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**Circles #3**

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

The drawing will make it clear but I can not get it in there.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

hi Agnishom,

Alternate segment theorem?

Haven't used this for years.

EDIT: earlier error corrected.

In any circle, centre A, EF a tangent at F, D some point on the circle:

Let EFD = x => FCD = x (5th angle property of a circle)

http://www.mathisfunforum.com/viewtopic … 88#p220488

Now consider the triangles EFC and EDF

angle E is common, and EFD = FCE = x

So EFC and EDF are similar

=> ED/EF = EF/EC => ED.EC = EF^2

Similarly ED.EC = EG^2

=> EF = EG

Thanks Phro.

Bob

*Last edited by bob bundy (2014-01-30 06:22:24)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

I think the diagram here does what you wanted in post 14. I need to think some more about whether it is ok or a cheat. You see as the equal property holds there's a danger with trying to construct a diagram that genuinely shows this as opposed to one that just looks right.

Anyway here's what I did.

(i) Join A to E, construct the midpoint and hence construct a circle with AE as diameter which makes EF a tangent to the original circle.

(ii) Do the same with the second circle to get G.

(iii) Find H the other point on the first circle which will make a tangent EH.

(iv) Bisect angle HEG.

(v) Find J where this bisector cuts BG.

(vi) Construct the circle with J as centre and with EG and EH as tangents.

Then EF = EH and EG = EH => EF = EG as required.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,394

No it doesn't work.

I repeated the construction but with E not on CB.

The diagram is somewhat complicated so I've tried to colour code it. The dotted circles are just there to fix the points F and G.

H is the second tangent point as before. The red circle does have both the line EH and the line EG as tangents but not both of G and H are the actual tangent contact points. I've chosen to make G one such point and you can see that H is not the other. Big red arrow.

This just shows how careful you have to be proving things. The result is true when E is on CB so that diagram looks ok. But this diagram shows that the construction is in-valid.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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That is not a big deal, we can change that line to: Draw a circle touching EH and EG and H and G only.

(It is still cheating, because we do not know if that can be done.)

What about bobbym's proof? He still did not tell me why those points must coincide.

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Hi Bob,

Ggrr. If only I had gone surfing. But the sea temperature is around 9 C. Ugh!

The water temperature is very bearable down here!

But the air temp isn't! We've had quite a few 40+ days lately, so the aircond's been working flat out (and thank goodness it's working!!)

The beach pics are ones I took of Bondi (Sydney) last November.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,705

Nice scenery, looks like Florida.

Agnishom wrote:

What about bobbym's proof? He still did not tell me why those points must coincide.

I think that is the easiest part, they are constructed to coincide. See the other thread.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Circles #3**