A Strange Wild Song (Poem by Lewis Carroll) and A topological juggle

           

A Strange Wild Song (Poem by Lewis Carroll) Written circa 1870

          

             A preparatory problem for the Kangaroo contest gave me a lot of trouble. I got it from Practice Materials, for Grade 11 and Grade 12. It's the first problem on the list, so it's rated as simple. And yet... not too simple.

Sorry if it's the French version.

          To avoid bothering myself with all sorts of topological juggling, I decided to try a solution with my own hand. I cut out the figure and tried to fit them together, finally obtaining the Answer: Option A is correct.

     Below are the photos that demonstrate this. It remains to be seen how a less...manufactured solution would have been possible.

I cropped the figure and arranged option A so that it would fit into the coverage by translation.

And here is the result

The other cases do not give the correct answer.

Option B

Option C

Option D

Option E

And this is what the 5 response options look like when returning by translation after the matching attempt.

          A coloring of the pentagons as they fit, to give us a more intuitive picture

Quadratic Fields in the "Revista Matematică (a Elevilor) din Timişoara

           Quadratic fiels are a very serious mathematical notion. A book accessible to any schoolchild is Mak TRIFKOVIĆ - Algebraic Theory of Quadratic Numbers (Springer Science+Business Media, New York, 2003).

           RMT is a magazine.. not too serious: don't try to use the electronic edition because you will be disappointed. You can take a look at the  magazine  here. In Issue 2 of 1975, the article on pages 3-8 contains a rather complicated statement about the radicals of these numbers.

If $\;\sqrt[n]{a+b\sqrt{d}}=x+y\sqrt{d}$ then $\;\sqrt[n]{a-b\sqrt{d}}=|x-y\sqrt{d}|$

and a sufficient condition for this to happen is

$$\sqrt[n]{a+b\sqrt{d}}=x+y\sqrt{d}\;\;\;\Rightarrow\;\;\sqrt[n]{a^2-b^2\cdot d} \in \mathbb{Q}\tag{N}$$

     The condition (N) is not sufficient, as their example shows

$\sqrt{18+2\sqrt{77}}=\sqrt{11}+\sqrt{7}\neq x+y\sqrt{77}$ although $\sqrt{18^2-2^2 \cdot 77}=4$

                    In my opinion, if $d$ is a prime number, then the condition (N) is indeed

 Necessary_ and_ Sufficient.

Mathematical Induction by Ear // Kulakla Matematiksel Tümevarım

               Forgive me, Turkish-speaking friends, if I have distorted the expression "a cânta după ureche" . Maybe, I don't know, it sounds like hell in English too.

             Everyone knows what Mathematical Induction is. Let $P(n)$ be a STATEMENT about the natural numbers. By STATEMENT we mean what in Logic is called PREDICATE. Predicates can be defined in first-order logic. I would express the axiom of induction like this:

$$P(0)\wedge \forall n(P(n)\rightarrow P(n+1))\Rightarrow \forall nP(n) \tag{I}$$

There are many debates on this topic. I understand that (I) is an axiom-schema. If we want to express (I) as a single axiom, that is,

$$\forall P(P(0)\wedge \forall n(P(n)\rightarrow P(n+1))\Rightarrow \forall nP(n))$$

then we go beyond first-order logic, because we apply the $\forall$ quantifier to a predicate.

The Integral $\int \frac{1}{ax^2+bx+c}dx$ in Problem 29046

          It was said in DidMath No. 1/2024 at page 17 that the arctangent function bothers us in the calculation of integrals. Such a situation just arose in Problem 29046, author Mihály BENCZE, Brașov.

              "29046    Let $a>1$. Determine $\int_{\frac{1}{a}}^a \frac{x^n\cdot \arctan x\;dx}{x^{2n+2}+x^{n+1}+1}\;.$"

ANSWER CiP

$$\frac{2\pi}{4(n+1)\sqrt{3}} \cdot \arctan \left( \frac{2}{\sqrt{3}} \cdot \frac {a^{2n+2}-1}{2a^{2n+2}+6a^{n+1}+1}\right )$$

 Solution CiP

                Lemma  Let $a \neq 0,\;\Delta=b^2-4ac<0$. Then

$$\int \frac{1}{ax^2+bx+c}dx=\frac{2}{\sqrt{-\Delta}}\cdot \arctan\left ( \frac{2ax+b}{\sqrt{-\Delta}}\right )+C \tag{I}$$

