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Two externally tangent circles,have a square between them,standing on the same base as the two circles.The circles have a radius of 1 unit each.The top two vertices of the square are touching one circle each.How do I find the area of the square?

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I forgot to add thay the top two vertices of the square are touching one circle each.And yeah,a picture was given along with the question in the original source.I had to describe it myself. –  rahul May 13 '13 at 4:59
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1 Answer

A picture would help. I hope it will be clear how to draw it.

Let the two circles have equations $(x+1)^2+(y-1)^2=1$ and $(x-1)^2+(y-1)^2=1$. So they are tangent to the $x$-axis, and symmetrical about the $y$-axis. The centre of the bottom side of the square is then at $(0,0)$.

Let the square have side $2t$. Then the point $(t,2t)$ is on the right-hand circle.

It follows that $(t-1)^2+(2t-1)^2=1$. This simplifies to $5t^2-6t+1=0$. The relevant root is $t=\frac{1}{5}$. So the square has area $\frac{4}{25}$.

Remark: We could strip away the coordinatization, and write the solution using only the Pythagorean Theorem. But coordinatization is a (provably) powerful tool in geometry, so one might as well use it.

If we really want to use Pythagorean Theorem only, let $O$ be the centre of the right-hand circle, and let $P$ be the point where the top right corner of the square meets the circle. Drop a perpendicular from $O$, towards the $X$-axis, and let $Q$ be the point where this perpendicular meets the top edge of the square, extended.

Then $PQ=1-t$ and $OQ=1-2t$. By the Pythagorean Theorem, $(PQ)^2+(OQ)^2=(OP)^2$, and therefore $(1-t)^2+(2-t)^2=1^2$. Expand, and continue as in the main solution.

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I can't understand how you did guess what "square between them" and "...the same basis as the two circles" actually mean, and why do you think it's middle section (segment parallel to a pair of sides through the diagonal's intersection) must be on the common tangent to both circle. I've taught high school mathematics and such a wording as the OP's problem's is beyond anything I could tolerate. –  DonAntonio May 13 '13 at 4:44
We will know after a while whether my interpretation is the intended one. But I am reasonably confident that it is. I thought that OP did a not too bad job of trying to describe a picture. –  André Nicolas May 13 '13 at 4:52
It would still be nice if you could write the solution using the pythagorean theorem,but thanks for this answer anyway,and I forgot to add that the top two vertices of the square are touching one circle each. –  rahul May 13 '13 at 5:06
Well, reading the OP's comment added a few minutes ago I realize I would never, ever have guessed the above: I thought that either the square's upper vertices are the circle's centers or else that the upper and lower sides both intersect each circle once...my adivinatory skills suck! –  DonAntonio May 13 '13 at 5:08
@rahul: Yes, I used that fact in the solution. The other top vertex (the one on the left) touches the left circle af the point $(-t,2t)$. –  André Nicolas May 13 '13 at 5:09
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