Easy lemma - Conjugate points

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Easy lemma - Conjugate points

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#1 h Apr 18, 2015, 5:39 AM • 2 Y

Y by Adventure10, Mango247

Denote $\mathcal{P}_X$ the power of $X$ with respect to $\omega$ (an circle). Prove that $\mathcal{P}_A+\mathcal{P}_B=AB^2 \Longleftrightarrow A$ and $B$ are conjugate points with respect to $\omega$

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#2 h Apr 18, 2015, 7:07 AM • 2 Y

Y by drmzjoseph, Adventure10

As it's not possible that both $A$ and $B$ lie inside of $\omega,$ then WLOG we assume that $A$ is outside of $\omega.$ $AM,AN$ are the tangents from $A$ to $\omega.$ $\mathcal{P}_A+\mathcal{P}_B=AB^2$ $\Longleftrightarrow$ $\mathcal{P}_B=AB^2-AM^2$ $\Longleftrightarrow$ $B$ has equal power WRT $\omega$ and $\odot(A,AM)$ $\Longleftrightarrow$ $B$ is on their radical axis $MN;$ the polar of $A$ WRT $\omega$ $\Longleftrightarrow$ $A$ and $B$ are conjugate points WRT $\omega.$

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suli

#3 h Apr 18, 2015, 10:56 PM • 1 Y

Y by Adventure10

I forgot what conjugate points were

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#4 h Apr 19, 2015, 12:03 AM • 1 Y

Y by Adventure10

suli wrote:

I forgot what conjugate points were

$A$ and $B$ are conjugate points with respect to $\omega \Longleftrightarrow A$ lies on polar of $B$ with respect to $\omega$
Or
$A$ and $B$ are conjugate points with respect to $\omega \Longleftrightarrow$ the circle of diameter $AB$ is orthogonal to $\omega$

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suli

#5 h Apr 19, 2015, 2:21 AM • 2 Y

Y by Adventure10, Mango247

Thanks for telling me that I need to get a book of projective and inversive geometry ASAP.

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#6 h Apr 19, 2015, 2:26 AM • 2 Y

Y by PXiomar, Adventure10

I have the book, in PDF of "Geometry of Conics" (Arseniy Akopyan) maybe you are interested

(Any user, too)

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#7 h May 1, 2015, 4:41 AM • 2 Y

Y by Adventure10, Mango247

My solution:

Denote $r$ the radius of $\omega$ and $O$ the center. If $p =\overrightarrow {OP}$ denotes the vector from $O$ .
$A$ and $B$ are conjugate points with respect to $\omega \Longleftrightarrow$ the circle of diameter $AB$ is orthogonal to $\omega$ i.e. $(\frac{a+b}{2})^2=(\frac{a-b}{2})^2+r^2$
$\Longleftrightarrow ab=r^2 \Longleftrightarrow a^2-r^2+b^2-r^2=(a-b)^2 \Longleftrightarrow \mathcal{P}_A+\mathcal{P}_B=AB^2$

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