Difference between revisions of "2016 USAMO Problems/Problem 3"
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Lines <math>I_B F</math> and <math>I_C E</math> meet at <math>P.</math> Prove that <math>\overline{PO}</math> and <math>\overline{YZ}</math> are perpendicular. | Lines <math>I_B F</math> and <math>I_C E</math> meet at <math>P.</math> Prove that <math>\overline{PO}</math> and <math>\overline{YZ}</math> are perpendicular. | ||
==Solution== | ==Solution== | ||
− | {{ | + | This problem can be proved in the following two steps. |
+ | |||
+ | 1. Let <math>I_A</math> be the <math>A</math>-excenter, then <math>I_A,O,</math> and <math>P</math> are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for <math>\triangle I_AI_BI_C.</math> | ||
+ | |||
+ | 2. Show that <math>I_AY^2-I_AZ^2=OY^2-OZ^2,</math> which implies <math>\overline{OI_A}\perp\overline{YZ}.</math> This can be proved by multiple applications of the Pythagorean Thm. | ||
+ | |||
+ | ==Solution 2== | ||
+ | [[File:2016 USAMO 3a.png|300px|right]] | ||
+ | We find point <math>T</math> on line <math>YZ,</math> we prove that <math>TY \perp OI_A</math> and state that <math>P</math> is the point <math>X(24)</math> from ENCYCLOPEDIA OF TRIANGLE, therefore <math>P \in OI_A.</math> | ||
+ | |||
+ | Let <math>\omega</math> be circumcircle of <math>\triangle ABC</math> centered at <math>O.</math> | ||
+ | Let <math>Y_1,</math> and <math>Z_1</math> be crosspoints of <math>\omega</math> and <math>BY,</math> and <math>CZ,</math> respectively. | ||
+ | Let <math>T</math> be crosspoint of <math>YZ</math> and <math>Y_1 Z_1.</math> | ||
+ | In accordance the Pascal theorem for pentagon <math>AZ_1BCY_1,</math> <math>AT</math> is tangent to <math>\omega</math> at <math>A.</math> | ||
+ | |||
+ | [[File:2016 USAMO 3b.png|500px|right]] | ||
+ | Let <math>I_A, I_B, I_C</math> be <math>A, B,</math> and <math>C</math>-excenters of <math>\triangle ABC.</math> | ||
+ | Denote <cmath>a = BC, b = AC, c = AB, 2\alpha = \angle CAB, 2\beta = \angle ABC, 2\gamma = \angle ACB,</cmath> | ||
+ | <cmath>\psi = 90^\circ – \gamma + \beta, X = AI_A \cap \omega, X_1 = BC \cap AI_A,</cmath> | ||
+ | <cmath>I = BI_B \cap CI_C, U= YZ \cap AI_A, W = Y_1Z_1 \cap AI_A,</cmath> | ||
+ | <math>V</math> is the foot ot perpendicular from <math>O</math> to <math>AI_A.</math> | ||
+ | |||
+ | <math>I</math> is ortocenter of <math>\triangle I_A I_B I_C</math> and incenter of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>\omega</math> is the Nine–point circle of <math>\triangle I_A I_B I_C.</math> | ||
+ | |||
+ | <math>Y_1</math> is the midpoint of <math>II_B, Z_1</math> is the midpoint of <math>II_C</math> in accordance with property of Nine–point circle <math>\implies</math> | ||
+ | <cmath>Y_1Z_1 || I_B AI_C || VO, IW = AW \implies TW \perp AI.</cmath> | ||
+ | <cmath>\angle AXC = 180 ^\circ – 2\gamma – \alpha = 90 ^\circ – \gamma + \beta = \psi.</cmath> | ||
+ | <cmath>\angle TAI = \angle VOA = 2\beta + \alpha = 90 ^\circ – \gamma + \beta = \psi.</cmath> | ||
+ | <cmath>I_A X_1 = IX_1 = BX_1 = 2R \sin \alpha \implies</cmath> | ||
+ | <cmath>\cot \angle OI_A A = \frac {VI_A}{VO} = \frac {R \sin \psi + 2R \sin \alpha}{R \cos \psi} = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}.</cmath> | ||
+ | <cmath>CX = \frac {ab}{b+c} \implies \frac {AI}{IX}= \frac {AC}{CX}= \frac {b+c}{a} \implies AI = AX \frac {b+c}{a+b+c},</cmath> | ||
+ | <cmath>AW = \frac {AI}{2}, UW = AU – AW,</cmath> | ||
+ | In <math>\triangle ABC</math> segment <math>YZ</math> cross segment <math>AX \implies \frac {AU}{UX} = \frac {m + nk}{k+1},</math> where <math>n = \frac {a}{b}, m = \frac{a}{c}, k=\frac {b}{c},</math> | ||
+ | <cmath>\frac {AU}{UX} = \frac{2a}{b+c} \implies AU = AX \cdot \frac {b+c}{2a +b +c}.</cmath> | ||
