Let $ABCDEF$ be a convex hexagon with $AB = CD = EF$, $BC =DE = FA$ and $\angle A+\angle B = \angle C +\angle D = \angle E +\angle F$. Prove that $\angle A=\angle C=\angle E$ and $\angle B=\angle D=\angle F$. Tran Quang Hung

Let $F$ be the boundary and $M,N$ be any interior points of a triangle $ABC$. Consider the function $f_{M,N}: F \to R$ defined by $f_{M,N}(P) = MP^2 +NP^2$ and let $\eta_{M,N}$ be the number of points $P$ for which $f{M,N}$ attains its minimum. (a) Prove that $1 \le \eta_{M,N} \le 3$. (b) If $M$ is fixed, find the locus of $N$ for which $\eta_{M,N} > 1$. (c) Prove that the locus of $M$ for which there are points $N$ such that $\eta_{M,N} = 3$ is the interior of a tangent hexagon.

Let $ABCDEF$ be a convex hexagon with $AB=DE$, $BC=EF$, $CD=FA$, and $\angle A-\angle D = \angle C -\angle F = \angle E -\angle B$. Prove that the diagonals $AD$, $BE$, and $CF$ are concurrent.

Let $ABCDEF$ be a convex hexagon satisfying $AC = DF, CE = FB$ and $EA = BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point.

Hexagon $ABCDEF$ has a circumscribed circle and an inscribed circle. If $AB = 9$, $BC = 6$, $CD = 2$, and $EF = 4$. Find $\{DE, FA\}$.

The surface of a soccer ball is made up of black pentagons and white hexagons together. On the sides of each pentagon are nothing but hexagons, while on the sides of each border of hexagons alternately pentagons and hexagons. Determine from this information about the soccer ball , the number of its pentagons and its hexagons.

Let $ABCDEF$ be a convex hexagon. In the hexagon there is a point $K$, such that $ABCK,DEFK$ are both parallelograms. Prove that the three lines connecting $A,B,C$ to the midpoints of segments $CE,DF,EA$ meet at one point.

Three identical spheres fit into a glass with rectangular sides and bottom and top in the form of regular hexagons such that every sphere touches every side of the glass. The glass has volume $108$ cm$^3$. What is the sidelength of the bottom? [img]https://1.bp.blogspot.com/-hBkYrORoBHk/XzcDt7B83AI/AAAAAAAAMXs/P5PGKTlNA7AvxkxMqG-qxqDVc9v9cU0VACLcBGAsYHQ/s0/2000%2BMohr%2Bp2.png[/img]

Let $P$ be a regular hexagon. For a point $A$, let $d_1\le d_2\le ...\le d_6$ the distances from $A$ to the six vertices of $P$, ordered by magnitude. Find the locus of all points $A$ in the interior or on the boundary of $P$ such that: (a) $d_3$ takes the smallest possible value. (b) $d_4$ takes the smallest possible value.

The hexagon $ABCDEF$ has all angles equal . We know that four consecutive sides of the hexagon have lengths $7, 6, 3$ and $5$ in this order. What is the sum of the lengths of the two remaining sides?

Someone wants to unscrew a square nut with side $a$, with a wrench whose hole has the form of a regular hexagon with side $b$. What condition should the lengths $a$ and $b$ meet to make this possible?

