Found problems: 200
Given a regular hexagon, whose sidelength is $ 1$ . What is the largest number of circles of radius $\frac{\sqrt3}{4}$ can be placed without overlapping inside such a hexagon? (Circles can touch each other and the sides of the hexagon.)
et $ABCDEF$ be a convex hexagon which has an inscribed circle and a circumcribed. Denote by $\omega_{A}, \omega_{B},\omega_{C},\omega_{D},\omega_{E}$ and $\omega_{F}$ the inscribed circles of the triangles $FAB, ABC, BCD, CDE, DEF$ and $EFA$, respecitively. Let $l_{AB}$, be the external of $\omega_{A}$ and $\omega_{B}$; lines $l_{BC}$, $l_{CD}$, $l_{DE}$, $l_{EF}$, $l_{FA}$ are analoguosly defined. Let $A_1$ be the intersection point of the lines $l_{FA}$ and $l_{AB}$, $B_1, C_1, D_1, E_1, F_1$ are analogously defined.
Prove that $A_1D_1, B_1E_1, C_1F_1$ are concurrent.
Let $ ABCDEF$ be a convex hexagon with $ AB \equal{} BC \equal{} CD$ and $ DE \equal{} EF \equal{} FA$, such that $ \angle BCD \equal{} \angle EFA \equal{} \frac {\pi}{3}$. Suppose $ G$ and $ H$ are points in the interior of the hexagon such that $ \angle AGB \equal{} \angle DHE \equal{} \frac {2\pi}{3}$. Prove that $ AG \plus{} GB \plus{} GH \plus{} DH \plus{} HE \geq CF$.
A regular hexagon is cut up into $N$ parallelograms of equal area. Prove that $N$ is divisible by three.
(V. Prasolov, I. Sharygin, Moscow)
In a convex cyclic hexagon $ABCDEF$, $BC=EF$ and $CD=AF$. Diagonals $AC$ and $BF$ intersect at point $Q,$ and diagonals $EC$ and $DF$ intersect at point $P.$ Points $R$ and $S$ are marked on the segments $DF$ and $BF$ respectively so that $FR=PD$ and $BQ=FS.$ [b]The segments[/b] $RQ$ and $PS$ intersect at point $T.$ Prove that the line $TC$ bisects the diagonal $DB$.
In the regular hexagon $ABCDEF$ on the line $AF$, the point $X$ is taken so that the angle $XCD$ is $45^o$. Find the angle $\angle FXE$.
(Kiev Olympiad)
Alice and Bob are playing in the forest. They have six sticks of length 1, 2, 3, 4, 5, 6 inches. Somehow, they have managed to arrange these sticks, such that they form the sides of an equiangular hexagon. Compute the sum of all possible values of the area of the hexagon.
Sheila is making a regular-hexagon-shaped sign with side length $ 1$. Let $ABCDEF$ be the regular hexagon, and let $R, S,T$ and U be the midpoints of $FA$, $BC$, $CD$ and $EF$, respectively. Sheila splits the hexagon into four regions of equal width: trapezoids $ABSR$, $RSCF$ , $FCTU$, and $UTDE$. She then paints the middle two regions gold. The fraction of the total hexagon that is gold can be written in the form $m/n$ , where m and n are relatively prime positive integers. Compute $m + n$.
[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS9lLzIwOTVmZmViZjU3OTMzZmRlMzFmMjM1ZWRmM2RkODMyMTA0ZjNlLnBuZw==&rn=MjAyMCBCTVQgSW5kaXZpZHVhbCAxMy5wbmc=[/img]
The midpoints of the opposite sides of the hexagon are connected by segments. It turned out that the points of pairwise intersection of these segments form an equilateral triangle. Prove that the drawn segments are equal.
(M. Volchkevich)
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.
The two circles $\Gamma_1$ and $\Gamma_2$ intersect at $P$ and $Q$. The common tangent that's on the same side as $P$, intersects the circles at $A$ and $B$,respectively. Let $C$ be the second intersection with $\Gamma_2$ of the tangent to $\Gamma_1$ at $P$, and let $D$ be the second intersection with $\Gamma_1$ of the tangent to $\Gamma_2$ at $Q$. Let $E$ be the intersection of $AP$ and $BC$, and let $F$ be the intersection of $BP$ and $AD$. Let $M$ be the image of $P$ under point reflection with respect to the midpoint of $AB$. Prove that $AMBEQF$ is a cyclic hexagon.
