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

Revision as of 11:12, 11 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.

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

Kimberling point X(24)

$1.$ Perspector of Triangle $ABC$ and Orthic Triangle of the Orthic Triangle. Denote $T_0$ obtuse or acute $\triangle ABC.$ Let $T_0$ be the base triangle, $T_1 = \triangle DEF$ be Orthic triangle of $T_0, T_2 = \triangle UVW$ be Orthic Triangle of the Orthic Triangle of $T_0$ . Let $O$ and $H$ be the circumcenter and orthocenter of $T_0.$

Then $\triangle T_0$ and $\triangle T_2$ are homothetic, the point $P,$ center of this homothety lies on Euler line $OH$ of $T_0.$

The ratio of the homothety is $k = \frac {\vec PH}{\vec OP}= 4 \cos A \cos B \cos C.$

Proof

WLOG, we use case $\angle A = \alpha > 90^\circ.$ Let $B'$ be reflection $H$ in $DE.$ In accordance with Claim, $\angle BVD = \angle HVE \implies B', V,$ and $B$ are collinear. Similarly, $C', W,$ and $C,$ were $C'$ is reflection $H$ in $DF,$ are collinear.

Denote $\angle ABC = \beta \angle CHD, \angle ACB = \gamma = \angle BHD \implies$

$\angle HDF = \angle HDE = \angle DHB' = \angle DHC' = 180^\circ - \alpha.$

$B'C' \perp HD, BC \perp HD \implies BC|| B'C' \implies AB ||HC', AC || HB' \implies$ $\triangle HB'C' \sim \triangle OBC, BB', CC'$ and $HO$ are collinear.

$HB' = H C' \implies \angle OCB = \angle OBC = \angle HB' C' = \angle HB' C' = \alpha - 90^\circ.$ $\angle HCD = 90^\circ - \beta \implies HB' = 4 HD \sin (\alpha - 90^\circ) \sin \alpha =$ $4 CD \tan(90^\circ- \beta) \cos \alpha \sin \alpha = 4 AC \sin (90^\circ - \gamma) = 4 BC \cos \alpha \cos \beta \cos \gamma \implies$ $\frac {\vec PH}{\vec OP}= 4 \cos A \cos B \cos C.$

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 75: / Line 75: @@
 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'  \implies AB ||HC', AC || HB' \implies</math>
+<math>\triangle HB'C' \sim \triangle OBC, BB', CC'</math> and <math>HO</math> are collinear.
+<math>HB' = H C' \implies \angle OCB =  \angle OBC = \angle HB' C' =  \angle HB' C' = \alpha - 90^\circ.</math>
+<math>\angle HCD = 90^\circ - \beta \implies HB' = 4 HD \sin (\alpha - 90^\circ)  \sin \alpha =</math>
+<math>4 CD \tan(90^\circ- \beta) \cos \alpha \sin \alpha = 4 AC \sin (90^\circ - \gamma) = 4 BC \cos \alpha \cos \beta \cos \gamma \implies</math>
+<math>\frac {\vec PH}{\vec OP}= 4 \cos A \cos B \cos C.</math>
 '''vladimir.shelomovskii@gmail.com, vvsss'''

2016 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search

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

Revision as of 11:12, 11 October 2022

Contents

Problem

Solution

Solution 2

See also