| |
| Autor |
 * Gleichseitige Dreiecke und Gerade |
|
Wario
Aktiv 
Dabei seit: 01.05.2020
Mitteilungen: 326
 |
Es seien $n\geq 2$ gleichseitige Dreiecke der Kantenlänge $s$ wie in folgender Abbildung nebeneinander angeordnet, die, wie gezeigt, von einer Geraden geschnitten werden:
![<math>
\def\allgm{1}
\def\bew{0}
\pagecolor{black!1}
% Gegebene Größen
\pgfmathsetmacro{\N}{5}
\pgfmathsetmacro{\s}{2}
\colorlet{unten}{orange!55}
\colorlet{oben}{yellow!55}
%\ifnum\bew=1
\pgfmathsetmacro{\N}{\bew==1 ? 7 : 6}
%\fi
\pgfmathsetmacro{\Ns}{int(\N+1)}
\pgfmathsetmacro{\Np}{int(\N-1)}
% Annotation - Kanten
\ifnum\bew=1
\tikzset{kante/.style={text=black} }
\else
\tikzset{kante/.style={text=black!1} }
\fi
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle,
Dreieck/.style={},
every label/.style={text=black!1},
]
\foreach[count=\nI] \n in {1,...,\N}{
\begin{scope}[xshift=-\nI*\s cm]
\coordinate[label=below:$A\n$] (A\n) at (0,0);
\coordinate[label=below:$B\n$] (B\n) at (\s,0);
\coordinate[label=$C\n$] (C\n) at (60:\s);
% unten füllen
\draw[Dreieck, name path global=dreieck\n] (A\n) -- (B\n) -- (C\n) --cycle;
% Annotation unten
\ifnum\bew=1
\path[] (A\n) -- (B\n) node[midway, below]{$s$};
\fi
\end{scope}
}
% Annotation rechts
\ifnum\bew=1
\path[] (B1) -- (C1) node[midway, right]{$s$};
\fi
% Gerade 1/2
\path[name path global=gerade] (C1) -- (A\N);
\foreach[count=\kNo] \k in {2,...,\N}{
\pgfmathtruncatemacro\K{\k+1}
\path[name intersections={of={dreieck\k} and gerade, name=Point}];
\ifnum\k=\N
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-2);
\coordinate[label=Y\k] (Y\k) at (Point-1);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$s$ %$y_{\k}$
};
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-1);
\coordinate[label=Y\k] (Y\k) at (Point-2);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$y_{\k}$};
\end{pgfonlayer}
\fi
% oben füllen
\ifnum\bew=1
\begin{pgfonlayer}{background}
\fill[oben] (X\k) -- (Y\k) -- (C\k) --cycle;
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\fill[unten] (X\k) -- (B\k) -- (A\k) -- (Y\k) --cycle;
\fill[unten] (A1) -- (B1) -- (C1) --cycle;
\end{pgfonlayer}
\fi
}
% Gerade 2/2
\draw[Dreieck] (C1) -- (A\N);
% Beweis ========================
% Annotation Strahlensatzfigur
\ifnum\bew=1%==========================
\draw[densely dashed] (C1) -- (C\N) -- +(-\s,0) coordinate[label=$C\Ns$] (C\Ns) node[midway, above]{$$} -- (A\N) node[midway, left]{$s$};
%
% Annotation oben
\foreach[evaluate={\K=int(\k+1)}] \k in {1,...,\N}{
\pgfmathsetmacro\text{\k==3 || \k==5 ? "" : "s"}
\path[] (C\k) -- (C\K) node[midway, above]{$\text$};
}
\foreach \k/\text in {\Ns/{n+1},\N/{n},\Np/{n-1}}{
\node[above] at (C\k) {$P_{\text}$};
}
\foreach \k in {1,2,3}{
\node[above] at (C\k) {$P_{\k}$};
}
\fi%============================
% Allgemeine Zeichnung
\ifnum\allgm=1
\pgfmathtruncatemacro{\Na}{\N-2}
\pgfmathtruncatemacro{\Nb}{3}
\fill[black!1] ([shift={(0,-2em)}]A\Na) rectangle ([shift={(-0.5*\s cm,2em)}]C\Nb) node[midway, text=black, font=\Huge]{$\boldsymbol\dots$};
\fi
\end{tikzpicture}
