miercuri, 8 iunie 2022
sâmbătă, 4 iunie 2022
Opublikowany problem matematyczny... za wcześnie // A published math problem ... too early
This is problem S:E22.89 from the GMB exercise supplement, from March 2022 (page 5; proposed for 5th grade).
Translated (thanks to DeepL):
"Let the natural numbers $\;a,\;b,\;c,\;x,\;y,\;z,\;$ be greater than $1$ such that
$$a^b\;=\;c,\;\;c^x\;=\;y,\;\;y^z\;=\;a^{2025}.$$
a) Calculate the multiplication $\;b\cdot x\cdot z\;.$
b) Give an example of numbers that check all the above conditions."
It is customary in the current year, for many problems, to contain the number 2022, as a kind of parameter, that can be in the problem any number. The author, Mr MUNTEANU Emanuel George, rushed three years earlier to publish this problem. Or is it "because" of the elegant decomposition of the number 2025 into prime factors : $2025=3^4 \cdot 5^2$ ?
Answer CiP
a) $b \cdot x\cdot z\;=\;2025;$
b) $a=7,\;b=15,\;c=7^{15},\;x=3,\;y=7^{45},\;z=45.$
Indeed, $a^b=7^{15}=c,\;c^x=(7^{15})^3=7^{45}=y,\;y^z=(7^{45})^{45}=7^{45\cdot 45}=7^{2025}=a^{2025}.$
A more general example is of the form:
$a\geqslant 2\;$ fixed natural number;
$z=3^{\alpha}\cdot 5^{\beta},\;\;0 \leqslant \alpha \leqslant 4,\;0 \leqslant \beta \leqslant 2$ but $(\alpha,\beta) \neq (0,0),\;(4,2)$ ; a divisor of the number 2025 ;
$b=3^{\gamma}\cdot 5^{\delta},\;\;0 \leqslant \gamma \leqslant 4-\alpha,\;0 \leqslant \delta \leqslant 2- \beta,$ but not to have both $\gamma =4-\alpha, \delta =2-\beta$; a divisor of the number $\frac{2025}{z};$
$x=3^{4-\alpha-\gamma} \cdot 5^{2-\beta-\delta};$
$y=a^{3^{4-\alpha} \cdot 5^{2-\beta}}.$
Solution CiP
a) $a^{2025}\underset{def}{=}\;y^z \;\underset{y=c^x}{=}\;\left ( c^x \right )^z=c^{x \cdot z}\;\underset{c=a^b}{=}\left ( a^b \right )^{x \cdot z}=a^{b \cdot x \cdot z}$, and from this it follows
$$b \cdot x \cdot z\;=\;2025. \tag{1}$$
b) We see from (1) that $b,x,z$ are divisors of 2025. We must have
$$z=3^{\alpha} \cdot 5^{\beta} \;,\tag{2}$$
with $(\alpha,\beta) \neq (0,0)$ since $z \neq 1$, and $(\alpha,\beta) \neq (4,2)$ since $b,x > 1$.
Beacause $b\cdot x=\frac{2025}{z}=3^{4-\alpha}\cdot 5^{2-\beta}$, we must have
$$b=3^{\gamma}\cdot 5^{\delta}$$
and we choose $\gamma,\; \delta$ so that $b$ and $x$ are greater than 1.
The base $a$ can be any natural number $\geqslant 2$.
In the concrete example we chose $a=7,\; \alpha =2,\; \beta=1,\;\gamma=\delta=1$.
We could count all possible cases for a given $a$, but we won't do it here.
$\blacksquare$
vineri, 3 iunie 2022
joi, 2 iunie 2022
SUPLIMENTUL cu EXERCIȚII al GMB N0 6/2009
Am rezolvat TOATE problemele S:E09.203-242, pe undeva pe niste foi....
La problema S:E09.220 probabil ca a fost o...ERATA...; am insemnat cu creionul.
Vezi in DRIVE
miercuri, 25 mai 2022
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.
$$\blacksquare$$
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