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**Gonzumzum****Member**- Registered: 2017-10-16
- Posts: 2

Hello all...

Anyone one has a clue how to prove this?????

I can´t even start doing anything....

https://i.pinimg.com/originals/7b/80/e4/7b80e4b0e4df83edbd06bccbbf8605df.jpg

Thanks

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

Hi Gonzumzum

Welcome to the forum.

The tangent to two circles creates a pair of similar triangles. Thus

EOdash/AOdash = K J/AJ

If you put in r2 r1 r3 and d, you will find r3 in the required form straight away. The other is similar.

Bob

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Gonzumzum****Member**- Registered: 2017-10-16
- Posts: 2

Hi.

Thanks.

It´s not clear how i can use d in that relation...besides, AOdash or AJ have some parts that cannot be defined with any of those suggested variables.

Could you specify the calculus?

Thanks,

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**alter ego****Member**- Registered: 2012-03-30
- Posts: 27

EOdash = r2

AOdash = d - r2

KJ = r3

AJ = 2r1 - r3

Alter

*Last edited by alter ego (2017-10-30 23:38:18)*

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