Find a point on the plane of a triangle such that the sum of its distances from three vertices is minimum.  I am guessing that it is the centroid but I can't prove that.
122064
Given a triangle in a plane with vertices A, B, C, find the point T that minimizes the sum of distances between A-T, B-T, and C-T.  I can experimentally determine this point by sampling the space and finding the minimum via an attractive, interactive web page:   Drag any vertex to see the gradient of value sums change and the point T be updated. I find it interesting to note that moving any vertex towards or away from T does not affect T, until the point where the vertex touches T and causes it to move along with it.  However, an experimental determination is neither efficient nor accurate, and prevents me from graphing other interesting properties such as the arcs of the vertices when a fixed sum is required.  Please help me find the equation for T (preferably in terms of vectors, as the answer should be the same in 2D or 3D).  Edit: Based on the answer, I've created an updated interactive using the calculated point and showing the construction:
20432
Give $a, b. c$ be the lengths of a triangle and $a+ b+ c= 3$. Prove that: $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\geq  \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}$$ My try  We have England MO inequality: $$\left ( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \right )^{2}\geq  \left ( a+ b+ c \right )\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )$$  and $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \geq  3=  a+ b+ c$$  But it cannot be used to prove the above inequality and how to use triangle inequality in this inequality. I need the help. Thanks!
601
I need to find the equation(s) of the plane containing the line  $$ \left\{  \begin{array}{c} x=1+k \\ y=1 \\ z = -2k \end{array} \right.  $$  lying at distance 1 from the point (-1,0,1).  I was trying to use "Distance from point to a plane" formula and compose a system, given that I know collinear points on the plane (from the line), however, I apparently miss something to finalize the solution.
37292
I have a polygon and the polygon's centroid. I want to select the most distant vertex from centroid (picture: point A). I used arcpy.FeatureVerticesToPoints_management to create a feature class containing points generated from polygon, but I don't know what to do next. Could you help me?  Also I want to find intersection point (point B) of the outline of polygon and line created by centroid and the most distant point (line OA).    I would be very grateful for any help.
41339
The following show you the whole question.     Find the distance d bewteen two planes   \begin{eqnarray} \\C1:x+y+2z=4 \space \space~~~ \text{and}~~~ \space \space C2:3x+3y+6z=18.\\ \end{eqnarray}   Find the other plane $C3\neq C1$ that has the distance d to the plane $C2$.   According to the example my teacher gave me, the answer should be :     Am I right? However, I do not know what is normal and why there are P(5) and Q($-\frac{1}{2}$).    Thank you for your attention
113789
I have 3 points in space A, B, and C all with (x,y,z) coordinates, therefore I know the distances between all these points. I wish to find point D(x,y,z) and I know the distances BD and CD, I do NOT know AD.  The method I have attempted to solve this using is first saying that there are two spheres known on points B and C with radius r (distance to point D). The third sphere is found by setting the law of cosines equal to the formula for distance between two vectors ((V1*V2)/(|V1||V2|)) = ((a^2+b^2-c^2)/2ab).   Now point D should be the intersection of these three spheres, but I have not been able to calculate this or find a way to. I either need help finding point D and I can give numeric points for an example, or I need to know if I need more information to solve (another point with a distance to D known).  Ok, now assuming that distance AD is known, how do I calculate point D?
50245
Suppose that $a,b,c$ are the lengths of the sides of a triangle. Prove that $$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$$   Attempt  It's hard to do much simplification here so we are going to have to use that $a &gt; b+c, b&gt;a+b, c&gt;a+c.$ Using that fact we obtain that $$b^2(c-b)+c^2(b-c) &gt; a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c).$$ Then should I use AM-GM to finish it or what inequality?
61135
Consider six distinct points in a plane. Let $m$ and $M$ denote respectively the minimum and the maximum distance between any pair of points.  Show that $M/m \geqslant \sqrt{3}$.
