Difference between revisions of "2016 USAMO Problems/Problem 3"

Revision as of 13:33, 12 October 2022

Problem

Let $\triangle ABC$ be an acute triangle, and let $I_B, I_C,$ and $O$ denote its $B$ -excenter, $C$ -excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\overline{AC}$ such that $\angle ABY = \angle CBY$ and $\overline{BE}\perp\overline{AC}.$ Similarly, points $F$ and $Z$ are selected on $\overline{AB}$ such that $\angle ACZ = \angle BCZ$ and $\overline{CF}\perp\overline{AB}.$

Lines $I_B F$ and $I_C E$ meet at $P.$ Prove that $\overline{PO}$ and $\overline{YZ}$ are perpendicular.

Solution

This problem can be proved in the following two steps.

1. Let $I_A$ be the $A$ -excenter, then $I_A,O,$ and $P$ are colinear. This can be proved by the Trigonometric Form of Ceva's Theorem for $\triangle I_AI_BI_C.$

2. Show that $I_AY^2-I_AZ^2=OY^2-OZ^2,$ which implies $\overline{OI_A}\perp\overline{YZ}.$ This can be proved by multiple applications of the Pythagorean Thm.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Solution 2

We find point $T$ on line $YZ,$ we prove that $TY \perp OI_A$ and state that $P$ is the point $X(24)$ from ENCYCLOPEDIA OF TRIANGLE, therefore $P \in OI_A.$

Let $\omega$ be circumcircle of $\triangle ABC$ centered at $O.$ Let $Y_1,$ and $Z_1$ be crosspoints of $\omega$ and $BY,$ and $CZ,$ respectively. Let $T$ be crosspoint of $YZ$ and $Y_1 Z_1.$ In accordance the Pascal theorem for pentagon $AZ_1BCY_1,$ $AT$ is tangent to $\omega$ at $A.$

Let $I_A, I_B, I_C$ be $A, B,$ and $C$ -excenters of $\triangle ABC.$ Denote $a = BC, b = AC, c = AB, 2\alpha = \angle CAB, 2\beta = \angle ABC, 2\gamma = \angle ACB,$ $\psi = 90^\circ – \gamma + \beta, X = AI_A \cap \omega, X_1 = BC \cap AI_A,$ $I = BI_B \cap CI_C, U= YZ \cap AI_A, W = Y_1Z_1 \cap AI_A,$ $V$ is the foot ot perpendicular from $O$ to $AI_A.$

$I$ is ortocenter of $\triangle I_A I_B I_C$ and incenter of $\triangle ABC.$

$\omega$ is the Nine–point circle of $\triangle I_A I_B I_C.$

$Y_1$ is the midpoint of $II_B, Z_1$ is the midpoint of $II_C$ in accordance with property of Nine–point circle $\implies$ $Y_1Z_1 || I_B AI_C || VO, IW = AW \implies TW \perp AI.$ $\angle AXC = 180 ^\circ – 2\gamma – \alpha = 90 ^\circ – \gamma + \beta = \psi.$ $\angle TAI = \angle VOA = 2\beta + \alpha = 90 ^\circ – \gamma + \beta = \psi.$ $I_A X_1 = IX_1 = BX_1 = 2R \sin \alpha \implies$ $\cot \angle OI_A A = \frac {VI_A}{VO} = \frac {R \sin \psi + 2R \sin \alpha}{R \cos \psi} = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}.$ $CX = \frac {ab}{b+c} \implies \frac {AI}{IX}= \frac {AC}{CX}= \frac {b+c}{a} \implies AI = AX \frac {b+c}{a+b+c},$ $AW = \frac {AI}{2}, UW = AU – AW,$ In $\triangle ABC$ segment $YZ$ cross segment $AX \implies \frac {AU}{UX} = \frac {m + nk}{k+1},$ where $n = \frac {a}{b}, m = \frac{a}{c}, k=\frac {b}{c},$ $\frac {AU}{UX} = \frac{2a}{b+c} \implies AU = AX \cdot \frac {b+c}{2a +b +c}.$ $\frac {AU – AW}{AW} = \frac {b+c} {2a + b + c}.$

