I am looking for a formal proof for the following problem: You travel from point A to point B on a right triangle only along its legs. For a 3 4 5 right triangle you would travel a distance of seven. Now imagine going in horizontal and vertical lines to point B while remaining within the right triangle (basically steps going from point A to B). You would again travel a distance of seven: . My question is once you start decreasing the size of your vertical and horizontal lines until they are infinitely small, when do you travel the length of your hypotenuse (for this example five) instead of the length of the legs (for this example seven)? Does this also mean at some point you travel a distance of in between the length of both legs and hypotenuse?
1302
What is wrong with this proof?    Is $\pi=4?$
34501
I would like a metric $d(\;)$ between pairs of (infinite) lines in $\mathbb{R}^3$, with these properties:   If two lines $L_1$ and $L_2$ are parallel and separated by distance $x$, then $d(L_1,L_2) = x$. $d(L_1,L_2)$ increases with the degree of skewness between the lines, i.e., the angle $\theta$ between their projections onto a plane orthogonal to the shortest segment connecting them (dashed below), where $\theta=0$ for parallel lines and $\theta=\pi/2$ for orthogonal lines. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;   Intuitively, I would like the metric to be related to the repulsive force between two electrically charged lines (but it need not be exactly the actual physical force).  Have such metrics been considered in the literature? If so, I'd appreciate descriptions and/or pointers&mdash;Thanks!  Update1. It turns out that the natural ad hoc definition (from the comments) fails to be a metric. For example, below, a sufficiently large value of $a$ ensures the triangle inequality will be violated: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; So this question may be more difficult than it initially appeared...  Update2.  The  have revealed that in fact there is no metric that satisfies my two conditions (and is continuous w.r.t. $\theta$)!
36077
Prove that for any piecewise smooth curve it is possible to find the parametrisation $\phi$ that is consistent with its length, ie.  length of a curve segment between $\phi(a)$ and $\phi(b)$ is equal to $|b-a|$.  It seems intuitive, but I don't have any idea how can I show this in the formal way.
70180
Assume that there are $n$ infinitely long straight lines lying on a plane in such a way that no two lines are parallel, and no three lines intersect at a single point. Prove that these lines divide the plane into $\frac{n^2+n+2}{2}$ regions. (Hint: Any two non-parallel straight lines on a plane must intersect at exactly one point.)  Can someone tell me where to start here? I know to prove the base where $n=0$, but don't know how to proceed.  Any help is greatly appreciated!
106503
I have been presented with an interesting problem. How can I decide whether a right triangle with given side lengths can be placed (with integer coordinate vertices) on a Cartesian plane so that the legs are not parallel to the x or y axes?  I'm a little stumped as to how the lengths come into play. I figured I would just be considering the slopes of the legs, and whether they were 0 or undefined. Also, try as I might, I can't make a sketch of a triangle that obeys this non-parallel alignment specification.  Any ideas?
56721
Imagine a two-dimensional grid consisting of 20 points along the x-axis and 10 points along the y-axis. Suppose the origin (0,0) is in the bottom-left corner and the point (20,10) is the top-right corner. A path on the grid consists of a series of moves in which each move is either one unit to the right or one unit up. Diagonal moves are not allowed. How many diﬀerent ways are there to construct a path starting at (0,0) and ending at (20,10)?  I'm a little stuck on this one. I feel I'm headed towards the right direction, but I'm not sure if I'm doing this right. For every move, there are 2 possible choices. If we want to get to (20,10), then there are 200 points from the origin to this point. And I think order matters here, so we would use permutations, so I come up with 200 P 2, which is 39,800.
120360
Is it possible to draw a triangle, if the length of its medians $(m_1, m_2, m_3)$ are given only?  Someone asked me this question, but I can not see it. Is it really possible?  UPDATE  Apart from the algebraic solution given by JimmyK4542, can anyone give me a direct construction?  I mean, it should sound like:     Draw a line segment sufficiently long. Cut the length of $m_1$ from it. Then $\ldots$
86302
I want to prove this and intuitively it makes sense. But I'm having a hard time coming up with a proof. So if a sequence converges, then we have a natural number for which the distance between all terms after it and the limit point get arbitrarily small. So how can I show that this also holds for every subsequence (which is like a subset of a sequence)?  (are subsequences always infinite?)  Could I suppose that there is a subsequence that doesn't converge to that limit, and find a contradiction? (and do the same for the other direction?)
11302
The question is quite straight forward, on $\mathbb{R}^2$ the maximum number of equidistant distinct points that can be layed (using the euclidean distance) is 3, forming thus an equilateral triangle. On $\mathbb{R}^3$ the maximum number of points is 4, forming thus a tetrahedon. My question is quite simple, is the maximum number of points that can be layed in $\mathbb{R}^n$ equal to $n+1$?
