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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 96,579

Hi Bob;

I did not even see yours until now. I was asking mathmatiKs. He seems to be having some problem with his posts.

I agree about the number of points of intersection.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,052

Now this is interesting.

I thought, maybe ABC give the last three points.

So I looked for similar triangles and found one. (see latest diagram)

But, annoyingly, GHI does not make another. ???

Bob

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

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

Hi Bob

In some cases like that are 6 points.

Bobbym

Triangles similars:

EHI FDI EFG

In yout picture

"... And as we let our own light shine, we unconsciously give other people permission to do the same. As we're liberated from our own fear, our presence automatically liberates others."

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

Hi Bob.

In your case the triangles would be:

JLG IKL HIJ

"... And as we let our own light shine, we unconsciously give other people permission to do the same. As we're liberated from our own fear, our presence automatically liberates others."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 96,579

Hi mathmatiKs;

Here is a an easier drawing:

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

AF = 6.30

BC=6.23

That will not be the error?

"... And as we let our own light shine, we unconsciously give other people permission to do the same. As we're liberated from our own fear, our presence automatically liberates others."

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

Similar triangles:

DGI FIE HFD

*Last edited by mathmatiKs (2012-08-04 09:46:05)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 96,579

Hi mathmatiKs;

Thanks, I will test them for AAA. Also, I will see if it works for some other triangles. See you later.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,052

I've just looked at the recent posts.

I have four similar triangles!

Colouring the circles helped.

See diagram.

AF = 6.30

BC=6.23

That will not be the error?

Yes. I am placing the points with the mouse and 'hand shake' causes small inaccuracies.

Bob

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

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

I did it in Kig (Interactive Geometry) and always works.

I just have to prove it.

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

bob bundy wrote:

I've just looked at the recent posts.

I have four similar triangles!

Colouring the circles helped.

See diagram.

AF = 6.30

BC=6.23

That will not be the error?Yes. I am placing the points with the mouse and 'hand shake' causes small inaccuracies.

Bob

Exactly. Are similar to the original triangle, that's what I say.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,052

It is quite late for me in the UK. And I need some sleep. Too much Olympic excitement!

I'll try a proof tomorrow.

Bob

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

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

If you are using Kig, you got an option of "intersection" that will draw the point of intersection. It is easier and error free.

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

Good bye Bob.

Good luck.

Really excited to discover new things, I'm always careful.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,052

hi mathmatiKs

Sleeping on this problem has given a result.

I wanted to construct the circles without having to 'make' the right radius.

So I used parallelograms.

By drawing lines parallel to the sides AB, BC and CA make three parallelograms

ABDC, BCFA and CAEB.

Opposite angles of a parallelogram are equal so

ABC = DFE, BCA = FED and CAB = EDF.

So DFE is similar to ABC (letters in the correct order)

To make the circles I can use D, E and F to fix the radius.

Circle on C: Radius = CD. ( note that this circle also goes through F because AB = CD and AB = FC )

Circle on B: Radius = BE. ( for the same reason this circle also goes through D )

Circle on A: Radius = AF ( for the same reason this circle also goes through E )

So three of the points from my earlier diagram are now indentified as D, E and F .

I have labelled the remaining intersections of circles as G, H and I

My diagram shows the same similar triangles as before but using the new letters.

To show the remaining triangles are also similar I will outline just one example.

Consider triangle EIG.

In this triangle IEG = BCA (opposite angles in a parallelogram)

Draw the line IG (shown in green)

angle IGF = IDF ( angles made at the circumference by the same chord IF )

=> IGE = EDF = CAB

So EIG is similar to CBA ( two angles are the same => all three must be the same )

Hope that helps,

Bob

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

I hope I have an avatar to appear

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

bob bundy wrote:

angle IGF = IDF ( angles made at the circumference by the same chord IF )

=> IGE = EDF = CAB

that means?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,052

hi mathmatiKs

In any circle, there are relationships between certain angles. These are called 'the angle properties of a circle'. I recently got asked to prove them all so you can look them up on:

http://www.mathisfunforum.com/viewtopic … 77#p220477

Look particularly at post #6.

I have made a simplified diagram for your question, erasing the bits that are not needed for this and marked the angles that are equal with a red dot.

Look at the chord IF. On the blue circle it makes the angle IDF and the angle IGF. So these are equal by property 2.

BAC is also equal because it is the opposite angle in the parallelogram ABDC.

Hope that helps you to understand this step in the proof.

There are three circles to 'play with' so you will be able to find lots of angles that are equal using the property. Look for a chord and a pair of angles made by the chord on the opposite side of the cirlce.

Bob

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**mathmatiKs****Member**- Registered: 2012-04-14
- Posts: 45

That property is still true even if the center of the circle has been moved out after the point D to point G?

mathmatiKs

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,052

Sorry. I do not understand your question.

Bob

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