Difference between revisions of "2010 IMO Problems/Problem 2"
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Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>. | Observation 1. D is the midpoint of arc <math>BDC</math> because it lies on angle bisector <math>AI</math>. | ||
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Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well. | Observation 2. <math>AI</math> bisects <math>\angle{FAE}</math> as well. | ||
Revision as of 12:01, 17 May 2015
Problem
Given a triangle 
, with 
as its incenter and 
as its circumcircle, 
intersects again at 
. Let 
be a point on arc 
, and 
a point on the segment 
, such that 
. If 
is the midpoint of 
, prove that the intersection of lines 
and 
lies on .
Authors: Tai Wai Ming and Wang Chongli, Hong Kong
Solution
Note that it suffices to prove alternatively that if meets the circle again at 
and 
meets at , then is the midpoint of .
Observation 1. D is the midpoint of arc because it lies on angle bisector .
Observation 2. bisects 
as well.
Key Lemma. Triangles 
and 
are similar.
Proof. Because triangles 
and 
are similar by AA Similarity (for 
and 
both intercept equally sized arcs), we have 
. But we know that triangle 
is isosceles (hint: prove 
), and so 
. Hence, by SAS Similarity, triangles and are similar, as desired.
Observation 3. As a result, we have 
.
Observation 4. 
.
Observation 5. If 
and intersect at 
, then 
is cyclic.
Observation 6. Because 
, we have 
.
Observation 7. 
is a parallelogram, so its diagonals bisect each other, so is the midpoint of 
, as desired.
See Also
| 2010 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
