This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 7

Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$? $\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$
Let be a natural number $ n\ge 2 $ and a real number $ r>1. $ Determine the natural numbers $ k $ having the property that the affixes of $ r^ke^{\pi ki/n} ,r^{k+1}e^{\pi (k+1)i/n} ,r^{k+n}e^{\pi (k+n)i/n} ,r^{k+n+1}e^{\pi (k+n+1) i/n} $ in the complex plane represent the vertices of a trapezoid.
Let $ABC$ be an acute, scalene triangle with circumcenter $O$, circumcircle $(O)$, orthocenter $H$. Line $AH$ meets $(O)$ again at $D \neq A$. Let $E, F$ be the midpoint of segments $AB, AC$ respectively. The line through $H$ and perpendicular to $HF$ meets line $BC$ at $K$. a) Line $DK$ meets $(O)$ again at $Y \neq D$. Prove that the intersection of line $BY$ and the perpendicular bisector of $BK$ lies on the circumcircle of triangle $OFY$. b) The line through $H$ and perpendicular to $HE$ meets line $BC$ at $L$. Line $DL$ meets $(O)$ again at $Z \neq D$. Let $M$ be the intersection of lines $BZ, OE$; $N$ be the intersection of lines $CY, OF$; $P$ be the intersection of lines $BY, CZ$. Let $T$ be the intersection of lines $YZ, MN$ and $d$ be the line through $T$ and perpendicular to $OA$. Prove that $d$ bisects $AP$.
In the complex plane, let $A$ be the set of solutions to $z^3 - 8 = 0$ and let $B$ be the set of solutions to $z^3 - 8z^2 - 8z + 64 = 0$. What is the greatest distance between a point of $A$ and a point of $B?$ $\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9 + \sqrt{3}$
Let $\Omega=\{z=x+iy~\in\mathbb{C}~:~|y|\leqslant 1\}$. If $f(z)=z^2+2$, then draw a sketch of $$f\Big(\Omega\Big)=\{f(z):z\in\Omega\}$$ Justify your answer.
$ABCD$ is a quadrilateral on complex plane whose four vertices satisfy $z^4+z^3+z^2+z+1=0$. Then $ABCD$ is a [list=1] [*] Rectangle [*] Rhombus [*] Isosceles Trapezium [*] Square [/list]
Consider a complex number whose affix in the complex plane is situated on the first quadrant of the unit circle centered at origin. Then, the following inequality holds. $$ \sqrt{2} +\sqrt{2+\sqrt{2}} \le |1+z|+|1+z^2|+|1+z^4|\le 6 $$ [i]Costică Ambrinoc[/i]