How prove that for every convex polygon there exist three consecutive vertices which their circumcircle include all poligon?
122040
Given a convex $n$-gon. The circumcircle is constructed for every triple of consecutive vertices of the polygon. We get the $n$ circles. Select the circle with the largest radius. Prove that the circle contains the polygon.  My work so far: $n=3 -$ triangle - obviously. $n=4 -$ If $\angle B = \max \left\{A,B,C,D \right\}$ then $ABCD \in \omega_{ABC}$  $n\ge 5$. I need help here.
70573
Consider a square of sidelength $n$ and $ (n+1)^{2} $ interior points. Prove that we can choose 3 of these points so that they determine a triangle (eventually degenerated) of area at most 1/2.  I couldn't get started on this. Any ideas?
101746
In arbitrary triangle $ABC$, let $G$ be centroid and $AH$ be a height.We extend $HG$ beyond $G$ to intersect circumcircle at $K$.Prove that $AK||BC$       I connected midpoint of $BC$ to $K$ and $A$ to show $MA=MK$ but failed...
3065
We consider in $\mathbb{R}^2$ the set of points $$\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}$$  Let $\Omega$ a rectangular structure consisting of the heads $\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}$  Let the two polygons $\sum_1 = \{M_4,M_5,M_2,M_1\}$ and $\sum_2 = \{M_5,M_6,M_3,M_2\}$   let the mesh $\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}$ such $$Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}$$  1- Prove that $(e_1,\sum_1,Q_1)$ and $(e_2,\sum_2,Q_1)$ are two finite elements of Lagrange.  2- Let $V$ the finite-dimensional space that corresponds to the mesh. Prove that $V \nsubseteqq H^1(\Omega)$  My problem is to prove 2.
108003
I have a four points in plane and need to test (based on point coordinates) whether they are able to form a convex quadrilateral:    Of course, the test should avoid configurations like these:      Given the diagonals, I can check whether the quadrilateral is convex (simply checking whether the intersection of diagonals is between both ends of both diagonals).  The real problem is how to label the four points and filter out all concave and degenerate configurations (like, for example: $A=B$).  If the labeling is possible (convex case found), the four points should be labeled such that $AC$ and $BD$ are diagonals of a convex quadrialteral.  I wonder if there is an elegant solution (rather than testing every possible permutation of $A, B, C$ and $D$).
92153
Graph $G$ has the diameter greater than $3$. Prove that the diameter of   $G'$ (the complement graph) is less than $3$.   What I was thinking about:  $G$ has the diameter greater than $3$ $\implies$ $G$ has not less than $4$ vertices.  If the $u$ and $v$ are vertices in $G$ that represents the diameter, then in $G'$ there is the edge $uv$. So, I was thinking that only first neighbors can be the diameter of $G'$. We need to prove now that between every two neighbors in $G$ there is only one other vertex in $G'$ which connects these two vertices (so the diameter is less than $3$).
87442
For an arbitrary convex quadrialteral $ABCD$:   $$AF=FC, BE=ED, GI=IH$$  Prove that points $F$, $E$ and $I$ are collinear.  I was able to solve the problem by using vectors and I think that it can be also solved analytically but I wanted some more elegant proof, more in the spirit of Euclid.
22092
Q) Let T be the line passing through the points P(–2, 7) and Q(2, –5). Let $F_{1}$ be the set of all pairs of circles $(S_{1}$, $S_{2}$) such that T is tangent to $S_{1}$ at P and tangent to $S_{2}$ at Q, and also such that $S_{1}$ and $S_{2}$ touch each other at a unique point, say, M. Let $E_{1}$ be the set representing the locus of M as the pair ($S_{1}$, $S_{2}$) varies in $F_{1}$. Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point R(1, 1) be $F_{2}$. Let $E_{2}$ be the set of the mid-points of the line segments in the set $F_{2}$. Let $C$ be the circle $x^2+y^2+6(2y+7x)=53$. The number of times $C$ intersects $E_{1}$ and $E_{2}$ is (are):
110883
What is the probability that the convex hull of $n+2$ random points on $n$-dimensional sphere contains sphere's center?