Six different points $A, B, C, D, E, F$ are marked on the plane, no four of them lie on one circle and no two segments with ends at these points lie on parallel lines. Let $P, Q,R$ be the points of intersection of the perpendicular bisectors to pairs of segments $(AD, BE)$, $(BE, CF)$ ,$(CF, DA)$ respectively, and $P', Q' ,R'$ are points the intersection of the perpendicular bisectors to the pairs of segments $(AE, BD)$, $(BF, CE)$ , $(CA, DF)$ respectively. Show that $P \ne P', Q \ne Q', R \ne R'$, and prove that the lines $PP', QQ'$ and $RR'$ intersect at one point or are parallel.
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 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 $AE \parallel MN$.
A convex hexagon is circumscribed about a circle of radius $1$. Consider the three segments joining the midpoints of its opposite sides. Find the greatest real number $r$ such that the length of at least one segment is at least $r.$
The two circles $\Gamma_1$ and $\Gamma_2$ intersect at $P$ and $Q$. The common tangent that's on the same side as $P$, intersects the circles at $A$ and $B$,respectively. Let $C$ be the second intersection with $\Gamma_2$ of the tangent to $\Gamma_1$ at $P$, and let $D$ be the second intersection with $\Gamma_1$ of the tangent to $\Gamma_2$ at $Q$. Let $E$ be the intersection of $AP$ and $BC$, and let $F$ be the intersection of $BP$ and $AD$. Let $M$ be the image of $P$ under point reflection with respect to the midpoint of $AB$. Prove that $AMBEQF$ is a cyclic hexagon.
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]
a) Given a convex hexagon $ABCDEF$ with all the equal angles. Prove that $$|AB|-|DE| = |EF|-|BC| = |CD|-|FA|$$
b) The opposite problem:
Prove that it is possible to construct a convex hexagon with equal angles of six segments $a_1,a_2,...,a_6$, whose lengths satisfy the condition $$a_1-a_4 = a_5-a_2 = a_3-a_6$$
The height of a truncated cone is equal to the radius of its base. The perimeter of a regular hexagon circumscribing its top is equal to the perimeter of an equilateral triangle inscribed in its base. Find the angle $\phi$ between the cone’s generating line and its base.
$AC' BA'C B'$ is a convex hexagon such that $AB' = AC'$, $BC' = BA'$, $CA' = CB'$ and $\angle A +\angle B + \angle C = \angle A' + \angle B' + \angle C'$. Prove that the area of the triangle $ABC$ is half the area of the hexagon.
(V Proizvolov)
In a convex hexagon $ABCDEF$ equalities $\angle ABC= \angle CDE= \angle EFA$ hold, and the angle bisectors of angles $ABC$, $CDE$ and $EFA$ intersect in one point. Rays $AB$ and $DC$ intersect at $P$, rays $BC$ and $ED$ - at $Q$, rays $CD$ and $FE$ - at $R$, rays $DE$ and $AF$ - at $S$.
Prove that $PR=QS$
[i]M. Zorka[/i]
Let us call a convex hexagon $ABCDEF$ [i]boring [/i] if $\angle A+ \angle C + \angle E = \angle B + \angle D + \angle F$.
a) Is every cyclic hexagon boring?
b) Is every boring hexagon cyclic?
You have a large number of congruent equilateral triangular tiles on a table and you want to fit $n$ of them together to make a convex equiangular hexagon (i.e. one whose interior angles are $120^o$) . Obviously, $n$ cannot be any positive integer. The first three feasible $n$ are $6, 10$ and $13$. Determine if $19$ and $20$ are feasible .
Let ${ABCDEF}$ be a convex hexagon, in which each of the diagonals ${AD, BE}$ , and ${CF}$ divides the hexagon into two quadrilaterals of equal area. Show that ${AD, BE}$ , and ${CF}$ are concurrent.
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]