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You are given a square with sidelength $x$. Draw $4$ circles $\Gamma_A$, $\Gamma_B$, $\Gamma_C$, $\Gamma_D$ in the square with radius $1$, such that no two circles touch/intersect. The circle can't intersect the borderline of the square aswell (it can touch though).

Find the minimal value of $x$ such that there doesn't exists points $A_1$, $B_1$, $C_1$ $D_1$ on $\Gamma_A$, $\Gamma_B$, $\Gamma_C$, $\Gamma_D$ respectively such that $A_1B_1C_1D_1$ is a cyclic quadrilateral (in short, find the smallest $x$ such that $A_1B_1C_1D_1$ never is cyclic).

My progress: I conjecture that the answer is $x=6$, and on the following image you find a construction:

four circles sit inside a square, with a fifth, larger circle crossing through three of the circles. A cloud of points surrounds the drawing. Notably, none of the points land on or within the fourth circle

How I tackled the problem: I let H,I,J be moving points on $\Gamma_A, \Gamma_B$ and $\Gamma_C$. Then I construct the circle and find all possible points in the plane it could reach. The blue dots/region is the reach of the circle. In this specific construction, it seems that the region never intersects $\Gamma_D$, and therefore I think $x=6$ is the minimum. However, this may not be an optimal construction, so I am in doubt.

edited: A second image to prove that conjecture in comment is false: In the image, we see that ELM is an equilateral triangle. We can show that there doesn't exist another circle in the triangle that can't be reached by three points on circles with circumcenters E,L,M and radius one. The "optimal circle" would've been the one centered at F. However, in the image, we see that there exists a circle intersecting our circle centered at F, proving that this construction isn't optimal.

edited x2: I think I found a better construction, this can lower the bound of x=6. No idea why I haven't tried this construction as it seems like a pretty easy idea: 3 circles, one in each corner. Circle in the middle can't be reached.

straight
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  • What’s your attempt, thanks! – user1176409 Jul 19 '23 at 10:47
  • Hello, see above. I edited my progress in the post. – straight Jul 19 '23 at 10:52
  • So wait, what you're doing is kind of ... an adversarial thing? you're looking for the smallest size of square where there is a way to draw the circles such that there is not a way to get a cyclic quadrilateral out of it? Is that the correct way of saying it? – Dan Uznanski Jul 19 '23 at 11:32
  • Wouldn't a more optimal solution be to have the unit circles arranged in an equilateral triangle, with only one of them in the corner of the square, and two touching the opposing two sides? I'm not sure of this. – Jaap Scherphuis Jul 19 '23 at 11:58
  • @DanUznanski exactly – straight Jul 19 '23 at 12:15
  • @JaapScherphuis I'll try it in geogebra – straight Jul 19 '23 at 12:15
  • @JaapScherphuis your conjecture is false. If x=6, then with your construction we can find a cyclic quadrilateral. Is it true that I can't add images to a comment? Is there any way I can show it to you? – straight Jul 19 '23 at 12:28
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    You can edit your post to include the new image. To avoid confusing new readers, it's helpful to make clear that it's an edit (e.g. "Edited to address Jaap's comment:..."). – TonyK Jul 19 '23 at 12:42
  • Thanks for checking out my suggestion. I was going to say that in your original configuration x=6 was not quite optimal because as you can see in this figure there is a gap between the central unit circle and the upper red circle suggesting that the square can be shrunk slightly. Your new configuration is much more promising, though. – Jaap Scherphuis Jul 19 '23 at 13:33

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