What's wrong with this picture? Impossible circles.

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 $$DA$$.

$$\text{C}_{AB}$$ 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.

• 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
Commented Nov 26, 2023 at 19:42
• Try it with four coins. Commented Nov 26, 2023 at 20:26
• @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) Commented Nov 27, 2023 at 6:01
• 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. Commented Nov 27, 2023 at 14:53
• If points š“,šµ,š¶,š· 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? Commented Nov 29, 2023 at 22:05

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.

• This is a fantastic answer. Commented Nov 28, 2023 at 4:09

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.

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.

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.