Difference between revisions of "2001 USAMO Problems/Problem 2"
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== Solution == | == Solution == | ||
− | {{ | + | It is well known that the excircle opposite <math>A</math> is tangent to <math>\overline{BC}</math> at the point <math>D_2</math>. (Proof: let the points of tangency of the excircle with the lines <math>BC, AB, AC</math> be <math>D_3, F,G</math> respectively. Then <math>AB+BD_3=AB + BF=AF = AG = AC + AG=AC + CD_3</math>. It follows that <math>2CD_3 = AB + BC - AC</math>, and <math>CD_3 = s-b = BD_1 = CD_2</math>, so <math>D_3 \equiv D_2</math>.) |
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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>. | ||
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+ | <center><asy></asy></center> | ||
+ | |||
+ | By [[Menelaus' Theorem]] 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> | ||
== See also == | == See also == |
Revision as of 20:22, 24 September 2008
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
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
.
![[asy][/asy]](http://latex.artofproblemsolving.com/0/6/1/061f07da08b2b0c2554b2543f4a362d2aedbcc4b.png)
By Menelaus' Theorem it follows that . It easily follows that
.
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 |