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?
[Math] area of square between tangent(externally) circles
geometry
Related Solutions
Let's get a figure in here, eh?
We can exploit the fact that the points of mutual tangency of the three inner circles (red points in the figure) form an equilateral triangle; we also know that the red points are the midpoints of the segments formed by joining any two of the three blue points (the centers of the inner circles).
We then deduce that the triangle formed by the red points is half the scale of of the triangle formed by the blue points, and find that the equilateral triangle formed by the blue points has a side length of 6 ft., and that the inner circles have a radius of 3 ft. Using the law of cosines, we reckon that the distance from a blue point to the center of the triangle formed by the blue points is $2\sqrt{3}$ ft. Adding to that the radius of an inner circle, we find that the radius of the outer circle is $3+2\sqrt{3}$ ft.
The area of the outer circle is $\approx$ 131.27 square feet.
Let $r$ be the radius of the smallest circle. Let $O_1$ be the center of the circle of radius 1, $O_2$ be the center of the circle of radius 2, $O_3$ be the center of the circle with radius 3, and $O_r$ be the center of the circle of radius $r$.
If you draw $\triangle O_1O_2O_r$, you can easily verify that it has side lengths 3, $1+r$, and $2+r$. In addition, you can also verify that the cevian $O_rO_3$ to the side with length 3 has length $3-r$. We now have enough information to solve for $r$. If we let $\theta=\angle O_1O_3O_r$, then by the Law of Cosines: \begin{align} 4+(3-r)^2+4(3-r)\cos\theta &= (1+r)^2\\ 1+(3-r)^2-2(3-r)\cos\theta &= (2+r)^2. \end{align} Adding the first equation to twice the second yields the equation $6+3(3-r)^2=(1+r)^2+2(2+r)^2$, which has the solution $r=\frac 67$.
Alternatively, you can use Descartes' Theorem, in particular, the following formula: $$\frac1r=\frac 11+\frac12-\frac13\pm2\sqrt{\frac1{1\cdot2}-\frac1{2\cdot3}-\frac1{3\cdot1}}=\frac 76,$$ so $r=\frac67$.
Best 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.