I like to look for mistakes in other people's works. So, I don't have time to find my own mistakes.
I was reading a wonderful book yesterday:
My Numbers, My Friends (Popular Lectures on Number Theory) by the author Paulo Ribenboim.
$$...\;if\;p\;is\;a\;prime\;dividing\;both\;P,\;Q,\;then...$$
$\blacksquare$
Those who wish can read as much as they want by clicking on the image of the book cover.
Click on the image to download.
For a wider collection click here. The password for opening encrypted files is ogeometrie
Click on the image to download.
For a wider collection click here. The password for opening encrypted files is ogeometrie
Nu este sigur daca voi reusi actualizarea ERATA la fiecare numar al revistei.
E.g $\{1,2,3,4,5\}=\{1,4\}\cup \{2,3\}\cup \{5\}$ ;
$\{1,2,3,4,5,6\}=\{1,6\}\cup \{2,5\} \cup \{3,4\}.$
The following is one of the old contest problems (22nd All Soviet Union Math Contest, 1988).
Let $m,\; n,\; k$ be positive integers with $m \geqslant n$ and $1+2+\cdots +n=m\cdot k$. Prove that the numbers $1,\;2,\;\cdots,n$ can be divided into $k$ groups in such a way that the sum of the numbers in each group equals $m$.
CiP example: $1+2+3+4+5+6+7+8+9=45$;
from $45=9\cdot 5$ the partition of $k=5$ groups is obtained, each with a sum of $m=9$
$$\{1,\;8\}\cup \{2,\;7\}\cup\{3,\;6\}\cup \{4,\;5\}\cup\{9\};$$
from $45=15 \cdot 3$ the partition of $k=3$ goups is obtained, each with a sum of $m=15$.
$$\{1,\;2,\;3,\;9\} \cup \{4,\;5,\;6\} \cup \{7,\;8\}.$$
CiP exampl$e^{bis}$: $1+2+3+4+5+6+7+8=36=12 \cdot 3$ and we have the partition of $k=3$ subsets, with the sum of $m=12$
$$\{1,\;2,\;3,\;6\} \cup \{4,\;8\} \cup \{5,\;7\}.$$
It would seem that the answer to the original question is $n=3\cdot p-1$ or $n=3\cdot p$. That is, in the case of $n=3\cdot p+1$, there is no such partition. (Indeed, if $n=3\cdot p+1$, then, because $$1+2+ \cdots +n=\frac{n(n+1)}{2}=\frac{(3p+1)(3p+2)}{2}=9\cdot \frac{p(p+1)}{2}+1,$$ this sum should be a multiple of 3, which is not true.)
If $n=3p$ then $1+2+ \cdots +n=\frac{3p(3p+1)}{2}=3 \cdot \frac{p(3p+1)}{2}=3\cdot \frac{p(p+1+2p)}{2}=3\cdot m$,
and $p$ or $p+1$ is divisible by $2$ so $m$ is integer.
If $n=3p-1$ then $\frac{n(n+1)}{2}=\frac{(3p-1) p}{2}=\frac {(2p+p-1)\cdot p}{2}=3 \cdot m$
with $m$ integer by the same arguments.
So in these cases we fall over the general problem given to the 22nd All Soviet Union Math Contest, 1988.
Another example in the case of $n=15$ is here : math.stackexchange.com/questions/31929/partition-of-equal-size-and-equal-sum
The proof is given by induction, resulting in a way of constructing the partitions.
The general problem has an object if $n \geqslant 3$.
For $n=3$ we have only one partition $\{1,\;2\} \cup \{3\}.$
For $n=4$ such partitions do not exsist.
For $n=5,\;6$ are just the examples presented previously.
For $n=7$ such partitions do not exist.
For $n=8$, because $1+2+\cdots +8=36=9 \cdot 4=12 \cdot 3=18 \cdot 2$ we have in addition to the presented previously and the partitions
$$\{1,\;2,\;3,\;4,\;8\}\cup \{5,\;6,\;7\},\;\{1,\;3,\;6,\;8\}\cup \{2,\;4,\;5,\;7\}$$
$$\{1,\;4,\;5,\;8\}\cup \{2,\;3,\;6,\;7\}\;,\;\{1,\;4,\;6,\;7\}\cup \{2,\;3,\;5,\;8\}$$
$$\{1,\;2,\;4,\;5,\;6\}\cup \{3,\;7,\;8\}\;,\;\{1,\;2,\;3,\;5,\;7\} \cup \{4,\;6,\;8\}$$
$$\{1,\;2,\;7,\;8\}\cup \{3,\;4,\;5,\;6\}.$$
For $n=9$, because $1+2+\cdots+9=45=9\cdot 5=15 \cdot 3$ are just the examples presented previously.