$\cot \angle UTW = \frac {TW}{UW} = \frac {AW \cdot \tan \psi}{AU – AW} = \tan \psi \cdot \frac {2a +b+c}{b+c} =$ $=\tan \psi \cdot \frac {2a}{b+c} + \tan \psi = \frac {2 \sin \alpha}{\sin \psi} \tan \psi + \tan \psi = \tan \psi + \frac{2 \sin\alpha}{\cos \psi}$ $\implies \angle UTW = \angle OI_AA .$ $TW \perp AI_A \implies TYZ \perp OI_A.$

Let $\triangle II_B I_C$ be the base triangle with orthocenter $I_A,$ center of Nine-points circle $O \implies OI_A$ be the Euler line of $\triangle II_B I_C.$

$\triangle ABC$ is orthic triangle of $\triangle II_B I_C,$

$\triangle DEF$ is orthic-of-orthic triangle.

$P$ is perspector of base triangle and orthic-of-orthic triangle.

Therefore $P$ is point $X(24)$ of ENCYCLOPEDIA OF TRIANGLE CENTERS which lies on Euler line of the base triangle. [[1]]

Claim $\frac {AU}{UX} = \frac {m + nk}{k + 1}.$ Proof $\frac {[AYZ]}{[ABC]} = \frac {AZ \cdot AY}{AB \cdot AC} = \frac {1}{(n + 1) \cdot (m+1)},$ $\frac {[BXZ]}{[ABC]} = \frac {BZ \cdot BX}{AB \cdot BC} = \frac {n}{(n + 1) \cdot (k+1)},$ $\frac {[CXY]}{[ABC]} = \frac {CY \cdot CX}{AC \cdot BC} = \frac {mk}{(m + 1) \cdot (k+1)},$ $\frac {[XYZ]}{[ABC]} = 1 - \frac {[AYZ]}{[ABC]} – \frac {[BXZ]}{[ABC]} – \frac {[CXY]}{[ABC]} = \frac {m+nk}{(m + 1) \cdot (k+1)\cdot (n+1)},$ $\frac {AU}{UX} = \frac {[AYZ]}{[XYZ]} = \frac {m + nk}{k + 1}.$

