## Monday Math 161

September 15, 2014

Find the infinite product $2^{\frac12}\cdot4^{\frac14}\cdot8^{\frac18}\cdots=\prod\limits_{n=1}^{\infty}\left(2^n\right)^{\frac1{2^n}}$

## Monday Math 160

September 8, 2014

Suppose we have four identical-looking coins. Three are fair, but one is biased, with a probability of coming up heads of 3/5. We select one of the four coins at random.

1. If we flip the selected coin twice, and it comes up heads both times, what is the probability that our coin is the biased one?

2. If we flip the selected coin three times, and it comes up heads all three times, what, then, is the probability that our coin is the biased one?

3. Generalize: We have m fair coins and one identical-looking biased coin with probability p of getting heads. If we select one coin at random, and obtain k heads in n flips, what is the probablility P(m,p,n,k) that we have the biased coin?

## Monday Math 159

August 18, 2014

Find $I=\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\ln\tan{x}}dx$

## Monday Math 158

August 11, 2014

Find a non-summation expression for the value of the sum $\cos{x}+2\cos{2x}+3\cos{3x}+\cdots+n\cos{nx}$.

## Monday Math 157

August 4, 2014

Consider a triangle, which we label ∆ABC, with circumcenter O and circumradius R=AO=BO=CO. Let us label the midpoints of the sides as MA, MB and MC, so that MA is the midpoint of BC (the side opposite A), and similarly, so that $\overline{AM_{A}}$, $\overline{BM_{B}}$ and $\overline{CM_{CA}}$ are the medians. Then $\overline{OM_{A}}$, <$\overline{OM_{B}}$ and $\overline{OM_{C}}$ are segments of the perpendicular bisectors of the sides. Let us also label the feet of the altitudes from A, B and C as HA, HB and HC, respectively, and let H be the orthocenter (the intersection of the altitudes $\overline{AH_{A}}$, $\overline{BH_{B}}$ and $\overline{CH_{C}}$).

Let us construct the point D on the circumcircle diametrically opposed to A; that is to say, the point D such that AD is a diameter of the circumcenter. Then AD=2R, and O is the midpoint of AD.

Now, by Thales’ theorem, ∠ABD and ∠ACD are both right angles. Now, since CD and the altitude $\overline{BH_{B}}$ are both perpendicular to AC, they are parallel to each other. Similarly the altitude $\overline{CH_{C}}$ and segment BD are parallel, both being perpendicular to AB. Thus, the quadrilateral BDCH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, we see that the midpoint MA of BC is also the midpoint of HD.

Now, let PA be the midpoint of the segment AH. Then $\overline{P_{A}M_{A}}$ is a midline of the triangle ∆AHD, and by the triangle midline theorem, $\stackrel{\longleftrightarrow}{P_{A}M_{A}}\parallel\stackrel{\longleftrightarrow}{AD}$ and PAMAAD=R.

Now, let N be the intersection of $\overline{P_{A}M_{A}}$ and HO. By the midline-median bisection theorem proven in this post, we see that, as HO is the median of ∆AHD that crosses midline $\overline{P_{A}M_{A}}$, N is the midpoint of both $\overline{P_{A}M_{A}}$ and HO. Thus, NMA=NPAPAMAR.

Now, consider the quadrilateral HOMAHA. Since $\overline{HH_{A}}$ and $\overline{OM_{A}}$ are both perpendicular to BC, HOMAHA is a right trapezoid. Letting QA be the midpoint of $\overline{H_{A}M_{A}}$, we see then that NQ is the median (or midline) of trapezoid HOMAHA.

By the first of the three items proven here, we see that $\overline{NQ}\parallel\overline{HH_{A}}\parallel\overline{OM_{A}}$, and so $\overline{NQ}\perp\overline{H_{A}M_{A}}$. Thus, we see that $\stackrel{\longleftrightarrow}{NQ}$ is the perpendicular bisector of $\overline{H_{A}M_{A}}$, and so, by the perpendicular bisector theorem, NHA=NMA, and so
NHA=NMA=NPAR.

