A more ...curvy... AFFINE PLAN

 

           In Jean Gallier, Basics of Affine Geometry (Ch. 2) shows an example of curved affine plan. He says: "this should dispell any idea that affine spaces are dull".

 

 

          Let's consider the set ( circular paraboloid) $$\mathfrak{P}=\{P=(x,y,z)\in \mathbb{R}\times \mathbb{R}\times \mathbb{R}\mid x^2+y^2=z\}.$$ Its elements are the points $P=(x_1,x_2,x_1^2+x_2^2).$ We will organize $\mathfrak{P}$ as a two dimensional affine space.

         The vector space associated to the affine space $\mathfrak{P}$ is $\mathbb{R}^2$. Tha action is the mapping

$$\mathfrak{P}\times \mathbb{R} \mapsto \mathfrak{P}$$

$$((x,y,x^2+y^2)\;,\;\begin{bmatrix} u\\v\end{bmatrix})\;\mapsto\;(x+u,y+v,(x+u)^2+(y+v)^2).$$

We also use the structural function of the affine space, $$\phi \;:\;\mathfrak{P}\times \mathfrak{P} \rightarrow \mathbb{R}^2,\;\;\phi(A,B)=\begin{bmatrix}b_1-a_1\\b_2-a_2 \end{bmatrix},$$ for $A=(a_1,a_2,a_1^2+a_2^2)\;,\;B=(b_1,b_2,b_1^2+b_2^2).$

In fact, $\mathfrak{P}$ is isomorphic to $\mathbb{A}_2$ (materialized by the plan $xOy$) by projection $pr_3$. 

          Let's take the example $A=(1,0,1) \in \mathfrak{P}\;,\;B=(0,1,1)\in \mathfrak{P}\;,\;\phi(A,B)=\begin{bmatrix}-1\\1 \end{bmatrix}.$ We choose the affine frame with the origin in point $A$ and a linear basis $\overrightarrow{e_1}, \overrightarrow{e_2}$:

$$\mathfrak{R}=(A=(1,0,1);\overrightarrow{e_1}=\begin{bmatrix}1\\0 \end{bmatrix},\;\overrightarrow{e_2}=\begin{bmatrix}0\\1 \end{bmatrix}).$$

For an arbitrary point $P=(x,y,x^2+y^2)\in \mathfrak{P}$, the vector $\phi(A,P)=\begin{bmatrix}x-1\\y\end{bmatrix}=(x-1)\overrightarrow{e_1}+y\overrightarrow{e_2}\;\;\Rightarrow\;\;P(x-1,y)$ are the affine Cartesian coordinates of $P$ in the frame $\mathfrak{R}$. In particular in the frame $\mathfrak{R}$ we have $A=(1,0,1)=A(1-1,0)=A(0,0)\;,\;B=(0,1,1)=B(0-1,1)=B(-1,1).$

Conversely, the point $C(c_1,c_2)$ that has the $(c_1,c_2)$ coordinates in the frame $\mathfrak{R}$ corresponds to the point $C=(c_1+1,c_2, c_1^2+c_2^2+2c_1+1) \in \mathfrak{P}.$

   Let us now investigate the colinear points with A and B. Points $A,B,C=(c_1,c_2,c_1^2+c_2^2)$ are affinely dependent iff the vectors $\overrightarrow{AC}\;,\;\overrightarrow{BC}$ are linerarly dependent. So

$$\overrightarrow{AC}=\phi(A,C)=\begin{bmatrix}c_1-a_1\\c_2-a_c \end{bmatrix}=\begin{bmatrix}c_1-1\\c_c \end{bmatrix}$$

$$\overrightarrow{BC}=\phi(B,C)=\begin{bmatrix}c_1-b_1\\c_2-b_2) \end{bmatrix}=\begin{bmatrix}c_1\\c_2-1 \end{bmatrix}$$

are linearly dependent. This means $\exists\;(\lambda,\mu)\neq (0,0)$ so that 

$\lambda \cdot \overrightarrow{AC}\;=\;\mu \cdot \overrightarrow{BC}$ $\Leftrightarrow \lambda \begin{bmatrix}c_1-1\\c_2 \end{bmatrix}=\mu \begin{bmatrix}c_1\\c_2-1 \end{bmatrix}\Leftrightarrow$

 $\begin{cases}-\lambda+(\lambda-\mu)c_1=0\\\mu+(\lambda-\mu)c_2=0 \end{cases} \tag{1}$