97323
Let L1 be the line passing through the point $P_1=(−11, 10, −1)$ with direction vector $\vec d_1=\begin{bmatrix}−2\\ 3\\ −1\end{bmatrix}t$, and let $L_2$ be the line passing through the point $P_2=(−8, 9, 10)$ with direction vector $\vec d_2=\begin{bmatrix}−1\\ 3\\1\end{bmatrix}t$. Find the shortest distance, $d$, between these two lines, and find a point $Q_1$ on $L_1$ and a point $Q_2$ on $L_2$ so that $d(Q_1,Q_2) = d$.   I dont really understand how to properly approach this question, how do i start?
56645
Find the minimum of the function      $$ f(x,y)=|ax-by+c|$$      where $a,b,c \in \mathbb N$ and $x,y \in \mathbb Z$.   The questions  and  are similar but they are in cases where $x, y$ are bounded.Taking the partial derivatives,etc. doesn't help. Is there a way to do this efficiently(for a computer program)?
46885
Find the minimum of $\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x+2)^2+(y+2)^2}+\sqrt{(x-1)^2+(y+1)^2}$ For real $x,y$. It is the distance of a point in a triangle from its vertexes which gives the minimum when $(x,y)$ is on the Fermat's point.  But now I am stuck in finding the location of Fermat's point.  $(1) \quad 4$ $(2) \quad 5$ $(3) \quad \sqrt{3}+2\sqrt{5}$ $(4) \quad 4\sqrt{2}$
35839
here's a question I got for homework (sorry if my translation is a bit unclear):  Let $X\sim‬G(p_1)$, $Y\sim ‬G(p_2)$, $X$ and $Y$ are independent. Prove that the minimum is also geometric, meaning: $\min(X,Y)\sim G(1-(1-p_1)(1-p_2))$. Instructions: first calculate the probability $P(\min(X,Y) &gt; k)$ and compare it to the parallel probability in (of?) a geometric random variable.  I have no idea where to start, even with the great clue that they've supplied. Any hints?
66865
Given four points $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$,$(x_4,y_4)$. How can we efficiently test them to make sure whether they are vertices of a parallelogram?  I think evaluating and comparing the distance between two points seems to be inefficient. What do you think? Do you have any better method?
3687
I have a mesh and in the region around each triangle, I want to compute an estimate of the principal curvature directions. I have never done this sort of thing before and  does not help a lot. Can you describe or point me to a simple algorithm that can help me compute this estimate?  Assume that I know positions and normals of all vertices.
99390
I came across this problem that seems a bit peculiar. Take  $$\sum_{1 \leq i &lt; j \leq n} |x_i-x_j|,$$ where $x_1,...,x_n \in [1,35].$ I want to figure out the maximum possible value of this sum in terms of $n$.  To me this, problem could possibly be bounded above by the triangle inequality, but I'm not sure what number I should use for triangulation. We can see that $$\sum_{1 \leq i &lt; j \leq n} |x_i-x_j|,$$ $$\leq \sum_{1 \leq i &lt; k \leq n} |x_i-35|+|x_j-35|,$$ although I am now having some troubles figuring out how this will help me to solve for a meaningful upper bound. Any recommendations?
95047
Let $a, b$ and $c$ three positive real numbers such that $a+b+c=3$. Find the minimum value of $$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}.$$  Here is my attempt. By symmetry we can assume that $a\leq b\leq c$. The function $f(x)=\frac{2-x^3}{x}$ is decreasing and convex on $]0,2^{1/3}]$. So if $c\leq 2^{1/3}$ then $3f(1)=3f(\frac{a+b+c}{3})\leq f(a)+f(b)+f(c)=A$ with equality if $a=b=c=1$.   If $c&gt;2^{1/3}$ I don't see how to proceed.
16759
Suppose there are $n$ points $(x_i, y_i)$ for $i = 1,\ldots,n$. Please find another point $(x, y)$ to minimize function: $$\sum_{i=1}^n \max(|x_i - x|, |y_i - y|)$$