</math>](/matheplanet/nuke/html/latexrender/pictures/6ae4734c2c879a887e3863e41a849fdf.png "\def\allgm{1}
\def\bew{0}
\pagecolor{black!1}
% Gegebene Größen
\pgfmathsetmacro{\N}{5}
\pgfmathsetmacro{\s}{2}
\colorlet{unten}{orange!55}
\colorlet{oben}{yellow!55}
%\ifnum\bew=1
\pgfmathsetmacro{\N}{\bew==1 ? 7 : 6}
%\fi
\pgfmathsetmacro{\Ns}{int(\N+1)}
\pgfmathsetmacro{\Np}{int(\N-1)}
% Annotation - Kanten
\ifnum\bew=1
\tikzset{kante/.style={text=black} }
\else
\tikzset{kante/.style={text=black!1} }
\fi
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle,
Dreieck/.style={},
every label/.style={text=black!1},
]
\foreach[count=\nI] \n in {1,...,\N}{
\begin{scope}[xshift=-\nI*\s cm]
\coordinate[label=below:$A\n$] (A\n) at (0,0);
\coordinate[label=below:$B\n$] (B\n) at (\s,0);
\coordinate[label=$C\n$] (C\n) at (60:\s);
% unten füllen
\draw[Dreieck, name path global=dreieck\n] (A\n) -- (B\n) -- (C\n) --cycle;
% Annotation unten
\ifnum\bew=1
\path[] (A\n) -- (B\n) node[midway, below]{$s$};
\fi
\end{scope}
}
% Annotation rechts
\ifnum\bew=1
\path[] (B1) -- (C1) node[midway, right]{$s$};
\fi
% Gerade 1/2
\path[name path global=gerade] (C1) -- (A\N);
\foreach[count=\kNo] \k in {2,...,\N}{
\pgfmathtruncatemacro\K{\k+1}
\path[name intersections={of={dreieck\k} and gerade, name=Point}];
\ifnum\k=\N
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-2);
\coordinate[label=Y\k] (Y\k) at (Point-1);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$s$ %$y_{\k}$
};
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-1);
\coordinate[label=Y\k] (Y\k) at (Point-2);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$y_{\k}$};
\end{pgfonlayer}
\fi
% oben füllen
\ifnum\bew=1
\begin{pgfonlayer}{background}
\fill[oben] (X\k) -- (Y\k) -- (C\k) --cycle;
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\fill[unten] (X\k) -- (B\k) -- (A\k) -- (Y\k) --cycle;
\fill[unten] (A1) -- (B1) -- (C1) --cycle;
\end{pgfonlayer}
\fi
}
% Gerade 2/2
\draw[Dreieck] (C1) -- (A\N);
% Beweis ========================
% Annotation Strahlensatzfigur
\ifnum\bew=1%==========================
\draw[densely dashed] (C1) -- (C\N) -- +(-\s,0) coordinate[label=$C\Ns$] (C\Ns) node[midway, above]{$$} -- (A\N) node[midway, left]{$s$};
%
% Annotation oben
\foreach[evaluate={\K=int(\k+1)}] \k in {1,...,\N}{
\pgfmathsetmacro\text{\k==3 || \k==5 ? \"\" : \"s\"}
\path[] (C\k) -- (C\K) node[midway, above]{$\text$};
}
\foreach \k/\text in {\Ns/{n+1},\N/{n},\Np/{n-1}}{
\node[above] at (C\k) {$P_{\text}$};
}
\foreach \k in {1,2,3}{
\node[above] at (C\k) {$P_{\k}$};
}
\fi%============================
% Allgemeine Zeichnung
\ifnum\allgm=1
\pgfmathtruncatemacro{\Na}{\N-2}
\pgfmathtruncatemacro{\Nb}{3}
\fill[black!1] ([shift={(0,-2em)}]A\Na) rectangle ([shift={(-0.5*\s cm,2em)}]C\Nb) node[midway, text=black, font=\Huge]{$\boldsymbol\dots$};
\fi
\end{tikzpicture}")
Berechne die hervorgehobene Gesamtfläche.
Vollständige Lösungen der Rätselaufgabe hier im Hide-Rahmen.
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Der Themensteller hat bestimmt, dass Du Lösungen oder Beiträge zur Lösung direkt im Forum posten darfst.
