Math Is Fun Forum

Gonzumzum

Member

Registered: 2017-10-16

Posts: 2

Sangaku Problem 525

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Sangaku Problem 525

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

Gonzumzum

Member

Registered: 2017-10-16

Posts: 2

Re: Sangaku Problem 525

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,

alter ego

Member

Registered: 2012-03-30

Posts: 29

Re: Sangaku Problem 525

EOdash = r2

AOdash = d - r2

KJ = r3

AJ = 2r1 - r3

Alter

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