answers

Tuesday, November 29, 2011

problem 10

the question was
Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal S$ are collinear. A windmill is a process that starts with a line $\ell$ going through a single point $P \in \mathcal S$ . The line rotates clockwise about the pivot $P$ until the first time that the line meets some other point belonging to $\mathcal S$ . This point, $Q$ , takes over as the new pivot, and the line now rotates clockwise about $Q$ , until it next meets a point of $\mathcal S$ . This process continues indefinitely.
Show that we can choose a point $P$ in $\mathcal S$ and a line $\ell$ going through $P$ such that the resulting windmill uses each point of $\mathcal S$ as a pivot infinitely many times.

now lets see its solution

This problem was very nice, creative and inspiring. Something a bit new which looks very nice. I will present a solution I found on AoPS, which I consider very interesting.

For any point $P$ , there is a line which cuts the remaining points in two groups, which differ with at most $1$ point from one another. This kind of a line works for the problem. Why? Imagine how such a line creates the windmill: it touches one point, then one side gains one point, and then this side loses the pivot, leaving the situation as before, i.e. with sides of the line differing by at most one point.

Pick now one point and a line which has the above property. At a given moment, the windmill will be parallel to this line, and by uniqueness, will coincide with this line. This reasoning shows that any point will be pivot in a $2\pi$ rotation of the windmill.

Thursday, November 24, 2011

question 8

Problem was this:

In the interior of triangle $P_1P_2P_3$ a point $P$ is given. Let $Q_1,Q_2,Q_3$ be the intersections of $PP_1, PP_2,PP_3$ with the opposing edges of triangle $P_1P_2P_3$ . Prove that among the ratios $\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}$ there exists one not larger than $2$ and one not smaller than $2$ .

Solution

Since triangles $P_1P_2P_3$ and $PP_2P_3$ share the base $P_2Q_2$ , we have $\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_1Q_1}{PQ_1}$ , where $[ABC]$ denotes the area of triangle $ABC$ . Similarly, $\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{P_2Q_2}{PQ_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{P_3Q_3}{PQ_3}$ . Adding all of these gives $\frac{[PP_1P_3]}{[P_1P_2P_3]}+\frac{[PP_1P_2]}{[P_1P_2P_3]}+\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{P...$ , or $\frac{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{PQ_3}+\frac{P_1Q_1}{PQ_1}=\frac{[PP_1P_2]+[PP_2P_3]+[PP_3P_1]}{[P_1P_2P_3]}=1$ We see that we must have at least one of the three fractions less than or equal to $\frac{1}{3}$ , and at least one greater than $\frac{1}{3}$ . These correspond to ratios $\frac{PP_i}{PQ_i}$ being less than or equal to $2$ , and greater than or equal to $2$ , respectively, so we are done.

Wednesday, November 23, 2011

question 7

question was

Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds

$f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor$

where $\left\lfloor a\right\rfloor$ is greatest integer not greater than $a.$

Solutions

Solution 1

Put $x=y=0$ . Then $f(0)=0$ or $\lfloor f(0) \rfloor=1$ .

$\bullet$ If $\lfloor f(0) \rfloor=1$ , putting $y=0$ we get $f(x)=f(0)$ , that is f is constant. Substituing in the original equation we find $f(x)=0, \ \forall x \in \mathbb{R}$ or $f(x)=a, \ \forall x \in \mathbb{R}$ , where $a \in [1,2)$ .

$\bullet$ If $f(0)=0$ , putting $x=y=1$ we get $f(1)=0$ or $\lfloor f(1) \rfloor=1$ .

For $f(1)=0$ , we set $x=1$ to find $f(y)=0 \ \forall y$ , which is a solution.

For $\lfloor f(1) \rfloor=1$ , setting $y=1$ yields $f(\lfloor x \rfloor)=f(x), \ (*)$ .

Putting $x=2, y=\frac{1}{2}$ to the original we get $f(1)=f(2)\lfloor f(\frac{1}{2}) \rfloor$ . However, from $(*)$ we have $f(\frac{1}{2})=f(0)=0$ , so $f(1)=0$ which contradicts the fact $\lfloor f(1) \rfloor=1$ .

So, $f(x)=0, \ \forall x$ or $f(x)=a, \ \forall x, \ a \in [1,2)$ .