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For a wider collection click here. The password for opening encrypted files is ogeometrie
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For a wider collection click here. The password for opening encrypted files is ogeometrie
The problem has been discussed here before.
I showed that $\left \{ cos\frac{\pi}{7},\;-cos\frac{2\pi}{7},\;cos\frac{3\pi}{7} \right \}=\left \{ sin\frac{5\pi}{14},\; -sin \frac{3\pi}{14},\; \sin\frac{\pi}{14} \right \}$
are the roots of the equation $8x^3-4x^2-4x+1=0$ (see Remark 2, relations (5)-(8) ). And identities are expressions of Vieta's relationships for this equation.
For 3.i), let $A=cos\frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}$.
Let's calculate $2sin\frac{\pi}{7} \cdot A=2sin\frac{\pi}{7}\cdot cos\frac{\pi}{7}-2sin\frac{\pi}{7}\cdot cos \frac{2\pi}{7}+2sin \frac{\pi}{7}\cdot cos \frac{5\pi}{7};$
with the formula of the double angle and the sum product formulas we get
$$2sin\frac{\pi}{7} \cdot A= sin\frac{2\pi}{7}-\left (sin \frac{3\pi}{7}-sin\frac{\pi}{7} \right )+ \left ( sin \frac{4\pi}{7}-sin\frac{2\pi}{7} \right ).$$
But the angles $\frac{4\pi}{7}$ and $\frac{3\pi}{7}$ are supplementary, so $sin \frac{4\pi}{7}=sin\frac{3\pi}{7}$, and we get $2sin\frac{\pi}{7} \cdot A= sin\frac{\pi}{7}$ from which 3.i) results.
For 3.ii), let's note that $\frac{\pi}{2}-\frac{\pi}{7}=\frac{5\pi}{14},\;\frac {\pi}{2}-\frac{2\pi}{7}=\frac{3\pi}{14},\;\frac{\pi}{2}-\frac{3\pi}{7}=\frac{\pi}{14}$
and with the complement formula this is obtained from 3.i).
For 3.iii), let $B=8\cdot cos\frac{\pi}{7}\cdot cos\frac{2\pi}{7} \cdot cos \frac{3\pi}{7}.$
We will calculate $sin\frac{\pi}{7}\cdot B$ applying several times the formula of the double angle and the formulas of the supplements.
$$sin\frac{\pi}{7}\cdot B=4\cdot \left ( 2sin \frac{\pi}{7} \cdot cos\frac{\pi}{7}\right )\cdot cos\frac{2\pi}{7}\cdot cos\frac{3\pi}{7}=4\cdot sin\frac{2\pi}{7}\cdot cos \frac{2\pi}{7} \cdot cos \frac{3\pi}{7}=$$
$$=2\cdot \left ( 2 \cdot sin\frac{2\pi}{7}\cdot cos \frac{2\pi}{7} \right ) \cdot cos \frac{3\pi}{7}=2\cdot sin\frac{4\pi}{7}\cdot cos\frac{3\pi}{7}=2\cdot sin\frac{3\pi}{7}\cdot cos \frac{3\pi}{7}=$$
$$=sin\frac{6\pi}{7}=sin\frac{\pi}{7}$$
hence $B=1.$
Now 3.iv) is obtained from 3.iii) with complement's formulae.
$\blacksquare$
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(Daneză)
Euclid showed in Book XIII, Proposition 10 of his Elements that the area of the square on the side of a regular pentagon inscribed in a circle is equal to the sum of the areas of the squares on the sides of the regular hexagon and the regular decagon inscribed in the same circle. In the language of modern trigonometry, this says:
Ptolemy used this proposition to compute some angles in his table of chords in Book I, chapter 11 of Almagest.
