Here is a surface that resembles a snowman:


Its equation has the form $f(x,y,z)=0$: if a point $(x,y,z)$ satisfies that equations it is shaded, otherwise it is "left blank". Your goal is to find $f$.

Some constraints:

  • The snowman is composed of three spheres: the "body", the "head" and the "hat".
  • The body and the head are tangent with respect to each other.
  • Half of the hat penetrates in the head.
  • The dimensions and position of the snowman in the space do not count...
  • ...as long as the relative dimensions between the body, the head and the hat are respected.

I suggest to use surfer to solve this puzzle. Surfer is an open-source program that allows you to draw a surface given its equation. Of course you can use another software (or paper-and-pencil) if you want.

Good luck and enjoy this puzzle!

  • 2
    $\begingroup$ Greetings, @melfnt, from one (of seemingly few) who also finds puzzling beauty in mathematical art. I wish there were more of these kinds of multilevel puzzles, where the poser (you, in this case) first solved a puzzle of how to represent something and subsequent solvers get to figure out how that was done. $\endgroup$ – humn Jun 4 '20 at 22:29
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    $\begingroup$ @humn thank you, posting many of these puzzle was my idea in the firs place. I think it is interesting to see how many different opinions this community have about this kind of puzzle: for now it gathered 3 upvotes, 2 downvotes and 1 closevote :) $\endgroup$ – melfnt Jun 5 '20 at 8:50
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    $\begingroup$ Interesting indeed, @melfnt, as would be any constructive comment from someone who finds fault with this kind of puzzle. I sure would like to be reassured that a vote of disapproval isn't simply an allergic or jaded response to mathematics. Thank you also for introducing surfer to those of us who weren't acquainted. $\endgroup$ – humn Jun 5 '20 at 15:25

Assume that the radii of the spheres $\Gamma_1, \Gamma_2, \Gamma_3 $ be $r_1,r_2,r_3$ both from bottom to top respectively. Now, let the sphere $\Gamma_1 $ be centered at the origin with the equation : $$x^2+y^2+z^2-r_1^2=0$$ Now, the sphere $\Gamma_2$ is tangent to the sphere $\Gamma_1$ hence it is centred at $(0,0,r_1+r_2)$. The equation of this sphere will be : $$x^2+y^2+(z-r_1-r_2)^2-r_2^2=0$$ The third sphere will then be centered at $(0,0,r_1+2r_2)$ and hence has the equation : $$x^2+y^2+(z-r_1-2r_2)^2-r_3^2=0$$ Hence the final equation will be $$(x^2 + y^2 + z^2 - r_1^2) (x^2 + y^2+(z - r_1 - r_2)^2 - r_2^2) (x^2 + y^2+(z - r_1 - 2 r_2)^2 - r_3^2)=0$$ The multiplication of all the terms in the final equation is an analogy from the combined equation of hyperbola. To be more clear, let the terms of the equations be $a,b$ and $c$ such that $abc=0$. For this to satisfy, at least one of them must be zero. This condition prevents any point outside the spheres represented by the equations because no sphere will be exits in the region other than their intended region. Contrary to this if we "add" the equations, it will NOT guarantee the existence of spheres only in the intended regions Now, consider a point $(0,r1+k,r1);k>0$. This point lies to the right of the tangent point of spheres $\Gamma_1$ and $\Gamma_2$. Now, no spheres exist here. So, this point will not satisfy the equation $abc=0$ and hence there is no existence of spheres here. If we assume addition of the terms, the other two terms can be adjusted to nullify the whole to zero which cannot and should not occur. Hence, the multiplication preserves the structure by forcing the existence of these spheres only in their intended regions

  • $\begingroup$ Well done! Can you just explain rot13(gur cebqhpg va gur svany rdhngvba)? $\endgroup$ – melfnt Jun 4 '20 at 20:07
  • $\begingroup$ Sorry I didn't understand your (deleted) comment. I meant, can you please update your answer to explain why did you rot13(zhygvcyl gbtrgure gur rdhngvbaf bs gur guerr fcurer gb bognva gur svany rdhngvba) ? $\endgroup$ – melfnt Jun 4 '20 at 20:18
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    $\begingroup$ It is not a mistake, think again about it. I just wanted you to explain it in the answer because maybe it is not completely obvious. $\endgroup$ – melfnt Jun 4 '20 at 20:21
  • $\begingroup$ I used the analogy of finding the pair of asymptotes to a hyperbola where the pair of those lines (asymptotes) are multiplied together to get both the asymptotes of the hyperbola. $\endgroup$ – John Brookfields Jun 4 '20 at 20:22
  • $\begingroup$ Oh, sure @humn. I'm really sorry for not appreciating your effort @[melfnt]. I was carried away solving the puzzle and it didn't cross my mind. It is really nice. $\endgroup$ – John Brookfields Jun 4 '20 at 23:01

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