Bei dieser Aufgabe kann ein öffentlicher Austausch über Lösungen, Lösungswege und Ansätze erfolgen. Hier musst Du keine private Nachricht schreiben! |
Wario
Aktiv 
Dabei seit: 01.05.2020
Mitteilungen: 326
|      Beitrag No.1, vom Themenstarter, eingetragen 2021-01-13
|
Zum Vergleich: Ist $n=10$ und die Fläche eines gleichseitigen Dreiecks gleich $6\text{cm}^2$, so ist die gesuchte Fläche gleich $41\text{cm}^2.$
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Squire
Senior
Dabei seit: 18.08.2015
Mitteilungen: 691
 | Beitrag No.2, eingetragen 2021-01-13
|
Wir berechnen zunächst die Summe der weißen Dreiecke.
Nach dem Strahlensatz sind die rechten Seiten der weißen Dreiecke
$s-\frac{s}{n},s-\frac{2s}{n},...s-\frac{(n-1)s}{n}$
und die linken Seiten der weißen Dreiecke
$s,s-\frac{s}{n-1},s-\frac{2s}{n-1},...s-\frac{(n-2)s}{n-1}$.
Die Flächensumme der weißen Dreiecke ist daher (Winkel an der Spitze ist immer 60°):
$\frac{1}{2}s(s-\frac{s}{n})\frac{\sqrt{3}}{2}+\frac{1}{2}(s-\frac{s}{n-1})(s-\frac{2s}{n})\frac{\sqrt{3}}{2}+...+\frac{1}{2}(s-\frac{(n-2)s}{n-1})(s-\frac{(n-1)s}{n})\frac{\sqrt{3}}{2}=$
$=\frac{s^2\sqrt{3}}{4}\cdot((1-\frac{1}{n})+(1-\frac{1}{n-1})(1-\frac{2}{n})+...+(1-\frac{n-2}{n-1})(1-\frac{n-1}{n})=$
$=\frac{s^2\sqrt{3}}{4}\cdot\sum_{k=1}^{n-1}(1-\frac{k-1}{n-1})(1-\frac{k}{n})=\frac{s^2\sqrt{3}}{4}\cdot(\frac{n}{3}-\frac{1}{6})$
Die orange Flächensumme ist schließlich
$\frac{s^2\sqrt{3}}{4}\cdot n-\frac{s^2\sqrt{3}}{4}\cdot(\frac{n}{3}-\frac{1}{6})=\frac{s^2\sqrt{3}}{4}\cdot(\frac{2n}{3}+\frac{1}{6})$
Grüße Squire
[Die Antwort wurde vor Beitrag No.1 begonnen.]
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haribo
Senior
Dabei seit: 25.10.2012
Mitteilungen: 2718
| Beitrag No.3, eingetragen 2021-01-13
|
schöne aufgabe,
ich komme für die hervorgehobene Gesamtfläche auf:
  
\ A*(4n+1)/6 oder noch etwas schöner A* (2/3 n+1/6) bei A=6 und n=10 also auch auf 6*(4*10+1)/6= 6*41/6= 41 auch ich ziehe die weissen teilflächen von n*A ab, die kleinste dreiecksfläche hat die fläche: A/(n*(n-1)) die zweitkleinste ist 2^2 mal grösser die drittkleinste 3^2... drum multipliziere ich für die abzugsfläche das kleinste weisse dreieck mit der summe aller quadratzahlen bis n-1: (n*(n-1)*(2n-1))/6 also gesuchte fläche =n*A-A/(n*(n-1))*(n*(n-1)*(2n-1))/6 gekürzt, ausgeklammert und umgestellt s.o.
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cramilu
Aktiv
Dabei seit: 09.06.2019
Mitteilungen: 683
Herkunft: Bierfranken
 | Beitrag No.4, eingetragen 2021-01-13
|
Ich komme [ebenfalls] auf...
\(A\: =\: s^2\,\cdot\,\frac{(4n+1)\,\cdot\,\sqrt{3}}{24}\)
Dazu habe ich Warios Grafik oben um eine Verbindungslinie
zwischen den Dreiecksspitzen erweitert.
Als "Gesamtfigur" ergibt sich ein gleichschenkliges Trapez.
Dann habe ich meine Betrachtung auf das linke Parallelogramm
eingeengt, welches zunächst in \((n-1)\) gleiche Rauten zerfällt.
Letztere werden durch die Schräge derart geteilt,
dass außer zwei gleichen Kolonnen von \((n-1)\) Dreiecksstümpfen
auch \((n-1)\) "Strahlensatzfiguren" entstehen - mit nach Strahlensatz
gleichmäßig wachsenden bzw. schrumpfenden parallelen Seiten und
Höhenlinien durch den jeweiligen "Kreuzungspunkt"...