In an acute triangle $ABC$ let $K,L,$ and $M$ be the midpoints of sides $AB,BC,$ and $CA,$ respectively. From each of $K,L,$ and $M$ drop two perpendiculars to the other two sides of the triangle; e.g., drop perpendiculars from $K$ to sides $BC$ and $CA,$ etc. The resulting $6$ perpendiculars intersect at points $Q,S,$ and $T$ as in the figure to form a hexagon $KQLSMT$ inside triangle $ABC.$ Prove that the area of this hexagon $KQLSMT$ is half of the area of the original triangle $ABC.$ [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra; diagram by adihaya*/ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 11.888712276357234, xmax = 17.841346447833423, ymin = 10.61620970860601, ymax = 15.470685507068502; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (12.488234161849352,12.833838721895551), B = (16.50823416184936,15.093838721895553), C = (16.28823416184936,11.353838721895551), K = (14.498234161849355,13.963838721895552), L = (16.39823416184936,13.223838721895552), M = (14.388234161849356,12.093838721895551), D = (13.615830174638527,13.467760858438725), F = (15.75135711740064,11.562938202365055), G = (15.625830174638523,14.597760858438724), H = (16.435061748056253,13.849907687412797), T = (14.02296781802369,12.74356027153236), Q = (16.032967818023693,13.873560271532357), O = (16.325061748056253,11.979907687412794); draw(A--B--C--cycle, zzttqq); draw((13.426050287639166,13.361068683160477)--(13.532742462917415,13.171288796161116)--(13.722522349916774,13.277980971439364)--D--cycle, qqwuqq); draw((14.054227993863618,12.223925334689998)--(14.133240861538676,12.426796211152979)--(13.930369985075695,12.505809078828037)--(13.851357117400637,12.302938202365056)--cycle, qqwuqq); draw((16.337846386707046,12.19724654447628)--(16.12050752964356,12.210031183127075)--(16.107722890992765,11.992692326063588)--O--cycle, qqwuqq); draw((15.830369985075697,11.765809078828037)--(15.627499108612716,11.844821946503092)--(15.54848624093766,11.641951070040111)--F--cycle, qqwuqq); draw((15.436050287639164,14.491068683160476)--(15.542742462917412,14.301288796161115)--(15.73252234991677,14.407980971439365)--G--cycle, qqwuqq); draw((16.217722890992764,13.86269232606359)--(16.20493825234197,13.645353469000101)--(16.42227710940546,13.63256883034931)--H--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(A--B, zzttqq); draw(B--C, zzttqq); draw(C--A, zzttqq); draw(M--D); draw(K--(13.851357117400637,12.302938202365056)); draw(F--L); draw(L--G); draw(K--H); draw(M--O); /* dots and labels */ dot(A,dotstyle); label("$A$", (12.52502834296331,12.93568440300881), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (16.548187989892043,15.193580123223922), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (16.332661580235147,11.457789022504372), NE * labelscalefactor); dot(K,linewidth(3.pt) + dotstyle); label("$K$", (14.536608166427676,14.02357961365791), NE * labelscalefactor); dot(L,linewidth(3.pt) + dotstyle); label("$L$", (16.43529320388129,13.28463192340569), NE * labelscalefactor); dot(M,linewidth(3.pt) + dotstyle); label("$M$", (14.433976542781535,12.155684063298134), NE * labelscalefactor); dot(D,linewidth(3.pt) + dotstyle); dot((13.851357117400637,12.302938202365056),linewidth(3.pt) + dotstyle); dot(F,linewidth(3.pt) + dotstyle); dot(G,linewidth(3.pt) + dotstyle); dot(H,linewidth(3.pt) + dotstyle); dot((15.922967818023695,12.003560271532354),linewidth(3.pt) + dotstyle); label("$S$", (15.96318773510904,12.063315602016607), NE * labelscalefactor); dot(T,linewidth(3.pt) + dotstyle); label("$T$", (14.064502697655428,12.802263292268826), NE * labelscalefactor); dot(Q,linewidth(3.pt) + dotstyle); label("$Q$", (16.076082521119794,13.931211152376383), NE * labelscalefactor); dot(O,linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

In a circle $O$, there are six points, $ A$, $ B$, $C$, $D$, $E$, $F$ in a counterclockwise order such that $BD \perp CF$ , and $CF$, $BE$, $AD$ are concurrent. Let the perpendicular from $B$ to $AC$ be $M$, and the perpendicular from $D$ to $CE$ be $N$. Prove that $AE \parallel MN$.

Consider the diagonals $A_1A_3, A_2A_4, A_3A_5, A_4A_6, A_5A_4$ and $A_6A_2$ of a convex hexagon $A_1A_2A_3A_4A_5A_6$. The hexagon whose vertices are the points of intersection of the diagonals is regular. Can we conclude that the hexagon $A_1A_2A_3A_4A_5A_6$ is also regular?