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 69: / Line 69: @@
 <cmath>\frac {AU}{UX} = \frac {[AYZ]}{[XYZ]} = \frac {m + nk}{k + 1}.</cmath>
-<i><b> Kimberling point X(24) </b></i>
-[[File:2016 USAMO 3g.png|450px|right]]
-Perspector of Triangle <math>ABC</math> and Orthic Triangle of the Orthic Triangle.
-<i><b>Theorem 1</b></i>
-Denote <math>T_0</math> obtuse or acute <math>\triangle ABC.</math> Let <math>T_0</math> be the base triangle, <math>T_1 = \triangle DEF</math> be Orthic triangle of <math>T_0, T_2 = \triangle UVW</math> be Orthic Triangle of the Orthic Triangle of <math>T_0</math>. Let <math>O</math> and <math>H</math> be the circumcenter and orthocenter of <math>T_0.</math>
-Then <math>\triangle T_0</math> and <math>\triangle T_2</math> are homothetic, the point <math>P,</math> center of this homothety lies on Euler line <math>OH</math> of <math>T_0.</math>
-The ratio of the homothety is <math>k = \frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C.</math>
-<i><b>Proof</b></i>
-WLOG, we use case <math>\angle A = \alpha > 90^\circ.</math>
-Let <math>B'</math> be reflection <math>H</math> in <math>DE.</math>
-In accordance with Claim, <math>\angle BVD = \angle HVE \implies B', V,</math> and <math>B</math> are collinear.
-Similarly, <math>C, W,</math> and <math>C',</math> were <math>C'</math> is reflection <math>H</math> in <math>DF,</math> are collinear.
-Denote  <math>\angle ABC = \beta = \angle CHD, \angle ACB = \gamma = \angle BHD \implies</math>
-<math>\angle HDF = \angle HDE = \angle DHB' = \angle DHC' = 180^\circ - \alpha.</math>
-<math>B'C' \perp HD, BC \perp HD \implies BC|| B'C'.</math>
-<math>OB = OC, HB' = HC', \angle BOC = \angle B'HC' = 360^\circ - 2 \alpha \implies OB ||HB', OC || HC' \implies</math>
-<math>\triangle HB'C' \sim \triangle OBC, BB', CC'</math> and <math>HO</math> are concurrent at point <math>P.</math>
-In accordance with Claim, <math>\angle HUF = \angle AUF \implies</math> points <math>H</math> and <math>P</math> are isogonal conjugate with respect <math>\triangle UVW.</math>
-<math>\angle HDE = \alpha - 90^\circ, \angle HCD = 90^\circ - \beta \implies</math>
-<math>HB' = 2 HD \sin (\alpha - 90^\circ) = - 2 CD \tan(90^\circ- \beta) \cos \alpha = - 2 AC \cos \gamma \frac {\cos \beta}{\sin \beta} \cos \alpha = - 4 OB \cos A \cos B \cos C.</math>
-<math>k = \frac {HB'}{OB} = \frac {HP}{OP}= - 4 \cos A \cos B \cos C \implies \frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C.</math>
-<i><b>Claim</b></i>
-[[File:2016 3 Lemma.png|400px|right]]
-Let <math>\triangle ABC</math> be an acute triangle, and let <math>AH, BD',</math> and <math>CD</math> denote its altitudes. Lines <math>DD'</math> and <math>BC</math> meet at <math>Q, HS \perp DD'.</math> Prove that <math>\angle BSH = \angle CSH.</math>
-<i><b>Proof</b></i>
-Let <math>\omega</math> be the circle <math>BCD'D</math> centered at <math>O (O</math> is midpoint <math>BC).</math>
-Let <math>\omega</math> meet <math>AH</math> at <math>P.</math>
-Let <math>\Omega</math> be the circle centered at <math>Q</math> with radius <math>QP.</math>
-Let <math>\Theta</math> be the circle with diameter <math>OQ.</math>
-We know that <math>OB = OP = OC = R, PH^2 = R^2 – OH^2 \implies</math>
-<math>QP^2 + R^2 = (QH+ HO)^2 \implies P \in \Theta,  \Omega \perp \omega.</math>
-Let <math>I_{\Omega}</math> be inversion with respect <math>\Omega, I_{\Omega}(B) = C.</math>
-Denote <math>I_{\Omega}(D) = D',  I_{\Omega}(S) = S',</math>
-<math>QH \cdot QO = QP^2 \implies I_{\Omega}(H) = O.</math>
-<math>HS \perp DD' \implies S'O \perp BC \implies BS' = CS' \implies \angle OCS' = \angle OBS'.</math>
-<math>\angle QSB = \angle QCS' = \angle OCS' = \angle OBS' = \angle CSS'.</math>
-<math>\angle BSH = 90 ^\circ  – \angle QSB = 90 ^\circ  – \angle CSS' =\angle CSH.</math>
-<i><b>Theorem 2</b></i>
-Let <math>T_0 = \triangle ABC</math> be the base triangle, <math>T_1 = \triangle DEF</math> be orthic triangle of <math>T_0, T_2 = \triangle KLM</math> be Kosnita triangle. Then <math>\triangle T_1</math> and <math>\triangle T_2</math> are homothetic, the point <math>P,</math> center of this homothety lies on Euler line of <math>T_0,</math> the ratio of the homothety is <math>k =  \frac {\vec PH}{\vec OP} = 4 \cos A \cos B \cos C.</math>
-We recall that vertex of Kosnita triangle are:  <math>K</math> is the circumcenter of <math>\triangle OBC, L</math> is the circumcenter of <math>\triangle OAB, M</math> is the circumcenter of <math>\triangle OAC,</math> where <math>O</math> is circumcenter of <math>T_0.</math>
 '''vladimir.shelomovskii@gmail.com, vvsss'''
 ==See also==
 {{USAMO newbox|year=2016|num-b=2|num-a=4}}

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Difference between revisions of "2016 USAMO Problems/Problem 3"

Revision as of 13:33, 12 October 2022

Contents

Problem

Solution

Solution 2

See also