POTW 3!

Points D and E are in fact the same point. (Or equivalently, the angle subtended by DEA is zero.) Thus there is no contradiction to having ADB and AEC both be right angles.

It would take me more than a blog comment to prove this, but that is what my intuition tells me.

Agree with #1: if everything were drawn correctly, both D and E would be symmetrical to A; i.e., they are both the bottom point where the two circles overlap, and ADE isn't a triangle, its a line.

In fact, were the diagram correctly drawn, it would have a bunch of very cool properties: XAY is a right angle with AX tangent to circle Y, and AY tangent to circle X; XY bisects the line A[DE], where [DE] is the correct intersection point of the circles symmetric to A; triangle ABC is similar, and exactly double in lengths, to triangle AXY; XY is parallel to BC; A[DE] is an altitude of triangle ABC; and so on.

What's most interesting to me is that this is all completely general: the two circles do not have to be the same size, so long as there's a true (non-degenerate) overlap between them.

Did I miss the POTW 2? I don't see the blog entry here.

The short answer to POTW2 is "divide by zero makes anything possible." X=10 is a solution, and 4x-40=0

If you had to ask exactly where things went wrong, it would either be to say that the diameters you thought you drew weren't actually diameters, or that the "line segment" BC wasn't actually a line segment but two with very small (but nonzero) slope.

I agree with the above posters; the only way for this to be true is if D and E are the same point, which means that the diagram must have one of the circles as slightly elliptical.

The simplest proof here I have is this: if AEC is a right angle perpendicular to BC, then AE BC at E. Similarly, AD bisects BC at D. Two bisectors of a line segment that run through the same point describe the same line, therefore AE = AD. And if AE = AD, then it follows that D = E; of course, if D = E then ADE is not a triangle but rather a line segment.

I have a feeling though that my proof may assume the thing I'm trying to show. Looking forward to cleverer answers (I did find a way to prove that AXE and ABE are similar triangles, but I couldn't seem to make anything with that).

If you had to ask exactly where things went wrong, it would either be to say that the diameters you thought you drew weren’t actually diameters

This is probably what I would say. If you look carefully, they don't go through the center of the dots representing the center of each circle.

Which I think means Jason wanted people to see that. He could have not drawn a dot at all or put the dot slightly off center and then had each line go through it 'correctly,' and I think that might have been more visually deceptive. By drawing it this way, he either intentionally or unintentionally left the reader an extra clue.

If you had to ask exactly where things went wrong, it would either be to say that the diameters you thought you drew weren’t actually diameters

In geometry proofs, the diagram as drawn has no bearing on the proof. The drawing is a representation of the scenario described in the proof, and the wording takes precedence over the drawing. Thus we are supposed to accept that AB and AC are in fact diameters, even if the drawing is slightly off, because that is what is explicitly postulated in this "proof".

@GAZZA: That proof assumes that the circles are of equal diameters. We can't assume that that is true (although the proof would be a lot easier if it were). In the general case, we can't conclude that AD/AE bisects BC. Whereas we can conclude that both AD and AE are both perpendicular to BC.

And I think the latter point is the key: in Euclidean geometry, given a line and a point not on the line, exactly one line is perpendicular to the given line. So if AD and AE are both perpendicular to BC, then D and E must be the same point.

In geometry proofs, the diagram as drawn has no bearing on the proof. The drawing is a representation of the scenario described in the proof, and the wording takes precedence over the drawing. Thus we are supposed to accept that AB and AC are in fact diameters, even if the drawing is slightly off, because that is what is explicitly postulated in this “proof”.

Something has to give, though, because the wording also includes the triangle AED. If we accept that AED is a triangle and not a line segment, then we have to say that AB and AC are not diameters after all, or that BC is not a line segment. There are reasons to prefer any of those three in some context or another. One reason might be that it's the illusion in the diagram that gives anyone an inkling that there is a triangle there in the first place. In that way it's a little like the missing square puzzle, which also relies on faulty intuitions about drawing, but in the latter there's only really one mistake, which is calling the large figure a "triangle" in the first place.

@Eric Lund

That proof assumes that the circles are of equal diameters. We can’t assume that that is true (although the proof would be a lot easier if it were).

BTW we also can't assume that the diameters AB and AC are perpendicular, as that's not stipulated in the text, though the diagram makes it look as if they are.

And I think the latter point is the key: in Euclidean geometry, given a line and a point not on the line, exactly one line is perpendicular to the given line. So if AD and AE are both perpendicular to BC, then D and E must be the same point.

Agreed, and that seems to work even if AB and AC are not perpendicular. But this result is based on the method of the text (showing that ADE and AED are right angles), and the diagram appears to show an inconsistency between that result and the "fact" that BC does not intersect the intersection of the two circles. So it would be nice to have an independent proof that BC does intersect that intersection. I guess that's what you had in mind when you said it would take you more than a blog comment to prove it. I'm terrible at geometry, so I won't even try!

The first thought that flashed in my head when I saw this problem was "Wait a second, Triangle AED DOES have two right angles, at the limit as length DE goes to zero." So it's not really a false proof, is it?

