UnivHypGeom2: Apollonius and harmonic conjugates

[EmptySpaceEnterprise]

polarity holds for general conics @03:58

geometers sketchpad in use @05:02

How to find the polar of a null point @05:50

Harmonic conjugates @08:58

discussion of various types of geometry @12:17

More on harmonic conjugates @16:37

examples of harmonic ranges and harmonic ranges theorem @24:00

Harmonic pencils and Harmonic bisectors theorem @28:34

Harmonic vector combinations theorem @32:37

Harmonic quadrangle theorem @34:34

[QuantumMaths]

For exercise 2.1, could one simply draw a chord through the desired point of tangency, select two points on the chord, find the polars of both of them, intersect them, take that point, and then draw a line through the desired point of tangency and the intersection of the two polars? Two points define a line, right? So the third point generated by two other polars would be redundant... right?

[footstep002]

I've tried several times to solve exercise2.1 using CaR program. But I can't plot the tangent line for gamma. As you know it is easy to find a null point starting 'from a pole outside circle'. It's just a result of first definition of pole, that includes a tangent line.

But in exercise2.1 null point is specified first, and I'm keep failing to get a proper tangent line for that null point.

Could you tell me how can I solve this problem?

[njwildberger]

@footstep002 Take any line through the null point, and find its pole. Then use the fact that if line A passes through point b, then the dual point a lies on the dual line B.

[alexzarhin]

Hello,

After several unsuccessful attempts to solve the stared question "how to find a center of a circle using only a straightedge", I looked in the internet and found out that it is impossible, according Poncelet-Steiner theorem.

Is this what you ment or I misunderstood something?

With Respect,

Alex.

[njwildberger]

@alexzarhin This is what I meant. It is useful in universal hyperbolic geometry to realize that the usual `center' of the null circle is not a valid concept in projective geometry.

[sstl517]

@njwildberger , if you are willing to forgo your requirement of "using a strait-edge only" and use a compass also, here is a good illustration of the technique:

wwwmakeitsolardotcom/science-f­air-ideas/90-find-circle-cente­r.htm

[Tr3xKuro]

@alexzarhin

LOL NOW I KNOW WHY HE HAS THAT SMUG GRIN ON HIS FACE WHILE HE'S TELLING US TO ATTEMPT IT.

[brangelito]

Thank you.

[danodet]

I love the way hierarchy of geometry is presented with respect to the toolkit allowed. I have a question about the carpenter’s square construction. Is it I or II ?

I. Given a line and a point on the line you can draw a line perpendicular to the given line that pass through the given point.

II. Given a line and a point in the plane you can draw a line perpendicular to the given line that pass through the given point.

It seems to me that II is stronger than I.

Thank you

[EclecticSceptic]

2) Out of interest, and this will be even more difficult to communicate here, is the following possible:

There's an arbitrary point a (let's say outside the circle). So let's draw 2 lines through it (L and K) that touch the circle at 2 points each. So L touches the circle at p, q. K touches at x, y. qx intersects py somewhere inside the circle. However, px is parallel to qy, and so there is no 2nd intersection, and the polar of 'a' cannot be drawn. Is this possible?

Thanks, and great lectures.

[EclecticSceptic]

@EclecticSceptic Sorry could you ignore this question please, it's invalid for many reasons.

[peterhi503]

Thank you, Wildberger, for another great lesson. I hope your ideas gain more currency.

[Waranle]

Thank you