Difference between revisions of "2011 USAMO Problems/Problem 3"
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| − | In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3\angle D</math>, <math>\angle C = 3\angle F</math>, and <math>\angle E = 3\angle B</math>. Furthermore <math>AB=DE</math>, <math>BC=EF</math>, and <math>CD=FA</math>. Prove that diagonals <math>\overline{AD}, < | + | In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3\angle D</math>, <math>\angle C = 3\angle F</math>, and <math>\angle E = 3\angle B</math>. Furthermore <math>AB=DE</math>, <math>BC=EF</math>, and <math>CD=FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent. |
==Solution== | ==Solution== | ||
==See Also== | ==See Also== | ||
{{USAMO newbox|year=2011|num-b=2|num-a=4}} | {{USAMO newbox|year=2011|num-b=2|num-a=4}} | ||
Revision as of 16:56, 15 May 2011
In hexagon 
, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy 
, 
, and 
. Furthermore 
, 
, and 
. Prove that diagonals 
, 
, and 
are concurrent.
Solution
See Also
| 2011 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