               Formula (I) can be verified with the definition, calculating for

$u=\frac{2}{\sqrt{-\Delta}}\cdot arctg \;\phi(x),\;\; \phi(x)=\frac{2ax+b}{\sqrt{-\Delta}}$

$$\frac {\mathrm{d}u}{\mathrm{d}x}=\frac{2}{\sqrt{-\Delta}}\cdot \frac{\phi^{'}}{\phi^2+1}=\frac{2}{\sqrt{-\Delta}}\cdot \frac{\frac{2a}{\sqrt{-\Delta}}}{\left ( \frac{2ax+b}{\sqrt{-\Delta}} \right ) ^2+1}=\frac{4a/(-\Delta)}{(4a^2x^2+4abx+b^2+4ac-b^2)/(-\Delta)}.$$

$\square$ end Lemma

For $a=1=b=c$ we obtain

              Corollary  $\int \frac{1}{x^2+x+1}dx=\frac{2}{\sqrt{3}} \cdot \arctan \left (\frac{2x+1}{\sqrt{3}} \right )+C \tag{1}$

 

          Solving the problem:

             We will not exaggerate with the claim to prove with derivatives that are equal the relationship

$$\arctan t +\arctan \frac{1}{t}=\frac{\pi}{2} \tag{2}$$

        Let $I:=\int_{1/a}^{a} \frac{x^n \cdot \arctan x}{x^{2n+2}+x^{n+1}+1}dx$. With substitution

$$x=\frac{1}{t},\;\;dx=-\frac{1}{t^2}\cdot dt,\;\;\;t=a\;for \;x=\frac{1}{a},\;\;t=1/a\;for\; x=a\;\;\;\;\Rightarrow$$

$I=\int_{a}^{1/a}\frac {\frac{1}{t^n} \cdot \arctan\left( \frac{1}{t} \right)}{\frac{1}{t^{2n+2}}+\frac{1}{t^{n+1}+1} }\cdot \left ( -\frac{1}{t^2} \right ) dt\overset{(2)}{=}\int_{1/a}^{a} \frac {t^n \cdot (\pi /2-\arctan t)}{t^{2n+2}+t^{n+1}+1}dt=\frac{\pi}{2} \cdot \int_{1/a}^{a} \frac{t^n}{t^{2n+2}+t^{n+1}+1}dt-$

$-\int_{1/a}^{a}\frac{t^n \cdot \arctan t}{t^{2n+2}+t^{n+1}+1}dt=\frac{\pi}{2}\cdot \int_{1/a}^{a} \frac{t^n}{t^{2n+2}+t^{n+1}+1}-I\;\;\;\Rightarrow I+I=\frac{\pi}{2}\cdot \dots$, so

$$I=\frac{\pi}{4} \cdot \int_{1/a}^{a}\frac{t^n}{t^{2n+2}+t^{n+1}+1}dt \tag {3}$$

To the integral in (3) we make the substitution

$$t^{n+1}=u,\;\;t^n \cdot dt=\frac{du}{n+1},\;\;\;u=1/a^{n+1}\;for\;t=1/a,\;\;u=a^{n+1}\;for\;t=a\;\;\;(b:=a^{n+1})$$

and we have  $I=\frac{\pi}{4} \cdot \int_{1/b}^{b} \frac{1}{u^2+u+1} \cdot \frac{du}{n+1}\overset{(1)}{=}\frac{\pi}{4(n+1)} \cdot \left ( \frac{2}{\sqrt{3}}\arctan (\frac{2u+1}{\sqrt{3}})\right)\Bigg\vert_{u=1/b}^{u=b}$

If we also consider the formula   $\arctan t-\arctan s=\arctan\frac{t-s}{1+t\cdot s}$   then ve obtain

$I=\frac{2\pi}{4(n+1)\sqrt{3}} \cdot \left (\arctan \frac{2b+1}{\sqrt{3}}-\arctan \frac{2/b+1}{\sqrt{3}} \right )=\frac{2\pi}{4(n+1) \sqrt{3}} \cdot \arctan \left (\frac{2}{\sqrt{3}}\cdot \frac{b^2-1}{2b^2+6b+1} \right )$

and puttin $b=a^{n+1}$ we get the answer.

$\blacksquare$

O Problemă cu cașu' rezolvată de CIOBANU // Problem 1300 - CMJ vol 56 issue 2

 The problem 1300 is compounded by my compatriots. 

The title of the post is intended to be an innocent joke.