Suppose we called the bottom circle intersection point F.

Note the vertical line AF, and draw the two horizontal lines perpendicular to AF which intersect A and F, respectively (AF bisects them). By symmetry, within a given circle, A,F, and the other two points where the horizontal lines intersect the circle define a circumscribed rectangle -- one each centered on X and Y. The diagonal of a circumscribed rectangle is a diameter; two of those diagonals are AC and AB. Therefore D,E, and F are all the same point.

@BDoyle

The conclusion was that "AED is a triangle with two right angles". I guess if you count a degenerate triangle as a triangle then you could say the proof is valid. If the accompanying text was phrased differently, we could take the proof as part of a reductio ad absurdum, where we assume a non-degenerate triangle, and then prove that it's degenerate. Here the author implies that the result is absurd (because he's talking about a non-degenerate triangle), but just revels in the absurdity instead of drawing an appropriate conclusion from it.

@Another Matt

Thanks for that proof. I couldn't immediately see quite how you got to the final sentence from the previous one, given that we have to be careful to distinguish between what follows from the original stipulations and what follows independently from your new stipulations. So can I take the liberty of adding some extra steps to spell it out in laborious detail, if only for my own satisfaction...

The diagonal of a circumscribed rectangle is a diameter. (I wasn't familiar with this theorem, but it seems intuitively obvious, and I'll take it for granted.) B and C were originally defined as the opposite ends of the diameters from A, and now it follows that they are also the opposite corners of the rectangles from A, i.e. the lower outer corners. By your stipulation, the bottom sides of the two rectangles constitute a single straight line intersecting F, so BFC is a straight line. Therefore, BFC is our original straight line BC, and BC intersects F.

One might raise a question as to just how independent your proof is from the original one, since they both make use of related facts about circumscribed angles. But it's probably as independent as we can reasonably expect.

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I see that POTW 4 is out. Since Jason hasn't put up a separate post about that, I will answer it here. The second step, where it is asserted that 2 ln cos α > ln cos α, assumes the very thing we are trying to prove. If 0 < cos α < 1, then ln cos α is negative, so the sense of the inequality will be incorrect. The inequality as stated is only true if cos α > 1.

Although POTW 4 is one of the easier ones so far, I do like the misdirection about trigonometry. It invites you to inspect shenanigans with the cos() function or the phase of the angle, where the crux is really the behavior of the natural log.

Also wikipedia has a nice writeup on the Bertrand Probability Paradox. One of the solutions really is better than the others. The problem makes you start with the circle, but if you start with a tangled mess of evenly-distributed intersecting lines and draw a circle anywhere on it, the bisecting radius solution can be deduced.

Yes, this one was easy; 2 ln cos alpha > ln cos alpha only if ln cos alpha >= 0. Since ln x <= 0 for any x < e, and since cos alpha is at most 1 which is indeed < e, then it follows that 2 ln cos alpha <= ln cos alpha. The rest proceeds ln cos^2 alpha <= ln cos alpha, and therefore cos^2 alpha <= cos alpha, and therefore cos alpha <= 1. This is of course true, but not particularly exciting.

I must admit to being stumped by the chord in a circle one though.

Ugh. I meant, ln x <= 0 for any x <= 1. Oops. Rest is still correct.

In POTW4, if you define a random chord as a line segment connecting two random points on a circle, then method one is correct.

In method two, the problem is that the point M is more likely to be the midpoint of one of these random chords the closer it is to point C. Basically, as M moves towards C in equal increments, the endpoints of two successive chords cover a larger arc of the circle, but for equal probability you want the increase in the arc to be constant. Sorry, lol, it is clear in my head but I don't explain well.

I haven't thought through #3 yet.

@19:

It turns out that defining "random chord" is hard. I'm sympathetic to Jaynes's analysis using maximal ignorance. Suppose on a Cartesian graph you had a vast collection of circles with random centers and radii, and another vast collection of lines with random slopes and intercepts, such that the average density of lines was the same everywhere. Then each circle will have a a set of "random chords", and because of the constant average density you'll end up with a similar distribution of chords in every circle, which is a virtue. Circles which are small enough that on average they have exactly one chord from the mesh of lines will have the same probability as chords in very large circles.

If you think about it that way, then the midpoints of the chords will end up concentrated toward the center for the reasons you mentioned, so the solution is much more like solution #2 (though there are probably other conceivable ways of getting a similar distribution, so solution #2 is not necessarily canonical). You don't have to think about it that way, of course, but there's something to guaranteeing equal chord distribution I find immensely satisfying.

I actually have a "Method 4" that gives a 4th answer. Let us assume that the radius of the circle is 1. Then by a bit of trigonometry we can prove that the sides of the triangle are of size sqrt(3). Now, a random chord of the circle can be of any size from 0 (at the tangents) to 2 (a diameter). Assuming random chords are uniformly distributed* that means that the probability they are of at least size sqrt(3) is (2 - sqrt(3))/2.

*Which I have by no means proven, and there is definitely reason to believe this is not the case. My point was not to solve the problem, merely to add a fourth confounding "reasonable sounding" way to calculate it.