If $\lambda=\mu$, then we get from (1), $-\lambda=\mu=0$, case excluded. If $\lambda \neq \mu$, then

$$c_1=\frac{\lambda}{\lambda-\mu}\;,\;c_2=\frac{-\mu}{\lambda-\mu}.$$

Designate $\alpha :=\frac{\lambda}{\lambda-\mu}\;\Rightarrow\;c_2=1-\alpha$, so points "colinear" with $A,\;B$ are
$$C=(\alpha,1-\alpha,1-2\alpha+2\alpha ^2) \tag{2}$$

which is expressed in the frame $\mathfrak{R}$:  $C(\alpha -1, 1-\alpha).$

In a particular case, we get the "middle" of the bipoint $\{A,B\}$ which is in the frame $\mathfrak{R}$:

$$\frac{1}{2}A+\frac{1}{2}B=\left (-\frac{1}{2},\frac{1}{2}\right )$$

to which corresponds thepoint $$C=\left (-\frac{1}{2}+1, \frac{1}{2},(-\frac{1}{2})^2+(\frac{1}{2})^2+2(-\frac{1}{2})+1\right )=C(\frac{1}{2},\frac{1}{2},\frac{1}{2}) \in \mathfrak{P}.$$

 

          We can obtain from the points (2) the equation of the "line" $AB$.

          In the frame $\mathfrak{R}$ it is   "x+y=0".(Attention, here x and y are the coordinates in $\mathfrak{R}$ of a generic point M(x,y).)

           In the frame $Oxyz$, we need to remove the parameter $\alpha$ from (2)
$$\begin{cases}x=\alpha\\y=1-\alpha\\z=1-2\alpha+2\alpha^2 .\end{cases}$$

We obtain the equivalent equations

$$\begin{cases}x+y=1\\x^2+y^2=z\end{cases}$$

that is, the intersection of paraboloid $\mathfrak{P}$ with the vertical plane $x+y=1$. 

If we choose in the $xOy$ plan the frame with the origin in $C'(\frac{1}{2},\frac{1}{2})$(projection of the midle of bipunct {A,B} above) and the "X" axis along the line $x+y=1$ we obtain(avoiding certain not too heavy calculations) the change of coordinates

$$\begin{cases}Y=z\\X=\frac{1}{\sqrt{2}}(1-2x) \end{cases}$$

in which the equation of the "line AB" is written $Y=\frac{1}{2}+X^2$ that is a parabola.

$$\blacksquare$$ 

           

 

 

 

 

      

 

 

 
 

 

ГЕОМЕТРИЯ Лобачевского

 Autor: Л. С. Атанасян

Titlu: ГЕОМЕТРИЯ Лобачевского
2-е издание, исправленное
(электронное)

 Editura : Москва
БИНОМ. Лаборатория знаний
2014

Descriere:

Эта книга выгодно отличается от других пособий по геометрии Лобачевского. Материал излагается на основе школьной аксиоматики абсолютной геометрии и аксиомы Лобачевского. Первая часть книги посвящена планиметрии Лобачевского, а вторая — стереометрии. В конце каждой главы даются задачи, в конце книги — ответы и указания к ним.

Cui se adreseaza:

Книга может с успехом использоваться студентами и преподавателями и физико-математических факультетов университетов, и педагогических вузов. Она также будет полезна учителям классов с углубленным изучением математики, для индивидуальной работы с учениками, интересующимися математикой.

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The Vieta-Jumping method. Three problems of number theory

      An article that sets out the method can be found here. It was published by YIMIN GE in the journal Mathematical Reflections, 5, (2007). See also here. And here for a book.

 

          PROBLEM #1 (IMO 1988, Problem 6) 

          Let $a,b$ be positive integers so that $ab+1$ divides $a^2+b^2$. Prove that $\frac{a^2+b^2}{ab+1}$ is a perfect square.

          PROBLEM #2

          Let $x,y$ be positive integers so that $xy$ divides $x^2+y^2+1$. Prove that $$\frac{x^2+y^2+1}{xy}=3.$$

 

           PROBLEM #3 (IMO 2007, Problem 5)

           Let $a,b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.

          

 

 

 

Problem med 7-gon diagonaler

RELATIONSHIP BETWEEN DIAGONALS in the REGULAR HEPTAGON        

           There is no respected geometry book that does not have the following problem:

 

See also

(In the book "Problems of Geometry" by Tzitzeica is problem 717.) This is the solution

     Another book also presents a personal solution of the authors (some academics)

In "Solution II" above, as in the Problem 717 solution, a Theorem of PTOLEMEU is used. Without this theorem, trying only reasonings with similarities of triangles, the authors tried in "Solutia I".

