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.
123801
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).
121436
Show $\rho (x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric on the metric space $X$, equipped with the Euclidean metric $d$.   I've already shown that the positivity $\rho(x,y)\geq 0$, the symmetry $\rho(x,x)=\rho(y,y)=0$.  I'm having trouble proving the last condition of a metric space though: the triangle inequality must be valid:  $\rho(x,y)\leq \rho(x,p)+\rho (p,y)$, for $p\in X$. So, $\dfrac{d(x,y)}{1+d(x,y)}\leq \dfrac{d(p,x)}{1+d(p,x)}+\dfrac{d(p,y)}{1+d(p,y)}$  Could someone provide a hint as to get started? No complete solutions please. Thanks in advance.
109903
I'm having problems understanding the excersice: E) Define $D_n$ as the group of symmetries of a regular n-gon. Name the vertices $V=\{V_0,V_1,...,V_{n-1}\}$ so that $$V_{k}=\exp({i\cdot\dfrac{2\pi k}{n})}$$ Let $S$ be the symmetry along the x-axis, and $R$ the rotation by $\dfrac{2\pi}{n}$. Show that $\{R,S\}$ generate $D_n$.  I tried to define something like $$R:\exp(i\cdot\dfrac{2\pi k}{n})\mapsto \exp(i\cdot\dfrac{2\pi (k+1)}{n}) $$   $$S:\exp(i\cdot\dfrac{2\pi k}{n})\mapsto \exp(i\cdot\dfrac{2\pi (-k)}{n}))$$ but these are not binary operation so $(V,R) ; (V,S)$ are not groups. To me it's clear that there are n rotations and n reflection so there are 2n symmetries on the n-gon; also fixing one reflection we can get the others by some combination of $r,s$.  I tried to solve the problem by defining $R,S$ as permutation of vertices and the vertices as coordinates in a lattice and it solves nicely, but I don't seem to understand this way of approaching the definitions.
46489
I'm studying for actuarial exams, but I always pick up mathematics books because I like to challenge myself and try to learn new branches.  Recently I've bought Topology by D. Kahn and am finding it difficult.  Here is a problem that I think I'm am answering sufficiently but any help would be great if I am off.  If $d$ is a metric on a set $S$, show that  $$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ is a metric on $S$.  The conditions for being a metric are $d(X,Y)\ge{0}, d(X,Y)=0$ iff $X=Y$, $d(X,Y)=d(Y,X)$, and $d(X,Y)\le{d(X,Z)+d(Z,Y)}$.  Thus, we simply go axiom by axiom.  1)  Since both $d(x,y)\ge{0}$ and $1+d(x,y)\ge{0},$ it is clear that $d_1(x,y)\ge{0}$.  (Is this a sufficient analysis?) 2) $d_1(x,x)=\frac{d(x,x)}{1+d(x,x)}=\frac{0}{1+0}=0$. 3) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=d_1(y,x).$ 4) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$  However, #4 is strictly less, not less than or equal to, according to my analysis, so where did I go wrong?
34191
Let $H$ be a subgroup of $G$. Prove that $gHg^{-1}=\{ghg^{-1}|h \in H\}$ is a subgroup of $G$, $g\in G$. Show that $o(gHg^{-1})=o(H)$.   Attempt:  As $e=geg^{-1}\in gHg^{-1}$ then $gHg^{-1}$ is non-empty. Let $x=gh_1g^{-1}$, $y=gh_2g^{-1}\in gHg^{-1}$, $h_1, h_2\in H$   then $xy^{-1}=\cdots=g(h_1h_2^{-1})g^{-1}\in gHg^{-1}$ as $h_1h_2^{-1}\in H$ ($H$ being the subgroup of $G$)  Therefore, $gHg^{-1}$ is a subgroup.   The problem is in 2nd part. I am unable to sgow that $o(gHg^{-1})=o(H)$. Please help me to solve the problem using the elementary tools of abstract algebra.
32131
I am having a lot of trouble understanding the solution to this problem.   $(gH)^n = g^nH \implies g^n \in H$  Why does $H^n$ just turn into the identity? I am very confused, any help is appreciated.
69904
I tried to prove it by proving that if $\frac GH$ is cyclic, there is $gH$ such that $(gH)^n=e_{(G/H)}=e_{(G)}*H$. Since $(gH)^n=g^nH$, $g^n=e_G$. So $G$ is cyclic.   I don't think it's correct since I didn't use the condition that $H$ is cyclic but I couldn't figure what's wrong. Would be great if someone can point out what I'm doing wrong and maybe point me in the right direction.   Thanks in advance.
40122
Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.  This time I have no approach. I haven't found any relation between the divisors of $n$ and the order of the subgroups of $G$. How would approach this?
90533
Question I found in "Introduction to Graph Theory" by Douglas B. West:     Let $G$ graph on $n$ vertices with connectivity $\kappa(G)=k \geq 1$. Prove that $$n \geq k(\operatorname{diam}(G)-1)+2$$ where $\operatorname{diam}(G)$ is the the maximal (edge) distance between a pair of vertices in $G$   Seems easy and I've seen the answer, too, which goes the same as thought: by taking the vertices the end vertices of the path of length $\operatorname{diam}(G)$, and applying Menger's Theorem. I'm having a hard time understanding one part - it seems like all the paths need to be of the same length for the $n \geq k(\operatorname{diam}(G)-1)+2$ to apply. Am I missing some part here?
77959
I have some properties which I do not understand how to form a proof for. Most of them are by induction, which is not my strong point Any help would be fantastic.   Looking at the chromatic polynomial $$P_G(k)=P_{G-e}(k) - P_{G/e}(k)$$  Properties   1.If G has n  vertices then $ P_g(k)$ has degree n and the coefficient of $k^n $ is 1   The signs of $P_g(k) $ alternate If G has q components the smalles non-zero term of $P_g(k)$ is the term in $k^q$
86362
Any rank 2 tensor would do the same trick, right?  Then, which is the motivation for choosing the metric one?    Also, if you help me to prove that $g^{kp}g_{ip}=\delta^k_i$, I would be thankful too ^^
32207
The only solution that I have been able to come up with is $d=1$, but because $d&gt;2$, I can't seem to figure out how to find a solution.
57844
Hello everyone interested. Here's a seemingly simple problem.  Suppose $G$ is a complete, simple graph with vertices $\{1,2,..,n\}$, $n\geq k\geq 3$.  How many cycles of length $k$ can we have on $G$?  The statement seems simple but I have trouble putting everything together.  So I say fix a vertex and start forming cycles with length k. But I get nowhere... What is the proper approach to this problem since counting is not easy? Is there a relevant theorem that I am unaware of?  Thanks for any help...
105893
A student sent me a question and I am stucked. I'd like to use classical geometry, instead of hard trigonometry if possible (can be trigonometric, but I'd like to not use scientific calculator; the only answer I got I need this), but I am in circles. The pentagon $ABCDE$ is regular. The answer is $48^\circ$ (I've constructed in Geogebra). Thank you in advance!
59785
$d_1(x_1,y_1)$ and  $d_2(x_2,y_2)$ are metric on $X$ and $d(x,y)$ is defined as: $$d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$ I am trying to show this is a metric. Can you give me some clue about proving the triangle inequality for $d$ ? Thank you for your help.
82274
I am not sure how to show the second part. Isn’t it obvious from the fact that    G and H are infinite?
34535
I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.  Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail. So I really hope someone could give me a hint on how to approach this problem.