Solution 1 CiP
We know the values $$sin 18^\circ=\frac{\sqrt{5}-1}{4},\;sin 30^\circ=\frac{1}{2},\;sin 36^\circ=\frac{\sqrt{10-2\sqrt{5}}}{4}.$$
The calculation is simple: $sin^2 18^\circ +sin^2 30^\circ =\left (\frac{\sqrt{5}-1}{4} \right )^2+\left (\frac{1}{2} \right )^2=\frac{5-2\sqrt{5}+1}{16}+\frac{1}{4}=\frac{6-2\sqrt{5}+4}{16}=\frac{10-2\sqrt{5}}{16}=sin^2 36^\circ .$
$\blacksquare$
Solution 2 CiP
First we show that
$cos 36^\circ -cos 72^\circ=\frac{1}{2}. \tag{1}$
For this, we denote $A:=cos 36^\circ -cos 72^\circ.$ We will calculate, with double-angle formula and product-to-sum identity
$$2sin 36^\circ \cdot A=2sin 36^\circ \cdot cos 36^\circ-2sin 36^\circ \cdot cos 72^\circ=sin (2\cdot 36^\circ)-(sin 108^\circ-sin36^\circ)=$$
$$=(sin 72^\circ-sin 108^\circ)+sin 36^\circ=sin 36^\circ,$$
(because of reflection in $90^\circ$ formulae, $sin108^\circ=sin(180^\circ-108^\circ)=sin 72^\circ)$, hence (1) follow.
Now we calculate with power-reduction formula
$sin^2 36^\circ-sin^2 18^\circ=\frac{1-cos (2\cdot 36^\circ)}{2}-\frac{1-cos(2\cdot 18^\circ)}{2}=\frac{1}{2}(cos 36^\circ-cos 72^\circ)\;\underset{(1)}{==}\;\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}=sin^2 30^\circ.$
$\blacksquare$
Remark CiP We also know $cos 36^\circ =\frac{\sqrt{5}+1}{4},\;cos 72^\circ =\frac {\sqrt{5}-1}{4}$ and (1) results easier.
<End Rem>
Let $A$ and $B$ be distinct points.
Proof CiP
(i)$\Rightarrow$(iii) As oriented segments, we have $\frac{\overline{XA}}{\overline{XB}}=\kappa\;\Rightarrow\;\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}\;\Rightarrow$
$$\Rightarrow\;\overrightarrow{OA}-\overrightarrow{OX}=\kappa(\overrightarrow{OB}-\overrightarrow{OX})\;\Rightarrow\;(1-\kappa)\overrightarrow{OX}=\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}$$
where $O$ is an arbitrary point. We get the conclusion.
(ii)$\Rightarrow$(i) $(1-\kappa)\overrightarrow{OX}=\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}\;\Rightarrow\;\overrightarrow{OA}-\overrightarrow{OX}=\kappa (\overrightarrow{OB}-\overrightarrow{OX}\;\Rightarrow\;\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}$
so the vectors $\overrightarrow{XA}$ and $\overrightarrow{XB}$ are collinear and their orientation is according to the sign of $\kappa$, as the figure shows. Hence $\overline {XA}=\kappa \cdot \overline{XB}$.
(iii)$\rightarrow$(ii) obvious.
The Lemma is proved. To prove the corollary we note that
$$\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}\;\Leftrightarrow\;\overrightarrow{AX}=-\kappa \cdot \overrightarrow{XB}=-\kappa(\overrightarrow{AB}-\overrightarrow{AX})\;\Leftrightarrow \;(1-\kappa)\overrightarrow{AX}=-\kappa \cdot \overrightarrow{AB}$$
hence $r=-\frac{\kappa}{1-\kappa}$.
$\blacksquare \blacksquare$
A young researcher thinks he has made a great discovery. A prestigious journal even published his research. Here are the links to the two articles:
In translation
"S:L23.322. Show that if $\;a,\;b,\;c\in (0,\;\infty)$ and $a+b+c=1$, then
$$\frac{a+1}{\sqrt{a+bc}}\geqslant 2."$$
Solution CiP
We have the known inequality
$\frac{x+y}{\sqrt{xy}}\geqslant 2 \tag{1}$
where $x,\;y$ are positive real numbers. The sign $"="$ occurs in (1) if and only if $x=y.$
(The deduction of (1) is immediate from the equivalents
$$(\sqrt{x}-\sqrt{y})^2 \geqslant 0\;\Leftrightarrow\;x-2\sqrt{x}\cdot \sqrt{y}+y \geqslant 0\;\Leftrightarrow\;x+y\geqslant 2\sqrt{xy}\;\Leftrightarrow\;(1)\;)$$
Back to the problem, it results from the given conditions that $\;a,b,c<1$ (because, for example $a=1-b-c<1-0-0$ and $-b,-c<0$).
