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

(Created page with "==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 triang...")
 
(Solution)
Line 8: Line 8:
 
==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
 +
 +
==See Also==
 +
 +
{{IMO box|year=1997|num-b=1|num-a=3}}
 +
[[Category:Olympiad Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Revision as of 00:00, 17 November 2023

Problem

The angle at $A$ is the smallest angle of triangle $ABD$. 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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

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