Given an equilateral triangle with sidelength $k$ cm. With lines parallel to it's sides, we split it into $k^2$ small equilateral triangles with sidelength $1$ cm. This way, a triangular grid is created. In every small triangle of sidelength $1$ cm, we place exactly one integer from $1$ to $k^2$ (included), such that there are no such triangles having the same numbers. With vertices the points of the grid, regular hexagons are defined of sidelengths $1$ cm. We shall name as [i]value [/i] of the hexagon, the sum of the numbers that lie on the $6$ small equilateral triangles that the hexagon consists of . Find (in terms of the integer $k>4$) the maximum and the minimum value of the sum of the values of all hexagons .

Let the regular hexagon $ABCDEF$ be inscribed in a circle with center $O$, $N$ be such a point Let $E-N-C$, $M$ a point such that $A- M-C$ and $R$ a point on the circumference, such that $D-N- R$. If $\angle EFR = 90^o$, $\frac{AM}{AC}=\frac{CN}{EC}$ and $AC=\sqrt3$, calculate $AM$. Notation: $A-B-C$ means than points $A,B,C$ are collinear in that order i.e. $ B$ lies between $ A$ and $C$.

The midpoints of alternative sides of a convex hexagon are connected by line segments. Prove that the intersection points of the medians of the two triangles obtained coincide.

The triangles $AB'C, CA'B$ and $BC'A$ are constructed on the sides of the equilateral triangle $ABC.$ In the resulting hexagon $AB'CA'BC'$ each of the angles $\angle A'BC',\angle C'AB'$ and $\angle B'CA'$ is greater than $120^\circ$ and the sides satisfy the equalities $AB' = AC',BC' = BA'$ and $CA' = CB'.$ Prove that the segments $AB',BC'$ and $CA'$ can form a triangle. [i]David Brodsky[/i]

Let $ABCDEF$ be a convex hexagon inscribed in a circle . Prove that the diagonals $AD, BE$ and $CF$ intersect at one point if and only if $$\frac{AB}{BC} \cdot \frac{CD}{DE}\cdot \frac{EF}{FA}=1$$

In the following, a rhombus is one with edge length $1$ and interior angles $60^o$ and $120^o$ . Now let $n$ be a natural number and $H$ a regular hexagon with edge length $n$, which is covered with rhombuses without overlapping has been. The rhombuses then appear in three different orientations. Prove that whatever the overlap looks exactly, each of these three orientations can be viewed at the same time.

Two perpendiculars are drawn from the midpoints of each side of the acute-angle triangle to two other sides. Those six segments make hexagon. Prove that the hexagon area is a half of the triangle area.

Let $A,B,C,D,E,F$ be the midpoints of consecutive sides of a hexagon with parallel opposite sides. Prove that the points $AB\cap DE$, $BC\cap EF$, $AC\cap DF$ lie on a line.

A plane passing through the center of a cube intersects the cube in a cyclic hexagon. Show that this hexagon is regular.

Let $O$ and $O'$ designate two dìagonally opposite vertices of a cube. Bisect those edges of the cube that contain neither of the points $O$ and $O'$. Prove that these midpoints of edges lie in a plane and form the vertices of a regular hexagon

Let $ABC$ be a triangle and $O$ be its circumcircle. Let $A', B', C'$ be the midpoints of minor arcs $AB$, $BC$ and $CA$ respectively. Let $I$ be the center of incircle of $ABC$. If $AB = 13$, $BC = 14$ and $AC = 15$, what is the area of the hexagon $AA'BB'CC'$? Suppose $m \angle BAC = \alpha$ , $m \angle CBA = \beta$, and $m \angle ACB = \gamma$. [b]p10.[/b] Let the incircle of $ABC$ be tangent to $AB, BC$, and $AC$ at $J, K, L$, respectively. Compute the angles of triangles $JKL$ and $A'B'C'$ in terms of $\alpha$, $\beta$, and $\gamma$, and conclude that these two triangles are similar. [b]p11.[/b] Show that triangle $AA'C'$ is congruent to triangle $IA'C'$. Show that $AA'BB'CC'$ has twice the area of $A'B'C'$. [b]p12.[/b] Let $r = JL/A'C'$ and the area of triangle $JKL$ be $S$. Using the previous parts, determine the area of hexagon $AA'BB'CC'$ in terms of $ r$ and $S$. [b]p13.[/b] Given that the circumradius of triangle $ABC$ is $65/8$ and that $S = 1344/65$, compute $ r$ and the exact value of the area of hexagon $AA'BB'CC'$. PS. You had better use hide for answers.