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I've been reading a bit about inversive geometry, particularly circle inversion. The following is a problem from Hartshorne's classical geometry, which I've been struggling with on and off for a few days.

enter image description here

I figured it would be helpful to show that $TU\perp OA$ first. At best, I tried to label angles according to which ones are congruent with each other. I know $\angle RPS$ and $\angle RQS$ are both right, as they subtend the diameter, so $\angle TPS=\angle UQS$ are both right as well. So $PTUQ$ is a cyclic quadrilateral, and thus $\angle RTQ=\angle PUQ$. Labeling $\angle SPQ$ as $3$ and $\angle PQS$ as $4$, I see that $1+2+3+4$ sum to two right angles.

That's about as far as my observations got me. My hunch is that $RTU$ is an isosceles triangle, and $PU$ is like a line of symmetry, but I'm not sure how to show it, and how to eventually conclude $TU$ meets $OA$ at $A'$, that is, $OA\cdot OA'=r^2$, where $r$ is the radius if $\Gamma$. Thanks for any ideas on how to solve this.

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    PU can't be a line of symmetry, since the setup is symmetric with respect to interchance of P/T and Q/U, so TQ would also have to be line of symmetry, which, from the diagram, it isn't.2011-04-27

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The three perpendiculars from the corners of a triangle to the sides meet in a common point, the orthocentre. You know that $UP$ is perpendicular to $RT$ and $TQ$ is perpendicular to $RU$; it follows that the line through $R$ and their intersection $S$ is perpendicular to $TU$.

[Update:] It turns out you can actually go on deducing all the angles much like you started. I'll use your angles $1$, $2$ and $3$. (I don't understand how you defined $4$; in the drawing it seems to mark the right angles $\angle SPT$ and $\angle SQU$ but in the text you defined it as $\angle PQS$.)

First, to make the symmetry of the situation manifest, let's also draw the lines $PA'$ and $QA'$. Here's an image (I'll be justifying the angles I filled in in a bit):

symmetrized diagram

The triangle $PQA'$ is the orthic triangle of the triangle $RTU$. If you take out the circle $\Gamma$, the diagram has $S_3$ symmetry, so the quadrilaterals $A'URP$ and $QRTA'$ are cyclic for the same reason as $PTUQ$; that justfies the angles I've filled in. Since $|OR|=|OP|=r$, triangle $ROP$ is isosceles, so $\angle OPR$ is 3, and hence $\angle OPA$ is 1. Thus the triangles $OPA$ and $OPA'$ have two angles in common (1 at $\angle OPA$ and $\angle OA'P$, and $\angle POA=\angle POA'$), and hence are similar. The inversion property then follows by taking the ratios of corresponding sides in these triangles.

P.S.: That the altitudes of $RTU$ are the bisectors of its orthic triangle is related to the fact (mentioned in the Wikipedia article) that the orthocentre is the incentre of the orthic triangle.

