2
$\begingroup$

Let's say we have two circles whose centers are spaced a fixed $x$ units apart from one another. Both circles have a radius $r$. Our goal is to identify the minimum value of $r$ so that the intersection of the two circles con entirely contain a square of sidelength $w$.

To illustrate this problem, I have attached a diagram below.

Intersecting circles and square

Note: In this illustration, the square's center is at the midpoint of the line connecting the two circle centers, and the squares sides are either parallel or perpendicular to that line.

Thus, if we are given a value of $w$ and $x$, how do we find the minimum value of $r$?

$\endgroup$
  • $\begingroup$ Note that per the illustration, the square's center is at the midpoint of the line connecting the two circle centers, and the squares sides are either parallel or perpendicular to that line. $\endgroup$ – Mark Fischler Apr 29 '16 at 23:20
  • $\begingroup$ Yes that this is correct, I will add this to the question to clarify. $\endgroup$ – ArKi Apr 29 '16 at 23:21
  • $\begingroup$ If we're not considering tilted squares... Let the intersection of the two circles be the points $P$ and $Q$. The center of the square lies in the midpoint of $PQ$, and its vertices lie on the circle arcs connecting $P$ and $Q$. Do you think you can take it form here? $\endgroup$ – Fimpellizieri Apr 29 '16 at 23:25
  • $\begingroup$ Since the position of the centres of the circles (relative to the square) is known, can you find the distance between a circle centre to the two furthest square corners? $\endgroup$ – peterwhy Apr 29 '16 at 23:26
1
$\begingroup$

Take $w<\frac12 x$. Observe that the optimal covering will come when the two cirles meet at the midpoints of the two horizontal sides of the square -- if you shring the radii further then those midpoints are no longer covered. Then we have a simple right triangle, from which $$r_\min = \frac12 \sqrt{w^2+x^2}$$.

Now tak $w>2x$. Then the natural right triangle has sides $(w/2, x/2+w/2,r$ frojm which $$ r_\min = \frac12 \sqrt{2w^2+2xw+x^2} $$

For $w$ roughly equal to $x$ we have to figure out which of these two caes holds.

$\endgroup$
  • $\begingroup$ Hi Mark, thanks for answering! I'm not sure if I entirely understand the first part of your answer - if the circles intersect at the midpoints of the horizontal sides of the square, then part of the square will be covered by the left circle, and part of the square will be covered by the right circle. However, I think the question is asking at what radius do both circles cover the square? Let me know if that's not clear. $\endgroup$ – ArKi Apr 29 '16 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.