Difference between revisions of "2001 USAMO Problems/Problem 2"
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Let <math>ABC</math> be a [[triangle]] and let <math>\omega</math> be its [[incircle]]. Denote by <math>D_1</math> and <math>E_1</math> the points where <math>\omega</math> is tangent to sides <math>BC</math> and <math>AC</math>, respectively. Denote by <math>D_2</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = AE_1</math>, and denote by <math>P</math> the point of intersection of segments <math>AD_2</math> and <math>BE_2</math>. Circle <math>\omega</math> intersects segment <math>AD_2</math> at two points, the closer of which to the vertex <math>A</math> is denoted by <math>Q</math>. Prove that <math>AQ = D_2P</math>. | Let <math>ABC</math> be a [[triangle]] and let <math>\omega</math> be its [[incircle]]. Denote by <math>D_1</math> and <math>E_1</math> the points where <math>\omega</math> is tangent to sides <math>BC</math> and <math>AC</math>, respectively. Denote by <math>D_2</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = AE_1</math>, and denote by <math>P</math> the point of intersection of segments <math>AD_2</math> and <math>BE_2</math>. Circle <math>\omega</math> intersects segment <math>AD_2</math> at two points, the closer of which to the vertex <math>A</math> is denoted by <math>Q</math>. Prove that <math>AQ = D_2P</math>. | ||
− | |||
− | |||
− | |||
− | + | 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 == | == See also == |
Revision as of 05:09, 22 September 2015
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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