Isogonal conjugate midpoint

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Isogonal conjugate midpoint

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Source: otis mock aime G44 generalization

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qwerty123456asdfgzxcvb

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qwerty123456asdfgzxcvb

#1 h Oct 19, 2024, 6:13 AM • 3 Y

Y by bryanguo, peppapig_, centslordm

Consider the set of all points $P$ such that their isogonal conjugate $Q$ in $\triangle ABC$ has $PQ$ parallel to $BC$ .

Prove that the midpoint of all $P,Q$ in this set lies on a rectangular hyperbola, and locate its center.

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DottedCaculator

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DottedCaculator

#2 h Oct 19, 2024, 2:32 PM • 3 Y

Y by qwerty123456asdfgzxcvb, soryn, centslordm

Let $p+q=x$ and $pq=y$ . Then, $a+b+c=x+abc\overline y$ . If $PQ\parallel BC$ , then $\frac{x^2-4y}{bc}$ is real, so $x^2-4y=b^2c^2(\overline x^2-4\overline y)$ . Substituting $\overline y=\frac{a+b+c-x}{abc}$ gives
$x^2-4(-abc\overline x+ab+bc+ca)=b^2c^2\overline x^2-4bc\frac{a+b+c-x}a.$ This hyperbola is clearly rectangular. Rewrite this as
$ax^2-ab^2c^2\overline x^2-4bcx+4a^2bc\overline x-4a^2b-4a^2c+4b^2c+4bc^2=0$ $(ax+abc\overline x-2a^2-2bc)(ax-abc\overline x+2a^2-2bc)=4b^2c^2-4a^4+4a^3b+4a^3c-4ab^2c-4abc^2.$
The asymptotes of this hyperbola are $x\pm b\overline x=\frac{2bc}a\pm2a$ , so the center of the hyperbola is $x=\frac{2bc}a$ . Therefore, the center of the rectangular hyperbola of the midpoints of $PQ$ is $\frac x2=\frac{bc}a$ , or the $A$ -isosceles trapezoid point.

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qwerty123456asdfgzxcvb

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qwerty123456asdfgzxcvb

#3 h Oct 19, 2024, 5:39 PM • 2 Y

Y by centslordm, TestX01

Consider the locus of all $P$ with $PQ$ parallel to $BC$ ; this is an isoptic cubic of $ABCA$ (with a weird limiting case where $A$ approaches parallel to $BC$ ), and thus the newton line $\tau$ is the $A$ -midline.

Then the poloconic of the point at infinity along the Newton line of an isoptic cubic is a rectangular hyperbola.

(Proof by dolphint: Consider the intersection of the line at infinity with the poloconic at $\infty_{\tau}, \infty_Y$ and the cubic at $\infty_{\tau}, I, J$ (circle points). Then we have $(I, J; \infty_{\tau}, \infty_Y) = -1$ which only happens if $\infty_{\tau}$ and $\infty_Y$ are perpendicular.)

(Proof of original problem with this fact: Let $\infty_{\tau}$ be this point at infinity, and let $\ell$ be a line through it. Let $\ell$ intersect the isoptic cubic not at the point at infinity at $P,Q$ (they are isogonal conjugates also, by definition), and let the line intersect the poloconic of the point at infinity again at $X$ . Then since by the definition of poloconic as $\frac{2}{\infty_{\tau}X} = \frac{1}{PX} + \frac{1}{QX}$ , we get that $(\infty_{\tau}, X; P, Q)$ is harmonic and thus $X$ is the midpoint of $PQ$ .)

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soryn

#4 h Oct 19, 2024, 8:21 PM

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Please,mr DottedCalculator, I do not understand how to find the affixes of the izogonal points. I do not understand your first equality a+b+c=..., and, how to conclude that the number (x^2-4y)/bc is real. Can explain me? Thx in advance

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soryn

#5 h Oct 20, 2024, 7:25 AM • 1 Y

Y by centslordm

Please,who can help me with the explanation? Exist an article with the approach the izogonal points in a triangle,using the complex numbers? Please.....If p+q=x, and pq=y, how obtain a+b+c=x+abc•(yconj)?

