How would I go about finding the formula for the complex number representation (affix) of the circumcenter of $\triangle abc$ given the affixes $a,b,c$?
120926
Find the center and radius of the circle which circumscribes the triangle with (complex) vertices $a_1,a_2,a_3$. Express the result in symmetric form.   I'm not sure where to start in this question. The circumcenter is the intersection of the perpendicular bisectors, but I don't know how to calculate the line equation of the perpendicular bisectors yet. As for the radius, after we find the circumcenter $c$ we can calculate it using $|c-a_1|$ (or $|c-a_2|$ or $|c-a_3|$; all three should be equal.) So how can I calculate the circumcenter?
105383
$a,b,c,d$ are complex numbers satisfying \begin{cases}   a+b+c+d=3 \\   a^2+ b^2+ c^2+ d^2=5 \\   a^3+ b^3+ c^3+ d^3=3 \\   a^4+ b^4+ c^4+ d^4=9 \end{cases}  Find the value of the following: $$a^{2015} + b^{2015} + c^{2015} + d^{2015}$$.
87954
Let  $$A_2 = \left[ \begin{array}{cc} 0 &amp; 1\\ 1 &amp; 0 \end{array}\right]$$  $$A_3 = \left[ \begin{array}{ccc} 0 &amp; 1 &amp; 1\\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0 \end{array}\right]$$  $$A_4 = \left[ \begin{array}{cccc} 0 &amp; 1 &amp; 1 &amp; 1\\ 1 &amp; 0 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 1 &amp; 0\end{array}\right]$$  and so on for $A_n$.  I was asked to calculate the determinant for $A_1, A_2, A_3, A_4$ and then guess about the determinant for $A_n$ in general. Of course the pattern is clear that  $$ \det A_n = (n-1)(-1)^{n-1} $$  but I was wondering as to what the proof of this is. I tried to be clever with cofactor expansions but I couldn't get anywhere.  Could someone explain it to me please?
113143
Suppose $a,b,c \in\mathbb N$, and the value of $c$ is known and fixed, while $a$ and $b$ are unknown and are both smaller than $c$. What is the total number of unique triangles possible with $a, b$ and $c$ as its sides?
106398
I am not sure I understand this I just to an explanation to this question:  If $a, b, c \in \Bbb{Z}^+$ and $a\mid bc$, does it follow that $a\mid b$ or $a\mid c$?
59304
I am finding a 4-digit number $abcd$ (in base-10 representation) satisfying the following property.  $$abcd=a^b c^d$$  I have been running my mind around this problem since a long time, but got no success.  I wrote a python program for a number with this property, and got the answer to be $2592$ since   $$2592=2^5\times 9^2$$  But I am looking for a purely mathematical way to solve this problem.   Thanks!
34259
I think I have found the final answer $\left(4 \left[\frac{2t \cosh(4t) - \sinh(4t)}{t^2}\right]\right)$ but need to verify it.
50174
Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$.  How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ?  I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$, but i can't go any further ...  If i expand it :   $2-ab-\frac{b}a-c-ac-\frac{a}c-b+2a+\frac{2}a-bc-\frac{c}b-a+2b+\frac{2}b+2c+\frac{2}c-4$  Which gives using $c=\frac{1}{ab}$ :   $5ab-\frac{b}a-\frac{1}{ab}-a^2b+\frac{1}a+b+\frac{1}b-2$
15100
Consider a plane $\mathbb{R}^2$ and an affine transformation $T: \mathbb{R}^2\to \mathbb{R}^2$ that maps points $A,B,C$ to points $A',B',C'$ correspondingly. We know that $A,B,C$ are not on the same straight line and so $T$ is determined uniquely. Given an arbitrary point $M$ how to find $T(M)$?  Basically, I'm trying to write a program that takes as input 14 numbers—the coordinates of the points $A,B,C, A',B',C',M$ and outputs two numbers—the coordinates of $M'=T(M)$, provided $A,B,C$ are not on the same straight line. How should I proceed?
70102
Is there a generalized way( a particular conic section of any shape,for instance an ellipse without determining its major/minor axis) to find the centre,vertices,foci,focal radii,letus rectum,eccentricity,etc when exists of any parabola,ellipse,hyperbola?(dealing separately)  $ax^2+by^2+cx+dy+exy+c=0$ ($a,b,c$ not all zero)
104046
I am very close to the solution I think. Since the circles cross the real axis, I want to find mappings from $z \to w$ such that $1/2 \to m$ and $0 \to -m$ and $-1 \to n$ and $1 \to -n$. Using $$ w = \frac{az+b}{cz+d}$$ we get $$ \frac{a+2b}{c+2d} = -\frac{b}{d}$$ and $$ \frac{b-a}{d-c} = - \frac{a+b}{c+d}$$ but where do we go from here? I am a little lost in the simplifying algebra here to solve for the $a,b,c,d$ in the transformation.
61620
Why does the complex equation $z=Ae^{it}+Be^{-it}$ represent an ellipse?, with $A,B \in \mathbb C$  How can it be described?
28642
In the triangle $ABC\,$ we denote by $O$ and $I$ the circumcenter and incenter respectively. The perpendicular bisectors of the line segments $IA$, $IB$ and $IC$ pairwise intersect thus defining the triangle $A_1B_1C_1$ Prove that $\vec {OI}=\vec {OA_1}+\vec {OB_1}+\vec {OC_1}$      Hi there. Im a high schooler new to olympiad geometry. Ive been unable to solve this problem despite repeated angle chasing could someone give a solution?Note I dont know stuff like projective geometry and hard lemmas so a simple proof would be most appreciated.
2724
The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.  So here is the puzzle:    The triangle $ABC$ is known to be isosceles, that is, $AC=BC$. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled "?".  The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution.
48006
Is it possible to get some help on this. I'm trying to help my friend solve this but it had been over 14 years since I did this type. All I have to work with is this.   Sorry if it is vague. Any help would be greatly appreciated!  Cosecant = cscx=1/sinx.
61841
I need a synthetic proof on this problem without the use of trigonometry.     Question: Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AF\perp BE$.    The proof is very easy using coordinate geometry.    I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
76111
Suppose $a,b,c$ lie in the unit circle of the complex plane and satisfy $a+b+c = 0$. Then, $a,b,c$ form the vertices of an equilateral triangle. Try So, I want to show that $|b-a|=|c-a|=|b-c|$. We are given $|a| = |b| = |c| = 1$ and $a+b+c=0$. Since $|z| = z \overline{z} $, we have $$ (b-a)( \overline{b}-\overline{a} ) = (c-a)( \overline{c} - \overline{a} )$$ which simplifies to $$ |b| - b \overline{a} - a \overline{b} - |a| = |c| - c \overline{a} - a \overline{c} - |a| $$ Hence, $$ c \overline{a} + a \overline{c} - b \overline{a} - a \overline{b} = 0$$ $$ a( \overline{c} - \overline{b} ) + \overline{a} ( c-b) = 0 $$ Here I am stuck. Perhaps this is the wrong approach?