I can show that GK = GJ and GI = GH by drawing the segments from the pentagon vertices to D and using the fact that DK is a diameter.  I'm having trouble showing that GK = GI = IH.
In Robert Dixon's Mathographics, a regular pentagon is constructed with straightedge and compass only.  It is the pentagon $ABCDE$ pictured below.   I am having trouble seeing why the central angles are all $72^\circ$, though. Can anyone provide the proof?  Also, does anyone know who this construction is due to? I haven't seen it anywhere, other than in Dixon's book; is it Dixon's result?     The result appears much more impressive without all the labels (which are slightly misplaced, please excuse this); however, I provided those so that answering would be easier. Also, it makes it easy to describe the steps in the construction:  1) Draw a circle (the red one)  with center $h$.  2) Draw the perpendicular lines $\ell_1$ and $\ell_2$ through $h$. Locate the points of intersection $f$, $B$, and $g$ with the red circle.  3) Bisect the line segment $gh$. Denote the center by $a$.  4) Draw the green circle with center $a$ and radius $ah$.  5) Draw the other green circle (as in (3) and 4)).  6) Draw the line segment through $f$ and $a$.  7) Locate the points of intersection $b$ and $c$ of the line segment with the circle constructed in step 4).  8) Draw the blue arcs (both have center at $f$ and the radii are $fb$ and $fc$).  9) Locate the points of intersection $A$, $C$, $D$, and $E$.   I actually have the solution to the first question, and will post it unless a more elegant explanation is provided (which is probably likely).  However, I find this construction particularly   beautiful, and would like to know who it is attributed to (Dixon doesn't say, explicitly).
I am running 12.04 LTS. It's a pretty default installation and is the only OS on this machine.  I have two users one has administrator rights the other does not. I accidentally removed administrator rights from the "admin" user.  I tried to  but that requires reaching GRUB during boot but I cannot get there because my /etc/default/grub has GRUB_HIDDEN_TIMEOUT=0 and changing that to a positive integer requires admin privileges to save the changes and run sudo update-grub.  Should I boot from LiveCD? The  discouraged my a bit. Right now all I have handy is an install DVD from 11.10, but that did not help me much.  Not sure what to do now.  Any suggestions?  NOTE: 12.04 does not have an admin group and I have not enabled root Installing Office 2010 on Ubuntu 15.04 using Wine I had installed Office 2010 32bit on Ubuntu 12.04 before using guides such as . However, due to package changes, I was not able to replicate the same on Ubuntu 15.04.  How does one install Office 10 using Wine? Evaluating and negotiating a start-up package I've been offered an academic position, and am now in the fortunate, but unfamiliar position of negotiating the offer.  I've seen a few blogs/comments/posts on this process.  However, I have no real idea how to evaluate the offer that's been made.  Coming from the position of a postdoc, the package seems impressive (given that, as a postdoc, my "package" consists of a very modest salary and health benefits).  However, I also realize the importance of maximizing (within reason) my initial offer, as this sets the baseline for, at least, the first few years of my faculty career and could provide support that would prove very important in the initial year or so while I'm trying to obtain grant funding.  In particular, I'm looking for two types of information.  First, what types of startup packages are standard for tenure-track assistant professors of Computer Science at a public RU/VH (i.e. Research) university?  This is just to give me some point of comparison.  Second, what types of changes are reasonable to negotiate?  Can I ask for more summer salary support while still conveying to them my desire to obtain grant funding as quickly as possible?  The offer already includes a reduced teaching load for the first few years --- is it also reasonable to ask for an extra semester off prior to a tenure review?  I understand these are fairly broad questions, but any input or suggestions would be very helpful.  I'm definitely interested in the job; I really liked the department and think it's a great opportunity.  I don't want to alienate anyone by pushing too hard, but at the same time, I want to make sure that I make every reasonable effort to maximize my chances for academic success, which includes maximizing my available resources by negotiating the best startup package I can. Uncountable well ordered set in $Z^-$ theory. I want to build an uncountable well-ordered set within the theory $Z^\textbf-$. So, I take $A=\omega$ (exists by infinity axiom) and define $$W:=\{(X,R)\in\mathcal{P}(A)\times\mathcal{P}(A\times A):\langle X,R\rangle\ \text{is a well-ordered set}\}$$  With this I consider the set $T:=W/\cong$ (where $\cong$ is isomorphism relation). Note that $W$ and $T$ are sets by power set axiom. So I define for each equivalence class $[x],[y]\in T$ the order  $$ [x]\leq_T[y]\Leftrightarrow \text{type}(x, R_x)\leq \text{type}(y,R_y) $$  Here, $R_x$ means the order of $x$ that make it belongs to $[x]$.   So, $T$ is well ordered by $\leq_T$ and it is uncountable.  Question: I don't pretty sure if I can build $\leq_ T$ without replacement axiom. Another doubt is, Can I take the $R_x$ without AC? What's the best way to protect your house from Endermen stealing your walls? I just started playing Dark Souls 2 without playing the first one.  Since I don't see any sell option in any merchant: Can I sell my old-unused-equipment for souls? If I cannot sell it, does it have any value to be keep until later? How do I make protocol foo hostname-aware? This is a  about hostname-awareness and proxying.   I know that some protocols are hostname-aware; that is, when I connect to the HTTP server at www.example.com it knows I want www.example.com's HTTP service, not www.example.net's, even though they are on the same IP address.  How can I do this for protocol foo?  (Temporary note: this question appears pursuant to .) How to install with Wubi when it says "Could not retrieve the required disk image files"? I tried to install Ubuntu 12.10 with the Wubi installer while running Windows, because my PC does not accept my BootCD as a bootable medium.  But Wubi showed me this error:  Ubuntu-Installationsprogramm (X) Ein Fehler ist aufgetreten: Could not retrieve the required disk image files Weiter Informationen finden Sie in der Protokolldatei: c:\users\username\appdata\local\temp\wubi-12.10-rev273.log   What can I do, because I want to get rid of my Windows 7 and use the way better Ubuntu 12.10?  Thanks for your huge answer, now my next problem after trying to install wubi with the .iso is, that the same error shows up again and i am not shure which part of the errorreport you need.  Can you tell me? Because i'm not that good at linux. Instant e-mail notifications of answers to questions I'd like to have Instant notifications by email of answers to my questions.  Can this be implemented? Is there a way to solve this integral?  $$\int \frac{dx}{\sqrt{-ax^2 + bx +c}}, \;\;\;\;\; \forall\; a,b,c \in (0,+\infty) \;in\;R$$  I have tried it by changing of variable and by parts, but no result. I don't know what else to do.  Thanks in advance. How does searching a file (locate) or autocompletion work? I am a Windows user and coming from Windows with all the indexing and eternal wait times when searching for files, I find it surprising how fast does locate work or the autocompletion (that I know) work in linux.  Is there any indexing being done in the background or how is this achieved? I still have a pretty clean installation, so maybe this goes bad with time, but in Windows the moment you want to search on a folder which is not indexed you have to wait for some seconds. Not enough space on /tmp I'm not able to run update manager as I get an error saying that there is not enough free space in the /tmp directory. I've practically cleaned out the tmp directory but the error persists.  here's df-h  /dev/loop0       13G   11G  952M  92% / udev            2.0G  4.0K  2.0G   1% /dev tmpfs           785M  920K  784M   1% /run none            5.0M     0  5.0M   0% /run/lock none            2.0G  584K  2.0G   1% /run/shm /dev/sda6        20G   14G  6.4G  68% /host  overflow        1.0M   16K 1008K   2% /tmp Bezout's identity in $F[x]$ Let $F$ be a field and $F[x]$ a polynomial ring. Let $p(x)$ be an irreducible polynomial.  Show that $\gcd(p(x),q(x)) = 1\Longrightarrow \exists r(x),s(x)$ such that $r(x)p(x)+s(x)q(x) = 1$.  I know the proof for Bezout's identity for , but this proof uses the notion of absolute value, which cannot be applied to a polynomial ring.   I am thinking about a way to prove the statement: Use the Euclidean algorithm to compute $\gcd(p(x),q(x))$ and show that the result is $1$, and somehow go backwards to obtain $r(x)p(x)+s(x)q(x) = 1$, but it is a bit messy, and I'm not yet able to see it completely.  Please offer some help. If you have some other methods, that would be great, too. Question  The test scores of three groups of people are saved as separate vectors in R.  set.seed(1) group1 &lt;- rnorm(100, mean = 75, sd = 10) group2 &lt;- rnorm(100, mean = 85, sd = 10) group3 &lt;- rnorm(100, mean = 95, sd = 10)   I want to know if there is a significant difference in the medians between these groups. I know that I could test group 1 versus group 2 using the Wilcoxon test, like so.  wilcox.test(group1, group2)   However, this compares only two groups at a time, and I would like to compare all three simultaneously. I would like a statistical test that yields a p value at the 0.05 significance level. Could someone please help?  Edit #1 - Mood's median test  Following user Hibernating's suggested answer, I tried Mood's median test.  median.test &lt;- function(x, y){     z &lt;- c(x, y)     g &lt;- rep(1:2, c(length(x), length(y)))     m &lt;- median(z)     fisher.test(z &lt; m, g)$p.value }  median.test(group1, group2)   However, this approach allows me to test for a significant difference between the medians of only two groups at a time. I am not sure of how to use it to compare the medians of all three simultaneously.  Edit #2 - Kruskal-Wallis test  User dmartin's suggested answer appears to be more or less what I need, and allows me to test all three groups simultaneously.  kruskal.test(list(group1, group2, group3))   Edit #3  User Greg Snow helpfully notes in his answer that the Kruskal-Wallis test is appropriate as long as it makes strict assumptions that make it also a test of means. How to add custom Modules in magento What is the justification for locking votes? Why am I not allowed to change my vote after some period of time? There are times when I go back to things I voted on, and decide they didn't deserve that specific vote. (I get wiser as I grow older) With powershell get exactly the same application list as in Add/Remove programms Question : $$\begin{align}a+b+c &amp;= 0\\ a^3 +b^3 +c^3 &amp;= 12\\ a^5 +b^5 +c^5 &amp;= 40\end{align}$$ Then , $a^4 + b^4 + c^4 = ?$  My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think $3abc = 12$ and got $abc = 4$ After that I am unable to link it with any other equation Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 $$   Applying d'Alembert criterion I have that $$ \lim_{n\to\infty} \frac{n^\alpha  \ln^\beta(n)}{(n+1)^\alpha  \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.  If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.  Should I study the rest of the cases (i.e. if $\alpha = 0, \beta &gt; 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.  Thanks in advance "There are a good many scholars ..." is it right? In "Seeing like a state" of James C.Scott there is a sentence at the beginning of a paragraph (Acknowledgements xi):     There are a good many scholars whose writings opened up new   perspectives for me or provided outstanding analyses of issues that I   could not have hoped to study so comprehensively on my own.   What worries me is a good many scholars. Is this some type of inversion or ancient English? Is this grammatically right? Why is a here? I thought that appropriate version would be There are many good scholars. Am I wrong?