33545
If $a_1,a_2,a_3,...,a_n$ are the side lengths of $A_1A_2A_3...A_n$ convex polygon,then$\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}$ is   $(A)&gt;\frac{1}{n-1}\hspace{1cm}(B)&lt;\frac{1}{n-1}\hspace{1cm}(C)&gt;\frac{1}{n}\hspace{1cm}(D)&lt;\frac{1}{n}$  $\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}=\frac{a^2_1}{a^2_n}+\frac{a^2_2}{a^2_n}+\frac{a^2_3}{a^2_n}+......+\frac{a^2_{n-1}}{a^2_n}&gt;(n-1)\frac{a^2_1}{a^2_n}\frac{a^2_2}{a^2_n}\frac{a^2_3}{a^2_n}......\frac{a^2_{n-1}}{a^2_n}$  and then no idea how to proceed.Any hints,please?
32308
Given a connected graph, how can we prove that the number of edge of its minimum edge cover plus its maximum matching is equal to the number of vertices?
113877
Need help with this proof using Ferrers' graph or otherwise.    Show that the number of partitions of $r+k$ into $k$ parts is equal to     The number of partitions of $r + {k+1 \choose 2}$ into $ k $ distinct parts   The number of partitions of $r$ into parts of size at most $k$
14290
Assume $G$ is a , i.e. a (finite) directed graph such that between any two vertices, $a$ and $b$, there is at least one edge in one of the two directions, $a\rightarrow b$ or $b\rightarrow a$.   Show that there is at least one , i.e. a path, following the direction of the edges and visiting all vertices exactly once. (This part is a classical exercise and it is here to prepare for the second part of the problem). Consider all possible losers, i.e. the set of all vertices that can be end-points of Hamiltonian paths of the tournament, see Part (1). Show that the subgraph of all losers has a , i.e. a cycle that follows the direction of the edges and passes through every vertex exactly once and returns to the initial vertex.   For example     these are all tournaments and the $\color{red}{\text{red}}$ dots are the losers. Notice how there is always a cycle joining them.
61671
How to prove that the intersection of any finite number of convex sets is a convex set?  I have no idea.
21669
I have $n$ points $x_1,\dots,x_n\in\Bbb R^d$, and I would like to check that some other point $y$ lies in their convex hull. How can I do this in some efficient way? I think that there was an algorithm based on checking the signs of pairwise inner products $x_i\cdot y$, however I was not able to find it.
30013
A circle is drawn that intersects all three sides of $\triangle PQR$ as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of $\triangle PQR$.     Designate the center of the circle $G$.  Thinking about it a bit, we realize if we could prove the incircle of the triangle and the circle given are concentric, we would know that the center of the circle is the incenter, for they have the same center.    If we shrink the circle given down to the incircle, the incircle should intersect the triangle at the midpoints of $AB$, $CD$, and $EF$.    We therefore denote the midpoint of $AB M$, the midpoint of $CD L$, andd the midpoint of $EF N$.  We know the incircle is concentric to the given circle if and only if $NG$ and $LG$ are congruent, because a circle is a circle if and only if the radii are congruent.  Thus, we must prove $NG\cong LG$.  When proving lengths equal, congruent triangles are always a good idea.  We see that $NGE\cong LGD$ seems likely, and it would also prove our answer if we could prove it.   We know that $GD\cong GE$ because they are both radii of the larger circle, and we also know $DL\cong EN$ because they are both half of two lines that are congruent.  We know $\angle GLD$ and $\angle GNE$ are both right angles, because that is where the smaller circle is tangent to the sides of the triangle.  Thus, $NGE\cong LGD$ by $HL$ congruency.  Because congruent parts of congruent triangles are congruent, we know $NG\cong LG$, which proves $G$ is the center of the incircle.  Is this proof sound?
101571
I understand how planar graphs and their dual graphs are connected, but I can't find a way to prove this because it just seems..obvious?