Found problems: 64
2020 Iranian Geometry Olympiad, 1
A trapezoid $ABCD$ is given where $AB$ and $CD$ are parallel. Let $M$ be the midpoint of the segment $AB$. Point $N$ is located on the segment $CD$ such that $\angle ADN = \frac{1}{2} \angle MNC$ and $\angle BCN = \frac{1}{2} \angle MND$. Prove that $N$ is the midpoint of the segment $CD$.
[i]Proposed by Alireza Dadgarnia[/i]
2019 Iranian Geometry Olympiad, 1
Circles $\omega_1$ and $\omega_2$ intersect each other at points $A$ and $B$. Point $C$ lies on the tangent line from $A$ to $\omega_1$ such that
$\angle ABC = 90^\circ$. Arbitrary line $\ell$ passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut $\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot of altitude from $A$ to $\ell$. Prove that points $X, Y$ and $Z$ are collinear.
[i]Proposed by Iman Maghsoudi[/i]
2015 Iran Geometry Olympiad, 2
let $ ABC $ an equilateral triangle with circum circle $ w $
let $ P $ a point on arc $ BC $ ( point $ A $ is on the other side )
pass a tangent line $ d $ through point $ P $ such that $ P \cap AB = F $ and $ AC \cap d = L $
let $ O $ the center of the circle $ w $
prove that $ \angle LOF > 90^{0} $
2017 Iranian Geometry Olympiad, 5
Sphere $S$ touches a plane. Let $A,B,C,D$ be four points on the plane such that no three of them are collinear. Consider the point $A'$ such that $S$ in tangent to the faces of tetrahedron $A'BCD$. Points $B',C',D'$ are defined similarly. Prove that $A',B',C',D'$ are coplanar and the plane $A'B'C'D'$ touches $S$.
[i]Proposed by Alexey Zaslavsky (Russia)[/i]
2021 Indonesia TST, G
let $ w_1 $ and $ w_2 $ two circles such that $ w_1 \cap w_2 = \{ A , B \} $
let $ X $ a point on $ w_2 $ and $ Y $ on $ w_1 $ such that $ BY \bot BX $
suppose that $ O $ is the center of $ w_1 $ and $ X' = w_2 \cap OX $
now if $ K = w_2 \cap X'Y $ prove $ X $ is the midpoint of arc $ AK $
2019 Iranian Geometry Olympiad, 2
Find all quadrilaterals $ABCD$ such that all four triangles $DAB$, $CDA$, $BCD$ and $ABC$ are similar to one-another.
[i]Proposed by Morteza Saghafian[/i]
2019 Iranian Geometry Olympiad, 4
Let $ABCD$ be a parallelogram and let $K$ be a point on line $AD$ such that $BK=AB$. Suppose that $P$ is an arbitrary point on $AB$, and the perpendicular bisector of $PC$ intersects the circumcircle of triangle $APD$ at points $X$, $Y$. Prove that the circumcircle of triangle $ABK$ passes through the orthocenter of triangle $AXY$.
[i]Proposed by Iman Maghsoudi[/i]
2017 Iranian Geometry Olympiad, 4
In the isosceles triangle $ABC$ ($AB=AC$), let $l$ be a line parallel to $BC$ through $A$. Let $D$ be an arbitrary point on $l$. Let $E,F$ be the feet of perpendiculars through $A$ to $BD,CD$ respectively. Suppose that $P,Q$ are the images of $E,F$ on $l$. Prove that $AP+AQ\le AB$
[i]Proposed by Morteza Saghafian[/i]
2022 Iranian Geometry Olympiad, 2
We are given an acute triangle $ABC$ with $AB\neq AC$. Let $D$ be a point of $BC$ such that $DA$ is tangent to the circumcircle of $ABC$. Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$, respectively, and let $M$ be the midpoints $EF$. Prove that the line tangent to the circumcircle of $AMD$ through $D$ is also tangent to the circumcircle of $ABC$.
[i]Proposed by Patrik Bak, Slovakia[/i]
2017 Iranian Geometry Olympiad, 5
Let $X,Y$ be two points on the side $BC$ of triangle $ABC$ such that $2XY=BC$ ($X$ is between $B,Y$). Let $AA'$ be the diameter of the circumcirle of triangle $AXY$. Let $P$ be the point where $AX$ meets the perpendicular from $B$ to $BC$, and $Q$ be the point where $AY$ meets the perpendicular from $C$ to $BC$. Prove that the tangent line from $A'$ to the circumcircle of $AXY$ passes through the circumcenter of triangle $APQ$.
[i]Proposed by Iman Maghsoudi[/i]
2018 Iranian Geometry Olympiad, 5
There are some segments on the plane such that no two of them intersect each other (even at the ending points). We say segment $AB$ [b]breaks[/b] segment $CD$ if the extension of $AB$ cuts $CD$ at some point between $C$ and $D$.
