What is an anti-semitism czar’s job?

In the month or so since she’s been appointed, there’s been a lot of controversy about Pres. Obama’s chosen envoy to the post of Special Envoy to Monitor and Combat Anti-Semitism (SEAS), Hannah Rosenthal.  The source of the controversy is mainly this article in Haaretz, in which she criticized Israeli Ambassador Michael Oren for making “most unfortunate” remarks against J Street, the liberal counterpart to AIPAC.  For example, Jeffrey Goldberg has pointed to a number of “non-neocon” Jews, such as Alan Solow, who are worried about Rosenthal using her platform to put her nose where it doesn’t belong.  Andrew Sullivan, on the other hand, defends Rosenthal, saying:

[S]he is allowed to answer questions in a press interview and because the ambassador should stay out of domestic American politics and not join in a campaign to torpedo a rival to AIPAC. The way in which Oren acted as a faction leader in American Jewish politics, rather than as a neutral representative of his country, open to all people of good will toward Israel, was obnoxious and undiplomatic.

This is actually an interesting point, and it gets to the heart of the sort of schizophrenic attitude we Jews have viz a viz Israel.  For example, is Israel representative of all Jews?  Does anti-Zionist = anti-semitic?  Etc. etc.

Looking at Rosenthal’s history, it is clear that she is a classic J Street persona.  She attracted the ire of the ADL last year when she recalled her experience at the National Israel Solidarity Rally in Washington in 2002 and stated that the list of speakers “espoused narrow, ultra-conservative views of what it means to be pro-Israel.”  IMHO, there’s nothing wrong with being a little to the left here, although I agree with Foxman on this one.  [This was the rally in which Paul Wolfowitz of all people was booed for considering the suffering of the Palestinians.  But I digress.]  But let’s look at this logically and figure out where all of this leads.

Are we to constantly conflate Israel and the Jewish diaspora?  The great majority of Jews love Israel and wish to see her defended, but they also have concerns closer to home.  And sometimes the interests of Israel and the interests of, say, the US, are not one and the same.  On the other hand, global anti-semitism can be correlated with Israeli actions.  So, what exactly is Rosenthal’s job then? And does she deserve all this criticism?

As Goldberg helpfully points out, the Office of the SEAS is a part of the US Dept of State.  It is a dept. of the Bureau of Democracy, Human Rights, and Labor led by Assistant Secretary Michael H. Posner, which is a part of the Under Secretary for Democracy and Global Affairs, led by Under Secretary of State for Democracy and Global Affairs Maria Otero.  So, this is pretty far down the food chain.  It attracts a lot of attention merely because, well, it’s about anti-semitism.  So, from a pure employment perspective, did Rosenthal step over the line?  Apparently, she did: concerning her remarks to Haaretz, the White House passed along this statement from Jeff Feltman,the Assistant Secretary for Near Eastern Affairs:

The Department of State values its close relationship with Ambassador Michael Oren and his staff at the Embassy of Israel in Washington. The United States and Israel enjoy extraordinarily close ties based on shared values, interests, and history, as well as the deep bonds between the Israeli people and the American people. Ambassador Oren plays an indispensible role in maintaining and strengthening our relationship through his day to day interaction with the Administration and Congress on issues of vital importance to both countries and his vigorous outreach to Americans of all origins and points of view.

While there was no specific criticism of Rosenthal, the message is clear: back off.

I know what Rosenthal is trying to do: she is trying to decouple Israeli actions from Jewish actions, and perhaps she feels that public criticism of Israel is as good a means for that as anything.  From a logical point of view, if your job is to combat anti-semitism, then it follows that part of your job is to protect Jews worldwide from hatred resulting from actions with which the have nothing to do.  That would include, for example, Israel’s action in Gaza last year, or the ongoing investigation of the bombing of the Buenos Aires JCC in 1994.  So, in this instance, Rosenthal seems like a good fit for the job.

