Chapter 5
Complex numbers from A to...Z · 166 exercises
Problem 30
On the sides of convex quadrilateral \(A B C D\) equilateral triangles \(A B M\), \(B C N, C D P\) and \(D A Q\) are drawn external to the figure. Prove that quadrilaterals \(A B C D\) and \(M N P Q\) have the same centroid.
4 step solution
Problem 30
Let a and \(z\) be complex numbers such that \(|z+a|=1 .\) Prove that $$ \left|z^{2}+a^{2}\right| \geq \frac{|1-2| a||}{\sqrt{2}} $$
6 step solution
Problem 31
Let \(A B C D\) be a quadrilateral and consider the rotations \(\mathcal{R}_{1}, \mathcal{R}_{2}, \mathcal{R}_{3}, \mathcal{R}_{4}\) with centers \(A, B, C, D\) through angle \(\alpha\) and of the same orientation. Points \(M, N, P, Q\) are the images of points \(A, B, C, D\) under the rotations \(\mathcal{R}_{2}, \mathcal{R}_{3}\), \(\mathcal{R}_{4}, \mathcal{R}_{1}\), respectively. Prove that the midpoints of the diagonals of the quadrilaterals \(A B C D\) and \(M N P Q\) are the vertices of a parallelogram.
3 step solution
Problem 31
Find the geometric images of the complex numbers \(z\) for which $$ z^{n} \cdot \operatorname{Re}(z)=\bar{z}^{n} \cdot \operatorname{Im}(z) $$ where \(n\) is an integer.
4 step solution
Problem 32
Let \(O_{9}, I, G\) be the 9 -point center, the incenter and the centroid, respectively, of a triangle \(A B C\). Prove that lines \(O_{9} G\) and \(A I\) are perpendicular if and only if \(\widehat{A}=\frac{\pi}{3}\)
4 step solution
Problem 32
$$ \text { Let } a, b \text { be real numbers with } a+b=1 \text { and let } z_{1}, z_{2} \text { be complex } $$ numbers with \(\left|z_{1}\right|=\left|z_{2}\right|=1\). Prove that $$ \left|a z_{1}+b z_{2}\right| \geq \frac{\left|z_{1}+z_{2}\right|}{2} $$
5 step solution
Problem 33
Let \(O_{9}, I, G\) be the 9 -point center, the incenter and the centroid, respectively, of a triangle \(A B C\). Prove that lines \(O_{9} G\) and \(A I\) are perpendicular if and only if \(\widehat{A}=\frac{\pi}{3}\)
5 step solution
Problem 33
$$ \text { Let } k, n \text { be positive integers and let } z_{1}, z_{2}, \ldots, z_{n} \text { be nonzero complex } $$ numbers with the same modulus such that $$ z_{1}^{k}+z_{2}^{k}+\cdots+z_{n}^{k}=0 . $$ Prove that $$ \frac{1}{z_{1}^{k}}+\frac{1}{z_{2}^{k}}+\cdots+\frac{1}{z_{n}^{k}}=0 $$
4 step solution
Problem 34
Two circles \(\omega_{1}\) and \(\omega_{2}\) are given in the plane, with centers \(O_{1}\) and \(O_{2}\), respectively. Let \(M_{1}^{\prime}\) and \(M_{2}^{\prime}\) be two points on \(\omega_{1}\) and \(\omega_{2}\), respectively, such that the lines \(O_{1} M_{1}^{\prime}\) and \(O_{2} M_{2}^{\prime}\) intersect. Let \(M_{1}\) and \(M_{2}\) be points on \(\omega_{1}\) and \(\omega_{2}\), respectively, such that when measured clockwise the angles \(\widehat{M_{1}^{\prime} O_{1} M_{1}}\) and \(M_{2}^{\prime} \widehat{O_{2} M}_{2}\) are equal. (a) Determine the locus of the midpoint of \(\left[M_{1} M_{2}\right]\). (b) Let \(P\) be the point of intersection of lines \(O_{1} M_{1}\) and \(O_{2} M_{2}\). The circumcircle of triangle \(M_{1} P M_{2}\) intersects the circumcircle of triangle \(O_{1} P O_{2}\) at \(P\) and another point \(Q\). Prove that \(Q\) is fixed, independent of the locations of \(M_{1}\) and \(M_{2}\).