Constructing diameter BE of the circumcircle gives us parallelogram CEAH, by a similar argument as above. Letting PB be the midpoint of the segment BH, analogous reasoning to the above shows that NHB=NMB=NPBR as well. Lastly, diameter CF, and midpoint PC of CH gives, by similar proof, that NHC=NMC=NPCR. Thus, the nine points MA, MB, MC, HA, HB, HC, PA, PB and MC are all equidistant from N.

Therefore, we see that for any triangle, the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter all lie on a single circle, which, for this reason, is usually known as the nine-point circle, and its center N the nine-point center. We also see that the radius of the the nine-point circle is half the radius of the circumcircle, and the nine-point center is the midpoint of the segment connecting the circumcenter and orthocenter. This latter tells us that for any non-equilateral triangle, the nine-point center lies on the Euler line (for an equilateral triangle, it coincides with the circumcenter, orthocenter, and centroid).

## Teen Girls and the “Selfie” Arms Race

July 30, 2014

So, I came across this article by schoolgirl Olympia Nelson, decrying the “sexual rat race” of selfies on social media. She complains that they all converge on a single formula, which obtain the most “likes” on Instagram and Faceboook: “Nothing with too much creativity but hip, titty and kiss. It’s the true scourge of the selfie.”

But why do they all converge on this? Miss Nelson goes on:

Why are we girls competing to be the Queen of Pouts? Why do we scour through photos of celebrities and all our ambitious friends to find out who is the new princess of prurient poses? Even demure girls are tempted to strike sexually suggestive poses. But they must be careful, not because parents are looking but because they might not score any ”likes” and might then feel a failure, unworthy among their peers.

How confident can you appear at being lascivious? How credible is your air of lewdness? A girl who is just a try-hard will lose credibility and become an outcast. So a lot depends on how much support you can get from other girls.

She further describes the multitude of techniques young women use to increase their popularity scores. She laments the “fake flattery” young women use “to get higher on the food chain”, and that these often too-intimate pictures are posted by girls who “are seeking some sort of approval from their friends.”

However, she then attempts to blame it on “boys’ tastes”, which she calls “not always sophisticated” and based on “what they see in pornography”:

Who do we blame for this moral mess? As feminists, we correctly blame patriarchy because boys are securely at the top of the status game. Boys end up with the authority. They have their cake and eat it.

First, lamentable as the rise of pornography in our degenerate, declining culture is, it’s role here is minor at best. Science has amassed quite a bit of evidence that men’s “tastes” on women’s looks are pretty uniform and to a great extent hardwired. So, despite how Miss Nelson may wish things were, it’s unlikely that what boys like will change.

Second, one should ask why young women like Miss Nelson place so much importance and status in interest and attention from the opposite sex? THe answer is the same reason why (heterosexual) men place so much importance and status in interest and attention from the opposite sex: not “patriarchy”1 or “misogyny”, but biology. From the Darwinian perspective, finding and attracting a mate, more specifically the best quality mate one can, is literally the Most Important Thing. Thus, any accounting of social status amongst post-pubescent human beings will have sexual attractiveness as a significant component. You can rail against human nature, but biology wins out in the long run. This also strikes at the first bit; even if male tastes were different, girls would just compete over those instead, just as viciously.

Most important, though, and which Miss Nelson somewhat acknowledges when not deflecting blame, is that the most important factor is not male approval, but peer approval. The great anxiety that drives this behavior is the fear of being unpopular with other girls, of becoming an outcast. When it comes to breaking down individuality and enforcing conformity and social hierarchy, boot camp has nothing on what high school girls do to one another.

How often does a teen girl demand the latest popular album not because they like the music, but because all their friends listen to it, so she has to have it to or she’ll just die? See this bit from Scientific American, where brain scans have not only confirmed this phenomenon, but shown that the primary emotional motivation is fear of failing to fit in.

