This article is specifically written to answer the geometry question: "What is the radius of a circle drawn in the space between a circle of radius 1 unit, and the corner of the enclosing square?" To better explain the question, and then answer it, I have drawn the diagram shown below. The question states that the radius of the larger circle is 1 unit of length, and that it is enclosed in a square. What is the radius of the smaller circle drawn in the space in the corner region? (The question was asked in a GCSE workbook, aimed for children aged 14-16, although the geometry and algebra became more complicated than I had expected.
By symmetry, the angle at A = 45 degrees, so triangles ACG and AEH are right-angled isosceles triangles.
AH and EH are equal to the radius of the larger circle = 1 unit
By Pythagoras' Theorem, length AE = √2
Length AD = AE - DE = √2 -1
However, length AD is not the diameter of the smaller circle. The diameter of the smaller circle is BD, not AD. We are still making progress, nonetheless.
Next, consider the ratio of the lengths AD:AF.
AD = √2 -1 as we showed earlier
AF = AD + DF = (√2 -1) + 2 (the diameter of the circle) = √2 + 1
So the ratio AD:AF = √2 -1 : √2 + 1
And the fraction AD/AF = √2 -1 / √2 + 1
Having combined all terms over the same denominator, we can now simplify (√2 -1)(√2 +1), since (a+b)(a-b) = a^2 - b^2
BD is the diameter of the smaller circle, BD = 0.216... Comparing this with the diameter of the larger circle, which is 2.00, we can see that the smaller circle is around 10% of the diameter of the larger one. This surprised me - I thought it was larger.
In future posts, I'll look at other arrangements of circles in corners - in particular the quarter-circle in the corner (which, as a repeating pattern, would lead to a smaller circle touching four larger circles in a square-like arrangement), and a third-of-a-circle in the corner of a hexagon. I'll then compare the two arrangements (in terms of space filled) and also check against any known alloys, looking at the ratios of diameters to see if I can find a real-life application.
The diagram isn't perfect, but I'm better at using a pencil and compasses than I am with drawing geometric shapes in Photoshop. The larger circle has centre E, and DE = EF = 1 unit. What is the radius of the smaller circle, with centre C (BC = CD = smaller radius).
By symmetry, the angle at A = 45 degrees, so triangles ACG and AEH are right-angled isosceles triangles.
AH and EH are equal to the radius of the larger circle = 1 unit
By Pythagoras' Theorem, length AE = √2
Length AD = AE - DE = √2 -1
However, length AD is not the diameter of the smaller circle. The diameter of the smaller circle is BD, not AD. We are still making progress, nonetheless.
Next, consider the ratio of the lengths AD:AF.
AD = √2 -1 as we showed earlier
AF = AD + DF = (√2 -1) + 2 (the diameter of the circle) = √2 + 1
So the ratio AD:AF = √2 -1 : √2 + 1
And the fraction AD/AF = √2 -1 / √2 + 1
What is the remaining distance between the circle and the origin?
Look again at the larger circle, and the ratio of the diameter to the distance from the corner to the furthest point on the circle?
The fraction AB/AD is equal to the fraction AD/AF. This fraction describes the relationship between the diameter of a circle and the additional distance to the corner of the enclosing square. The diameter of the circle is not important, the ratio is fixed.
So we can divide the shorter length AD in the ratio AD:AF, and this will give us the length AB and (as we already know AD) the diameter of the smaller circle, BD.
To express it more simply and mathematically: AD/AF = AB/AD
Substituting known values for AD and AF, this gives:
AD/AF = AB/AD
(AD^2) / AF = AB
(√2 -1)^2 / (√2 + 1) = AB
Evaluating:
(√2 -1)^2 = 3 - 2 √2 = 0.17157...
and:
(√2 + 1) = 2.4142...
Now: BD (diameter or circle) = AD ('corner' of larger circle) - AB ('corner' of small circle)
Substituting values for AD and AB, and then combining terms over the same denominator, we get:
So we can divide the shorter length AD in the ratio AD:AF, and this will give us the length AB and (as we already know AD) the diameter of the smaller circle, BD.
