Let A, B, C, D and E be five points marked, in clockwise order, on the unit circle in the plane (with centre at origin). Let α and β be real numbers and set f(P) = αx + βy where P is a point whose coordinates are (x, y). Assume that f(A) = 10, f(B) = 5, f(C) = 4 and f(D) = 10. Which of the following are impossible?  (a) f(E) = 2  (b) f(E) = 4  (c) f(E) = 5  I have try this question many times , but from my point of view   the answer  is f(E) = 5 becoz it is moving in clockwise direction and after f(E) =5 then f(E)=10, so my answer is f(E)=5. I don't know   whether answer is correct or not  if anbody help  me , i would be very thankful to him...
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Let $A,B,C,D,E$ be five points marked in  clockwise direction on the unit circle in the plane with centre at origin. If $f(P)=\alpha x+\beta y$ for $P=(x,y)$   If $f(A)=10,f(B)=5,f(C)=4,f(D)=10$. What is $f(E)$?  Please help on how to start
117852
Find the number of integer points lying in or inside a circle of radius $n\in \mathbb N$ centered at the origin.   The problem asks for all $(a,b)\in \mathbb Z^2$ such that $a^2+b^2\leq n^2$. Looking at such points lying strictly in the North-East quadrant, pick an $a\in \{1,\ldots,n\}$ and let $b$ run through $\{1,\ldots,\lfloor \sqrt{n^2-a^2}\rfloor\}$ : there are exactly $\displaystyle \sum_{k=1}^n \lfloor \sqrt{n^2-k^2} \rfloor$ integer points strictly in the quadrant.  Multiply that by $4$ to account for all $4$ quadrants, add the origin and points on the $x$ and $y$ axis and the total number is $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor$$  I'm interested in asymptotics of this sum. A trivial estimate is the following:$$\begin{align}\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor &amp;= \sum_{k=0}^n \sqrt{n^2-k^2} + O(n)\\ &amp;= n^2\cdot \left(\frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}}\right) + O(n) \\ &amp;=n^2 \left( \int_0^1 \sqrt{1-t^2}dt + o(1)\right) + O(n) \\ &amp;= \frac \pi 4n^2 + o(n^2)\end{align} $$  Hence $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + o(n^2)$$     Is there a way to refine the estimate above and get $\pi n^2 + O(n)$ instead (or something better) ?
120028
BdMO 2014     There are $7$ points on a circle.Any 2 consecutive points are at equal distance from one another.How many acute angled triangles can you form taking any 3 of these points?   I believe there are 5 possible triangles that we can form using any three points.Then we can find out which of these are acute-angled.Then we can compute the number one by one and sum them up.Am I right?
117503
Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is  a. a circle if $ T $ is a circle. b. a circle minus (0, 0, 1) if $ T $ is a line.  Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$  To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:  $$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$  We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:  $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.   I really am at a  loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?
39523
If $A=\left\{1,2,3,4,5\right\}$ Then     Find Number of functions $f : A \to A$ such that $f(f(x))=f(x)$   Case $1.$ if $f$ is injective then $f(f(x))=f(x)$ $\implies$ $f(x)=x$, hence there is only one injective function which is an identity function.  Case $2.$ When $f$ is many to one function  Let us assume that $f(1)=f(2)=f(3)=f(4)=f(5)=k$ Then clearly for any $k$ from the set $A$, $f(f(x))=f(x)$  hence there are five such functions.  Are there any other possibilities?
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This is from Putnam:     If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.   Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.  Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.  Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.  Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.
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Please don't sign this as a duplicate of , they aren't. I am interested about the actual algorithm, not a proof of the existence.  Does a such mapping exists? I think, it must, because both of them have the same cardinality ($\aleph_1$).  My actual suggestion for a such mapping were simply merge the digits of the real numbers. For example, from $\pi$ and $e$ we could got  $3.1415...$ and $2.7182...$ would lead to $32.17411852...$ .  Although I am not sure it were ok. Maybe a better, clearer solution also exists?
11395
I have two function $f$ and $r$, such that: $y = f(x)$, $r(y) = x$. I read that it means that: $f \circ r = I_{U}$, where $x \in U, y \in U$, where $U$ is a finite set. What does $f \circ r = I_{U}$ mean? What does this circle mean and how could i read all this writing $f \circ r = I_{U}$ ?
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Suppose we are given two circles, one inside the other, that are tangent at a point $z_0$. I'm trying to map the region between these circles to the unit disc, and my thought process is the following:  I feel like we can map $z_0$ to $\infty$, but I'm not really sure about this. If it works, then I get a strip in the complex plane, and I know how to handle strips via rotation, translation, logarithm, power, and then I'm in the upper half-plane (if I've thought this through somewhat properly). My problem really lies in what point $z_0$ can go to, because I thought the point symmetric to $z_0$ (which is $z_0$ in this case) had to be mapped to $0$. Is this the right idea, and if what other details should I make sure I have? Thanks!
100073
$f(z)= a\frac{z-z_0}{\overline{z_0}\cdot z-1}$, where $\left|a\right| = 1$ and  $\left|z_0\right| &lt; 1$ .  If $\left|z\right|= 1$, it is obvious that $\left|f(z)\right| = 1$, thus  $f(z)$ maps unit circle to unit circle.  If $z= z_0$, $f(z)= 0$.   So why does this transformation send  $\left|z\right|&lt;1$ to itself?
