# How can I calculate the radius of a circle that touches AD and DC and goes through point B of a square

I'm having a lot of trouble with an exercise in which I have to calculate the radius of a circle that only touches AD and CD of a square with the side 1 length and goes through point B.
Heres my sketch:

After thinking about it for about 20 minutes, I just can't find an approach to solve this.
I just don't have any given parameters.
Does anybody have any idea on how to solve this?

• My idea: 20 minutes weren't enough time of thinking. – Michael Hoppe Dec 17 '16 at 22:40
• When did 20 minutes become a long time to think about a math problem? – Sasho Nikolov Dec 18 '16 at 1:56

Let $I$ and $J$ be the tangent points.

So, $OI=JO=OB=r$ and then

$$OD=r\sqrt{2} \quad \text{(Pythagoras Theorem)}$$

$$BD=BO+OD=\sqrt{2}=r+r\sqrt{2} \Rightarrow r=\frac{\sqrt{2}}{1+\sqrt{2}}$$

• Very nice, but...why would the square diagonal $\;BD\;$ pass through the circle's center? The other diagonal doesn't... – DonAntonio Dec 17 '16 at 21:11
• One argument can be simetry. Other is if we a point $O$ on $BD$ such that $OI=OJ=\frac{\sqrt{2}}{1+\sqrt{2}}$ (perpendicular) then we get that $OB$ also has the same size. So there is a circle going through $I,J,B$ . – Arnaldo Dec 17 '16 at 21:21
• Thanks, that was very helpfull and i understood it incredibly fast, probably beacuse I should have been able to solve it myself in the first place, thanks a lot – user2741831 Dec 17 '16 at 21:24
• You are very welcome! – Arnaldo Dec 17 '16 at 21:24

The circle only touches the sides, that means that the sides are tangent to the circle. Say the center of the circle is O, the touching point on DC is M, the touching point on AD is N. You can easily prove that O is on the diagonal BD (two right angle triangles DNO and DMO). Then, say the radius of the circle is denoted by $R$. The length ON=OM=$R$, and since O is on the diagonal, DNO and DMO are right angle isosceles triangles. Then ON=ND=$R$, so OD=$R\sqrt{2}$. The length of BD=$\sqrt{2}$=OB+OD=$R+R\sqrt{2}$. Therefore $R=\frac{\sqrt{2}}{1+\sqrt{2}}$

• Did you mean "the diagonal $\;BD\;$ ...? – DonAntonio Dec 17 '16 at 21:12
• Ooops. Thanks. I'll fix that – Andrei Dec 17 '16 at 21:12

This problem is much easier in reverse. Start with a circle of radius $1$ centred at $(0,0)$ on cartesian coordinates.

Then we know

$CD$ has the equation $y=1$

$AD$ has the equation $x=-1$

$B$ is at $\left(\sqrt{\frac 12},-\sqrt{\frac 12}\right)$.

Thus the sidelength of the square is $\sqrt{\frac 12}+1$ .

Now by scale arguments, we can say that for a square of sidelength $1$,

$R= \frac1{\sqrt{\frac 12}+1} = \frac{\sqrt2}{1+\sqrt2}$.

• Just a suggestion: your posts will be much better received if you would bother to use mathjax yourself to format them. – Paul Sinclair Dec 18 '16 at 2:24
• @PaulSinclair Thanks very much for the edit. I'm not a career mathemetician, I'm a chemical engineer. I would love to learn Mathjax but I don´t have the time right now. – Level River St Dec 18 '16 at 2:29
• Sorry, but that is obviously false. One look at your profile is enough to see that you put in plenty of recreational time on various S.E. sites. It took me maybe 15 minutes to figure out enough Latex and mathjax as I used in editing this post. And before I joined, I had almost no exposure to Latex, and had never seen mathjax before. – Paul Sinclair Dec 18 '16 at 2:56