What's wrong with this picture? Impossible circles.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.