On the other hand, in criticizing an Israeli Ambassador, who has the right to speak with whom he wishes, she stepped on the toes of her State Dept. colleagues.  So, the reprimand from Feltman is justified: Rosenthal stepped out of line.  This is not to say that what she said was necessarily wrong.  I think the truth is a little more complex than some bloggers portray it.  That said, if Rosenthal is to survive in her job, then she is going to have to accept the constraints of her position and figure out a way to do her job without attacking the Israeli government.

Top Ten of the Decade: Movies

In alphabetical order:

  • Best In Show:  Hands down the funniest movie of the decade, by the folks that brought you Spinal Tap.  Seriously, anyone who has ever raised a colicky baby cannot have at least pissed their pants watching the yuppie couple try to locate and replace the bee toy for their substitute baby.  And Fred Willard is in rare form as the stupid color guy, awesome enough to have been imitated several times since.
  • Casino Royale:  Daniel Craig is Bond.  The tension between Craig and Judy Dench is worth the price of admission alone.  But ultimately, it’s the respect shown to Fleming’s original work, albeit updated for the 00’s rather than the 50’s, is really what makes this a great film.
  • Downfall: I remember the horror when this movie was about to come out: Hitler, as a human?  How awful!  Humbug.  It’s Hitler as the pathetic, reckless, shell of a man in his last days holed up in his bunker, from the point of view of a young secretary.  This is the best portrayal of the monster I have seen in film and after reading Kershaw’s biography of Hitler, quite accurate and well done.  Not to be missed.
  • The Incredibles: I put up two Pixar movies, they were that good.  This one is great for the exasperation shown to those who would bring down those whose gifts will benefit us all.  It is a libertarian movie, showing the effects of a fear of excess success.  And best of all, it is chock full of great lines and stylish animation.  [“No kepps!”  “I’m the greatest good you’re ever gonna get!”]
  • Kill Bill vol. 1: A total guilty pleasure.  Sorry, but the extended scene in the teahouse with the Crazy 88’s has to rank as amongst the most singularly entertaining, ever.  And that jar of Vaseline!
  • Michael Clayton: A powerful movie about the moral compromises we make when we’re in over our heads.  Clooney as the title character is extremely likable and sympathetic, and I really enjoyed the ending and the way the story unwound.  Tilda Swinton as Karen Crowder, the general counsel of the baddies, is also in over her head.  She’s the opposite: school smart, kiss-ass, very unlikable, she falls victim to her moral failures in the end.  And Tom Wilkinson as Arthur Edens, a man who loses it and ceases his compromises, nearly steals the show.  Possibly my favorite.
  • Minority Report:  This one was pure excellence all around, asking some very disturbing questions about the nature of justice, all while presenting a very disturbing vision of the future without sinking into a dystopia.  How insane would you go having your retinas scanned dozens of times a day for personalized marketing?  And, as much as I hate to admit it, Cruise was very very good in this.  [And who could resist Max von Sydow?]
  • There Will Be Blood: Daniel Day Lewis gives a chilling portrayal of a man who is out to make a buck doing what he does best, and is driven to inhuman acts in the process of becoming rich.  The moral foundations of our nation – religion and business – are sickeningly amoral.  You will want to wash your hands after this movie, but no matter.  It is intense and just good cinema.
  • V for Vendetta: The slew of movies based on graphic novels in this decade provided great entertainment.  I think this one is the best of the bunch, but I am by no means backed up by the critics, who largely panned it.  Fuck them.  This is a straightforward vision of a fascist society in the making, in its execution, and its destruction.  The incredibly talented Hugo Weaving has to give a performance of feeling behind a mask that betrays none, and he is terrific.  Natalie Portman proves herself worthy of a good script and does the shaved head thing with dignity and righteous anger.  And John Hurt plays the mirror image of his Winston Smith character in 1984 as the fascist dictator, mostly seen on a giant screen yelling at his subordinates.
  • WALL-E: This is a cartoon.  I had to keep reminding myself of this as I watched likely the most incredible animation ever depicted on the big screen.  O, and this was accomplished with a mostly mute robot that used mechanical pieces to convey emotional signals.  I guess this too is another libertarian movie in a way, as it warns about the consequences of giving into our lazy instincts and letting other people clean up our messes.