8 step solution
Problem 35
Isosceles triangles \(A_{3} A_{1} O_{2}\) and \(A_{1} A_{2} O_{3}\) are constructed externally along the sides of a triangle \(A_{1} A_{2} A_{3}\) with \(O_{2} A_{3}=O_{2} A_{1}\) and \(O_{3} A_{1}=O_{3} A_{2} .\) Let \(O_{1}\) be a point on the opposite side of line \(A_{2} A_{3}\) from \(A_{1}\), with \(\widehat{O_{1} A_{3} A}_{2}=\frac{1}{2} \widehat{A_{1} O_{3} A}_{2}\) and \(\widehat{O_{1} A_{2} A_{3}}=\frac{1}{2} \widehat{A_{1} O_{2} A}_{3}\), and let \(T\) be the foot of the perpendicular from \(O_{1}\) to \(A_{2} A_{3} .\) Prove that \(A_{1} O_{1} \perp O_{2} O_{3}\) and that
4 step solution
Problem 36
A triangle \(A_{1} A_{2} A_{3}\) and a point \(P_{0}\) are given in the plane. We define \(A_{s}=A_{s-3}\) for all \(s \geq 4\). We construct a sequence of points \(P_{0}, P_{1}, P_{2}, \ldots\) such that \(P_{k+1}\) is the image of \(P_{k}\) under rotation with center \(A_{k+1}\) through angle \(120^{\circ}\) clockwise \((k=0,1,2, \ldots)\). Prove that if \(P_{1986}=P_{0}\) then the triangle \(A_{1} A_{2} A_{3}\) is equilateral.
4 step solution
Problem 37
Two circles in a plane intersect. Let \(A\) be one of the points of intersection. Starting simultaneously from \(A\) two points move with constant speeds, each point travelling along its own circle in the same direction. After one revolution the two points return simultaneously to \(A\). Prove that there exists a fixed point \(P\) in the plane such that, at any time, the distances from \(P\) to the moving points are equal.
6 step solution
Problem 38
Inside the square \(A B C D\), the equilateral triangles \(A B K, B C L, C D M\) \(D A N\) are inscribed. Prove that the midpoints of the segments \(K L, L M, M N, N K\) and the midpoints of the segments \(A K, B K, B L, C L, C M, D M, D N, A N\) are the vertices of a regular dodecagon.
5 step solution
Problem 39
Let \(A B C\) be an equilateral triangle and let \(M\) be a point in the interior of angle \(\widehat{B A C}\). Points \(D\) and \(E\) are the images of points \(B\) and \(C\) under the rotations with center \(M\) and angle \(120^{\circ}\), counterclockwise and clockwise, respectively. Prove that the fourth vertex of the parallelogram with sides \(M D\) and \(M E\) is the reflection of point \(A\) across point \(M\).
3 step solution
Problem 41
Let \(A B C\) be a triangle, \(H\) its orthocenter, \(O\) its circumcenter, and \(R\) its circumradius. Let \(D\) be the reflection of \(A\) across \(B C\), let \(E\) be that of \(B\) across \(C A\), and \(F\) that of \(C\) across \(A B\). Prove that \(D, E\) and \(F\) are collinear if and only if \(O H=2 R\)
2 step solution
Problem 42
Let \(A B C\) be a triangle such that \(\widehat{A C B}=2 \widehat{A B C}\). Let \(D\) be the point on the side \(B C\) such that \(C D=2 B D .\) The segment \(A D\) is extended to \(E\) so that \(A D=D E .\) Prove that $$ \widehat{E C B}+180^{\circ}=2 \widehat{E B C} $$
5 step solution