Then the numbers $x=1-b$ and $y=1-c$ are positive. But we have
$$a+1=1-b-c+1=(1-b)+(1-c)=x+y$$
and
$$a+bc=1-b-c+bc=(1-b)(1-c)=xy$$
and then the inequality (1) expresses exactly what the statement requires.
$\blacksquare$
Remark CiP The statement does not ask when the $"="$ sign occurs. We will answer, as a bonus.
The sign $"="$ occurs if and only if $(a,b,c)=(1-2t,t,t)$ with $t\in (0, \frac{1}{2}).$
Even the values $t=0$ (when $a=1,\;b=c=0$) and $t=\frac{1}{2}$ (when $a=0,\;b=c=\frac{1}{2}$) are admissible for the statement.
<end Rem>
..................................................................................................................................
As Mr. Konstantine ZELATOR used to us, a problem never appears alone. I am referring to his posts related to problems in the Crux Mathematicorum magazine (although he also has concerns about UFO-logy). As I announced in the title, we will also try to solve the other problems on the photographed page.
" S:L23.321. Let $ABCD$ be the convex quadrilateral whose diagonals
intersect at $O$. If $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AO}=\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{OC}$, show that $ABCD$
is a parallelogram."
Solution CiP
It seems that no figure would be needed.
The relationship given between the six vectors is written, replacing some of them with the sum of two others, in the form
$(\overrightarrow{AO}+\overrightarrow{OB})+(\overrightarrow{AO}+\overrightarrow{OD})+\overrightarrow{AO}=(\overrightarrow{BO}+\overrightarrow{OC})+(\overrightarrow{DO}+\overrightarrow{OC})+\overrightarrow{OC}\;\Leftrightarrow$
$\Leftrightarrow\;3\overrightarrow{AO}+\overrightarrow{OB}+\overrightarrow{OD}=\overrightarrow{BO}+3\overrightarrow{OC}+\overrightarrow{DO}\;\Leftrightarrow$
$\Leftrightarrow \;3(\overrightarrow{AO}-\overrightarrow{OC})=2\overrightarrow{BO}+2\overrightarrow{DO}\;\Leftrightarrow\;$
$\Leftrightarrow\;3(\overrightarrow{AO}+\overrightarrow{CO})=2(\overrightarrow{BO}+\overrightarrow{DO}).\;\tag{1}$
On the left side of the relation (1) we have a sum of two vectors collinear with $\overrightarrow{AC}$, so the result will also be a vector collinear with $\overrightarrow{AC}$. On the right side of the relation (1) we have a sum of two vectors collinear with $\overrightarrow{BD}$, so the result will also be a vector collinear with $\overrightarrow{BD}$. Thus, equality (1) is possible only if
$$\overrightarrow{AO}+\overrightarrow{CO}=\overrightarrow{0}=\overrightarrow{BO}+\overrightarrow{DO}\;\Leftrightarrow$$
$\Leftrightarrow\;\overrightarrow{AO}=-\overrightarrow{CO}=\overrightarrow{OC}$ and $\overrightarrow{BO}=-\overrightarrow{DO}=\overrightarrow{OD}.$ But this means that point $O$ is both the midpoint of segment $[AC]$ and the midpoint of $[BD]$, thus $ABCD$ is a parallelogram.
$\blacksquare$
I saw, flipping through a list of problems for the Mathematics Olympiad for Juniors, the following question:
"If $0<x<1$ prove that $x-x^3\;<\frac{1}{2}$."
I would be curious what the simplest solution would be. I had another post like this here. And I mean a solution that does not call for advanced mathematical knowledge, for example Calculus.
Anyone with even a little Calculus experience realizes that the statement of the problem hides an insufficiency: does the expression $x-x^3$ have a maximum or a minimum on the interval $0<x<1$ ?; and what are the values of $x$ that achieve them ?
Following this line, Calculus teaches us to find out the largest and smallest value taken by the function
$$f\;:\;[0;1]\rightarrow \mathbb{R},\;\;f(x)=x-x^3$$
(or the extension of this function to the entire real axis $-\infty <x<+\infty$).
The derivative ${f}'=1-3x^2$ has critical (stationary) points $x_\pm=\pm \frac{1}{\sqrt{3}}$.
