Given a convex quadrilateral $ABCD$. The medians of the triangle $ABC$ intersect at point $M$, and the medians of the triangle $ACD$ at point$ N$. The circle, circumscibed around the triangle $ACM$, intersects the segment $BD$ at the point $K$ lying inside the triangle $AMB$ . It is known that $\angle MAN = \angle ANC = 90^o$. Prove that $\angle AKD = \angle MKC$.
Problem
Source: St. Petersburg 2019 10.4
Tags: geometry, circumcircle, Centroid, equal angles
Buffaloesarecute
30.10.2019 12:33
\bump.....
math_pi_rate
30.10.2019 13:23
Hi all! Kindly, if possible, do not give a solution to this problem till the 15th of November, as this is also being used in the Indian Postal Set - 2019. Thanks.
math_pi_rate
18.11.2019 08:04
Hey guys! Thanks for agreeing to the request above. Anyway, here's a solution to this problem, as the contest time has ended now. You all can chime in too . SOLUTION: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.23, xmax = 10.49, ymin = -6.33, ymax = 4.89; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-2.68,0.47)--(-0.26,4.45), linewidth(2) + wrwrwr); draw((-0.26,4.45)--(2.26,-0.24), linewidth(2) + wrwrwr); draw((2.26,-0.24)--(-2.68,0.47), linewidth(2) + wrwrwr); draw(circle((-0.4161625988880266,-1.319427096488521), 2.885655750833899), linewidth(2) + wrwrwr); draw((-3.0110100700199145,-2.5818753979572193)--(2.26,-0.24), linewidth(2) + wrwrwr); draw((-2.68,0.47)--(1,2.105), linewidth(2) + wrwrwr); draw((-4.3865151050298685,-6.0978130969358295)--(-0.26,4.45), linewidth(2) + wrwrwr); draw((-0.21,0.115)--(-4.3865151050298685,-6.0978130969358295), linewidth(2) + wrwrwr); draw((-0.19333333333333363,-1.33)--(-0.26,4.45), linewidth(2) + wrwrwr); draw((-2.68,0.47)--(-1.6021717016766228,-1.9559376989786097), linewidth(2) + wrwrwr); draw((-0.22666666666666682,1.56)--(-1.6021717016766228,-1.9559376989786097), linewidth(2) + wrwrwr); /* dots and labels */ dot((-2.68,0.47),dotstyle); label("$A$", (-3.17,0.53), NE * labelscalefactor); dot((-0.26,4.45),dotstyle); label("$B$", (-0.71,4.43), NE * labelscalefactor); dot((2.26,-0.24),dotstyle); label("$C$", (2.35,-0.03), NE * labelscalefactor); dot((1,2.105),linewidth(4pt) + dotstyle); label("$P$", (1.09,2.27), NE * labelscalefactor); dot((-0.21,0.115),linewidth(4pt) + dotstyle); label("$Q$", (0.03,0.27), NE * labelscalefactor); dot((-0.22666666666666682,1.56),linewidth(4pt) + dotstyle); label("$M$", (-0.71,1.83), NE * labelscalefactor); dot((-0.19333333333333363,-1.33),dotstyle); label("$Z$", (-0.07,-1.91), NE * labelscalefactor); dot((-3.0110100700199145,-2.5818753979572193),linewidth(4pt) + dotstyle); label("$T$", (-3.57,-2.45), NE * labelscalefactor); dot((-1.6021717016766228,-1.9559376989786097),linewidth(4pt) + dotstyle); label("$N$", (-1.41,-2.43), NE * labelscalefactor); dot((-4.3865151050298685,-6.0978130969358295),linewidth(4pt) + dotstyle); label("$D$", (-4.85,-6.07), NE * labelscalefactor); dot((-1.4654708966091958,1.36868770921719),linewidth(4pt) + dotstyle); label("$K$", (-1.79,1.57), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $T$ be the point such that $AMCT$ is an isosceles trapezoid (with $AM \parallel CT$). Also let $Z$ be the reflection of $B$ in $M$, and $P,Q$ be the midpoints of $BC,AC$. Then by midpoint theorem, $MP \parallel CZ$. But, by the given hypothesis in problem, $AM \parallel CN$. Since $A,M,P$ are collinear, this gives that $Z \in CN$. As $MZ$ passes through midpoint of $AC$, and $CZ \parallel AM$, so we get that $AMCZ$ is a parallelogram. Thus, $AZ=CM=AT$, i.e. $A$ lies on the perpendicular bisector of $TZ$. Also, $CT \parallel AM$ means that $C,Z,N,T$ are all collinear. As $N$ is the foot of perpendicular from $A$ to $TZ$, we get that in fact $N$ is the midpoint of $TZ$. Consider $\triangle BTZ$. Then $M,N$ are midpoints of $BZ$ and $TZ$, which gives $BT \parallel MN$. Also, in $\triangle BQD$, we have $BM=2MQ$ and $DN=2NQ$, which means that $MN \parallel BD$. Combining the above two informations, we have $T \in BD$. As $T \in \odot (AKMC)$ (By its definition), so we get $$\measuredangle AKD=\measuredangle AKT=\measuredangle ACT=\measuredangle MAC=\measuredangle MKC$$where $\measuredangle$ represents directed angle modulo $180^{\circ}$. Hence, done.