This is one of the sangaku problems (see pic).

Pretty much all the information given can be read from the picture.

Can someone point me in the right direction? How do you approach such a problem?

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- Feb 22nd 2013, 10:57 AMmetlxSangaku problem
This is one of the sangaku problems (see pic).

Pretty much all the information given can be read from the picture.

Can someone point me in the right direction? How do you approach such a problem? - Feb 22nd 2013, 11:45 AMHallsofIvyRe: Sangaku problem
Here's how I would approach it (the

**hard**way- someone cleverer may be able to find a simpler way):

Start by setting up a coordinate system such that the origin is at the bottom of the large (radius 3) circle and the y axis is a diameter. Then that circle has equation $\displaystyle x^2+ (y- 3)^2= 9$ or $\displaystyle x^2+ y^2- 6y= 0$. The next circle, with radius 2, is tangent to that first circle at (0, 6) and so must have center at (0, 4) and equation $\displaystyle x^2+ (y- 4)^2= 4$. Finally, the circle with radius 1 has center at (0, 1) and so equation $\displaystyle x^2+ (y- 1)^2= 4$.

In terms of that coordinate system, the problem is to find a, b, and r so that $\displaystyle (x- a)^2+ (y- b)^2= r^2$ is tangent to all three of those circles. - Feb 22nd 2013, 12:41 PMmetlxRe: Sangaku problem
Hmm.

So we can assume that$\displaystyle a>0 $ and $\displaystyle b>0 $ since the circle is in the first quadrant, given that (0,0) is at the bottom of the big circle.

From your pointers I was thinking to determine where each of the "known" circles intersects with the "r-circle". I turned the equations a bit and I ended up going in circles (no pun intended).

Intersection between the biggest and the r-circle:

$\displaystyle x^2 + y^2 - 6y = x^2 - 2ax + a^2 + y^2 - 2by + b^2 - r^2$

$\displaystyle -2ax + a^2 - 2by + b^2 - r^2 + 6y = 0$

Intersection between circle with radius 2 and the r-circle:

$\displaystyle x^2 + y^2 - 8y + 12 = x^2 - 2ax + a^2 + y^2 - 2by + b^2 - r^2$

$\displaystyle -2ax + a^2 - 2by + b^2 - r^2 + 8y - 12 = 0$

As far as I know this gives me no more information on how to isolate a, b and r. Or am I wrong?

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Another idea I had in mind is to construct a right triangle by connecting the radiuses:

$\displaystyle (1+r)^2 + (2+r)^2 = (1+2)^2$

$\displaystyle 1 + 2r + r^2 + 4 + 4r + r^2 = 9$

$\displaystyle r = \frac{-3 + \sqrt{17}}{2} $

Would that be the correct value of r?

(also, i'm interested in your way of doing it, so i'd very much like to see how to proceed from there). - Feb 22nd 2013, 01:06 PMmetlxRe: Sangaku problem
Never mind for the second part. There is no way of knowing for sure if the triangle would be a right-angled triangle.

- Feb 22nd 2013, 05:06 PMjohngRe: Sangaku problem
Here's a solution. The answer is r = 6/7.

Attachment 27196 - Feb 28th 2013, 05:37 AMMINOANMANRe: Sangaku problem
Problems of that type can be solved easily only if you apply inversive geometry.

Few years ago I solved a similar problem with a slight difference the two circles inside were equal having as r the 1/4 of the biger circle.

the interesting point is that the general term of the sequence of the circles inscribed and touching each other can be calculated.

The nth term of the sequence is Rn=R/[(n-1)^2+2]

and expresses the radius of the nth circle of the sequence inscribed incide the circle where R is the radius of the big circle.

MINOAS - Feb 28th 2013, 05:51 AMMINOANMANRe: Sangaku problem
here is a similar one in the cut the knot website

1 + 27 = 12 + 16 Sangaku