And being cast out from the pack means becoming prey to it. All the worst incidents of bullying I’ve heard or read about, in terms of psychological cruelty and viciousness, the perpetrator and victim have been teenage girls. There’s a reason the “alpha bitch” and “girl posse” tropes have such frequency and cross-cultural resonance. Or see the work of Rachel Simmons. And Paul Graham’s “Why Nerds Are Unpopular” helps further explain why high school amplifies this:

Why is the real world more hospitable to nerds? It might seem that the answer is simply that it’s populated by adults, who are too mature to pick on one another. But I don’t think this is true. Adults in prison certainly pick on one another. And so, apparently, do society wives; in some parts of Manhattan, life for women sounds like a continuation of high school, with all the same petty intrigues.

I think the important thing about the real world is not that it’s populated by adults, but that it’s very large, and the things you do have real effects. That’s what school, prison, and ladies-who-lunch all lack. The inhabitants of all those worlds are trapped in little bubbles where nothing they do can have more than a local effect. Naturally these societies degenerate into savagery. They have no function for their form to follow.

When the things you do have real effects, it’s no longer enough just to be pleasing. It starts to be important to get the right answers, and that’s where nerds show to advantage. Bill Gates will of course come to mind. Though notoriously lacking in social skills, he gets the right answers, at least as measured in revenue.

.

Given that human nature doesn’t change except on evolutionary time scales, and the prison-like nature of modern high school isn’t likely to be transformed anytime soon (save for its end with the total collapse of industrial civilization), not much can really be done. And the best thing to ameliorate the situation, parental involvement to limit the “selfie arms race”, Miss Nelson rejects in her concluding paragraph. She wants an end to girls being “compelled to act the stereotype, because those who opt out commit themselves to social leprosy”, but it is her fellow girls who provide the compulsion, and treat the non-conforming like lepers.

Miss Nelson identifies a real (if intractable) problem, but then spends the rest of her article deflecting all responsibility away from herself and her peers, blaiming “patriarchy”, and then rejecting any solution that might require trade-offs or any change in behavior on her part, instead wanting a vague, magical solution that doesn’t require any action on her part or any limits on her behavior. She wants fried ice; she is the one who wants to have her cake and eat it too. In other words, she’s a teenage girl.

1. In fact, under an actual patriarchy, the problem would be much less, if not non-existant, because patriarchs would not tolerate, and would curb, the female bad behavior and status games that drive this issue.

## Monday Math 156

July 28, 2014

For a trapezoid, the median (also known as the midline or mid-segment) is the segment connecting the midpoints of the legs of the trapezoid. This post presents a proof that:

1. the median of a trapezoid is parallel to the bases;
2. the length of the median is half the sum of the lengths of the bases;
3. the midpoints of the diagonals of a trapezoid also lie on its midline.

## Monday Math 155

July 21, 2014

The triangle midline theorem, also called the midsegment theorem, states that the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. This may be trivially proven via triangle similarity (SAS similarity condition) and the corresponding angles postulate. More interesting, however, is to prove it using triangle congruence.

Let D and E be the midpoints of sides AB and AC, respectively, of ∆ABC. Let us extend segment DE past E to point F such that DE=EF, and let us draw CF.

Since DE=EF, AE=EC, and vertical angles ∠AED and ∠CEF are congruent, we see by the SAS condition that ∆ADE≅∆CFE. Thus, CF=AD=BD. Also, ∠FCE≅∠DAE; but since these are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AD}$ and $\stackrel{\longleftrightarrow}{CF}$ cut by transversal $\stackrel{\longleftrightarrow}{AC}$, we see that $\stackrel{\longleftrightarrow}{AD}\parallel\stackrel{\longleftrightarrow}{CF}$. But then the quadilateral BCFD has a pair of opposite sides, BD and CF, which are of equal length and parallel, so it is therefore a parallelogram, and so $\stackrel{\longleftrightarrow}{DE}\parallel\stackrel{\longleftrightarrow}{BC}$. And since opposite sides of a parallelogram have equal length, DF=BC, and so DEDFBC.

