[superprismatic]

What conditions must be put on 4 concentric

circles so that it is possible to draw a

rectangle having one corner on each circle?

[curr3nt]

R_x = radii of circle x

R₂² - R₁² = R₄² - R₃²

Where R₄ > R₃ and R₂ > R₁ R₂ and R₃ have to be one of the inner circles.

edit - example I was playing with had one of the rectangle edges along the diameter of the smallest circle.

Edited March 9, 2011 by curr3nt

[Guest]

Can you elaborate a bit...As it's worded, I would think that one could just draw four circles..in a square configuration and overlay a rectangle?!

[benjer3]

Colbizzy: By concentric, it means they all have to have the same center point, like a target.

[curr3nt]

Slight update

R_x = radii of circle x

R₄² + R₁² = R₃² + R₂²

Where R₄ > R₃ > R₂ > R₁

[Guest]

I apologize in advance if I am proceeding on erroneous assumptions here. I’m assuming that 1) Concentric means only “sharing a common center” and not necessarily that the circles graduate in size, and 2) That we can draw our circles in more than one plane.

In this case, as long as the circles are concentric, on separate planes, and the same size, the circles themselves would all intersect at two points, like lines of longitude. So any rectangle with a corner in the intersection would meet the criteria. I’m sorry if this isn’t what you meant. I can probably whip up a more geometric answer.

[superprismatic]

I apologize in advance if I am proceeding on erroneous assumptions here. I’m assuming that 1) Concentric means only “sharing a common center” and not necessarily that the circles graduate in size, and 2) That we can draw our circles in more than one plane.

In this case, as long as the circles are concentric, on separate planes, and the same size, the circles themselves would all intersect at two points, like lines of longitude. So any rectangle with a corner in the intersection would meet the criteria. I’m sorry if this isn’t what you meant. I can probably whip up a more geometric answer.

The problem intent is that there are 4 differently sized concentric circles in the plane. I hope you enjoy working on this problem.

Welcome to the Den! Your out-of-the-box way of looking at things will help you with many of the puzzles here.

[curr3nt]

Radius values of 7, 15, 20 and 24

Using

c² = a² + b²

r₁² = x² + x²

r₂² = x² + y²

r₃² = x² + z²

r₄² = y² + z²

Since

x² + x² + y² + z² = x² + y² + x² + z²

then

r₁² + r₄² = r₂² + r₃²

is this not a condition you are looking for?

[superprismatic]

Radius values of 7, 15, 20 and 24

Using

c² = a² + b²

r₁² = x² + x²

r₂² = x² + y²

r₃² = x² + z²

r₄² = y² + z²

Since

x² + x² + y² + z² = x² + y² + x² + z²

then

r₁² + r₄² = r₂² + r₃²

is this not a condition you are looking for?

Nice picture! Sorry I didn't respond to you yet, but I'm not sure

that you solved the general problem. I'm working to see if your

solution is general, but I haven't finished yet. For what it's

worth, I tried and failed to find a counterexample. I even tried

examples where the sides of the rectangle had unusual slopes.

So, I think you are correct, but I don't have an actual proof.

But, I am a little short on time right now, so I'll work on it when

I can. It would help if you could send me your proof in more detail.

But, even if you don't, I'll eventually get back to you. Thanks

for your interest in the problem!

[curr3nt]

Radius values of 7, 15, 20 and 24

***change the left x to a w***

Using

c² = a² + b²

r₁² = w² + x²

r₂² = w² + y²

r₃² = x² + z²

r₄² = y² + z²

Since

w² + x² + y² + z² = w² + y² + x² + z²

then

r₁² + r₄² = r₂² + r₃²

Same proof works for if the center isn't inside the rectangle. Look at the picture and slide either the left or bottom line to the opposite side of the center. In those cases w or x still exist between the rectangle and the center.

[benjer3]

Very nice Curr3nt. It even works for a square with only a corner touching the inner circle and two on two congruent circles.

Odd slopes don't matter because you can always turn the picture so the rectangle is at right angles to the frame.

[curr3nt]

Shows different lengths for w, x, y and z on the blue lines (see first picture for labeling, had to use MS Word to make this so labeling...meh)

Also shows the three types of rectangles, center inside rectangle, rectangle partially in smallest circle and only corner on smallest circle.

Equations with w, x, y and z hold up for all three types since the center of the circle is the center of the rectangle contained by the smallest circle.

edit - wording

Edited March 10, 2011 by curr3nt

[superprismatic]

curr3nt's solution is indeed general. And he did it quickly!

Nice work!