          We have to show that: $e^{\frac{\pi \imath}{7}}$ is a solution of the equation

$$\imath \sqrt{7}\cdot z^6-z^5+z^4+z^3+z^2+z-1 =0\tag{E}$$

                 Solution CiP

               We have equality [(1) for $n=3$]:

$$\sin \frac{\pi}{7} \cdot \sin \frac {2\pi}{7} \cdot \sin \frac{3\pi}{7}=\frac{\sqrt{7}}{8} \tag {1}$$

     Let $z=e^{\frac{\pi \imath}{7}}$. According to the De Moivre's and Euler's formulas we have

$z^7=e^{\pi \imath}=-1, \;\;|z|=1,\;\bar z=\frac{1}{z},\;\;\sin \frac{k\pi}{7}=\frac{z^k-\bar z ^k}{2\imath}=\frac{z^{2k}-1}{2\imath \cdot z^k},\;k=1,\;2,\;3$  so

$$z^7+1=0\quad ; \quad z^6-z^5+z^4-z^3+z^2-z+1=0 \tag{C}$$

$$\sin \frac{\pi}{7}=\frac{z^2-1}{2\imath \cdot z},\; \quad \sin \frac{2\pi}{7}=\frac{z^4-1}{2\imath \cdot z^2},\; \quad \sin \frac{3\pi}{7}=\frac{z^6-1}{2\imath \cdot z^3}.\tag{S}$$

Substituting the values ​​(S) into (1) we obtain, after multiplying by $(2\imath)^3=-8\imath$

$$-\sqrt{7} \imath \cdot z^6=(z^2-1)(z^4-1)(z^6-1) \Leftrightarrow$$

$$\Leftrightarrow\;\;-\sqrt{7}\imath \cdot z^6=z^{12}-z^{10}-z^8+z^4+z^2-1 \tag{2}$$

          We will apply the Long Division Algorithm to the polynomial $P=X^{12}-X^{10}-X^8+X^4+X^2-1$  to both $Q=X^7+1$ , $Q_1=X^6-X^5+X^4-X^3+X^2-X+1$.

$\begin{array}{c|c}X^{12}&+0X^{11}&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1&Q=X^7+1\\\hline\\-X^{12}&&&&&&&-X^5 &&&&&&D=X^5-X^3-X\\\hline \\&&-X^{10}&&-X^8&&&-X^5&+X^4&&+X^2&&-1&\\\hline &&+X^{10}&&&&&&&+X^3&&&&&\\\hline &&&&-X^8&&&-X^5&+X^4&+X^3&+X^2&&-1\\ \hline\\&&&&&&&-X^5&+X^4&+X^3&+X^2&+X&-1&=R\\\end{array}$

In the first case we obtain

$$P=Q \cdot D+R \tag{3}$$

and in the second case

$\begin{array}{c|c}X^{12}&+0X^{11}&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1&Q_1=X^6-X^5+X^4-X^3+X^2-X+1\\\hline -X^{12}&+X^{11}&-X^{10}&+X^9&-X^8&+X^7&-X^6&&&&&&&D_1=X^6+X^5-X^4-X^3-X^2-X\\\hline &X^{11}&-2X^{10}&+X^9&-2X^8&+X^7&-X^6&+0X^5&+X^4&+0X^3&+X^2&+0X&-1\\\hline &-X^{11}&+X^{10}&-X^9&+X^8&-X^7&+X^6&-X^5\\\hline &&-X^{10}&+0X^9&-X^8&+0X^7&+0X^6&-X^5&+X^4&+0X^3&+X^2&+0X&-1\\\hline &&+X^{10}&-X^9&+X^8&-X^7&+X^6&-X^5&+X^4\\\hline &&&-X^9&+0X^8&-X^7&+X^6&-2X^5&+2X^4&+0X^3&+X^2&+0X&-1\\\hline &&&+X^9&-X^8&+X^7&-X^6&+X^5&-X^4&+X^3\\\hline &&&&-X^8&+0X^7&+0X^6&-X^5&+X^4&+X^3&+X^2&+0X&-1\\\hline &&&&+X^8&-X^7&+X^6&-X^5&+X^4&-X^3&+X^2\\\hline &&&&&-X^7&+X^6&-2X^5&+2X^4&+0X^3&+2X^2&+0X&-1\\\hline &&&&&+X^7&-X^6&+X^5&-X^4&+X^3&-X^2&+X\\ \hline &&&&&&&-X^5&+X^4&+X^3&+X^2&+X&-1&=R_1 \end{array}$

$$P=Q_1 \cdot D_1+R_1 \tag{4}$$

I got the same result $R=R_1=-X^5+X^4+X^3+X^2+X-1$. The equality (2) is written

$$-\sqrt{7} \imath \cdot z^6=P(z)=Q(z) \cdot D(z)+R(z)$$

but according to (C) we have $Q(z)=0=Q_1(z)$ and so 

$$-\imath \sqrt{7} \cdot z^6=-z^5+z^4+z^3+z^2+z-1$$

which is equivalent to the equation (E)

$\blacksquare$

A Problem in Litigation : by Cristian OLTEANU

          The author of Problem #4 in the image is a diligent GMB contributor. The problem was given to 8th graders during the "Nicolae POPESCU" Mathematics Competition...