Solution 2

Substituting $y=0$ we have $f(0) = f(x) [f(0)]$ . If $[f0)] \ne 0$ then $f(x) = \frac{f(0)}{[f(0)]}$ . Then $f(x)$ is constant. Let $f(x)=c$ . Then substituting that in (1) we have $c=c[c] \Rightarrow c(1-[c])=0 \Rightarrow c=0$ , or $[c]=1$ . Therefore $f(x)=c$ where $c=0$ or $c \in [1,2)$

If $[f(0)] = 0$ then $f(0)=0$ . Now substituting $x=1$ we have $f(y)=f(1)[f(y)]$ . If $f(1) \ne 0$ then $[f(y)] = \frac{f(y)}{f(1)}$ and substituting this in (1) we have $f([x]y)=\frac{f(x)f(y)}{f(1)}$ . Then $f([x]y)=f(x[y])$ . Substituting $x=1/2, y=2$ we get $f(0)=f(1)$ . Then $f(1)=0$ , which is a contradiction Therefore $f(1)=0$ . and then $f(y)=0$ for all $R$

Then the only solutions are $f(x)=0$ or $f(x)=c$ where $c \in [1,2)$ .( By m.candales [2])

Solution 3

Let $y=0$ , then $f(0)=f(x)\left\lfloor f(0)\right\rfloor$ .

Case 1: $\left\lfloor f(0)\right\rfloor\neq 0$

Then $f(x)=\frac{f(0)}{\left\lfloor f(0)\right\rfloor}$ is a constant. Let $f(x)=k$ , then $k=k\left\lfloor k \right\rfloor \Leftrightarrow k=0 \vee 1\leq k<2$ . It is easy to check that this are solutions.

Case 2: $\left\lfloor f(0)\right\rfloor= 0$

In this case we conclude that $\left\lfloor f(0)\right\rfloor= 0\Rightarrow f(0)=0$

Lemma:If $y$ is such that $0\leq f(y)<1$ , $f(y)=0$

Proof of the Lemma: If $x=1$ we have that $f(\left\lfloor x\right\rfloor y)=f(y)=f(x)\left\lfloor f(y)\right\rfloor =0$ , as desired.

Let $0\leq x<1$ , so that we have: $0=f(0)=f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor\Rightarrow$ $\Rightarrow f(x)=0 \vee 0\leq f(y)<1 \Rightarrow f(x)=0 \vee f(y)=0$ , using the lemma.

If $f$ is not constant and equal to $0$ , letting $y$ be such that $f(y)\neq 0$ implies that $f(x)=0, \forall 0\leq x<1$ .

Now it's enough to notice that any real number $x$ is equal to $ky$ , where $k\in \mathbb{Z}$ and $0\leq y< 1$ , so that $f(x)=f(ky)=f(\left\lfloor k\right\rfloor y)=f(k)\left\lfloor f(y)\right\rfloor=0$ . Since $x$ was arbitrary, we have that $f$ is constant and equal to $0$ .

We conclude that the solutions are $f(x)=k$ , where $k=0 \vee 1\leq k<2$ .( By Jorge Miranda

Solution 4

Clearly $f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\...$ , so $(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0$ for all $x,y\in\mathbb{R}$ .

If $\left\lfloor f(y)\right\rfloor = 0$ for all $y \in \mathbb{R}$ , then by taking $x=1$ we get $f(y)=f(1)\left\lfloor f(y)\right\rfloor = 0$ , so $f$ is identically null (which checks).

If, contrariwise, $\left\lfloor f(y_0)\right\rfloor \neq 0$ for some $y_0 \in \mathbb{R}$ , it follows $f(x) = f(\lfloor x \rfloor)$ for all $x \in \mathbb{R}$ .

Now it immediately follows $f(x) = f(\lfloor x \rfloor \cdot 1) = f(x)\lfloor f(1) \rfloor$ , hence $f(x)(1 - \lfloor f(1) \rfloor) = 0$ .

For $x=y_0$ this implies $\lfloor f(1) \rfloor = 1$ . Assume $\lfloor f(0) \rfloor=0$ ; then $1 \leq f(1) = f\left ( 2\cdot \dfrac {1} {2} \right ) = f(2)\left \lfloor f \left ( \dfrac {1} {2} \right ) \right \rfloor = ...$ , absurd.

Therefore $\lfloor f(0) \rfloor \neq 0$ , and now $y=0$ in the given functional equation yields $f(0) = f(x)\lfloor f(0) \rfloor$ for all $x \in \mathbb{R}$ , therefore $f(x) = c \neq 0$ constant, with $\lfloor c \rfloor = \lfloor f(1) \rfloor = 1$ , i.e. $c \in [1,2)$ (which obviously checks).

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