We can also prove a similar theorem: that the line through the midpoint of one side of a triangle parallel to a second side of the triangle bisects the third side, and that the segment of that line inside the triangle is one-half as long as the parallel side. As with the midpoint theorem, it is trivial to prove with triangle similarity (this time the AA condition) and the corresponding angles postulate. So, instead, we use a similar construction as above.

Let D be the midpoint of side AB of ∆ABC, and let E be the point where the line through D parallel to BC intersects AC. Construct the line through C parallel to AB, and let F be the point where it intersects $\stackrel{\longleftrightarrow}{DE}$

Then BCFD is a parallelogram, and since opposite sides of a parallelogram have the same length, BD=CF and BC=DF. And so CF=BD=AD. And since ∠ADE and ∠CFE are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AD}$ and $\stackrel{\longleftrightarrow}{CF}$ cut by transversal $\stackrel{\longleftrightarrow}{DF}$, they are congruent. Similarly, ∠DAE≅∠CFE, and so, by the ASA condition, ∆ADE≅∆CFE. Thus, AE=EC, and E is thus the midpoint of AC. We see also that DE=EF, and since DF=BC, thus DEDFBC.

Using both of these theorems together, we can prove a third: that a midline (mid-segment) of a triangle and the triangle median that intersects it bisect each other.

Let MA, MB and MC be the midpoints of sides BC, AC and AB, respectively, of ∆ABC. Thus, $\overline{M_{B}M_{C}}$ is a midline of ∆ABC, and $\overline{AM_{A}}$ a median. Let P be the point where they intersect.

By the midline theorem, $\stackrel{\longleftrightarrow}{M_{B}M_{C}}\parallel\stackrel{\longleftrightarrow}{BC}$ and MBMCBC.

This means, then, that $\stackrel{\longleftrightarrow}{M_{C}P}\parallel\stackrel{\longleftrightarrow}{BM_{A}}$, and so by our second theorem above with regards to ∆ABMA, we see that P must be the midpoint of $\overline{AM_{A}}$, and MCPBMA.

Similarly, our second theorem applied to triangle AMAC establishes that MBPCMA. But BMA=CMA, and so MBP=MCP, and P is the midpoint of $\overline{M_{B}M_{C}}$ as well.

## Monday Math 154

July 14, 2014

Continuing the series on triangle centers, let us consider ∆ABC, with circumcenter O and centroid G. Construct the line segment OG, and extend it out from G to the point H such that GH=2OG.

Next, construct the median from vertex A to the midpoint M of side BC. Then G lies on AM, with AG=2GM, as we showed here. Recalling that the circumcenter is the intersection of the perpendicular bisectors of the sides, we see that OMBC.

Now, since AG=2GM, GH=2OG, and ∠AGH≅∠MGO, we see (by the SAS similarity condition) that ∆AGH~∆MGO. And since these triangles are similar, corresponding angles ∠HAG and ∠OMG are congruent. However, these are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AH}$ and $\stackrel{\longleftrightarrow}{OM}$ cut by transversal $\stackrel{\longleftrightarrow}{AM}$, and therefore $\stackrel{\longleftrightarrow}{AH}\parallel\stackrel{\longleftrightarrow}{OM}$. And since OMBC, we see $\stackrel{\longleftrightarrow}{AH}\perp\overline{BC}$, and $\stackrel{\longleftrightarrow}{AH}$ is the triangle altitude from A to BC.

Analogous constructions show that H must also be on the triangle altitudes from B and C:

Thus, we see that H is the orthocenter of ∆ABC. So, we see that for any non-equilateral triangle, the circumcenter O, centroid G and orthocenter H are collinear, with GH=2OG; the line through these triangle centers is known as the Euler line of the triangle. (For an equilateral triangle, O, G and H are all the same point.)

## Monday Math 153

July 7, 2014

A few weeks ago, I demonstrated a proof that all three medians of any triangle are concurrent; in other words, that the centroid exists.

Now, how about demonstrating that the circumcenter and orthocenter exist; that is to say, showing that the three perpendicular bisectors of the sides are concurrent, and that the three altitudes are concurrent, for any triangle.