113070
When $f$ and $g$ are invertible functions we have  $$(g\circ f)^{-1}=f^{-1}\circ g^{-1}$$  I think we have this in every Group $(A,*,u,')$  $(a*b)=c$   $(a*b)*b'=a=c*b'$  $c'*a=c'*(c*b')=b'$   $(c'*a)*a'=b'*a'$      $$c'=(a*b)'=b'*a'$$   and in general we have   $(a_1*a_2*...*a_n)'=a_n'*...*a_2'*a_1'$  1)What is the specific name of this theorem?   2)There is something wrong with my derivation?  EDIT Can we see this as a property of a function $f:A\rightarrow A$ on a structure $(A,\circ)$:  $f(x\circ y)=f(y)\circ f(x)$  ?
11282
(a) Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.  (b) Let $P_1 P_2 \dotsb P_{18}$ be a regular 18-gon. Show that $P_1 P_{10}$, $P_2 P_{13}$, and $P_3 P_{15}$ are concurrent.  I have gotten nowhere on this problem, but I have a hint:  One of those three segments is more interesting than the other two. Which one, and why? And how can you use that fact to make part (a) relevant?  Any help is appreciated!
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$f: \mathbb{R}^5 \rightarrow \mathcal{M}_{2}(\mathbb{R}) $  $ (x_1,x_2,x_3,x_4,x_5) \rightarrow \begin{pmatrix} x_1 &amp; x_2 \\ x_1 + x_3 &amp; x_5 \end{pmatrix}$   Determine the basis of $Kerf$ Determine a basis of $Imf$ Determine the matrix $M_1 = M(f', \mathcal{B}, \mathcal{B '})$ of $f$      1 . $x = (x_1,x_2,x_3,x_4,x_5) \in Kerf \iff f(x_1,x_2,x_3,x_4,x_5) = \begin{pmatrix} x_1 &amp; x_2 \\ x_1 + x_3 &amp; x_5 \end{pmatrix} =\begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 0\end{pmatrix}$  I got $(x_1,x_2,x_3,x_4,x_5) = (0,0,0,x_4,0)$.   A base of $Kerf$ is $(0,0,0,1,0)$.    $Imf^= Vect &lt;f(e_1), f(e_2), f(e_3), f(e_4), f(e_5) &gt; = &lt;E_{11}, E_{12}, E_{21}, E_{22}&gt;$   So the family $(E_{11}, E_{12}, E_{21}, E_{22})$ is basis.  with $E_{ij} $ is the elementary matrix.   The coordinate $x_4$ is not on the matrix.   Are my answers correct?   I need help with question 3, too. Thank you.
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Theorem 2.1 : Suppose that $f$ is an integrable function on the circle with $\hat f(n)=0$ for all $n \in \Bbb Z$. Then $f(\theta_0)=0$ whenever $f$ is continuous at the point $\theta_0$.      Proof : We suppose first that $f$ is real-valued, and argue by contradiction. Assume, without loss of generality, that $f$ is defined on $[-\pi,\pi]$, that $\theta_0=0$, and $f(0) \gt 0$.      Since $f$ is continuous at $0$, we can choose $ 0\lt \delta \le \frac \pi2$, so that $f(\theta) \gt \frac {f(0)}2$ whenever $|\theta| \lt \delta$. Let $$p(\theta)=\epsilon + \cos\theta,$$   Where $\epsilon \gt 0$ is chosen so small that $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$. Then, choose a positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1 + \frac \epsilon2$, for $|\theta| \lt \eta$. Finally, let $p_k(\theta)=|p(\theta)|^k$, and select $B$ so that $|f(\theta)| \le B$ for all $\theta$. This is possible since $f$ is integrable, hence bounded.      By construction, each $p_k$ is a trigonometric polynomial, and since $\hat f(n)=$ for all $n$, we must have $\int_{-\pi}^{\pi} f(\theta)p_k(\theta)\,d\theta=0$ for all $k$.   I understood the first paragraph clearly. But the rest is not making it's way into my head.   In the beginning of second paragraph, how does the given range works for choosing $\delta$? If the continuity is used to get the range, then how? How can we choose $\epsilon$ so small such that, $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$? How can we choose positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1+ \frac \epsilon2$, for $|\theta| \lt \eta$. Why do we must have $\int_{-\pi}^{\pi}f(\theta)p_k(\theta)\,d\theta=0$ for all $k$?
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Given $P(z)=z^6+6z+10$, find how many roots are in each quadrant    I have already seen that $P(z)$ has six different roots, and that none of them are real or of the form $ki$, $k\in \Bbb R$. Since the coefficients of $P(z)$ are real, then it's roots are 3 conjugate pairs, and therefore the possibilities are limited to two.  Defining $f(z)=z^6$, $g(z)=z^6+6z+10$ and applying Rouche in $|z|&lt;2$, $|f(z)-g(z)|&lt;|f(z)|$ I get that all roots are contained in the disc, but how do I go about splitting them into the four quadrants? From what I know about this polynomial, finding zeros in one quadrant is enough to determine the rest, but I'm stuck trying to apply rouche to a quarter circle (basically I'm having trouble finding a suitable f and g)
90179
I've been reading some proofs regarding finite projective planes of order n, and often they start out by assuming that each line contains n+1 points. Is this a fact that follows from the axioms for finite projective planes? Or does this relate to the order of the finite field being projected? (for example in the Fano plane, does the fact that each line contains 3 points follow from the fact that there are 3 elements {0,1,2} in Z_2?) I haven't been able to find a clear explanation of this (clearly I do not yet have a solid understanding of how such planes are constructed). If someone could enlighten me that would be excellent.
10003
For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?  I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly  $$ f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right) $$ but I can't seem to finish it, any help?