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The planar diagram shows points $A,B,C,D$ (they don't have to be the vertices of any particular kind of quadrilateral) and four circles: $\text{C}_{AB}$ with diameter $AB$, $\text{C}_{BC}$ with diameter $BC$, $\text{C}_{CD}$ with diameter $CD$ and $\text{C}_{DA}$ with diameter $DA$.

enter image description here

$\text{C}_{AB}$ and $\text{C}_{CD}$ are disjoint (no points in common; neither is inside the other).
$\text{C}_{BC}$ and $\text{C}_{DA}$ are disjoint.

Is this possible?

I don't think it's possible. The diagram actually shows four ellipses. When I draw four circles, I am unable to make the pairs of circles disjoint. But I don't know how to prove that this is impossible. I've been trying proof by contradiction, but contradiction eludes me.

Context: I've been trying to crack another question, and my effort led to this question.

Dan
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    Note that there is nothing in your pictures that indicate that AB, BC, CD and DA are diameters (as opposed to just chords) of their respective circles. It's only your description that gives this information. If we remove this constraint then the picture becomes possible, and it stays possible even if we add the constraint that ABCD is a square. – Stef Nov 26 '23 at 19:42
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    Try it with four coins. – Andrew Morton Nov 26 '23 at 20:26
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    @Stef it's quite normal for diagrams to not fully give out all information in the text, and may even not correspond to the actual truth, since it's only for illustration (the non-truth-ness may be due to intentional fudging as not to give out the answer directly, e.g., proving points coincidence, or collinearity, etc., or maybe unintentional) – justhalf Nov 27 '23 at 06:01
  • If $C_{AB}$ and $C_{BC}$ intersect at $B$ and $B'$ then you will find that if $D$ is the same side of $BB'$ as $A$ then $C_{AD}$ contains $B'$, while if $D$ is the same side of $BB'$ as $C$ then $C_{CD}$ contains $B'$ (and if $D$ is on $BB'$ then all four circles intersect at $B'$). So at least one of the two circles involving $D$ circles must intersect both of the other two circles. – Henry Nov 27 '23 at 14:53
  • If points ,,, don't have to be the vertices of any particular kind of quadrilateral, does that not imply that lengths AB and DC, or AB and DC don't matter? – Robbie Goodwin Nov 29 '23 at 22:05

4 Answers4

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Suppose $P$ is a point in the small area in the middle of the drawing. Then $P$ lies outside of the circles, hence $\angle BPA < 90^\circ$, $\angle CPB < 90^\circ$, $\angle DPC < 90^\circ$, and $\angle APD < 90^\circ$. Summing up we obtain $\angle BPA + \angle CPB + \angle DPC + \angle APD < 360^\circ$. On the other hand, note that the angles $BPA, CPB, DPC, APD$ sum up to full angle, thus $\angle BPA + \angle CPB + \angle DPC + \angle APD = 360^\circ$, which gives a contradiction.

timon92
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If $C_{AB}\cap C_{CD}=\emptyset$ then the radius of $C_{BC}$ is greater than the sum of radius of $C_{AB}$ and $C_{CD}$ and, because $\overline{BC}$ is diameter of $C_{BC}$, the center of this circumference is situated in the vertical $BC$. It follows, very clearly, that $C_{BC}\cap C_{AD}$ cannot be empty.

user541686
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Piquito
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Impossible. (Damn I got nerd snipped by this^^).

Imagine a square with the point A, B, D, D as its corners. When the diameters of each circle should be equal to one line of the square, the center of each circle would need to lay on the line of the square.

Thus, the radiuses of the circles would meet in the center of the image.

You can actually see with your eyes, that the line A-B is not through the center of the circle, and thus it is not the diameter.

Valentin Seehausen
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Nov 27 '23 at 18:23
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    You got 'nerd sniped'. Getting 'snipped' is an invasive procedure followed by sitting on bags of ice. – JimmyJames Nov 27 '23 at 19:35
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Similar to Valentin Seehausen's answer, consider a square ABCD. As each circle has a radius equal to half the side length of this square, all of the circles will share a point at the center of this square. If any one side is shortened from this state, the circles adjacent to that side (for AB those would be $\text{C}_{BC}$ and $\text{C}_{AD}$) necessarily overlap because their centers get closer without their radii decreasing. Their radii cannot be decreasing because that would require their diameter endpoints getting closer together.

The same principle applies for shortening both opposite sides: the midpoint of that side, and thus the center of that circle, moves closer to the center of the opposite circle without the endpoints of that side getting closer together, and thus without the radius decreasing.

This proves that the scenario is impossible for any arrangement of A,B,C,D which has them as the vertices of a quadrilateral with at least one pair of parallel sides which are shorter than or congruent with the shortest possible line segment that connects the lines each of those sides are on. At this time, I cannot further this proof. Unless someone beats me to it, I will further it soon.

Disdyakis triacontahedron
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