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soryn

#6 h Oct 20, 2024, 6:57 PM

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Really,nobody? At least with an article..

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soryn

#7 h Oct 23, 2024, 9:40 PM

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Bump....

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soryn

#10 h Nov 11, 2024, 3:38 AM

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A good man?

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#11 h Nov 11, 2024, 5:12 AM • 1 Y

Y by qwerty123456asdfgzxcvb

Condition of parallel is useless.Just notice that the midpoint lies on a fixed rectangular hylerbola passes through IIaIbIc as long asdirection of line connecting a pair of Isogonal conjugate is fixed.My solution is to construct external triangle IaIbIc of ABC and change reference triangle.

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soryn

#12 h Nov 11, 2024, 6:00 AM

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Thx. I do not understand the beginning part ,the first equation of solution of Dotted Calculator. Who can explain me,how obtain the equations? Thx in advance

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kooooo

#13 h Nov 11, 2024, 1:55 PM • 2 Y

Y by qwerty123456asdfgzxcvb, centslordm

This problem is the particular case of the following well-known property.

Given a $\triangle ABC$ , let $U, V$ be the isogonal conjugate WRT $\triangle ABC$ lying on the pivotal isogonal cubic with pivot $P$ . Then the harmonic conjugate of $P$ WRT $U, V$ lies on a fixed conic.

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qwerty123456asdfgzxcvb

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qwerty123456asdfgzxcvb

#14 h Dec 1, 2024, 7:48 PM • 1 Y

Y by centslordm

kooooo wrote:

This problem is the particular case of the following well-known property.

Given a $\triangle ABC$ , let $U, V$ be the isogonal conjugate WRT $\triangle ABC$ lying on the pivotal isogonal cubic with pivot $P$ . Then the harmonic conjugate of $P$ WRT $U, V$ lies on a fixed conic.

This is known as the poloconic of P in the pivotal isogonal cubic

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polishedhardwoodtable

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polishedhardwoodtable

#15 h Dec 4, 2024, 9:40 PM • 1 Y

Y by qwerty123456asdfgzxcvb

Let $B=(0,0)$ , $C=(1,0)$ , and $A=(a,b)$ , let $PQ$ intersect $AB$ and $AC$ at points $X$ and $Y$ respectively. One can see that $X=(ak,bk)$ and $Y=(ak+1-k,bk)$ . Finally observe that $\angle XPB=\angle XBQ$ so triangles $XPB$ and $XBQ$ are similar, and similarly we have $YQC$ and $YCP$ are similar. Thus we obtain the equalities $XB^2=XP*XQ$ and $YC^2=YQ*YP$ . From here letting $P=(p,bk)$ and $Q=(q,bk)$ one can solve a linear system of equations for $p$ and $q$ , and obtain that the midpoint $M$ is the point $(ak+\frac{1-k}{2}+\frac{k^2(2a-1)}{2(1-k)},bk)$ . This is a parameterization of all points $M$ in terms of the changing value $k$ , $M=(X(k),Y(k))$ . We can turn this parameterization into an equation for the hyperbola in $x$ and $y$ by observing that $k=\frac{Y(k)}{b}=\frac{y}{b}$ thus we know $x=X(\frac{y}{b})=\frac{ay}{b}+\frac{1-\frac{y}{b}}{2}+\frac{\frac{y}{b}^2(2a-1)}{2(1-\frac{y}{b})}$ . Simplifying this equation leads to a ton of canceling and the final equation is just $\frac{b}{2}-ab=(x+a-1)(y-b)$ , which is clearly rectangular and clearly has center $A'=(1-a, b)$ which is just $A$ reflected over the perp. bisector of $BC$ .

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soryn

#16 h Apr 28, 2025, 3:39 PM

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Nice......
Thx

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