Buffer polygon inward by percentage of area I would like to buffer a polygon inward by a percentage of its area, using the polygons below as an example.    How could I shrink each polygon inward resulting in a 35% reduction of area? I am using the term "reverse buffer" because I want the newly created polygons to have the same shape as the originals and be centered and within the originals, like the schematic below.    I am on ArcGIS 10.2 and have SA and ET Geotools.
Creating polygons as % of original area using ArcGIS Desktop? I have a polygon shapefile and need to create new polygons, within the originals, as % of original area.  I used buffer by field (Field= new area as %) but the buffers are created in circles around the polygons.  What am I doing wrong?  I am using ArcGIS Desktop 10.
Given: $a = qb + r$. Then it holds that $\gcd(a,b)=\gcd(b,r)$. That doesn't sound logical to me. Why is this so?    Addendum by LePressentiment on 11/29/2013: (in the interest of  and averting a duplicate)  What's the intuition behind this result? I only recognise the proof and examples solely due to algebraic properties and formal definitions; I'd like to apprehend the result naturally. Why would Louise telling the Chinese commander about his wife's last words change his mind? In the future in Arrival, the commander says that he changed his mind about going to war with the aliens because Louise called his private number and told him his wife's dying words. My question is what did Louise say to him (like in English) and why did it change his mind? Topical example:  'There was a scuffle outside the local pub. There appears to be no connection with religious extremists.'  ...but now you're wondering if there was. It's a nasty tactic used to raise fear and doubt - deliberately calling attention to something by explicitly denying it.  What's the term for this? How do you print the current month, day and year using awk and date? I can't seem to grab the month, day and year using   awk '{print $2, $3, $6}' date    Does anyone know how to? Finding an angle within an 80-80-20 isosceles triangle The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.  So here is the puzzle:    The triangle $ABC$ is known to be isosceles, that is, $AC=BC$. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled "?".  The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution. I use time manager for a polygon layers with dates and it works. I would like to see the cummulative addition of my polygons. So I have 4 periods I want the poylgons from 1900-1950 to be added to the new ones in 1950-2000 etc in the next time step. How can I perform that or does another open source tool exist for that? More than once I've accidentally run a number of commands and polluted my bash history. How do I close my terminal without saving my bash history? I'm using Fedora. Why is the direct product of a finite number of nilpotent groups nilpotent? Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and  $$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$ for all $n\gt 0$.  Show that the sequence is  upper bounded.  My idea: since  $$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$ then $$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$ then $$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$ so $$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$ since $$a_{n+1}&gt;a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}&lt;\dfrac{1}{a_{1}+i^2}$$ so $$\dfrac{1}{a_{n+1}}&gt;\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)$$ But the RHS might be $\lt0$ for a sufficiently large starting value; for instance, with $a_{1}=\dfrac{99}{100}$ then $$\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)&lt;0,n\to\infty$$  see:   so this method won't let me bound the series and I don't know what else to do. Matryoshka words Since it's Friday and others are playing games, here's another one to take us into the weekend.  Everyone knows about those Russian nesting dolls (called matryoshka dolls, I believe). A small doll nests inside a larger, which itself is inside a larger doll, and so on.   This game does the same with words. You start with a three-letter word, add a letter to make another word, and so on. Longest word wins. The catch? Each additional letter has to form a legitimate word.  Here's one to start you off:     bur   burr   burro   burrow   burrows     Good luck. How to remove some pages from the navigation bullets in Beamer? I'm making a beamer presentation using the miniframes outer theme, with navigation bullets, and the subsection=false option, meaning the bullets are all on a single row. The problem I have is that, if one section has too many pages, the bullets overflow to the next section:  Section 1 Section 2 oooooooooooooooooooo   I tend to have many frames, some of them just asking a question, leading to the next detailed frame. I don't really need those frames to be represented by navigation bullets, and removing them would fix the problem. Any idea how to do that?  