Geometry sangaku puzzle, incribed circle circle/triangle/square

Question

Hello I am trying to solve a geometry puzzle, its been 30 years since I was in school and I struggled with maths! I would love to get some help to find out what the radius of the bigger circle is if the radius of the smaller circle "乙" is 3.06. Are you clever enough to figure this one out? What formula do you need? What is the radius?

Answer 1

Better answer.

$$q = 2 p \sqrt{2} \qquad\stackrel{p=3.06}{\to}\qquad q \approx 8.65$$

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Previous answer.

$$q = 2 r - r \sqrt{2} = \sqrt{2}\left( \sqrt{2} r - r \right) = \sqrt{2}\cdot 2 p \quad\to\quad q = 2 p \sqrt{2} \quad\stackrel{p=3.06}{\to}\quad q \approx 8.65$$

Answer 2

Lets call the radius of the big circle $R.$

The diagonal of the green square is the diameter of the circle. $2R$

We can find the side length by the pythagorean theorem.

$s^2 + s^2 = (2R)^2\\ 2s^2 = 4r^2\\ s = \sqrt 2 R$

The distance from the center of the big circle then to the side is $\frac 12$ the side length $\frac {\sqrt 2}{2} R$

The shortest distance from the center of the circle to the side of the green square (and the edge of circle Z) is $\sqrt \frac {1}{2} R$

And the diameter of the blue circle Z is the radius of the big circle less this distance to the edge of the square. $(1-\frac {\sqrt 2}{2}) R$

We are almost there.

$2Z = (1-\frac {\sqrt 2}{2})R\\ R = \frac {2}{1-\frac {\sqrt 2}{2}} Z$

We can do some prettying up first, or we can jam it through the calculator like it is. I am not sure I want to walk through the steps to do the simplification.

But you would get.

$R = (4 + 2\sqrt 2) Z\\ R = 6.83\cdot 3.06 = 20.89$

Answer 3

If $x$ is the radius of the small blue circle and $y$ the radius of the bigger red circle, then $y = \sqrt 8 \cdot x$.

To see this, denote with $a$ and $d$ the side length and the length of the diagonal of the green square respectively. By Pythagoras theorem $$d = \sqrt{2}a.$$

Note that $d$ is also the diameter of the biggest circle. Since this fits exactly two small circles of radius $2x$ next to the square we have $$ 4x+a=d $$ and thus $$x=(d-a)/4 = (\sqrt 2 -1 )/4 \cdot a $$ Now consider five points $C$,$P$, $Q$ and $A$ and $B$ in the plane:

• $C$ is the centre of the red circle

• $P$ is the point where the red circle touches the diagonal of the green square

• $Q$ is the point where the red circle touches the right side of the green square
• $A$ is the upper right corner of the green square
• $B$ is the lower right corner of the green square

The triangles $PCA$ and $QCA$ are similar because they share same angles, thus also their side lengths coincide, i.p. $$d/2 =\vert PA\vert = \vert QA\vert$$ The triangle $CQB$ has a right angle at $Q$ and a $45$°-angle at $B$, thus it is isosceles, i.e $$ y = \vert QC \vert=\vert QB \vert $$
Moreover $$ \vert AQ \vert + \vert QB \vert = a.$$ Combining the three last equations yields: $$ y = a - d/2 = (1-1/\sqrt 2)\cdot a.$$ Compare the two equations for $x$ and $y$ to see that $$ y = 4 \cdot \frac{1-1/\sqrt{2}}{\sqrt 2 -1} \cdot x = \sqrt 8 \cdot x $$