Difference between revisions of "2001 USAMO Problems/Problem 2"
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Now consider the homothety that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>. | Now consider the homothety that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>. | ||
− | + | <asy> | |
+ | pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); </asy> | ||
By Menelaus' Theorem on <math>\triangle ACD_2</math> with segment <math>\overline{BE_2}</math>, it follows that <math>\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math> | By Menelaus' Theorem on <math>\triangle ACD_2</math> with segment <math>\overline{BE_2}</math>, it follows that <math>\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math> | ||
Revision as of 21:21, 23 September 2015
Contents
[hide]Problem
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 .
Solution
Solution 1
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 .
Solution 2
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 .
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.