Difference between revisions of "2001 USAMO Problems/Problem 2"

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(Problem)
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== Problem ==
 
== Problem ==
 
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 06:16, 22 September 2015

Problem

Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$.

See also

2001 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions

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