Thu 6 Jul 2006

Geometry Puzzle Solution

A week ago I posted an innocent looking geometry puzzle. Today, I give the solution.

This problem turned out to be a little harder then I remembered (at least, the solution turned out to be longer than I hoped when I wrote it out). I have another couple of problems to post at some point; they should be easier than this one.

To save a bit of space, I won't repeat the problem. Check the older post if you need a reminder. Further, to avoid having to using full-on XHTML and MathML, I will be giving a bit of a verbose explanation, rather than a dozen lines of symbology.

Is There A Solution?

For many mathematics problems, it doesn't hurt to ask whether a solution exists at all, particularly if you cannot immediately see what the solution would be. For this problem, we can fairly easily show that there should be one point satisfying the equal angle requirement on a line connecting the centers of the two circles. Let's address that concern briefly before actually finding the solution. Since the rest of this post kind of puts the "existence" issue to rest by constructing a solution, this may seem superfluous to you. If so, skipping ahead will not lose you anything.

Imagine a point that is really close to the larger of the two circles and on the line connecting the centers. Then the angle between the tangents to the larger circle will be much greater than the angle between the tangents to the other circle (draw a diagram if you don't believe me). Now imagine the same point all the way over near the smaller circle. Unless the two circles are massively different in size, the opposite will now be true: the angle between the tangents to the nearest (smaller) circle will be larger than the angle to the larger circle. So if we slide the point along the line between the two circles' centers, the two angles we care about will switch from having the lefthand one being larger to having the righthand one being larger. This will happen smoothly and so there will be a point just as they are about to switch where the two angles are equal.

[Yes, if you're really keen about your mathematics rigour and have been paying way too much attention in University or College calculus or analysis classes, you are shouting out that I haven't proven this change function is continuous and therefore I can't apply the Mean Value Theorem. Push off! This is my soapbox and I can wave my hands if I want to. If you care about that, you can work out the details for yourself. No, no... come back! If you are in this category, you are probably the only three people reading this and I value your attention. Good point; well raised.]

Constructing The (Second) Simplest Solution

Let's try to work some features about this point we are trying to find. Have a look at the first diagram:

This shows the first simplification we can introduce. Although the problem talks about the angle between the two tangents from P to each circle, because of the symmetry of the whole situation, we could just as easily talk about the angle between one tangent and a line to the centre of the circle. That angle will always be exactly half of the angle between the two tangents. Henceforth, we will therefore deal with one tangent and a line to the center(s) of the circles.

So, looking at figure 1, we want to select P on the central line such that the two marked angles are equal. I have also marked in the centers of the circles and A and B represent the places where the tangents intersect with the circles (so lines AP and BP are at right angles to the radii AX and BX).

Because of the special nature of P (the two marked angles are equal), the two triangles XAP and YBP are similar (the corresponding angles on each triangle are the same). So the corresponding sides on the two triangles are in the same ratio to each other. That means XA:BY is the same as AP:BP and XP:YP.

This is good news. We know the lengths XA and BY from the circles we are given. So we need to find a point P on the line XY such that XP:YP = XA:BY. This is not too difficult. Best demonstrated with the aid of another diagram, I think...

This time, we construct the two radii that are perpendicular to the line XY, one pointing up, one pointing down. Constructing a perpendicular line with a straightedge and compass is not too difficult. Here is one good set of explanations that Google turned up.

The line connecting A' and B' in figure 2 will divide line XY into the required ratio. To see this, we notice that the two angles at P formed by XY intersection A'B' are equal and so triangles A'XP and B'YP are similar again. Therefore XA':YB' = XP:YP, which is what we want.

We have now found a point P that solves the problem.

A quick comment on how this solution "works": In the first diagram, we assumed P existed and worked out some useful properties (inspired guessing, or selective editing of my thought processes, led us to work out the "right" useful properties immediately). In the second diagram, we took some known information (the circles and X and Y) and constructed the point P that satisfied the special properties we wanted.

It is important to realise that A and B in figure 1 are not the same as A' and B' in figure 2. This should be obvious, because the lines A'P and B'P are clearly not tangent to the circles. This slight "change the points" trick makes geometry problems hard sometimes: we worked out something about P, A and B and then we threw away A and B and magically introduced A' and B'. However, this was necessary, because we could construct A' and B', whereas A and B were mysteries, until we knew P (given P, A and B in figure 1 are unique and all the calculations we did will be true for them; but they do not help us find P in the first place).

Finding All Solutions (Including The Simplest One)

It does not take a lot of imagination to realise that there should be other solutions to this problem, although they will not lie on the line XY. If we know P, it stands to reason that we should be able to move a little bit away from P and still keep the angles between the tangents (or between one tangent and the center of the respective circle) equal.

Unfortunately, I cannot remember (or work out again, even more mortifyingly) how to prove this geometrically. Back in January, 1989, just after I had finished high school, I attended a summer mathematics school for Australian students (I had been invited back as part of the smaller "second year" group after getting a lot out of the post-year-11 version). The late Esther Szekeres, a talented mathematician and also an extremely good speaker and educator, gave a memorable series of lectures on geometrical problem solving and this was one of the after-class problems she set. I am fairly sure I was able to find the set of all points geometrically back then, but I have apparently become less intelligent over the years.

Anyway, it is possible to construct the set of all solutions (the locus, to use the correct term) using algebra and a lot of paper (an analytic solution, rather than a geometric one). I won't bore you with the details, but, rather, just give the answer. If you really care and need some hints, drop me an email and I'll help out. But it isn't too difficult if you draw a diagram similar to figure 1 with a point Q away from the line XY.

Assuming the two circles are not the same size, the locus of points satisfying the condition we want will be a circle surrounding the smaller circle. The point P we found in the previous section will obviously be on the circle. The only other solution on the line XY from figures 1 and 2 is on the same diameter as P:

The dotted line in figure 3 is the set of all points we seek (I have intentionally cut off the top and the bottom of the circle to make the image fit on the page). The line that is tangent to both circles will intersect the central line at some point on the far side of the smaller circle.

This is why the previous section only found the second simplest solution: this far intersection point is also obviously a solution, since it uses the same tangents for both circles. No thinking required to see that one!

Constructing the common tangent line is also something I am not sure how to do purely with a straightedge and compass at the moment (it is not at all rigorous to just "do it by eye", since determining the precise contact points on the circles is the hard part). So, I'm not very happy with my solution here, when judged by beauty alone.

By the way, if the two circles are the same size, figure 3 will obviously not apply (the common tangent will be parallel to the line XY, so no intersection). In that case, though, P will be precisely halfway between X and Y and the locus will be a line perpendicular to XY, passing through P (which is constructable with a straightedge and compass, as shown in the link I posted earlier).

The Three Circle Problem

The second "bonus" part of the problem I posted (and this was the problem Esther originally gave us back in 1989) asked how to find a point like P if there were three circles involved.

The solution here is easy if you know the solution for two circles. It's a very common technique in geometric problems: consider the three circles in pairs. Construct the locus for, say, circle #1 and circle #2. Now construct the locus for circles #2 and #3. These two loci will intersect in one or more places. Those intersection points satisfy the condition that the angles between the tangents to each of the three circles will be equal. Everything on the first locus has equal angles for circles #1 and #2, everything on the second has equal angles for #2 and #3. So at the intersection "angle to #1" = "angle to #2" = "angle to #3" (because "angle to #2" is a property of both loci).

Topics: mathematics/puzzles