114924
Why if the absolute value is the distance from $0$ on the number line, is $\lvert 3-\pi \rvert = \pi - 3$ and $\left\lvert\sqrt{2}-1\right\rvert= \sqrt{2}-1$
84903
Which rectangle has the largest perimeter inscribed in a right-angled triangle of legs $3$ and $4$, where one of the sides lies on the hypotenuse?  I've done the calculation for the largest area and found the value of the sides of the rectangle, using a derivative, and found the values of 6/5 and 5/2 but I don't know how to answer the final part of the question.
118215
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 &lt; 2(xy + yz + xz)$        My solution: Consider  $$x^2 + y^2 + z^2 &lt; 2(xy + yz + xz)$$  Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$  Hence  $$(x+y+z)^2-2(xy+yz+xz) &lt; 2(xy + yz + xz)$$  $$ (x+y+z)^2 - 4(xy+yz+xz) &lt; 0 $$  As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 &gt;0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$  Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
41973
Every day you walk from point A to point B, which are $3$ miles apart. There is a $50$% chance each walk that there is an invisible wall somewhere strictly between the two points (never at A or B). The wall extends $1$ mile in each direction perpendicular to the line segment (direct path) between A and B, and its position can be at any random location between the two points. That is, it can be any distance between A and B such as $0.1$ miles away from A, $1.5$ miles away, $2.9$ miles away.... You don't know if the wall is present or where it is until you walk into it.  You must walk around the wall if present as there is no other way to circumvent it.  Assume the wall is negligible thickness (like a force field), all ground is perfectly flat, and the y coordinates at both A and B are $0$ (although I don't think the optimal path will change much if they weren't).  What strategy minimizes the average expected walk distance between A and B?  How do we know for certain this strategy yields the shortest average walking distance?  To get the $100$ bounty points, I would like a convincing argument from you as to why you feel your solution is optimal and cannot be beaten.  For those of you using computer simulation, anything reasonably close to optimal is a candidate for the bounty.   Can anyone beat the optimal solutions submitted so far?  There are still a few days to try.  Even a slight beat (if not due to roundoff error or some other error) would help prove previous solutions are non-optimal.  You can just use the table of $31$ points (x-y coordinates) and compute that path length and then if you can find better then I will accept that answer.  I think by Sunday night I may have to award the bounty otherwise it may get auto-awarded.
61042
Last year, I took physics and over the summer I have started to wonder about why many phenomena work the way they do (such as why is $\mathrm{KE} =\frac{1}{2}mv^2$, etc.). I have found answers to all of my questions, except for this one:  Why are we allowed to use trigonometry to get vector components? I understand that if we draw a right triangle, then we can express the adjacent side with $\text{length of hypotenus} \times \cos(\theta)$. But why does this work with velocity and with forces? I have tried to reason through this many times, but I am not able to figure out why we are able to do this. Any ideas?   BTW: My attempts for reasoning through velocity were as follows: If we have a ball moving in two dimensions, and we are given that the ball has $\sqrt{3}$ the speed in the x-direction than the y-direction, then we can conclude that the ball will move in a 30-degree angle. I am not sure how to apply this to forces or acceleration though. For example, why is it that in the case of a ball spinning horizontally while attached to a string, that the y component of the strings tension gets smaller as the centripetal force increases? I can see how we can prove this by using trigonometry, but why does this trig work in the first place?
30849
at point no. 2 in the diagram, let the object accelerates to a velocity v1 (just along the direction of the slope) due to rolling down the ramp. Now, v1 has a x-component as well as y-component.  Now, my question is:  In the diagram, In between the points no. 2 and 3, the object passes over a horzontal distance.  During this part of motion, which statement of the following is true:  i)  the magnitude of the velocity remains constant, i.e. equal to v1, and only the direction changes.  or, ii) the magnitude of the velocity changes, in such a way, that only the x-component of v1 remains still; whereas, the y-component of v1 completely vanishes,  as soon as the object leaves the ramp and gets onto the horizontal trajectory(between the points marked 2 and 3).
58439
Someone posted a question on the notice board at my University's library. I've been thinking about it for a while, but fail to see how it is possible. Could someone verify that this is a valid question and point me in the right direction?  'Given an infinite set of points in a plane, if the distance between any two points is an integer, prove that all these points lie on a straight line.'
67151
I am not sure how to approach this problem.  Every day, a dance student walks from her home to dance-school, which is located $12$ blocks east and $16$ blocks north from home. She always takes the shortest walk of $28$ blocks.  $a$) How many different walks are possible?  $b$) Suppose that $5$ blocks east and $6$ blocks north of her home lives her dance friend, whom she meets each day on her way to dance school. Now how many walks are possible?  For $a$ I was thinking $P(28,12)\cdot P(28,16)$, but not sure if that would work because you can't walk the furthest block north before the closest block north.  And for $b$ I'm not really sure where to start.