That's awesome. She wishes her first daddy male parental figure could see this.
"Well, the whole thing is defined by the two points and the three distances, with nothing else going in any other particular direction, so the result has to be round. A circle or a ball or a bunch of circles or something.
If you think of each point-and-distance pair as defining a hollow ball, the result is that the sparkles follow the rules for both balls. Depending on the sizes and positions of the balls, that might be nothing, a point, a circle, or a hollow ball.
This circle is special: it's the biggest circle you can have using the first ball. It cuts the first ball in half at its widest point. That's special because it makes the top slice of the ball the same as the bottom slice. If you put the second point above the first point and kept the rest the same, you'd get the same circle.
Let's imagine that you move your illusion so the two points are on the wall - a smooth vertical part of the wall. You can actually do it if you want. Remember, the shape has to be round, so we don't need to worry about all of it. Whatever points are exactly on the wall already tell us everything about the shape.
So, here's the first point, the second point, a point on the circle to their left, a point on the circle to their right, and we could have gotten this by putting the second point above the first, so let's imagine another 'second point' too. Now, there are four triangles which are all identical in shape although some of them are flipped around. See how they all have a squarish angle, where they meet at the first point? Since they're all the same shape, that angle has to be the same in all of them. Four of them make a complete turn around the first point, so the angle is exactly a fourth of a turn. That's what makes this work.
Any triangle with a square angle at the first point would have worked. The fun part is that a triangle with lengths of 3, 4, and 5 of whatever - fourths of the radius - has a square angle and has lengths that are easy to measure out.
To prove that, we're going to need several triangles like this of different sizes. The shape of all of them is the same, just the size is different. The first one is five times bigger: its sides are 15-20-25 instead of 3-4-5, but it's still a 3-4-5 triangle. The 'something' in 'three somethings' is five. Five something-smallers.
Now, we draw a line from the square point of the big triangle to the other side, so that it makes two square angles against it. That gives us two smaller triangles.
This one has the narrower angle of the big triangle and it has a square angle, so it has the same shape over all.
This one has the wider angle of the big triangle and it has a square angle, so it also has the same shape.
How big are each of these smaller triangles? This one has its 'five side' pressed against the big one's 'four side' which was 20. Twenty is five fours, so it must be measuring in fours.
The other one has its 'five side' pressed against the big one's 'three side', which is 15, five threes, so it's measuring in threes.
The question is, does this diagram actually work? There's two lines which we can measure two ways. Do we get the same length both ways?
The inner line is four threes, which is twelve, and also three fours, which is twelve again.
The 'five line' of the big triangle is five fives, which is twenty-five, and also four fours plus three threes, sixteen plus nine, which add to twenty-five."
If Nico points out the circular argument, he can probably follow a more rigorous proof, and if not, that's also fine.