Difference between revisions of "1997 IMO Problems/Problem 2"

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==Problem==
 
==Problem==
  
The angle at <math>A</math> is the smallest angle of triangle <math>ABD</math>. The points <math>B</math> and <math>C</math> divide the circumcircle of the triangle into two arcs.  Let <math>U</math> be an interior point of the arc between <math>B</math> and <math>C</math> which does not contain <math>A</math>.  The perpendicular bisectors of <math>AB</math> and <math>AC</math> meet the line <math>AU</math> and <math>V</math> and <math>W</math>, respectively.  The lines <math>BV</math> and <math>CW</math> meet at <math>T</math>.  Show that.
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The angle at <math>A</math> is the smallest angle of triangle <math>ABC</math>. The points <math>B</math> and <math>C</math> divide the circumcircle of the triangle into two arcs.  Let <math>U</math> be an interior point of the arc between <math>B</math> and <math>C</math> which does not contain <math>A</math>.  The perpendicular bisectors of <math>AB</math> and <math>AC</math> meet the line <math>AU</math> and <math>V</math> and <math>W</math>, respectively.  The lines <math>BV</math> and <math>CW</math> meet at <math>T</math>.  Show that.
  
 
<math>AU=TB+TC</math>
 
<math>AU=TB+TC</math>
 
  
 
==Solution==
 
==Solution==

Latest revision as of 22:58, 28 March 2024

Problem

The angle at $A$ is the smallest angle of triangle $ABC$. The points $B$ and $C$ divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$. The perpendicular bisectors of $AB$ and $AC$ meet the line $AU$ and $V$ and $W$, respectively. The lines $BV$ and $CW$ meet at $T$. Show that.

$AU=TB+TC$

Solution

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See Also

1997 IMO (Problems) • Resources
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions