2001 USAMO Problems/Problem 2
Let be a triangle and let be its incircle. Denote by and the points where is tangent to sides and , respectively. Denote by and the points on sides and , respectively, such that and , and denote by the point of intersection of segments and . Circle intersects segment at two points, the closer of which to the vertex is denoted by . Prove that .
It is well known that the excircle opposite is tangent to at the point . (Proof: let the points of tangency of the excircle with the lines be respectively. Then . It follows that , and , so .)
Now consider the homothety that carries the incircle of to its excircle. The homothety also carries to (since are collinear), and carries the tangency points to . It follows that .
By Menelaus' Theorem on with segment , it follows that . It easily follows that .
The key observation is the following lemma.
Lemma: Segment is a diameter of circle .
Proof: Let be the center of circle , i.e., is the incenter of triangle . Extend segment through to intersect circle again at , and extend segment through to intersect segment at . We show that , which in turn implies that , that is, is a diameter of .
Let be the line tangent to circle at , and let intersect the segments and at and , respectively. Then is an excircle of triangle . Let denote the dilation with its center at and ratio . Since and , . Hence . Thus , , and . It also follows that an excircle of triangle is tangent to the side at .
It is well known that We compute . Let and denote the points of tangency of circle with rays and , respectively. Then by equal tangents, , , and . Hence It follows that Combining these two equations yields . Thus that is, , as desired.
Now we prove our main result. Let and be the respective midpoints of segments and . Then is also the midpoint of segment , from which it follows that is the midline of triangle . Hence and . Similarly, we can prove that .
2001usamo2-2.png Let be the centroid of triangle . Thus segments and intersect at . Define transformation as the dilation with its center at and ratio . Then and . Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since and , maps lines and to lines and , respectively. It also follows that and or This yields as desired.
Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle .
Here is a rather nice solution using barycentric coordinates:
Let be , be , and be . Let the side lengths of the triangle be and the semi-perimeter .
Therefore, and Clearly then, the non-normalized coordinates of
Normalizing, we have that
Now, we find the point inside the triangle on the line such that . It is then sufficient to show that this point lies on the incircle.
is the fraction of the way "up" the line segment from to . Thus, we are looking for the point that is of the way "down" the line segment from to , or, the fraction of the way "up".
Thus, has normalized -coordinate .
As the line has equation , it can easily be found that
Recalling that the equation of the incircle is We must show that this equation is true for 's values of .
Plugging in our values, this means showing that Dividing by , this is just
Plugging in the value of The first bracket is just and the second bracket is Dividing everything by gives which is , as desired.
As lies on the incircle and , , and our proof is complete.
We again use Barycentric coordinates. As before, let be , be , and be . Also
Now, consider a point for which . Then working component-vise, we get from which we can easily get the coordinates of as;
It suffices to show that .
Let be the incenter of triangle . We claim that is the midpoint of . Indeed,
Hence the claim has been proved.
Since is the center of and the midpoint of , thus is the point diametrically opposite to , and hence it lies on and closer to off the 2 points. Thus as desired.
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