2001 USAMO Problems/Problem 2

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 .

we use barycentric coordinates. It’s obvious that un-normalized, D1 = (0 : s − c : s − b) ⇒ D2 = (0 : s − b : s − c), so we get a normalized D2 =

0, s−b a , s−c a � . Similarly, E2 =

s−a b , 0, s−c b � . Now we obtain the points P =

s−a s , s−b s , s−c s � by intersecting the lines AD2 : (s−c)y = (s−b)z and BE2 : (s − c)x = (s − a)z. Let Q0 be such that AQ0 = P D2. It’s obvious that Q0 y + Py = Ay + D2y, so we find that Q0 y = s−b a − s−b s = (s−a)(s−b) sa . Also, since it lies on the line AD2, we get that Q0 z = s−c s−b · Q0 y = (s−a)(s−c) sa . Hence, Q 0 x = 1 − ((s − b) + (s − c))(s − a) sa = sa − a(s − a) sa = a s Hence, Q 0 = � a s , (s − a)(s − b) sa , (s − a)(s − c) sa � 16 A B C I D1 E1 D2 E2 Q P Figure 5.1: USAMO 2001/2 Let I =

a 2s , b 2s , c 2s � . We claim that, in fact, I is the midpoint of Q0D1. Indeed, 1 2 � 0 + a s � = a 2s 1 2 � (s − a)(s − b) sa + s − c a � = (s − a)(s − b) + s(s − c) 2sa = ab 2sa = b 2s 1 2 � (s − a)(s − c) sa + s − b a � = c 2s Implying that Q lies on the circle; in particular, diametrically opposite from D1, so it is the closer of the two points. Hence, Q = Q0 , so we’re done.

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

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