I could cheat by using overlays to have the question and the answer on the same frame, but that would be a mess...  Code example to answer percusse's question: this should give you one circle per frame, not one per subsection. If you add many frames in one subsection, the circles will overflow over the next section.  \documentclass[compress]{beamer} \useoutertheme[subsection=false]{miniframes} \begin{document} \section{S1} \subsection{SS1} \frame{1} \frame{2} \frame{3} \subsection{SS2} \frame{1} \frame{2} \frame{3} ... \end{document} Gravatar links on "Filtered Questions" page are not HTTPS Finite groups with exactly one maximal subgroup I was recently reading a proof in which the following property is used (and left as an exercise that I could not prove so far). Here is exactly how it is stated.     Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.   Now, does this mean exactly one conjugacy class of maximal subgroups? If it really means exactly one maximal subgroup, then what can we say about a finite group that possesses exactly one class of (at least two) maximal subgroups that are all conjugate to each other? Something I did in the first Animal Crossing was make a house on an alternate character solely to showcase bugs/fish I felt were too cool to only leave in the museum. However bugs can't be mailed or dropped, so I'm at a loss as to how to do this.  Is there a way I can give bugs/fish to a second character or will that character have to find them themselves? How to create a sequence with leading zeroes using brace expansion When I use the following, I get a result as expected:  $ echo {8..10} 8 9 10   How can I use this brace expansion in an easy way, to get the following output?   $ echo {8..10} 08 09 10   I now that this may be obtained using seq (didn't try), but that is not what I am looking for.  Useful info may be that I am restricted to this bash version. (If you have a zsh solution, but no bash solution, please share as well)  $ bash -version GNU bash, version 3.2.51(1)-release (x86_64-suse-linux-gnu) If every pair of congruence equations admits solutions, then the entire system admits solutions Let a system of three linear congruence equations in integers be given;      \begin{cases}x\equiv b_1\mod c_1\\ x\equiv b_2\mod c_2\\ x\equiv b_3\mod c_3\\ \end{cases}   with $c_1,c_2,c_3\in\mathbb Z_{+}$. Show that, if every pair of congruence equations admits solutions, then the entire system admits solutions.    Now, if every pairs admits solutions, then for example $b_1\equiv b_2\equiv a\mod \gcd(c_1,c_2)$ for some $a$ then;   $x-a\equiv b_1-a\mod c_1$  $x-a\equiv b_2-a\mod c_2$  replacing $x$ by $x':=\frac{x-y}{\gcd(c_1,c_2)}$, we obtain:  $x'\equiv b_1'\mod c_1'$  $x'\equiv b_2'\mod c_2'$  and since $\gcd(c_1,c_2)=1$ by Chinese Remainder Theorem we have a unique solution $\mod{c_1'c_2'}$, so $x\equiv (\gcd(c_1,c_2)x'+a)\mod{\operatorname{lcm(c_1',c_2')}}\quad(1)$  We can do it with another pair (take $2$nd and $3$rd, since we know that such a pair admits solutions;  $x\equiv(\gcd(c_2,c_3)x''+a')\mod{\operatorname{lcm(c_2',c_3')}}\quad(2)$  Now how can I show that for the pair $(1)$ and $(2)$ also solutions exist ? What is wrong with this fake proof $e^i = 1$? $$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$  Obviously, one of my algebraic manipulations is not valid. Autocovariance of an ARMA(2,1) process - derivation of analytical model for $\gamma( k)$ $GL_n(\mathbb F_q)$ has an element of order $q^n-1$ For fixed prime power $q$ show that the general linear group $GL_n(\mathbb F_q)$ of invertible matrices with entries in the finite field $\mathbb F_q$ has an element of order $q^n-1$.   I tried to show this question with showing diagonal matrix but i can't find element directly  competible with order i think i am on wrong way please give me clue ? I am looking for the combinatorial argument for the identity:  \begin{equation} k\binom{n}{k} = n\binom{n-1}{k-1} \end{equation}  This is easy to show algebraically as:  \begin{equation} \binom{n}{k} = \dfrac{n(n-1)(n-2)(n-k+1)}{k(k-1)!} \end{equation}   What is the combinatorial argument? What are some general ideas to get started?   Here is a clarification of 2. From what I have seen so far, proving (combinatorially) an identity with an addition sign usually implies that we need to partition a set (this makes sense because of the addition rule and provides a nice visual). On the contrary, the previous observation leads me to believe that multiplication in identities can be resolved with the multiplication principle, but what is the "visual/interpretation" for this? Could someone provide such an interpretation for the example identity given above?