This problem was sent to Einstein in 1930s, when he was at the height of his fame. At the time, general relativity was being widely dismissed in the media as the bizarre work of a madman. People evidently didn't appreciate his genius, as they considered this puzzle worthy of his time and yet it was easy enough for little old me to solve. Einstein, of course, was never too proud to approach a problem, even though such simple geometry that to him must have seemed like child's play after the Riemannian geometry he invoked in the theory of relativity. He duly solved the problem and returned the answer.
Anyway, enough of the historical background. I read about this problem in a biography of the great man, but no solution was presented. I thought it was worthy of solution and you might be interested too. I'll pose the problem here and the solution in a seperate node, in time-honoured fashion. Have a go yourself before spoiling the fun!
Here we go: You have two mutually-tangential (just touching) circles of radius 8 and 2. Find the equation of the non-trivial tangent between them. Examine the poorly-rendered ASCII art below for clarification.
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Have a go yourself, if you like. It may help you understand the solution, which I have generously presented below.
Note that there are two non-trivial tangents. Ultimately, you can choose either. Symmetry ensures that the problem is much the same.
This is my solution. I'm sure it isn't The Ultimate SolutionTM but it gives you a nice elegant answer and it's easy to follow. So, let's go.
I'll be considering the general case for two circles (named A and B) of radius a and b respectively. WLOG, assume a > b (just to make things neat). Assume also (in a classic case of WLOG) that circle A is centered at the origin.
Imagine a trapezium formed from the following four points:
- The centre of A, at (0,0).
- The center of B, at (a + b, 0).
- The point of contact of the tangent line with A.
- The point of contact of the tangent line with B.
It should look (something) like this:
W ____ X
| | b
a | / Y
|θ/ a + b
I've lettered the points, and marked in the appropriate lengths:
WZ: The line joining the centre of A to the tangent line clearly has length a, since it is a radius.
XY: Likewise the line from the centre of B to the tangent line has length b.
YZ: The line joining the two centres has length a + b.
WX: We don't know the length of the top (although it's staightforward to find), but it's not important. This side is part of the tangent line, remember.
Note I also added the angle θ. Since the tangent to a circle forms a right-angle with its radius, WZ is perpendicular to the tangent line. Therefore, finding θ allows us to find the gradient of the tangent line. Let's find θ now.
Form a right-angled triangle by drawing a line perpendicular to WZ that meets Y. This point on WZ will be called D. The triangle should look like this:
| \ a + b
a - b
As before, YZ is a + b. The reason for DZ having length a - b should be obvious from the first diagram.
Now, if you managed to decipher my more-than-clumsy diagrams so far, you're in clear waters now. We can easily find θ because
cos(θ) = a - b
a + b
from the above diagram. Now let us denote the equation of the tangent line we've been looking for all along. It's a straight line of course so we'll say: y = mx + c. m is of course the gradient, and c the y-intercept. The gradient of WZ is then tan(θ), so the gradient of WX (perpendicular to WZ) must then be -1 / tan(&theta); more simply, -cot(θ). WX is part of the tangent line, so
m = -cot(θ).
Substituting in this value of m, the equation of the tangent is:
y = -x · cot(θ) + c
c = y + x · cot(θ)
So if we could just find a specific point on the tangent line, we can solve the above equation and find c. Let's find that point then!
Both W and X actually lie on the tangent line, but let's use W because it's easiest. From the first figure, it's easy to form the following right-angled triangle:
a / |
The base of this triangle is of length a · cos(θ); it's height is a · sin(θ). Remember we initially set Z = (0,0), so W = (a · cos(θ), a · sin(θ) ). There's our point!
Return to our equation for the tangent line:
c = y + x · cot(θ)
Substitute in our values for x and y from the point W:
c = a·sin(θ) + a·cos(θ) · cot(&theta)
= a( sin(θ) + cos(θ) / tan(θ) )
Examine the last term here: cos(θ) / tan(θ). Recall that, by definition, tan(θ) = sin(θ) / cos(θ). Therefore:
-------- ≡ ---------
Substituting this back into our equation for c:
c = ------ ( sin²(θ) + cos²(θ) )
Recall the famous trigonometric identity:
sin²(θ) + cos²(θ) ≡ 1.
Therefore, the brackets in the above equation vanish, leaving
c = a / sin(θ) = a · csc(θ).
And there we have it. Elegant equations for θ, m and c. We have the equation of the tangent. We can easily plug in the original values of a = 8 and b = 2 and get the specific equation requested. Alas, I have no calculator to hand.
I like to think Einstein himself solved it in this manner, but I haven't been able to track down his original solution. Feel free to /msg me if any of this is unclear and I'll do what I can to resolve it.
This has been another HTML Nightmare. Please be sympathetic.