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...") |
m (→Problem) |
||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | The angle at <math>A</math> is the smallest angle of triangle <math> | + | 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== | ||
{{solution}} | {{solution}} | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1997|num-b=1|num-a=3}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 23:58, 28 March 2024
Problem
The angle at is the smallest angle of triangle
. The points
and
divide the circumcircle of the triangle into two arcs. Let
be an interior point of the arc between
and
which does not contain
. The perpendicular bisectors of
and
meet the line
and
and
, respectively. The lines
and
meet at
. Show that.
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 |