To express it more simply and mathematically: AD/AF = AB/AD
Substituting known values for AD and AF, this gives:
AD/AF = AB/AD
(AD^2) / AF = AB
(√2 -1)^2 / (√2 + 1) = AB
Evaluating:
(√2 -1)^2 = 3 - 2 √2 = 0.17157...
and:
(√2 + 1) = 2.4142...
Now: BD (diameter or circle) = AD ('corner' of larger circle) - AB ('corner' of small circle)
Substituting values for AD and AB, and then combining terms over the same denominator, we get:
Having combined all terms over the same denominator, we can now simplify (√2 -1)(√2 +1), since (a+b)(a-b) = a^2 - b^2
BD is the diameter of the smaller circle, BD = 0.216... Comparing this with the diameter of the larger circle, which is 2.00, we can see that the smaller circle is around 10% of the diameter of the larger one. This surprised me - I thought it was larger.
In future posts, I'll look at other arrangements of circles in corners - in particular the quarter-circle in the corner (which, as a repeating pattern, would lead to a smaller circle touching four larger circles in a square-like arrangement), and a third-of-a-circle in the corner of a hexagon. I'll then compare the two arrangements (in terms of space filled) and also check against any known alloys, looking at the ratios of diameters to see if I can find a real-life application.
Nice post, and an an elegant solution.
ReplyDeleteI can see another way to work this out which is quite fun. If you first imagine putting an identical circle in the upper right corner of the unit square, then adding two smaller circles in the remaining spaces in these corners, then two more, and two more, and so on, the infinite sum of the diameters of all these ever decreasing circles should be equal to the diagonal of the square. Since every circle is in a constant ratio to the next largest, this sum is easy to calculate.
In your example, with unit radius for the large circle and radius r for the smaller one in your diagram, this gives us:
2 + 2(2r) + 2(2r^2) + 2(2r^3) + … = 2(sqrt(2))
1 + 2r(1 + r + r^2 + r^3 + …) = sqrt(2)
2r/(1-r) = sqrt(2) - 1
This can be rearranged to give you the answer:
r = (sqrt(2) - 1)/(sqrt(2) + 1)
Nice example, but isn't the solution overtly complex? If we denote the small radii with r, all it takes is to notice that AC = r*sqrt(2), CD = r, DE = 1, AE = sqrt(2). Therefore, on one hand we have AD = AE - DE = sqrt(2) - 1, and on the other: AD = AC + CD = r(sqrt(2)+1). So r(sqrt(2)+1) = sqrt(2)-1 and r = (sqrt(2)-1)/(sqrt(2)+1).
ReplyDeleteI like it. Short, simple and straightforward - I like the way you've broken down the lengths of the diagonal instead of comparing their ratios.
ReplyDeleteHi David - I get the result being 0.343... not 0.216 - so not so small as you thought?
ReplyDeleteHi David, I get 0.343... too. Thanks for sharing the problem. it's a good one.
ReplyDeleteThanks, John - I will revisit my calculations and check back with you!
ReplyDeleteWhat happens in 3-D? N-D? Bear in mind that the length of the longest diagonal of an 'N-cube', ie a 4-D cube, 5-D cube etc. is given by the square root of the dimensionality. (So a 4D hypercube has a longest diagonal of exactly 2 units, always a bit of a surprise!). You'll find that although the diameter of the big circle and its higher dimensional remains constant at 2 units, that of the smaller circle increases without limit as the number of dimensions increases. This gets very weird & counterintuitive!
ReplyDeleteAlso think about the infinite number of circles you get as you approach the corner of the big square as you go 'up the dimensions'!
I say to my student: Always draw the task first. Messure ecsact the little and the big cirkle.Its easy to se that about 10% is very wrong. He should start all over again when he got that answer. He should compare the answer with the reality.
ReplyDeleteI just draw two circles and the difference is about 5,9 times bigger diameter on the big compared to the smaler one. 5.9 : 1
ReplyDelete