If $0<x<1$ then $x-x^3 \leqslant \frac{2}{3\sqrt{3}}. \;\;\;\;\;(1)$
Equality occur at $x_+=\frac{1}{\sqrt{3}}$.
Remark CiP Before I had time to complete the Variation Table of the function, Mr. George Stoica already posted his comment with an elementary solution to both the initial version (obtaining a slight strengthening) and a solution to the "optimal" version (1).
The title of the post is inspired by the similar title of a book, we could call it famous, by Ivan NIVEN, (The MATHEMATICAL ASSOCIATION of AMERICA, 1981). Even after decades, this book is still quoted with interest; see Berkeley Math Circle : Tom Rike, Maxima and Minima Problems WITHOUT Calculus, 2002
We will give a demonstration based on the maximum product principle. It is stated as follows
If the sum of $n$ non-negative real numbers is constant, then their
product is maximum when the numbers are equal.
Let be the product of three factors $P=x\cdot (1-x) \cdot (1+x).$
The sum of the three factors is $x+(1-x)+(1+x)=2+x$. Unfortunately, it is not constant, so we will resort to a "trick". We consider the product
$$P_1=(\sqrt{3}+1)(2+\sqrt{3}) \cdot P=$$
$$=[(\sqrt{3}+1)x]\cdot [(2+\sqrt{3})(1-x)] \cdot [1+x].$$
The sum of the three factors is now
$[x\sqrt{3}+x]+[2+\sqrt{3}-2x-x\sqrt{3}]+[1+x]=3+\sqrt{3}$
and is now a constant. So the maximum of $P_1$ occurs when
$$(\sqrt{3}+1)x=(2+\sqrt{3})(1-x)=1+x.$$
But the three conditions above are satisfied exactly when $x=\frac{1}{\sqrt{3}}.$
Now, $P_1$ and $P$ differ by a positive factor, so they are maximum at the same time.
$\blacksquare$
Things don't end there. Some might say that this solution is like pulling a rabbit out of a hat. I applied a method known in the French mathematical literature as belonging to M. Grillet. Unfortunately, in the "Éléments d'algèbre, à l'usage des candidats au baccalauréat ès sciences et aux écoles spéciales" par Eugène Rouché (Paris, MALLET-BACHELIER , IMPRIMEUR-LIBRAIRE, 1857) on page 181, only the Nouvelles Annales are mentioned, without any other specification.
We start by finding three numbers $\alpha,\;\beta,\;\gamma\;$ so that the product
$$P_1=\alpha \cdot \beta \cdot \gamma \cdot P=[\alpha \cdot x]\cdot [\beta \dot (1-x)] \cdot [\gamma \cdot (1+x)]$$
has a constant sum of factors. The sum of these factors is
$$(\alpha x)+(\beta-\beta x)+(\gamma +\gamma x)=(\alpha -\beta+\gamma)\cdot x+(\beta+\gamma).$$
If we choose $\alpha -\beta +\gamma =0$, then $\alpha = \beta-\gamma$ and
$$P_1=[(\beta-\gamma)x]\cdot [\beta(1-x)]\cdot [\gamma(1+x)]$$
has a constant sum of factors. $P_1$ is maximum for
$$(\beta-\gamma) \cdot x =\beta-\beta\cdot x=\gamma+\gamma \cdot x .\tag{2}$$
From the first equality we get $x=\frac{\beta}{2\beta-\gamma}$ and from the second equality we get $x=\frac{\beta-\gamma}{\beta+\gamma}.$ In order for relations (2) to be compatible, we must have
$$\frac{\beta-\gamma}{\beta+\gamma}=\frac{\beta}{2\beta-\gamma}\;\Leftrightarrow\;\beta^2-4\beta \gamma+\gamma^2=0.$$
We choose $\frac{\beta}{\gamma}>1$ to have $\beta-\gamma >0$, so $\frac{\beta}{\gamma}=2+\sqrt{3}.$ In our trick we took $\gamma=1,\;\beta=2+\sqrt{3},\;\alpha=1+\sqrt{3}.$
(Remark. In the case of more than three variables, conditions like those in (2) lead to equations of degree higher than 2, so they are more difficult to handle.)
!! Finally, I found exactly. Grillet, H. Méthode élémentaire pour résoudre quelques questions sur les maximums. Nouvelles annales de mathématiques : journal des candidats aux écoles polytechnique et normale, Serie 1, Volume 9 (1850), pp. 70-73.
