For those of you who don’t want to try to answer, try the question at the bottom.

(Assume everything is snug. No overlaps.)

1. Which is a better fit – a small circle in a small square or a large circle in a large square?
2. Which is a better fit – a circle in a square, or a square in a circle?
3. Where can you produce the tighter fit – with two congruent circles in a rectangle, or two non-congruent circles in a triangle?

How can you compare two “fits” to decide which is better?

1. January 18, 2007 am31 9:45 am 9:45 am

I would use the ratio of the difference between inner and outer figure area to the total area as a measure of the fit. The lower this ratio is, the better. You could also think of it as a percent, the percent of the larger figure not filled by the smaller one.

1. These are equal. (All circles are similar and all squares are similar.)

2. Circle in square fits better. It’s (4-pi) / 4 versus (pi – 2) / pi.

3. For congruent circles, the more sides you have to circumscribe around the circles, the better fit you can get. This is true even if you don’t allow concave polygons. So rectangle beats triangle.

Allowing non-congruent circles in the triangle changes things significantly. I guess that the best case is an equilateral triangle with an inscribed circle and then a small circle in one of the points, but I have no proof at all for that.

2. January 18, 2007 am31 10:13 am 10:13 am

I think #3 is the real question for teachers. I suspect you are right, but haven’t solved it myself. Getting the ratio will be a challenge.

Oh, and there are more possible answers to the last question (unnumbered) than just ratios, although that’s what they all really boil down to.

Jonathan

January 19, 2007 am31 12:46 am 12:46 am

#3 is a really cool problem.

I tried it out for the two circles in a triangle case, and, barring any mistakes, (which are quite possible), these are the results:
(1) With an equilateral triangle, taking a central circle and another small circle to one side, the ratio of the area of the circles to the triangle turns out to be pi*(1+1/9)/(3*sqrt(3)), which is about 0.67
(2) Alternatively, you can take an isosceles triangle, and try to fit two circles within it, and solve the optimization problem to maximize the ratio of the two areas. This gives a maximum ratio of about 0.69 (I cheated and resorted to a numerical optimization after formulating the problem). This is, of course, greater than for the equilateral triangle using the central circle, and the triangle obtained this way is not equilateral.
(3) Either way, the ratio is less than what you can get for two congruent circles in a rectangle. For this case, trivially, you can at least get pi/4.

4. January 19, 2007 am31 8:40 am 8:40 am

Wow. I play “how about?” and generate little problems for myself all the time. #3 was one of those. Sometimes they are trivial, sometimes they are way too hard, and sometimes they are in-between, and challenging. I think #3 hit the in-between category.

Thanks for the work. How did you optimize this? I did not even consider the isosceles triangle with cogruent circles case, just assumed the O inscribed in an equilateral triangle, with a second O tucked into the corner was the best we could do.

Jonathan

January 19, 2007 pm31 6:25 pm 6:25 pm

Hi, If it wasn’t too clear from my message, I consider two non-congruent circles in an isosceles triangle.

To formulate the problem, I used a bit of analytical geometry. Consider two circles, first one with radius unity, and center (1+d), second one with radius r and center (d+2+r). Thus, the two circles are tangential, and the triangle will be such that one side is tangent to both circles. I assumed that this line would pass through the origin. The other side is a reflection of this line, and the third side is a vertical tangent to the second circle at x=d+2+2r. Now, r is a function of d, and you can also find the slope of the tangent line, and thus the length of the vertical tangent, all are functions just of d. Then, the area of the circles and the area of the triangle can be found as functions of d, and the ratio can can be maximized. The expression has radicals and I did not want to take the derivative and set to 0, so I just plotted it and found the maximum.

Keep the problems coming. In general, it is much more difficult to formulate good problems than to solve them.

Thanks,

Clueless

6. January 21, 2007 pm31 11:04 pm 11:04 pm

Sometimes formulating good problems is more about altering a problem that is suitable for one audience to make it challenging, but not overwhelming, for another. Part of what you see here (three problems of increasing difficulty), is my attempt to show some of that “altering process” (and also to engage a variety of readers).

I have taught lessons on doing this, and will think about writing up something for this blog.

Thanks for the encouragement (and the nice solution!)

Jonathan