The Expanding Rail Problem

Problem:

A railroad track is a mile in length. On a really hot day, the track expanded so that, while its endpoints were fixed in place, the track bowed into a circular arc. If the track expanded by a foot, about how far did the track deviate from its original position?

Solution:

rail

This problem is a classic illustration in how to do estimates and accurate computation.  I got it out of Forman Acton’s book, “Real Computing Made Real”.  This old chestnut is one that is simple to set up, but difficult to execute.

I take a different approach to Acton, who uses this to challenge the reader in his understanding of preserving significant figures.  His approach is entirely numerical.  I choose, rather, to illustrate the power of analysis to preserve the sig figs and to help feel my way around while doing so.

First, consider a crude estimate to help us gain some perspective.  Instead of the circular arc, imagine the track, originally of length \(a=5280\) is stretched like a violin string at the center to a length \(a+\varepsilon=5281\); it would then look the the straight line segments as in the dotted lines in the figure.  This is not the correct answer, but it puts us in the ballpark.  To find the deviation \(d\), we simply use Pythagoras:

\(d^2+(\frac{a}{2})^2=[\frac{1}{2}(a+\varepsilon)]^2\)

We can then solve for \(d\) as usual.  Nevertheless, note that, as stated, \(a\) is 3 orders of magnitude larger than \(\varepsilon\).  We can therefore neglect the term \((\frac{\epsilon}{2})^2\) and write, with great [numerical] accuracy:

\(d \approx \sqrt{\frac{a \varepsilon}{2}}=51.38093 \ldots\)

The take away here is that \(\varepsilon \ll d\), while at the same time, \(d \ll a\).  More accurately stated, \(d=O[\sqrt{a \varepsilon}]\).  Those who guessed that \(d\) would be about 1 ft would be very wrong.  The neat thing about this problem is that such a small change results in a big result.

That said, we have not solved the correct problem and now need to get on with it.  To do this, we write down the geometric relations algebraically:

\(\theta=\frac{a+\varepsilon}{r}\),
\(\sin \frac{\theta}{2}=\frac{a}{2 r}\),
and
\((\frac{a}{2})^2+(r-d)^2=r^2\).

Now that we are guided by orders of magnitude, we can scale our quantities appropriately. Let \(x=\frac{a}{2 r}\), \(y=\frac{d}{r}\), and \(\delta=\frac{\varepsilon}{a}\). We can then rewrite the above equations in a more compact form, eliminating \(\theta\):

\(\sin{x(1+\delta)}=x\),
and
\(y^2-2 y+x^2=0\).

We do not know a lot about the magnitude of the quantities \(x\) and \(y\), but we do know from the above analysis that \(y \ll x\), so even though we could solve the second equation exactly in terms of \(x\), we need not even do that. All we need to do is ignore the \(y^2\) term and we get, in a first approximation,

\(y=\frac{x^2}{2}+O[x^4]\).

The quantity \(x\) is determined through the first equation and depends on the value of the small quantity \(\delta\). If we think about what to expect as a solution, note that this equation is transcendental and we cannot expect an exact solution. However, if we look at the limiting equation at \(\delta=0\), which is \(\sin{x}=x\). The only solution to that equation lies at \(x=0\). We then conclude that when \(\delta\) is small, so is \(x\).

This is very useful information, because it tells us how to go about approximating the solution to the first, transcendental equation. Because we now know that \(x\) is small, we can Taylor expand the sine term about \(x=0\) and get the following approximate equation:

\(x=x(1+\delta)-\frac{1}{6}x^3 (1+\delta)^3+O[x^5]\)

This can be solved easily for \(x\):

\(x=\sqrt{\frac{6 \delta}{(1+\delta)^3}}(1+O[\delta])\)

[That last factor \(O[\delta]\) comes from considering an expansion to a quadratic equation that would be solved if we considered the \(O[x^5]\) term. More on this later.]