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    Thanks joriki, I'll try to use this to conclude $A'$ is indeed the inverse of $A$.2011-04-27
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    I apologize for the second ping, but I still haven't been able to figure out the meat of my original question. I know at best now that $PTA'S$ and $UQSA'$ are cyclic, but I don't see how that relates to showing that $TU$ really meets $OA$ at $A'$. Do you have any insight as to why this is? Thank you.2011-04-27
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    @yunone: I was thinking along different lines -- in analogy to the standard tangent construction of the inversion, you could conclude $OA\cdot OA'=r^2$ from similar triangles if you could show that $\angle A'PO = \angle OAP$, but I don't see how to show that.2011-04-27
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    @yunone: I'm thinking it might have something to do with $OPR$ and $OQR$ being isosceles and $PQ$ forming equal opposite angles at $A$.2011-04-27
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    Thanks for taking the time to think about it. I'll try to get further based on your suggestion, but please let me know if you figure out, which is more likely to happen before I do.2011-04-27
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    @yunone: I updated my post with a complete answer.2011-04-27
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    Wonderful, thank you, joriki. One question, you mention $S_3$ symmetry? I intuitively see that $RPA'U$ and $RTA'Q$ are cyclic, since each cyclic quadrilateral is in a sense obtained by chopping off a smaller triangle above the orthic triangle. As in, $TPQU$ is obtained by chopping off triangle $RPQ$ that extends above the orthic triangle, and like wise we could chop off $PTA'$ or $QUA'$ to get the others, and it would make sense that the resulting quadrilaterals would be cyclic. Forgive my sloppy speaking, but how does $S_3$ symmetry make this rigorous?2011-04-27
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    @yunone: Well, I think it's I who should be making apologies for sloppy speaking :-) By $S_3$ symmetry, I meant that if you take out the circle, all known relations (linear equations among the angles) in the remaining diagram are invariant with respect to permuting the corners (and correspondingly permuting the remaining points). We only used the circle to establish that $\angle TPS$ and $\angle UQS$ are right angles, and from that we deduced that $RA'U$ is a right angle; once we know this, we can dispense with the circle, and then everything we know follows from these three right angles...2011-04-27
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    ... so, forgetting about the circle, we're left with just an arbitrary triangle, its three altitudes and the three lines connecting their base points, and there's nothing to break the $S_3$ symmetry, so whatever you'd deduced for $PTUQ$ must hold *mutatis mutandis* for the other two corresponding quadrilaterals, since everything you used in that derivation is invariant under the permutations. But actually all this abstraction isn't saying much more than that you can deduce the cyclicity of $RPA′U$ and $RTA′Q$ the same way you deduced that of $PTUQ$, from the right angles at the diagonals.2011-04-27
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    Oh I see, and once you have that $PQA'$ is the orthic triangle, it's simple enough to see that the top angles of $A'URP$ and $QRTA'$ are both right, so they are cyclic. Thanks for clarifying.2011-04-27
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i will follow up on jorikis answer. first we will show that A and A' cut the diagonal RS harmonically. that is the cross ratio of four collinear points $(R, S; A, A^\prime)$ defined by $\frac{RA}{SA} / \frac{RA^\prime}{SA^\prime}$ is unity. this will imply that $A$ and $A^\prime$ are conjugates. cross ratio $(R, S; A, A^\prime)$ is equal to the cross ratio of the lines $(RP, SP; AP, A^\prime P)$ which is $\frac{\sin(\angle RPA) \sin(\angle S P A^\prime)}{\sin(\angle SPA) \sin(\angle RPA^\prime)}$ by law of sine. this is unity because $\angle RPS = 90^\circ$ and the line $PS$ bisects $\angle AP A^\prime.$

p.s. i got this idea by picking $P$ so that $AP$ is orthogonal to $RS.$ in this special case the result follows easily. now generalize.

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    This looks nice. But there seem to be some typos in it. $\angle RAS$ is $180^\circ$, not $90^\circ$. Also the ratio contains the same angle twice, and it isn't unity. I suspect that what you meant to write was $\angle RAS = 90^\circ$ and $\frac{\sin(\angle RPA) \sin(\angle S P A^\prime)}{\sin(\angle APS) \sin(\angle RP A^\prime)}$? This is indeed unity and corresponds to the cross ratio of the points.2011-04-27
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    Do you have a separate argument for $PS$ bisecting $\angle APA'$, or is this where you're following up on my answer?2011-04-27
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    the argument i used to show $PS$ bisecting $\angle APA^\prime$ is that $QUTP, SA^\prime TP$ are cyclic quadrilaterals so $\angle QRS = \angle QPS = \angle QTU = \angle QPU.$ all these angles are made by equal arcs in their respective circles.2011-04-27
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    I think you need $\angle SPA'$ in there somewhere?2011-04-27
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    yes. $\angle QRS = \angle QTS = \angle QPU = \angle QTU = \angle A^\prime PS.$ hope i got all angles right. when i am editing i cant see the figure. so i am copying from my notes. that is my excuse for the typos.2011-04-27