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This one has many solutions, a few of which are rather amazing in my opinion.
Suppose O,V, and W are three circles in the plane, none of which intersect each other and all of which have different radii.
Given any two of the three circles, there will be exactly four lines tangent to both circles. Two of those lines will not intersect the segment connecting the centers of the two circles. Those two lines are called the common external tangents of the two circles.
Suppose the external tangents of O and V meet at a point P.
Suppose the external tangents of O and W meet at a point Q.
Suppose the external tangents of V and W meet at a point R.
Prove that P,Q, and R all lie on one line.
By the way: Would people like me to occasionally post nice elementary problems, like this one and the ones Slithe has been posting, or should this blog focus on more advanced mathematics? What subjects are most interesting to you?
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I wanna see the solution 
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I proved it using AutoCAD, does that count? 
Meh, imageshack is being buggy and won't let me upload the picture.
Anyways, this gave me an excuse to refresh my CAD memory, so I guess it wasn't a total loss.
I now patiently await the correct answer.
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I'll wait until tomorrow afternoon and then post the answer if no one gets it by then
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I think I'm close to a pure algebra plug and chug solution, but I'm much more interested in seeing if there is an elegant solution.
Unfortunately I must now go to school, so I will continue this later. I'm sure by the time I get back polemarch will have found at least 3 different ways to solve it, thus making the problem look ridiculously easy.
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Yea I was working on cranking out some algebra too...boring
Trying to find a nicer solution but I really should do my calc homeowork T_T
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Analytical way of solution is the most obvious but not very elegant. I will post it later.
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It is easy to show that point of intersection of 2 external tangents lies on the line connecting centers of circles and the distance between center of larger circe and intersection point is
where l_{12}
is the distance between centers of circles, r_1 and r_2 are radii of circles. Using this we can find the coordinates of the intersection point
where x_1 and x_2 are coordinates of the centers of circles. Similarly
Let us consider 3 circles now, 1st is the largest and 3rd is the smallest.
P - point of intersection of external tangets of 1st and 2nd circles
Q - point of intersection of external tangets of 1st and 3rd circles
R - point of intersection of external tangets of 2nd and 2rd circles
The coordinates of these points are as follows (we always can choose coordinate system so that x_1=0, y_1=0, x_2=0)
Let us consider vectors PQ and Pr and show that they have the same direction (it means these points lie on 1 line). The coordinates of these vectors are
Last equations are equal.
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Yeah... once you realize that WQ is a multiple of WO based on the radii of those circles, it's pretty easy to churn out an algebraic solution. I started to do one with Cartesian coordinates then gave up out of tediousness.
Something slightly nicer (at least in terms of notation) is that you can use vectors to show that PR and RQ have the same direction by expressing them both as multiples of WV and WO.
There must be some nice solutions involving reflections or something... but I don't see any. Sorry to disappoint you, Slithe. Looking forward to seeing other solutions.
edit: Nice, 15vs1 - I'm impressed at the writeup.
Muirhead: I think this difficulty level is reasonable, with harder ones occasionally mixed in if you're intending to do it regularly. As for the types of problems, go with your tastes. I personally am not really a fan of geometry problems for the most part because I'm not very good at visualizing them, so I tend to resort to analytic geometry, which I don't find very satisfying. (Judging by your sig, I guess you disagree!) Of course, there are some problems that are really nice; or it'll all be worth it if you or someone else posts some beautiful solution to this one.
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I don't have much of an intuition for geometry either, although I do find it interesting. I believe the answer I got via brute force was pretty much the same as 15vs1.
Currently, my intuition says that there's some way to use some crazy triangle properties to solve this elegantly. I could just be way off base though...
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More advanced mathematics would be interesting.
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I admire your tenacity in working out the coordinates, 15vs1 . I wouldn't have the strength to carry it through.
+ Show Spoiler +Perhaps the simplest way to do this is with Desargues' theorem, but I doubt everyone here knows that theorem. I leave it as an exercise to whoever knows it to fill in that proof.
Here is the elegant way:
+ Show Spoiler +Imagine the plane containing the three circles as lying in three dimensional space. Suppose we are looking down on the plane from above, as we usually do. Construct three cones with bases equal to the three circles. Make sure that each cone is pointing up, towards us. Make sure each cone is right and that the height of each cone is equal to the radius of its base. This means that all three cones are similar. Denote the tip of the cone with base O by A. Denote the tip of the cone with base V by B. Denote the tip of the cone with base W by C. Now, we claim that the line through A and B passes through P. Why is this true? Well, notice that there is a dilation centered at P which takes circle O into circle V. Because all the cones are similar, this dilation also takes the cone with base O into the cone with base V. Thus, it takes A, the tip of the cone above O, into B, the tip of the cone above V. Similarly, we have that the line through A and C passes through Q, and the line through B and C passes through R. Thus, P,Q, and R all lie in the plane that contains the points A,B, and C. They also lie in the plane that contains the circles. The intersection of two planes is a line
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OK I think I have a cleaner method then brute force coordinates for solving this that uses more triangles. I didn't check whether my solution works yet, but have a look.
+ Show Spoiler +So we have the three circles O, V, W, and the three intersection points P, Q, R. I'm going to make one assumption that I should prove first but do not feel like doing right now. The assumption is that I know the lengths of OP, OQ, and VR, via the ratios of the circle radii. For example, |PO| = |VO| * (radius of O) / (radius of O - radius of V). Now for the next part, here's a picture to help the visualization. If and only if P, Q, and R are on a straight line, then the areas of the triangles OPQ and OQR have to add up to equal the area of the triangle OPR. Given this, we just need to calculate the areas of the triangles, which is not difficult because we can extrapolate most of the information. Right off that bat we should know the lengths of OV, VW, OW. I also assume that my above equation is correct, in which case we should know OP, OQ, VR. From here on out, it's just simple trigonometric calculations, i.e. law of sines and cosines, heron's formula, to show that the areas add up correctly.
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Slithe that works
I'm not sure it is nicer than coordinates though.
I don't like causing people pain, so all problems I post should have short, hopefully cool/interesting solutions. I just posted the answer to this one under spoilers.
Also, thanks for the feedback everyone.
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Great solution. I would never find it but anyway i think you should give more time. Elementary math is not worse than advanced if the problem is interesting so i would be glad to see here any types of problems.
On January 30 2008 07:41 Muirhead wrote:
I don't like causing people pain,
I wish all professors did not like it too.
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How about by Menelaus it's obvious
Eh I'm late but I was just reading your other blog post
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