     The notations in the figure above correspond to this solution. First of all, notice the congruence of the angles:

$$\angle BAC \equiv \angle CAD \equiv \angle CDB \equiv \angle ADB =\frac{\pi}{7},$$

$$\angle ABD \equiv \angle ACD =\frac{4\pi}{7}.$$

Note now that the quadrilateral $BCDN$ is rhombus ($CD \parallel BE\;,BC \parallel AD\;$and $DB$ bisect $\angle CDN$). It follows from here

$$\angle DNM = \angle DCM=\frac{4\pi}{7}\;,\;ND=BC=a.$$ 

      We have the similarity of triangles $\Delta ABM \sim \Delta ACD$ where do we get

$\frac{BM}{CD}=\frac{AM}{AD}=\frac{AB}{AC}\Leftrightarrow$

$$\frac{BM}{a}=\frac{AM}{c}=\frac{a}{b}. \tag{1}$$

 As the third angle in a triangle, we have $\angle DMN \equiv \angle DMC=\frac{2\pi}{7}$ so $\Delta DMN \equiv \Delta DMC$, so $ND=DC=a$. 

       Further $AM=AN\;$ (from angle calculations we have $\angle ANM=\pi-\angle DNM=\pi-\frac{4\pi}{7}=\frac{3\pi}{7},\;\angle AMN=\pi-2\cdot \frac{2\pi}{7}=\frac{3\pi}{7}$) so 

$AM=AN=AD-ND=c-a$. Then the second equation of (1) is written

$$\frac{c-a}{c}=\frac{a}{b}$$

which immediately results the relationship $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$.

$\blacksquare$ 

           A demonstration without words is here. It is a true demonstration without words.

          Let's reproduce the hexagon, starting with the side $CD$. Obviously we have $CD \parallel IK$.

          We observe from angle calculations that the points $G,D,K$ and $G,C,I$ are respectively collinear: $\angle GCD+\angle DCI=\frac{3\pi}{7}+\frac{4\pi}{7}=\pi$. Obviously we have the similarity indicated on the figure, hence the conclusion.

 

$\blacksquare\;\blacksquare$

 

A nonomogeneous Conditioned Algebraic Identity

                In Romanian Journal of Mathematics "Gazeta Matematica - Seria B", no 2/2022,  on page 60, there is an article on Problem 28 003, which appeared in GMB 2/2021 (page 107) and was solved in GMB 9/2021 (page 404). Here we demonstrate LEMA 4, which the author states "results by direct calculation"

In translation, for which we thank DeepL, reads as follows;

         " For any positive numbers $a$, $b$, $c$ there is equivalence

$$a^2+b^2+c^2+abc=4\;\;\Leftrightarrow\;\;\frac{a}{2a+bc}+\frac{b}{2b+ca}+\frac{c}{2c+ab}=1." \tag{E}$$

     

     Solution CiP

               To obtain homogeneous algebraic expressions, we use the notation $abc :=d^2$. Then the condition on the left-hand side of the (E) is expressed

$$a^2+b^2+c^2+d^2=4. \tag{1}$$

Further we have $ab=\frac{d^2}{c}$ and the analogues, so the expression on the right-hand side of (E) is

$$R:=\frac{a}{2a+bc}+\dots \;=\frac{a}{2a+\frac{d^2}{a}}+\frac{b}{2b+\frac{d^2}{b}}+\frac{c}{2c+\frac{d^2}{c}}$$

so

$$R=\frac{a^2}{d^2+2a^2}+\frac{b^2}{d^2+2b^2}+\frac{c^2}{d^2+2c^2} \tag{2}.$$

Both (1) and (2) are now homogeneous expressions. (1) is  symmetric, while (2) is symmetric only with respect to $\{a,b,c\}$, so we will consider $d$ as a parameter.

          Let's calculate the  expression $R$ by bringing it in (2) to the same common denominator. The calculation will push itself forward. We'll substitute everywhere possible $abc \rightarrow d;\;a^2b^2c^2 \rightarrow \;d^2$.

     The denominator of $R$ is

$$Q:=(2a^2+d^2)(2b^2+d^2)(2c^2+d^2)=$$

$=(4a^2b^2+2a^2d^2+2b^2d^2+d^4)(2c^2+d^2)=8a^2b^2c^2+4a^2c^2d^2+\dots +2b^2d^4+d^6.$

Finally we get

$$Q=d^2[d^4+2(a^2+b^2+c^2+4)d^2+4(a^2b^2+b^2c^2+c^2a^2)].$$

     The numerator of $R$ is, more thickly

$$P:=a^2(2b^2+d^2)(2c^2+d^2)+b^2(2c^2+d^2)(2a^2+d^2)+c^2(2a^2+d^2)(2b^2+d^2)$$

but in the end we find

$$Q=d^2[(a^2+b^2+c^2+12)d^2+4(a^2b^2+b^2c^2+c^2a^2)].$$

In conclusion

$$R=\frac{P}{Q}=\frac{(a^2+b^2+c^2+12)d^2+4(a^2b^2+b^2c^2+c^2a^2)}{d^4+2(a^2+b^2+c^2+4)d^2+4(a^2b^2+b^2c^2+c^2a^2)}.\;\tag{3}$$

From relation (3) we obtain that (where the sum $\sum$ extends to $a,b,c$)

$$R=1\;\Leftrightarrow\;(\sum a^2+12)d^2=d^4+2(\sum a^2+4)d^2$$
 $$\Leftrightarrow\;\sum a^2+12=d^2+2\sum a^2+8$$

$$\Leftrightarrow\;\;\sum a^2+d^2=4$$

so we got precisely condition (1). This demonstrates the required equivalence.

