Suppose you have 2 circles that intersect each other in such a way that each circle passes through the other's center. What is the area between the circle(or common area) i.e. area between the centres of the circles?
If we pick randomly two points inside a circle centred at $O$ with radius $R$, and draw two circles centred at the two points with radius equal to the distance between them, what is the expected area of the intersection of the two cirlces that contain the origin $O$.
How can window snap be enabled in Openbox? I would like to have window snap behaviour in Openbox. For me, this would mean that dragging a window to the edge of the display using the mouse would change the size of the window such that it is docked in half or quarter of the display, as can be done in Unity7. How can this be done in Openbox? Problem. Use Tutte's 1-factor theorem to prove that every 3-regular graph with no bridges has a 1-factor.  I honestly have no idea how to approach this problem. I have drawn a couple of graphs to gain some insight to no avail. Any hints or tips will be greatly appreciated. Functional equations leading to sine and cosine Search with Auto-Complete (like Gmail / Intellisense) Does centrifugal force exist? Casting vs using the 'as' keyword in the CLR When programming interfaces, I've found I'm doing a lot of casting or object type conversion.  Is there a difference between these two methods of conversion?  If so, is there a cost difference or how does this affect my program?  public interface IMyInterface {     void AMethod(); }  public class MyClass : IMyInterface {     public void AMethod()     {        //Do work     }      // Other helper methods.... }  public class Implementation {     IMyInterface _MyObj;     MyClass _myCls1;     MyClass _myCls2;      public Implementation()     {         _MyObj = new MyClass();          // What is the difference here:         _myCls1 = (MyClass)_MyObj;         _myCls2 = (_MyObj as MyClass);     } }   Also, what is "in general" the preferred method? D+ and D- on USB I'm making a small project which will take +5v from USB, What should I do with the D+ and D-, keep it open or short them to ground or do I have to add some components?  I want to run ATMEGA168 and 16x2 LCD with back light. How to assign symbols to functions Is there a better alternative to ambiguous for a word with a wide range of meanings, difficult to find, because they become different in connection with person's opinion. If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$ given that $x^2~,~y^2~,~z^2\ne1$  I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully): Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality. This means that we are saying that $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$ Consider $\tan (A+B+C)$: $$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$ So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$. Now consider $\tan (2A+2B+2C)$: $$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$ $$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$ $$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$ $$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$ as required.  My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;) Thank you for your help. Total number paths between two nodes in a complete graph I gave a scammer my telephone number, address, and name. How bad is it? Updated question because old one was embarassing: I saw a spam email from "PayPal" and I didn't realize that it was a phishing attempt. I gave the phishing site my name, telephone number, and email. What could he do with them?  Old question: Well, I saw a spam email, but I didn't realize that it's really a scam at first. I gave him my telephone number, address, and name. What would he do with them? Could he ruin ALL of my reputation with it? A composition of a periodic function that is somehow not periodic. solving diophantine equations using congruences So, a question in Elementary Number Theory by David M. Burton:     Problems 4.4      2(a) Using congruences, solve the diophantine equation 4x + 51y = 9       Hint: $4x \equiv 9(mod \ 51) $ gives $ x = 15 +51t$, whereas $ \ 51y \equiv 9 (mod \ 4)$ gives $y = 3+4s $. Find the relation between $s$ and $t$ From this hint, I substitute $ x = 15 +51t$ and $y = 3+4s$ in the given equation $4x + 51y = 9$ to get the relation $s = -t-1$ and using which I can get the solutions $ x = 15 + 51t, \ y =-1-4t$ But we know that the solutions of $ax \equiv b (mod \ n)$ is equivatent to the solutions of the diophantine equation $ax -ny = b$, so, solutions of $4x \equiv 9(mod \ 51) $ is equivalent to solutions of the equation $4x-51y=9$ whereas, solutions of $51y \equiv 9(mod \ 4) $ is equivalent to solutions of the equation $-4x+51y=9$ Now, my question is: as, the above equations $4x-51y=9$ and $-4x+51y=9$ are different, then what is the point of substituting $ x = 15 +51t$ (solution of $4x-51y=9$) and $y = 3+4s $ (solution of $-4x+51y=9$) in the given equation $4x+51y=9$ to find a relation between $t$ and $s$ ? Isn't it absurd as $ x = 15 +51t$ and $y = 3+4s $ are solutions of two different equations? Can anyone please explain this? PS: I did it like this $4x \equiv 9(mod \ 51) $ gives $ x = 15 +51t$ and by substituting it in $4x+51y=9$ we get $y=-1-4t$ Custom date format with jQuery validation plugin How can I specify a custom date formate to be validated with the  for jQuery? I created some linestring features in EPSG:32632 (WGS 84 / UTM zone 32N) exactly 10 000 m long.  QGIS calculates all line lengths (horizontal, vertical, diagonal) as 10 001.9 m. The measure tool explains "Project CRS transformation is turned on and ellipsoidal calculation is selected. The coordinates are transformed to the chosen ellipsoid (WGS84), and the distance is calculated in meters."  When I set the ellipsoid to "None / Planimetric" in project properties, I get the expected results (10 000 m).  I don't understand the phrase "coordinates are transformed to the chosen ellipsoid (WGS84)", is it a transformation to unprojected degrees (4326?)? Since 32632 already is based on WGS84, is there anything to transform? If there is a kind of "great circle" calculation involved, I would have expected lengths can only get smaller.  Is QGIS's calculation correct/meaningful, is it plain wrong or do I see rounding errors?  This are the geometries I tested:  LineString (370000 5615000, 370000 5625000) LineString (366464.46609406732022762 5616464.46609406732022762, 373535.53390593267977238 5623535.53390593267977238) LineString (365000 5620000, 375000 5620000) LineString (373535.53390593267977238 5616464.46609406638890505, 366464.46609406638890505 5623535.5339059317484498)   Edit as addition to MappaGnosis's answer: I forgot that in UTM , this seems to be the simple answer for the title question. I was surprised recently to find that it's possible to have a return statement in a finally block in Java.  It seems like lots of people think it's a bad thing to do as described in ''. Scratching a little deeper, I also found '' which shows some pretty horrible examples of other types of flow control in finally blocks.  So, my question is, can anyone give me an example where a return statement (or other flow control) in a finally block produces better / more readable code? How is my laptop ram running in dual channel? Order of the product of two commuting elements with coprime orders in a group. I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.    We have that  $$ (gh)^{mn}=g^{mn}h^{mn}=1 $$  using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|&lt;mn$ and arriving at a contradiction, but it did not take me anywhere. Short story about survivors on Earth after the atmosphere has frozen