            With the data in the figure, the problem has no solution. I modified it to my taste, obtaining:

           4.    Calculate the minimum of the expression $\frac{1}{ab}+9ab$, knowing that

                  $(3a+2)(b+1)=6,\;a,b \in (0,+\infty)$, and find the values ​​of the

                  numbers $a$ and $b$ for which this minimum is obtained.

ANSWER CiP

$$\frac{1}{ab}+9ab\geqslant 6;\;\;Equality\;for\;a=\frac{1}{3},\;b=1\;\;or\;\;a=\frac{2}{3},\;b=\frac{1}{2}$$

                     Solution CiP

             $\frac{1}{ab}+9ab=3 \cdot \left( \frac{1}{3ab}+3ab\right )\geqslant 3 \cdot 2=6.$ We used that

 $x+\frac{1}{x} \geqslant 2,\;x>0$, with equality if and only if $x=1$. In our case we have equality for $3ab=1$.

          To find the numbers that achieve the minimum:

$$(3a+2)(b+1)=6\Leftrightarrow 3ab+3a+2b+2=6\underset{3ab=1}{\Leftrightarrow} 3a+2b=3.$$

So we have the conditions

$$3a+2b=3\quad 3a\cdot 2b=2$$

so $3a$ and $2b$ are the roots of the quadratic equation $y^2-3y+2=0.$ It results that

$$\{3a,\;2b\}=\{1,\;2\}$$

from which we obtain the answer.

$\blacksquare$

By-products of PRODUCT // Ürün yan ürünleri

               Here we will demonstrate the formulas

$$\sin \frac{\pi}{2n+1}\cdot\sin \frac{2\pi}{2n+1}\cdot_{\dots} \cdot\sin \frac{n\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^n} \tag{1}$$

$$\sin\frac{\pi}{2n}\cdot\sin \frac{2\pi}{2n}\cdot_{\dots}\cdot\sin \frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}} \tag{2}$$

              We will start from a previously demonstrated formula, written for $m\in \mathbb{N} \setminus \{0,\;1\}$ instead of $n$:

$$\sin \frac{\pi}{m}\cdot\sin \frac{2\pi}{m}\cdot_{\dots} \cdot \sin \frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}} \tag{3}$$

          Let's put in (3): $m$-odd, i.e. $m=2n+1,\;n\in \mathbb{N}\setminus\{0\}$. We will now write down a few more of the $m-1=2n$ factors of (3):

$\sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1}\cdot_{\dots} \cdot \sin \frac{n\pi}{2n+1}  \cdot$

$\cdot \sin \frac{(n+1)\pi}{2n+1} \cdot_{\dots}\cdot \sin \frac{(2n-1)\pi}{2n+1} \cdot \sin \frac{2n\pi}{2n+1}=\frac{2n+1}{2^{2n}} \tag{4}$

Considering the formula $\sin (\pi-\alpha)=\sin \alpha$ and noting that

$$\pi-\frac{2n\pi}{2n+1}=\frac{\pi}{2n+1} \quad \pi-\frac{(2n-1)\pi}{2n+1}=\frac{2\pi}{2n+1} \quad \dots \quad \pi-\frac{(n+1)\pi}{2n+1}=\frac{n\pi}{2n+1}$$

we see that in (4) the factors are two by two equal, one from each row. Then

$$\left (  \sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1} \cdot_{\dots} \cdot \sin \frac{n\pi}{2n+1}\right )^2=\frac{2n+1}{2^{2n}}$$

and taking the square root of both members results (1).

$\blacksquare$

          Let's put in (3) $m$-even, i.e. $m=2n,\;n\in \mathbb{N}\setminus \{0\}$. We now have

$\sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n}\cdot_{\dots} \cdot \sin \frac{(n-1)\pi}{2n} \cdot \overset{=1}{\overbrace{\sin \frac{n\pi}{2n}}} \cdot$

$\cdot \sin \frac{(n+1)\pi}{2n} \cdot_{\dots} \cdot \sin \frac{(2n-2)\pi}{2n} \cdot \sin \frac{(2n-1)\pi}{2n}=\frac{2n}{2^{2n-1}} \tag{5}$

Noting that 

$$\pi-\frac{(2n-1)\pi}{2n}=\frac{\pi}{2n}\quad \pi-\frac{(2n-2)\pi}{2n}=\frac{2\pi}{2n}\quad \dots \quad \pi-\frac{(n+1)\pi}{2n}=\frac{(n-1)\pi}{2n}$$

and considering again the formula  $\sin (\pi-\alpha)=\sin \alpha$ we have

$$\left ( \sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n}\cdot_{\dots} \cdot \sin \frac{(n-1)\pi}{2n}\right )^2=\frac{n}{2^{2n-2}}.$$

     Taking the square root of both members results (2).

$\blacksquare\;\blacksquare$