# Circle inscribed between quarter circles - proving its center point

Square $$ABCD$$ has a side length of $$1$$. $$BGD$$ and $$AFC$$ are quarter circles, and there is an inscribed circle between the quarter circles. Two points of tangency $$F$$ and $$G$$ are labeled.

How can I prove $$O$$ is the center of the inscribed circle?

I was thinking to show $$EFGH$$ is a rectangle (or equivalently an inscribed parallelogram). It is easy to see $$FG$$ and $$EH$$ are parallel to $$AD$$ by symmetry, but I cannot justify why $$EF$$ and $$HG$$ are parallel to $$AB$$ (well I thought it was obvious, but upon further thinking it didn't seem so obvious!).

The problem of solving for the radius of the circle has two solutions on Quora. However, I was having issues justifying the steps of the proofs. For example, the first proof seems to assume $$O$$ is the center of the circle. And the second states the three points of tangency in the circle form an equilateral triangle (which doesn't seem obvious to me).

It is definitely true $$O$$ is the center, but I'm at a mental block for proving this obvious thing. Any suggestions would be much appreciated, thanks!

Disclosure: I run the YouTube channel MindYourDecisions. I will give credit by linking to this thread in the "sources" for the video/blog post.

• Given two tangent circles $\bigcirc A$ and $\bigcirc O$, consider the line tangent to both of them at their point of tangency, $G$. Necessarily, $\overline{AG}$ and $\overline{OG}$ are perpendicular to that line; but there is only one such perpendicular, so $A$, $O$, $G$ are collinear.
– Blue
Dec 24, 2019 at 6:55

As mentioned in a comment, the centers of tangent circles lie on the perpendicular to the tangent line at the point of tangency. So, in OP's diagram, $$O$$, $$A$$, $$F$$ are collinear, as are $$O$$, $$B$$, $$G$$.

The length of the radius of $$\bigcirc O$$ follows from computing the power of point $$A$$ with respect to that circle in two ways:

$$|AM|^2 = |AE||AF| \quad\to\quad s^2 = 2(s-r)\cdot 2s \quad\to\quad 4r = 3s \quad\square$$

It's perhaps worth noting that $$\triangle AOM$$ is of the $$3$$-$$4$$-$$5$$ variety.

• Elegant answer. I came across this problem in a commentary on Euclid I.1 that asked for the construction presuming no knowledge of circle properties, which made me realize we can also just use the Pythagorean Theorem at $O$ to solve: $(2s-r)^2=r^2+s^2 \rightarrow 4r=3s$. Mar 13 at 16:27

$$\overline{AG}$$ is perpendicular to $$\bigcirc O$$ (at $$G$$) and hence $$\overline{EG}$$ is one of its diameters. Likewise for $$\overline{DF}$$. Two distinct diameters always meet at the center. Q.E.D.

Note: FareedAbiFarraj's demonstration that $$\overline{OF} = \overline{OG}$$ does not prove that $$O$$ is the center:

• Short and perfect. Thanks for illustrating the counter-example too: I kept finding such flaws in my attempts at a proof. Dec 24, 2019 at 8:11

I had earlier solved this problem and found out the radius of the inscribed circle to be 3/8 of side of the square. If side of the square is taken as "a" then the co-ordinates of centre is (a/2,3a/8). I am taking point A as the origin. We can get the co-ordinates F and G by solving the equations of the inscribed and quarter circles. Then we can prove the colinearity of AOG and DOF.

• The square side is given as $1$. Regardless, you haven't solved the problem at all. Do you see that? Dec 24, 2019 at 17:32