$\blacksquare$

          REMARK

          As Mr. Leo Giugiuc stated, in a personal message, the implication $\Rightarrow$ in (E) can be proved using trigonometric substitutions.

     Indeed, from $a^2+b^2+c^2+abc=4$ it result $\;a^2,b^2,c^2\; \leq 4$, so let's put

$$a=2\sin \alpha,\;b=2 \sin \beta,\;\;\alpha, \beta\;\in (0, \frac{\pi}{2}].$$

Then $c$ is the positive solution of the second degree equation

$$c^2+4c \sin \alpha \cdot \sin \beta+4\sin^2 \alpha+4\sin^2 \beta-4=0,$$

for which we have $\Delta^{'}_c=4\sin^2\alpha \cdot \sin^2 \beta -4\sin^2 \alpha-4\sin^2 \beta+4=4(1-\sin^2\alpha)(1-\sin^2 \beta)=4 \cos^2 \alpha \cdot\cos^2 \beta),$

$$c=\frac{-2 \sin \alpha \cdot \sin \beta +\sqrt{\Delta^{'}_c}}{1}=2\cos \alpha \cdot \cos \beta - 2 \sin \alpha \cdot \sin \beta=2\cos (\alpha+\beta).$$

     Further we have $\frac{a}{2a+bc}=\frac{2\sin \alpha}{4\sin \alpha+2\sin \beta \cdot 2\cos(\alpha+\beta)}=\frac{\sin \alpha}{2(\sin\alpha+\sin \beta \cdot \cos(\alpha+\beta))}=$

$=\frac{\sin \alpha}{2(\sin \alpha +\cos \alpha \cdot \cos \beta \cdot \sin \beta-\sin \alpha \cdot \sin^2 \beta)}=\frac{\sin \alpha}{2(\cos \alpha \cdot \cos \beta \cdot \sin \beta+\sin \alpha(1-\sin^2 \beta))}=\frac{\sin \alpha}{2\cos \beta(\cos \alpha \cdot \sin \beta+\sin \alpha \cdot \cos \beta}.$

So, repeating the same kind of calculations, we got 

$$\frac{a}{2a+bc}=\frac{\sin \alpha}{2\cos \beta \cdot \sin (\alpha+\beta)},\;\frac{b}{2b+ac}=\frac{\sin \beta}{2\cos \alpha \cdot \sin(\alpha+\beta)},\;\frac{c}{2c+ab}=\frac{\cos(\alpha+\beta)}{2\cos\alpha \cdot \cos \beta}.$$

     Then

$R=\frac{\sin \alpha}{2\cos \beta \cdot \sin(\alpha+\beta)}+\frac{\sin \beta}{2 \cos \alpha \cdot \sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{\sin \alpha \cdot \cos \alpha+\sin \beta \cdot \cos \beta}{2\cos \alpha \cdot \cos \beta \cdot \sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=$

$=\frac{\sin 2\alpha+\sin 2\beta}{4\cos \alpha \cdot \cos \beta \cdot \sin(\alpha+\beta) }+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{2\sin (\alpha+\beta)\cdot \cos(\alpha-\beta)}{4\cos \alpha \cdot\cos \beta \cdot \sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{\cos(\alpha-\beta)}{2\cos \alpha \cdot \cos \beta}+\frac{\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=$

$=\frac{\cos(\alpha -\beta)+\cos(\alpha+\beta)}{2\cos \alpha \cdot \cos \beta}=\frac{2\cos \alpha \cdot \cos \beta}{2\cos \alpha \cdot \cos\beta}=1$.

END REM

BIRO ISTVAN makes substitutions $x=\frac{bc}{a},\;y=\frac{ca}{b},\;z=\frac{ab}{c}$ and has a simpler calculation to make.

LEO GIUGIUC points to another inequality that starts from the same hypothesis.

Kunihiko Chikaya makes an equally difficult calculation but proves both implications.

 

GAZETA MATEMATICĂ Seria B N0 2/2022

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SUPLIMENTUL cu EXERCIȚII al GMB N0 2/2022

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DIDACTICA MATEMATICĂ N0 2/2021

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