendicular from </math>C<math> on </math>AB<math>. If </math>E<math> and </math>F<math> are the symmetrical points of </math>K<math> with respect to the lines </math>PQ<math> and </math>BC<math>, prove that the points </math>A, E<math> and </math>F$ are collinear.
 
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Latest revision as of 14:01, 14 December 2023

Problem

The circles $\Gamma_1$ and $\Gamma_2$ intersect at two different points $A$ and $K$. The tangent common to $\Gamma_1$ and $\Gamma_2$ closest to $K$ touches $\Gamma_1$ at $B$ and $\Gamma_2$ at $C$. Let $P$ be the foot of the perpendicular from $B$ on $AC$, and $Q$ the foot of the perpendicular from $C$ on $AB$. If $E$ and $F$ are the symmetrical points of $K$ with respect to the lines $PQ$ and $BC$, prove that the points $A, E$ and $F$ are collinear.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

OIM Problems and Solutions

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