+ | <cmath>\frac {AU – AW}{AW} = \frac {b+c} {2a + b + c}.</cmath> | ||
+ | [[File:2016 USAMO 3c.png|400px|right]] | ||
+ | <cmath>\cot \angle UTW = \frac {TW}{UW} = \frac {AW \cdot \tan \psi}{AU – AW} = \tan \psi \cdot \frac {2a +b+c}{b+c} =</cmath> | ||
+ | <cmath>=\tan \psi \cdot \frac {2a}{b+c} + \tan \psi = \frac {2 \sin \alpha}{\sin \psi} \tan \psi + \tan \psi = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}</cmath> | ||
+ | <cmath>\implies \angle UTW = \angle OI_AA .</cmath> | ||
+ | <cmath>TW \perp AI_A \implies TYZ \perp OI_A.</cmath> | ||
+ | |||
+ | Let <math>\triangle II_B I_C</math> be the base triangle with orthocenter <math>I_A,</math> center of Nine-points circle <math>O \implies OI_A</math> be the Euler line of <math>\triangle II_B I_C.</math> | ||
+ | |||
+ | <math>\triangle ABC</math> is orthic triangle of <math>\triangle II_B I_C,</math> | ||
+ | |||
+ | <math>\triangle DEF</math> is orthic-of-orthic triangle. | ||
+ | |||
+ | <math>P</math> is perspector of base triangle and orthic-of-orthic triangle. | ||
+ | |||
+ | Therefore <math>P</math> is point <math>X(24)</math> of ENCYCLOPEDIA OF TRIANGLE CENTERS which lies on Euler line of the base triangle. | ||
+ | [[https://artofproblemsolving.com/wiki/index.php/Kimberling%E2%80%99s_point_X(24)]] | ||
+ | [[File:2016 USAMO 3d.png|300px|right]] | ||
+ | |||
+ | <i><b>Claim</b></i> <cmath>\frac {AU}{UX} = \frac {m + nk}{k + 1}.</cmath> | ||
+ | <i><b>Proof</b></i> | ||
+ | <cmath>\frac {[AYZ]}{[ABC]} = \frac {AZ \cdot AY}{AB \cdot AC} = \frac {1}{(n + 1) \cdot (m+1)}, </cmath> | ||
+ | <cmath>\frac {[BXZ]}{[ABC]} = \frac {BZ \cdot BX}{AB \cdot BC} = \frac {n}{(n + 1) \cdot (k+1)}, </cmath> | ||
+ | <cmath>\frac {[CXY]}{[ABC]} = \frac {CY \cdot CX}{AC \cdot BC} = \frac {mk}{(m + 1) \cdot (k+1)},</cmath> | ||
+ | <cmath>\frac {[XYZ]}{[ABC]} = 1 - \frac {[AYZ]}{[ABC]} – \frac {[BXZ]}{[ABC]} – \frac {[CXY]}{[ABC]} = \frac {m+nk}{(m + 1) \cdot (k+1)\cdot (n+1)}, </cmath> | ||
+ | <cmath>\frac {AU}{UX} = \frac {[AYZ]}{[XYZ]} = \frac {m + nk}{k + 1}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | |||
==See also== | ==See also== | ||
{{USAMO newbox|year=2016|num-b=2|num-a=4}} | {{USAMO newbox|year=2016|num-b=2|num-a=4}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 21:36, 17 October 2022
Contents
Problem
Let be an acute triangle, and let
and
denote its
-excenter,
-excenter, and circumcenter, respectively. Points
and
are selected on
such that
and
Similarly, points
and
are selected on
such that
and
Lines and
meet at
Prove that
and
are perpendicular.
Solution
This problem can be proved in the following two steps.
1. Let be the
-excenter, then
and
are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for
2. Show that which implies
This can be proved by multiple applications of the Pythagorean Thm.
Solution 2
We find point on line
we prove that
and state that
is the point
from ENCYCLOPEDIA OF TRIANGLE, therefore
Let be circumcircle of
centered at
Let
and
be crosspoints of
and
and
respectively.
Let
be crosspoint of
and
In accordance the Pascal theorem for pentagon
is tangent to
at
Let be
and
-excenters of
Denote
is the foot ot perpendicular from
to
is ortocenter of
and incenter of
is the Nine–point circle of
is the midpoint of
is the midpoint of
in accordance with property of Nine–point circle
In
segment
cross segment
where
Let be the base triangle with orthocenter
center of Nine-points circle
be the Euler line of
is orthic triangle of
is orthic-of-orthic triangle.
is perspector of base triangle and orthic-of-orthic triangle.
Therefore is point
of ENCYCLOPEDIA OF TRIANGLE CENTERS which lies on Euler line of the base triangle.
[[1]]
Claim
Proof
vladimir.shelomovskii@gmail.com, vvsss
See also
2016 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.