# Constructing the circle inversion inverse of a point with ruler only

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.

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. – joriki Apr 27 '11 at 4:53

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):

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 $A$. – yunone Apr 27 '11 at 5:09
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. – yunone Apr 27 '11 at 6:02
@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. – joriki Apr 27 '11 at 6:06
@yunone: I'm thinking it might have something to do with $OPR$ and $OQR$ being isosceles and $PQ$ forming equal opposite angles at $A$. – joriki Apr 27 '11 at 6:30
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. – yunone Apr 27 '11 at 6:47

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. – joriki Apr 27 '11 at 15:54
Do you have a separate argument for $PS$ bisecting $\angle APA'$, or is this where you're following up on my answer? – joriki Apr 27 '11 at 15:56
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. – abel Apr 27 '11 at 17:01
I think you need $\angle SPA'$ in there somewhere? – joriki Apr 27 '11 at 18:25
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. – abel Apr 27 '11 at 20:04