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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

In the adjoining figure, AB is tangent at A to the circle with center O , point D is interior to the circle, and DB intersects the circle at C. If BC = DC = 3,OD = 2 , and AB = 6, then find the radius of the circle.

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,434

Hi cooljackiec;

**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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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

thanks. can you help me on this?

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,382

hi

There's a theorem that is true for all circles. For any four points on the circumference, see diagram

AE x EC = DE x EB

That should be enough for you to do this one.

Bob

ps. to prove it look for equal angles using the circle theorems and then similar triangles.

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

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

hmm.. that is the power of a point theorem right?? i think it works for secants/tagents too

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,382

I've not met it called that, but Wiki agrees with you:

http://en.wikipedia.org/wiki/Power_of_a_point

And yes it does apply to tangents.

Bob

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

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

still stuck

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,434

Hi;

I got an answer of 31 by another method. Can you check it?

**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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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

seems correct. how did you do that?

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,434

I have been experimenting with the team of bobbym and geogebra, trying to be more than an amazed bystander. The great gAr would describe it as a "mathod!"

**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: 6,382

hi cooljackiec

still stuck

I'm assuming you mean with the question in post 3.

I put x = XP and y = QY.

Then applied the theorem twice:

5 x 6 = x(27+y)

7 x 12 = (x+27)y

I subtracted to eliminate xy.

Then made y the subject of that and substituted back into one equation to make a quadratic.

It factorised easily with one positive and one negative root. Rejected the negative and then substituted back to find y.

Got the same answer as bobbym.

Bob

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

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