In fact, we can even neglect the factor of \((1+\delta)^3\) in the denominator because it, too, adds a factor of \(O[\delta]\). Then we are simply left with

\(x=\sqrt{6 \delta}(1+O[\delta])\).

If we substitute back into the above equation for \(y\), and use the original definitions of the dimensionless parameters, we find, after a little rearrangement, the following first approximation for the desired deviation:

\(d=\frac{\sqrt{6}}{4}\sqrt{a \varepsilon}(1+O[\frac{\varepsilon}{a}])\).

Using the numbers provided, the deviation resulting from a 1 ft addition in length to the track is about 44.49719 ft, to within some factor of a constant times 0.019% error.

Is this good enough? Acton provides an answer good to 7 significant figures: 44.49845, which means the approximation is good for an error of 0.0028%, or 4 sig figs. This is fine in most cases, but let’s assume we want to reproduce Acton’s result. How do we do this?

We go back to the transcendental equation and Taylor expand out an additional term:

\(x=x(1+\delta)-\frac{1}{6}x^3 (1+\delta)^3+\frac{1}{120}x^5 (1+\delta)^5+O[x^7]\)

Now we can write that \(x=\sqrt{6 \delta}(1+B \delta+O[\delta^2])\) for some \(B\), for which we will solve by plugging into the above equation. When we do this, we note that all terms in \(x\) are of even powers, and terms of \(O[\delta]\) all cancel, so we are left with a coefficient of \(\delta^2\) equaling zero; this gives \(B=-\frac{27}{20}\). This result gets put into the quadratic equation above, where we now assume that \(y\) takes the form \(y=3 \delta (1+C \delta+O[\delta^2])\). Note that we include the \(y^2\) term now. Verifying that the terms of \(O[\delta]\) cancel, and equating the coefficient of \(\delta^2\) to zero, we get that \(C=-\frac{6}{5}\). When we go back to the original parameters, we find that

\(d=y r\),
\(\Rightarrow d=\frac{a}{2}\frac{3\delta(1+C\delta+O[\delta^2])}{\sqrt{6 \delta}(1+B\delta+O[\delta^2])}\),
\(\Rightarrow d=\frac{\sqrt{6}}{4}a\sqrt{\delta}(1+(C-B)\delta+O[\delta^2])\),

which gives us our correction:

\(d=\frac{\sqrt{6}}{4}\sqrt{a \varepsilon}(1+\frac{3}{20}\frac{\varepsilon}{a}+O[(\frac{\varepsilon}{a})^2])\).

This gives us \(d=44.498455 \ldots\), which is not too shabby.

Where Local Control Goes Awry, Cont’d

A few days ago, I posted a story about the E Ramapo, NY school board.  The Board has a majority Hasidim and has used their clout to replace the long-serving attorney to the Board with one 4X as expensive and for the seeming purpose of trying to transfer tax dollars meant for public schools to the special needs programs at the yeshivot that the Hasidic children attend.  The VP of the Board and leader of the meeting in which the attorney was voted in over howls of protest from the non-Hasidic community was one Aron Wieder, who spent the 20 Nov meeting looking sullen and annoyed that he had to deal with angry people as he virtually ignored them.

While I thought Mr. Wieder lacked class and basic manners, I do not fault the Hasidim for their actions.  They are only doing what is in their interest and exercising the fruits of majority rule and the political process.

Mr. Wieder seems to have seen what we did in his performance that night.  Here is a speech to address the polarizing issue of the lawyer before the community:

A few thoughts:

  • “Change we can believe in…”?  Huh?  I didn’t think the frum went so much for Obama.
  • He basically accuses the non-Hasidic board members who claimed that they were never invited to interview the new lawyer of lying.  Nice way of uniting the community.
  • No proof offered in that accusation, BTW.
  • “God Bless America”: the hooting and hollering when he says this were just creepy.  Who was in the audience here?