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(4cm);
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 = -5.267474904743955, xmax = 11.572179069738377, ymin = -10.642621257034536, ymax = 4.543526642434019; /* image dimensions */
/* draw figures */
draw((-4,-2)--(1.08,-2.03), linewidth(2));
draw(shift((-2.1866176795507295,-2.0107089507113147))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
draw((-0.16981767035094117,3.225314210196242)--(-2.1866176795507295,-2.0107089507113147), linewidth(2) + linetype("4 4"));
draw((-0.16981767035094117,3.225314210196242)--(-0.8194002739586808,1.538865607509914), linewidth(2));
label("$A$",(-1.2684397405642523,3.860690076971137),SE*labelscalefactor,fontsize(16));
label("$B$",(-1.9211395070170559,2.002590777612728),SE*labelscalefactor,fontsize(16));
label("$C$",(-4.971261820527631,-1.6571211388676117),SE*labelscalefactor,fontsize(16));
label("$D$",(1.08925640451367566,-1.6571211388676117),SE*labelscalefactor,fontsize(16));
/* dots and labels */
dot((-4,-2),dotstyle);
dot((1.08,-2.03),dotstyle);
dot((-0.16981767035094117,3.225314210196242),dotstyle);
dot((-0.8194002739586808,1.538865607509914),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
$a)$ Is it possible that each segment when extended from both ends, breaks exactly one other segment from each way?
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(4cm);
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 = -6.8, xmax = 8.68, ymin = -10.32, ymax = 3.64; /* image dimensions */
/* draw figures */
draw((-2.56,1.24)--(-0.36,1.4), linewidth(2));
draw((-3.32,-2.68)--(-1.24,-3.08), linewidth(2));
draw(shift((-2.551651190956802,-2.8277593863544612))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
draw(shift((-0.8889576602618603,1.3615303519809556))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
draw((-2.551651190956802,-2.8277593863544612)--(-0.8889576602618603,1.3615303519809556), linewidth(2) + linetype("4 4"));
draw((-1.4097008194020806,0.049476186483185636)--(-1.8514772275312024,-1.0636149148218605), linewidth(2));
/* dots and labels */
dot((-2.56,1.24),dotstyle);
dot((-0.36,1.4),dotstyle);
dot((-3.32,-2.68),dotstyle);
dot((-1.24,-3.08),dotstyle);
dot((-1.4097008194020806,0.049476186483185636),dotstyle);
dot((-1.8514772275312024,-1.0636149148218605),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
$b)$ A segment is called [b]surrounded[/b] if from both sides of it, there is exactly one segment that breaks it.\\
([i]e.g.[/i] segment $AB$ in the figure.) Is it possible to have all segments to be surrounded?
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(7cm);
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 = -10.70976151557872, xmax = 18.64292748469251, ymin = -16.354300717041443, ymax = 9.136192362141452; /* image dimensions */
/* draw figures */
draw((1.0313140845297686,0.748205038977829)--(-1.3,-4), linewidth(2.8));
draw((-5.780195085389632,-2.13088646583346)--(-2.549994860479401,-2.13088646583346), linewidth(2.8));
draw((4.121070821400425,-3.816208322308361)--(1.78,-1.88), linewidth(2.8));
draw(shift((-0.38228674372374466,-2.13088646583346))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
draw((-2.549994860479401,-2.13088646583346)--(-0.38228674372374466,-2.13088646583346), linewidth(2.8) + linetype("4 4"));
draw(shift((0.32979226045261084,-0.6805897691262632))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */
draw((4.121070821400425,-3.816208322308361)--(0.32979226045261084,-0.6805897691262632), linewidth(2.8) + linetype("4 4"));
draw((-3.6313140845297687,-8.74820503897783)--(3.600422205681574,5.980726991931396), linewidth(2.8) + linetype("2 2"));
label("$A$",(-0.397698406272906,1.754593418658662),SE*labelscalefactor,fontsize(16));
label("$B$",(-2.6377720405041316,-3.266261278756151),SE*labelscalefactor,fontsize(16));
/* dots and labels */
dot((1.0313140845297686,0.748205038977829),linewidth(6pt) + dotstyle);
dot((-1.3,-4),linewidth(6pt) + dotstyle);
dot((-5.780195085389632,-2.13088646583346),linewidth(6pt) + dotstyle);
dot((-2.549994860479401,-2.13088646583346),linewidth(6pt) + dotstyle);
dot((4.121070821400425,-3.816208322308361),linewidth(6pt) + dotstyle);
dot((1.78,-1.88),linewidth(6pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
[i]Proposed by Morteza Saghafian[/i]
2021 Iranian Geometry Olympiad, 2
Two circles $\Gamma_1$ and $\Gamma_2$ meet at two distinct points $A$ and $B$. A line passing through $A$ meets $\Gamma_1$ and $\Gamma_2$ again at $C$ and $D$ respectively, such that $A$ lies between $C$ and $D$. The tangent at $A$ to $\Gamma_2$ meets $\Gamma_1$ again at $E$. Let $F$ be a point on $\Gamma_2$ such that $F$ and $A$ lie on different sides of $BD$, and $2\angle AFC=\angle ABC$. Prove that the tangent at $F$ to $\Gamma_2$, and lines $BD$ and $CE$ are concurrent.
2014 Iran Geometry Olympiad (senior), 1:
ABC is a triangle with A=90 and C=30.Let M be the midpoint of BC. Let W be a circle passing through A tangent in M to BC. Let P be the circumcircle of ABC. W is intersecting AC in N and P in M. prove that MN is perpendicular to BC.
2020 Iranian Geometry Olympiad, 3
Assume three circles mutually outside each other with the property that every line separating two of them have intersection with the interior of the third one. Prove that the sum of pairwise distances between their centers is at most $2\sqrt{2}$ times the sum of their radii.
(A line separates two circles, whenever the circles do not have intersection with the line and are on different sides of it.)
[color=#45818E]Note.[/color] Weaker results with $2\sqrt{2}$ replaced by some other $c$ may be awarded points depending on the value of $c>2\sqrt{2}$
[i]Proposed by Morteza Saghafian[/i]