# 3 square in a circle (geometric question)

There is three square in a circle like the picture attached. How can find radius of circle ?
I know that's not a hard problem,but I was away from geometry for years. If possible give me hint or idea to start to solving. Thanks in advance. the smallest side is $$6$$
and medium square side is $$6+6$$ and big side is $$6+6+6$$
But I got stuck here and not to go over there ...

• Draw the radius from the centre of the circle to either of the vertices of the big square touching the circle. Then apply Pythagoras. – Nelli Kuukeri Mar 26 at 15:39
• @NelliKuukeri: where is the centre according to you? – Rahul Verma Mar 26 at 15:52

It is only the Pythagorean theorem: • I am in doubt that you see the picture! please look again,it is not a symmetric sahpe. – Khosrotash Mar 26 at 16:20
• @Khosrotash, You can move the middle square to the center as you wish. Notice that the other two squares are touching the circle and it is indeed a symmetric shape. – Seyed Mar 26 at 16:31
• @Khosrotash, khaahesh mikonam. – Seyed Mar 26 at 16:35
• What theorem allows you to state “$(6+12+18)\times X = 6\times12$”? – gen-z ready to perish Mar 26 at 17:45
• @gen-zreadytoperish, it is called Intersecting Chord Theorem. You can read about it here: mathopenref.com/chordsintersecting.html – Seyed Mar 26 at 18:44

Coordinate Geometry works.

Draw the horizontal diameter of the circle.
Let the center of the line segment between the squares be the origin $$(0,0)$$.
Let the center of the circle be $$(x, 0 )$$.

What is the coordinate of the top right blue point?



What is the coordinate of the top left blue point?



Hence, what is an equation that we can write?



Hence, what is the radius of the circle?

• By the way I just saw your appearance on Numberphile. Cool stuff! – Maximilian Janisch Mar 27 at 14:28
• Thanks!    – Calvin Lin Mar 27 at 14:44 Equations: $$a + b = 36$$ $$a^2 + 9^2 = r^2$$ $$b^2 + 3^2 = r^2$$

Using 2 and 3: $$b^2 - a^2 = 72$$ Factoring: $$(b-a)(b+a) = 72$$ $$(b-a)36 = 72$$ $$b-a = 2$$ Adding the first: $$2b = 38$$ $$b = 19$$ $$a = 17$$ Adding 2 eq y 3 eq: $$19^2 + 17^2 +9^2 + 3^2 = 2r^2$$ $$370 = r^2$$ $$r = \sqrt{370} = 19.23....$$

The circumradius of a triangle is $$R=\frac{abc}{4A}$$ where $$a,b,c$$ are the sides of the triangle and $$A$$ is its area. The sides are $$18$$, $$\sqrt{36^2+6^2}=6\sqrt{37}$$, and $$\sqrt{36^2+12^2}=12\sqrt{10}$$ and the area is $$\frac12\,18\cdot36=324$$. Therefore, the circumradius is $$R=\frac{18\cdot6\sqrt{37}\cdot12\sqrt{10}}{4\cdot324}=\sqrt{370}$$

• :+1 nice ,very nice idea – Khosrotash yesterday Apply the Pythagorean theorem to the triangles ADO and CBO to match the overall length AB = 36,

$$\sqrt{r^2-9^2} + \sqrt{r^2-3^2} =36$$

Solve to obtain the radius $$r=\sqrt{370}$$.