No matter anyway: Mr. Wieder has been offered a position as Administrative Assistant to Spring Valley’s [Democratic, non-Jewish] Mayor, and has taken it.  Good for him.  But I hope that the inclusion of the Hasidim in their larger community brings folks closer together.

Integrating binomial coefficients

Problem: Let \(C(z)\) = coefficient of \(x^{2010}\) in the expansion of \((1+x)^z\) about \(x=0\). What then is the value of

\(\int_{0}^{1}dy\) \(C(-y-1)\) \(\sum_{m=1}^{2010}\frac{1}{y+m}\) ?

Solution:

Let \(k=2010\).  The value of the binomial coefficient desired is

\(C(z)=\binom{z}{k}=\frac{z(z-1)(z-2) \ldots (z-k+1)}{k!}\).

It then follows that

\(C(-y-1)=\frac{(-1)^k}{k!}(y+1)(y+2) \ldots (y+k)\).

If we let \(\frac{(-1)^k}{k!}p(y)=C(-y-1)\), then it turns out that

\(\sum_{m=1}^{k}\frac{1}{y+m}=\frac{p'(y)}{p(y)}\).

Therefore

\(I(k)=\int_{0}^{1}dy \: C(-y-1)\sum_{m=1}^{k}\frac{1}{y+m}=\int_{0}^{1}dy \: p'(y)\). \(I(k)=\frac{(-1)^k}{k!}(p(1)-p(0))=\frac{(-1)^k}{k!}(2)(3) \ldots (k+1)\) \(I(k)=\frac{(-1)^k}{k!}(k+1)!=(-1)^k (k+1)\)

The answer is therefore 2011.

Where local control goes awry

When we lived in Poughkeepsie and kept kosher, we would occasionally travel down to Monsey in Rockland County for shopping.  There is a strip mall in Monsey in which every shop is exclusively kosher; it was there where we could do our one-stop grocery shopping and get some kosher Chinese.  Such a mall can only be supported by a strong and orthodox Jewish community, which Monsey is certainly.  Monsey, NY, a hamlet of the town of Spring Valley and part of the E Ramapo school district, is one of a handful of upstate NY villages that have gone majority Hasidic over the years.  And, as the taxpayers of E Ramapo school district have discovered, such a situation brings with it a special clash of interests that seems to be upending a social contract that has bonded Americans of all backgrounds for decades.

The story that has touched off this observation has to do with a seemingly boring meeting of the E Ramapo school board on Nov 20.  However, this meeting, held at 12:40 AM while most taxpayers were sleeping, was designed to slip a very controversial move through with minimal fuss.  That move was the changing of the Board’s legal representation, from the local attorney who had represented the board for the past 33 years, to a new one.  The new one, it was revealed, not only would charge 4 times as much for an appearance at a meeting, but also had problems with the Attorney General.  And the leader of the Board, Aron Wieder, wanted the change to be confirmed then and there, with no consideration of a transition plan or any other sort of risk management.  And this new attorney would be confirmed no matter what that evening.  How is this possible?

It was possible because, out of 9 members, the Board has 6 Hasidim.  As a rule, the Hasidim [nominally called “ultra-orthodox” in the news – I find that moniker meaningless] do not send their children to the public schoold, but to private yeshivot.  They do, however, pay taxes.  And, to their dismay, they were seeing their school taxes rise even as the local school population was dropping.  So the Hasidim did what we in America applaud: they took action, used their numbers, and got a majority to the school board.  This of course is disastrous news to the people who actually use the public schools, now that a majority of the board only has an interest in minimizing their taxes.  Now, supporters of the Hasidic community point out that they have not been all about slashing taxes and have, for example, expanded the full-time kindergarten.  Further, they, like anybody else, have an interest in quality public schools.

