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Difference between revisions of "2009 OIM Problems/Problem 3"

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== Problem ==
 
== Problem ==
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Let <math>C_1</math> and <math>C_2</math> be two circles with centers <math>O_1</math> and <math>O_2</math> with the same radius, which intersect at <math>A</math> and <math>B</math>. Let <math>P</math> be a point on the arc <math>AB</math> of <math>C_2</math> that is inside <math>C_1</math>. Line <math>AP</math> intersects <math>C_1</math> at <math>C</math>, line <math>CB</math> intersects <math>C_2</math> at <math>D</math> and the bisector of <math>\angle CAD</math> intersects <math>C_1</math> at <math>E</math> and <math>C_2</math> at <math>L</math>. Let <math>F</math> be the point symmetrical to <math>D</math> with respect to the midpoint of <math>PE</math>. Show that there exists a point <math>X</math> satisfying <math>\angle ∠XFL = \angle XCD = 30^{^circ}</math> and <math>CX = O_1O_2</math>.
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Let <math>C_1</math> and <math>C_2</math> be two circles with centers <math>O_1</math> and <math>O_2</math> with the same radius, which intersect at <math>A</math> and <math>B</math>. Let <math>P</math> be a point on the arc <math>AB</math> of <math>C_2</math> that is inside <math>C_1</math>. Line <math>AP</math> intersects <math>C_1</math> at <math>C</math>, line <math>CB</math> intersects <math>C_2</math> at <math>D</math> and the bisector of <math>\angle CAD</math> intersects <math>C_1</math> at <math>E</math> and <math>C_2</math> at <math>L</math>. Let <math>F</math> be the point symmetrical to <math>D</math> with respect to the midpoint of <math>PE</math>. Show that there exists a point <math>X</math> satisfying <math>\angle XFL = \angle XCD = 30^{\circ}</math> and <math>CX = O_1O_2</math>.
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Latest revision as of 15:21, 14 December 2023

Problem

Let $C_1$ and $C_2$ be two circles with centers $O_1$ and $O_2$ with the same radius, which intersect at $A$ and $B$. Let $P$ be a point on the arc $AB$ of $C_2$ that is inside $C_1$. Line $AP$ intersects $C_1$ at $C$, line $CB$ intersects $C_2$ at $D$ and the bisector of $\angle CAD$ intersects $C_1$ at $E$ and $C_2$ at $L$. Let $F$ be the point symmetrical to $D$ with respect to the midpoint of $PE$. Show that there exists a point $X$ satisfying $\angle XFL = \angle XCD = 30^{\circ}$ and $CX = O_1O_2$.

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

Solution

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See also

OIM Problems and Solutions

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