OH = r^2/ OE, line exterior to a circle related

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Source: 2019 France JBMO TST 1.2

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#1 h Jul 20, 2020, 10:15 PM

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Let $\Gamma$ be a circle with center $O$ and radius $r$ , and a straight line $\ell$ that does not intersect. We note with $E$ the intersection point of $\ell$ and the perpendicular from $O$ on $\ell$ . Let $M$ be a point of $\ell$ different from $E$ . The tangents from $M$ at $\Gamma$ intersect $\Gamma$ in $A$ and $B$ . Finally, let $H$ be the intersection point of the lines $AB$ and $OE$ . Prove that $OH =\frac{r^2}{OE}$ .

This post has been edited 1 time. Last edited by parmenides51, Jun 21, 2022, 1:52 AM

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#2 h Jul 21, 2020, 12:40 AM

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Pretty straightforward using projective geometry.
Clearly $AB$ is the polar of $M$ with respect to $\Gamma$ as it is the line through the tangency points from $M$ to $\Gamma$ . Then $H$ lies in the polar of $M$ . By La Hire's theorem, $M$ lies in the polar of $H$ . As the polar of $H$ is perpendicular to $OH$ and passes through $M$ , it means that $l$ is the polar of $H$ with respect to $\Gamma$ . Let $X$ and $Y$ be the intersections of $OH$ with $\Gamma$ , then $(X,Y;H,E)$ is a harmonic bundle where $O$ is the midpoint of $XY$ . Then it is known that $OH\cdot OE=OX^2=r^2$ as desired.
A fun corollary of the problem is that if we move $M$ over $l$ , $H$ remains fixed.

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#3 h Jul 21, 2020, 3:33 AM

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Supertinito wrote:

Pretty straightforward using projective geometry.
Clearly $AB$ is the polar of $M$ with respect to $\Gamma$ as it is the line through the tangency points from $M$ to $\Gamma$ . Then $H$ lies in the polar of $M$ . By La Hire's theorem, $M$ lies in the polar of $H$ . As the polar of $H$ is perpendicular to $OH$ and passes through $M$ , it means that $l$ is the polar of $H$ with respect to $\Gamma$ . Let $X$ and $Y$ be the intersections of $OH$ with $\Gamma$ , then $(X,Y;H,E)$ is a harmonic bundle where $O$ is the midpoint of $XY$ . Then it is known that $OH\cdot OE=OX^2=r^2$ as desired.
A fun corollary of the problem is that if we move $M$ over $l$ , $H$ remains fixed.

Once you know $\ell$ is the polar of $H$ , by the very definition of polar, we have $OH \cdot OE = r^2$ (at least by how I've seen polar defined). I don't think we need the extra step you mentioned.

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#4 h Jul 21, 2020, 1:26 PM

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dgreenb801 wrote:

Once you know $\ell$ is the polar of $H$ , by the very definition of polar, we have $OH \cdot OE = r^2$ (at least by how I've seen polar defined). I don't think we need the extra step you mentioned.

Yeah you're right. I suppose that you use the inversion thingy to define polar lines, then as you said the end of the proof is obvious. I'm used to the definition using harmonic bundles, that's why I used the extra step at the end. But at the end of the day everything is more or less equivalent. Thanks for reminding me that fact (I kind of forgot about it :oops:

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