The fifth Problem of the Week has now been posted. This one is probably my favorite of the term. I think it's fairly challenging. It will have to hold you for a while, though, since POTW will be taking next week off.

I've also posted a solution to POTW 4. Enjoy!

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I've just posted the new Problem of the Week, along with an official solution to last week's problem. But this one will have to hold you for a while, since I'm taking next week off.

The fifth Problem Of the Week has now been posted at the big website. I've also posted an “official” solution to Problem Four. POTW will be taking two weeks off after this one, so you will have to make this last. (Spring break is almost upon us, which seems incredible considering how cold it is…

Occasionally I rant about the general awfulness of mathematics textbooks. If I were to express my major objection in the most charitable possible way, it is that most textbooks are written like reference books. They are usually very good at recording the basic facts of a subject and proving them…

As you might have noticed, Sunday Chess Problem had the week off. If you really need to get your fix, though, you can have a look at this web page I made for my chess problems. You'll recognize a few of them from the Sunday Chess Problem series.
I did, however, manage to get the new POTW up.…

Since 0 <= cos(x) <= 1 in the first quadrant (acute angle x), ln cos(x) < 0. Hence 2 ln cos(x) < ln cos(x).

Sorry, wrong page. I was reading POTW 4 (which I mised entirely last week), and there's no way to go back and delete a post. Jason, if you care, feel free to delete both #1 and this #2.

I believe I can prove that point O doesn't exist. Start with an isosceles triangle ABC with AB and AC the equal sides and draw the bisector AD for angle BAC, which will be the same as the bisector of the side BC. Now extend side AC until angle ABC is a right angle (call the new vertex E) and extend AD, to the side of the new triangle at point F; now we have a right triangle, ABF, equivalent to the one in your initial figure. Finally, draw a line from C parallel to ADF to contact side BF at point G. Triangles BDF and BCG are similar, with a side ratio of 1:2. Thus BF is equal to FG and less than FE. The perpendicular bisector of BE will thus be somewhere on FE, and will not contact AF.

The hole in the proof is the claim that triangles OFC and OEA are congruent, on the basis that they have two sides of equal length and an equal angle. However, that theorem, the Side-Angle-Side (SAS) theorem, only holds when the angle in question is the one between the two given sides. That's why the elements are stated in that order. When the equal angle is not the one between the two equal sides, as here, there may be two triangles satisfying the given conditions. Or as my junior high school geometry teacher put it, there is no such theorem as Angle-Side-Side. (Angle-Angle-Side and Angle-Side-Angle, however, are both valid.)

Extend BO to where it intersects with AC; call that point G. Then using half-angle identities we should be able to discover the range of possible ratios between AC and GC.

Without going into deriving the half-angle formulas let's use this one:

$latex \tan\frac{\theta}{2} = \frac{\sin\theta}{1+\cos\theta}$

We also know that

$latex \tan\theta = \frac{\sin\theta}{\cos\theta}$

Then for 0° < $latex \theta$ < 90° (forgive me for using degrees instead of radians):

$latex \frac{AG}{AC} = \frac{\sin\theta}{1+\cos\theta} \cdot \frac{\cos\theta}{\sin\theta} = \frac{\cos\theta}{1+\cos\theta}$

So now we can take limits (assume that lim of 90° is from below and lim of 0° is from above):

$latex \lim_{\theta \to 90} \frac{\cos\theta}{1+\cos\theta} = 0$

and

$latex \lim_{\theta \to 0} \frac{\cos\theta}{1+\cos\theta} = \frac{1}{2}$

This means that $latex 0 < \frac{AG}{AC} < \frac{1}{2}$, so the bisector of angle ABC intersects with line segment AC between the right angle and the bisector of line segment AC, so the point O where BO and DO meet can't ever be inside the triangle.

The rest of the proof falls apart from there. (LaTeX gods, let my markup be pure)

Not too bad. Limits are badly formatted, but it's readable.

You can visualize it in this little app:

http://www.mathopenref.com/triangleincenter.html

You can make a right angle at B and then move C back and forth to see how the limits above work.

Wait,

SB takes LaTeX input?I mean, the PNGs are crude, but whatever.How is it delimited?

* Looks at WP documentation*

Just dollar signs? Hang on:

$\dfrac{1}{2}$

Oh, wait. Please pardon my excitement.

$latex \dfrac{1}{2}$

^ I've been remiss: Many thanks to Another Matt. I'm now wondering just how much I can get away with.

End thread derailment.

My pleasure. I asked Jason about this about a year ago, and I've always been pleased to put it to use. It's so nice to be able to read formatted math stuff. I'm astounded at the general superior quality of LaTeX, too – even in a music dissertation I'd have been totally lost without it. It really is one of the great triumphs of open-source software in my opinion.

Here's where I asked:

http://scienceblogs.com/evolutionblog/2014/10/12/potw-6/

The point O lies outside the triangle. To get F, we will need to extend BC, whereas E will be within AB. So we get BF = BC+CF = BE = BA-EA

Eric Lund:

In the general case this may be true, but these are right triangles and so two legs or a leg and a hypotenuse are enough to determine the other side via the Pythagorean theorem; so the congruence would hold if the rest of the proof were well-argued.

Another Matt: Nice proof, but it seems simpler to note that $latex \tan(x) $ is convex on in the interval $latex (0, pi/2) $, so that $latex \tan \left( \frac{\theta}{2} \right) \leq \frac{1}{2} \tan \theta$ for all acute $\latex \theta$. (Indeed, for all convex functions $\latex f$, we have $latex f(x) + f(y) \leq f(x + y)$.) In particular, taking $latex \theta$ to be angle B, we get the same conclusion you came to (the angle bisector intersects the opposite side between the right angle and the midpoint of AC), without considering limits.

Thanks. I had forgotten about the properties of convex functions! That is indeed simpler.

My last math course was in 1997... I study what I need for musical purposes, but I'm perpetually rusty otherwise. :)

POTW 6: The assumption is that there is at least one pair of numbers x and y which satisfy the two equations simultaneously. This is not the case. This can be seen by solving the second equation for y and substituting the result (y = x - 4) into the first equation. But when you attempt to simplify the result, x drops out, and you are left with 16 = 0. Alternatively, plot the two equations, and you will find that they describe parallel lines.

Indeed, this holds if you change the second equation to x - y = a for any a != 0 (replace 16 in the above derivation with a^2). If a == 0, then the first equation reduces to the trivial 2 = 2; i.e., they are the same line, and the number of solutions is infinite.

Right, the first equation takes the form a + 1/a = 2. This can only be the case when a=1, that is when x/y=1, or x=y. So x-y=4 is simply impossible. In fact, this problem was confusing at first because x-y=4 seemed like such a non sequitur I thought I was misreading something.