But this latest dealing with the attorney has rightfully set off the anger and mistrust of the non-Hasidic taxpayers.  Why would the Hasidic members, who ostensibly only wish to control costs, abruptly switch legal counsel to one that is 4 times more expensive?  The answer, it was revealed, was even more disturbing.  The new counsel is also the counsel for the school district of Lawrence, NY, which is also controlled by Hasidim, and has successfully diverted public funds for the special needs programs at their yeshivot.  There was no consideration of any other attorney.  One non-Hasidic board member stated for the record that he never met or interviewed the new counsel: the whole concept of the attorney switch was done behind the backs of the non-Hasidic board members.

The meeting was recorded and put on Youtube; a short contentious part can be seen here:

[The entire meeting, which is about an hour, is split into 6 parts: 1, 2, 3, 4, 5, 6.  It is worth watching it to see the entire context.]

As awful as the actions of Mr. Wieder are, I cannot fault him nor the Hasidic community.  What is happening is purely legal, albeit procedurally improper.  But in the grand scheme of things, the Hasidim have it right.  They were unhappy with a situation, they used the political process and their numbers to make things how they wanted it.  And now they are using their power to divert their tax dollars into causes they want.  You and I may not like what happened at that meeting, nor do we like the way Mr. Wieder runs things.

What is insane is the way schools are funded and boards are run.  The problem is that Mr. Wieder and his 5 cronies on the board were driven to be there because of what they felt were unfair expropriations of his tax dollars.  And now he’s turned it around on his community.  There needs to be more oversight from the states on these boards so that arbitrary decisions like those taken by Mr. Wieder could not be possible.  Mr. Wieder, through his arrogance and a sense of entitlement, put his board in major legal risk, as the superintendent so clearly points out.

The story did not end there.  An interim attorney was hired because of all the mess that Mr. Wieder never thought through, and the decision to hire Mr. Wieder’s attorney is still pending.  But the problem remains. What do you do when the majority of your school board, who are backed by a majority of the voters who send their kids to private schools, has no interest in the school?  This will be an interesting story to follow.

The Swiss Minaret Ban, or why I love America, part 13,207

You may have heard of the controversy over the recent ban on new construction of minarets in Switzerland.  22 of 26 cantons, representing 57.5% of the popular vote, voted for the ban.  Multiculturalists and religious officials were quite upset with the ban.

To me, this ban is interesting in not only what it says about Switzerland as a nation, but also as a nation of law in contrast to the US.

To be perfectly honest, I don’t think I would much care for a minaret in my neighborhood.  This is not due to any distaste for Islam, but rather distaste for so much of my home going to something which will only annoy me.  I imagine this is what many Swiss voters felt.

But what I feel as a single voter is of little consequence.  Rather, it’s how the legal infrastructure is set up to deal with my feelings.  And clearly, in the US we have a huge advantage.  Recall the First Amendment to the Constitution:

Congress shall make no law respecting an establishment of religion, or prohibiting the free exercise thereof…

If we wanted a national ban on minarets, as the Swiss have done, we’d have to amend the Constitution.  Good luck with that.  But, such a national ban shouldn’t be done in the first place.  Minarets should be allowed to be built where there is local approval.  If there is an area chock-a-block full of devout Muslims, then they should get a minaret.  If a few Muslims want to build a minaret in a place that will annoy the local populace, then local regulations will prevent the building.  The system we have is, of course, not perfect: if it were, then the Mormon temple serving the Boston area would not be dominating a residential neighborhood in Belmont.  But, at the very least, we cannot embarrass ourselves as the Swiss apparently have by making a local issue first a national, and then an international, one.

Nonlinear system producing periodic functions

Consider the following system:

\(y’\) = \(-z^3\); \(z’\) = \(y^3\); \(y[0]=1\), \(z[0]=0\).

Show that the solutions \(y=f[x]\) and \(z=g[x]\) are periodic functions, and evaluate the period.

Solution:

Clearly, \(y^3 y’\) + \(z^3 z’=0\), so that the quantity \(y^4+z^4\) is constant.  Given the boundary conditions, \(y^4+z^4=1\) for all values of \(x\).  When plotted in phase space, this is a closed curve and is therefore representative of a periodic system.

