I have this problem: Two circles shown in the figure, of centers $O$ and $O'$, both have radius equal to $r$, are externally tangent to each other and are further tangent to the line $t$. The square $ABCD$ has side $AB$ on line $t$ and the other two vertices, $C$ and $D$, which belong respectively to the circumferences bounding the two circles of centers $O '$ and $O$. What is the measure of the side of the square $ABCD$?

enter image description here

I have made this assumption with this figure: I have drawn from $O$ and $O'$ the perpendiculars to $t$ and joined $O$ with $D$ and $O'$ with $C$. I have denoted by $H$ and $H'$ the orthogonal projections of the centers $O$ and $O'$. From $P$ I have drawn the perpendicular to the line $t$. This line $t$ cut the square in two parts. Named $\overline{DC}=x$. The rectangular trapezoids $ADOH$ and $BCO'H'$ are equivalents.

enter image description here

$\overline{BH'}=\overline{AH}=r-x/2$, $\overline{PF}=r-x$, hence $$\text{area}(DCO'O)=\frac{(2r+x)\cdot (r-x)}{2}$$ $$\text{area}(OO'H'H)=2r\cdot r=2r^2$$ $$\text{area}(AHOD)=\frac{(x+r)\cdot(r-\frac x2)}{2}=\text{area}(BH'O'C)$$ Definitely it must be:

$$2r^2-(x+r)\left(r-\frac x2\right)-\frac{(2r+x)(r-x)}{2}=x^2 \tag 1$$

Solving the $(1)$ I obtained true for all $x$ and not $\frac 25r$. Where did I make a mistake?

  • 2
    $\begingroup$ You did not make a mistake. Also, you did not get true for all $x$, you were just not able to find anything new about $x$ $\endgroup$
    – D S
    Mar 1 at 20:21
  • $\begingroup$ @DS I can no think of simple to solve this problem :-( $\endgroup$
    – Sebastiano
    Mar 1 at 20:25
  • $\begingroup$ The given answer is simple, no? $\endgroup$
    – D S
    Mar 1 at 20:25
  • $\begingroup$ Consider also that $C$ and $D$ are on the circles, not just the corners of any centred square on $t$. $\endgroup$
    – peterwhy
    Mar 1 at 20:26
  • 2
    $\begingroup$ In @RobinSparrow's diagram, if $r=5$ then $r-x=4$ and $r-2x=3$, giving you that triangle. $\endgroup$
    – Neil
    Mar 2 at 12:40

2 Answers 2


In your diagram, make the sides of the square $2x$.

Then $r^2=(r-x)^2+(r-2x)^2$



Leaving $x=\frac{r}{5}$ so the square has side $\frac{2r}{5}$.


To show why sides of $2x$ are convenient

enter image description here

  • $\begingroup$ Thank you very much. Can you, please, explain me the reason that the side of the square is $2x$? Can you please extend your answer also with a drawing? Thank you very much. $\endgroup$
    – Sebastiano
    Mar 1 at 20:24
  • $\begingroup$ Just make $DC=2x$ and your diagram is unchanged, forget the rest of it, and draw the right triangle from either center to the nearest corner of the box. Then you will see the sides of the right triangle are $r-x$ and $r-2x$, I chose $2x$ for the sides seeing/knowing that these would be the sides of the triangle, so I don't have to deal with fractions. $\endgroup$ Mar 1 at 20:26
  • $\begingroup$ You could make the sides of the square length $s$, substitute in $x=\frac{s}{2}$, go through those lines of algebra, and you'll find the same answer, you'll just have a few fractions to go through first. $\endgroup$ Mar 1 at 20:28
  • 1
    $\begingroup$ Sure, added a picture, hopefully this makes it clear. But try that substitution, work it out, you'll see you end up with the same thing either way. $\endgroup$ Mar 1 at 20:37
  • 1
    $\begingroup$ Drawn in Keynote, it’s free on Macs. You will get the same solution, in terms of r, no matter how you represent the side $\endgroup$ Mar 1 at 23:28

Consider this argument:

I am given some constants $a$,$b$,$c$, and I want to find the value of variable $d$ for which some relation $X$ holds based on the given. I somehow know that this value must be unique.
I start solving this question. I find that in this case, $d = f(a,d) + f(b,d) + f(c,d) - g(a,b,c,d)$ for some functions $f,g$. Then I evaluate those functions by keeping $d$ as it is on both sides. Eventually, after cancelling, I get $d=d \iff 0=0$.

Did I make any mistake? No. I just put the givens into some equations, without deriving something of use. Maybe I missed using some specific property or failed to notice that adding another function $h(a,b,c,d)$ to RHS changes nothing, i.e., $h(a,b,c,d) = 0$ which could have been enough to determine $d$ based on $a,b,c$.

Does this mean $d$ can take all real (or whatever set it was defined on) values? No. It means that $d=d$ holds for my unique value of $d$, which is not of much consequence. It can also be understood as saying that I used a subset of information about $X$, and found a result which is a direct implication of that subset only. However, when I add the missing piece, I get the unique solution.

Note: This does not mean the missing piece is sufficient alone to get the value of $d$.

After this rant, I will proceed to answer the original question.
Here, in your solution have used the following facts about this diagram:

  1. Both circles have the same radius $r$ and are tangent to each other
  2. The yellow region is a square of side $x$
  3. Two vertices of the square lie on a common tangent of the circle.

Now, try to draw a figure armed with this information.

You will understand that there are infinite such squares since you haven't used the fact that:
4. The two other vertices of the square lie one on each circle.

Now, here is a simple solution (actually the same as RobinSparrow):
Draw $DM \perp OO'$, where $M$ lies on $OO'$. Then, by Pythagoras on $\Delta DMO$, $$r^2 = (r-\frac x2)^2+(r-x)^2$$ Verify that this equation relies on all four facts.

Solving, $$r^2 +\frac 54x^2 - 3xr =0 $$ $$\iff (r-\frac 52 x)(r-\frac x2) = 0$$ Obviously, $r \neq \frac x2$, so $\boxed{x = \frac 25 r}$.

  • $\begingroup$ +1 also for you. Can you do a drawing, please? $\endgroup$
    – Sebastiano
    Mar 1 at 20:45
  • 1
    $\begingroup$ I can't do that right now. I only made one additional construction: Drop a perpendicular from $D$ onto $OP$. $\endgroup$
    – D S
    Mar 1 at 20:49

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