To compute the period \(p\), we begin with the obvious observation that

\(p=\int_{0}^{p}dx=4 \int_{0}^{p/4}dx\).

Note that

\(dx=dz/y^3=dz(1-z^4)^{-\frac{3}{4}}\)

so that

\(p=4 \int_{0}^{1}dz(1-z^4)^{-\frac{3}{4}}\)

This integral is doable by using the substitution \(w=1-z^4\) to reveal that

\(p=\int_{0}^{1}dw[w(1-w)]^{-\frac{3}{4}}\)

The true nerd will recognize the above integral is in the form of a Beta integral and is thus equal to

\(p\) = \(\frac{\Gamma[\frac{1}{4}]^2}{\sqrt{\pi}}\) \(\approx 7.4163\)

Solution to the irregular hexagon problem

Three equilateral triangles share a common point which is the center of a circle on which the three sides opposite the point become three sides of a cyclic hexagon [a hexagon inscribed in the circle] ABCDEF, with, say, AB, CD, and EF as the sides of the equilateral triangles. Consider the triangle formed from the midpoints of sides BC, DE, and FA [G, H, and J, respectively]. Prove that triangle GHJ is equilateral.

hextri

Solution:

Let O be the origin of the circle.  Let \(\alpha =\angle BOC\), \(\beta =\angle DOE\), and \(\gamma =\angle FOA\).  Note that \(\alpha +\beta +\gamma =\pi \).

The length of a side of \(\Delta GHJ\), say, \(\overline{GH}\), is determined by the length of the bisectors, in this case \(\left|\overline{OG}\right|\) and \(\left|\overline{OH}\right|\); and the angle \(\angle GOH\) between the bisectors, as follows:

\(\left|\overline{GH}\right|^2\) = \(\left|\overline{OG}\right|^2 \) + \(\left|\overline{OH}\right|^2\) – \(2 \left|\overline{OG}\right|\) \( \left|\overline{OH}\right|\) \(\cos \angle GOH \)

where \(\left|\overline{OG}\right|\) = \(\cos \frac{\alpha}{2}\) , \(\left|\overline{OH}\right|\)= \(\cos \frac{\beta}{2}\), and \(\angle GOH=\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3}\).

Then in terms of the parameters \(\alpha\) and \(\beta\), the length of the side \(\overline{GH}\) is

\(\left|\overline{GH}\right|^2\) = \(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\beta}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos(\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3})\)

The same reasoning applies to the other sides, e.g.,

\(\left|\overline{JG}\right|^2\) = \(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\gamma}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\gamma}{2} \cos(\frac{\alpha }{2}+\frac{\gamma}{2}+\frac{\pi }{3})\)

To prove that \(\Delta GHJ\) is equilateral, we must show that the lengths of its sides are equal.  The problem then reduces to showing that

\(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\beta}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos(\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3})\) = \(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\gamma}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\gamma}{2} \cos(\frac{\alpha }{2}+\frac{\gamma}{2}+\frac{\pi }{3})\)

where, again,  \(\alpha +\beta +\gamma =\pi \).

Suffice it to say that this is an exercise in manipulating trig identities beyond the patience of most sane people, even those that consider math puzzles for recreation.  I will leave most of the details to the reader, assuring said reader that the identity to be proven does in fact hold true.  Your trusty problem poser has performed said manipulations himself, and can offer the following road map.

First, observe that \(2 \cos(z + \frac{\pi}{3})\) = \(\cos z\) – \(\sqrt{3}\) \(\sin z\).  Note that the term associated with the \(\sqrt{3}\) will be treated separately.

Also note that \(\cos \frac{\gamma }{2}\) = \(\sin \frac{\alpha +\beta }{2}\) and \(\sin \frac{\gamma }{2}\)=\(\cos \frac{\alpha +\beta }{2}\).

Keep organized, and enjoy!