Leider habe ich es mir hernach unnötig kompliziert gemacht,
weil ich zunächst den Flächeninhalt \(A_P\) des Parallelogrammes
ermittelt habe, dann selbigen halbiert und erst am Ende den \(A_K\)
der "Kolonne" von Strahlensatzdreiecken abgezogen...
Die genaue Rechnerei liefere ich bei Gelegenheit nach;
"zwischendurch" benötigt man die Summenformeln nach Gauß
und jene nach ?Brahmagupta? (Quadratzahl-Summe)...
... und dann Obacht beim Verrechnen[!] und "Kürzen" der \(n\)! 🙄
-----------------
ADMIRATIONIS SUI SATISFACTIONIS SACRA SITIS
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Wario
Aktiv
Dabei seit: 01.05.2020
Mitteilungen: 326
| Beitrag No.5, vom Themenstarter, eingetragen 2021-01-14
|
2021-01-13 04:47 - Wario im Themenstart schreibt:
Es seien $n\geq 2$ gleichseitige Dreiecke der Kantenlänge $s$ wie in folgender Abbildung nebeneinander angeordnet, die, wie gezeigt, von einer Geraden geschnitten werden:
Berechne die hervorgehobene Gesamtfläche.
Lösung.
![<math>
\def\allgm{1}
\def\bew{1}
\pagecolor{black!1}
% Gegebene Größen
\pgfmathsetmacro{\N}{7}
\pgfmathsetmacro{\s}{2}
\colorlet{unten}{orange!66}
\colorlet{oben}{yellow!66}
\pgfmathsetmacro{\N}{\bew==1 ? 7 : \N}
\pgfmathsetmacro{\Ns}{int(\N+1)}
\pgfmathsetmacro{\Np}{int(\N-1)}
% Annotation - Kanten
\ifnum\bew=1
\tikzset{kante/.style={text=black} }
\else
\tikzset{kante/.style={text=black!1} }
\fi
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle,
Dreieck/.style={},
every label/.style={text=black!1},
]
\foreach[count=\nI] \n in {1,...,\N}{
\begin{scope}[xshift=-\nI*\s cm]
\coordinate[label=below:$A\n$] (A\n) at (0,0);
\coordinate[label=below:$B\n$] (B\n) at (\s,0);
\coordinate[label=$C\n$] (C\n) at (60:\s);
% unten füllen
\draw[Dreieck, name path global=dreieck\n] (A\n) -- (B\n) -- (C\n) --cycle;
% Annotation unten
\ifnum\bew=1
\path[] (A\n) -- (B\n) node[midway, below]{$s$};
\fi
\end{scope}
}
% Annotation rechts
\ifnum\bew=1
\path[] (B1) -- (C1) node[midway, right]{$s$};
\fi
% Gerade 1/2
\path[name path global=gerade] (C1) -- (A\N);
\foreach[count=\kNo] \k in {2,...,\N}{
\pgfmathtruncatemacro\K{\k+1}
\path[name intersections={of={dreieck\k} and gerade, name=Point}];
\ifnum\k=\N
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-2);
\coordinate[label=Y\k] (Y\k) at (Point-1);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{n}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$s$ %$y_{\k}$
};
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-1);
\coordinate[label=Y\k] (Y\k) at (Point-2);
\ifnum\k=\Np
\path[] (X\k) -- (C\k) node[kante, pos=0.7, right]{$x_{n-1}$};
\path[] (Y\k) -- (C\k) node[kante, pos=0.7, left]{$y_{n-1}$};
\else
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$y_{\k}$};
\fi
\end{pgfonlayer}
\fi
% oben füllen
\ifnum\bew=1
\begin{pgfonlayer}{background}
\fill[oben] (X\k) -- (Y\k) -- (C\k) --cycle;
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\fill[unten] (X\k) -- (B\k) -- (A\k) -- (Y\k) --cycle;
\fill[unten] (A1) -- (B1) -- (C1) --cycle;
\end{pgfonlayer}
\fi
}
% Gerade 2/2
\draw[Dreieck] (C1) -- (A\N);
% Beweis ========================
% Annotation Strahlensatzfigur
\ifnum\bew=1%==========================
\draw[densely dashed] (C1) -- (C\N) -- +(-\s,0) coordinate[label=$C\Ns$] (C\Ns) node[midway, above]{$$} -- (A\N) node[midway, left]{$s$};
%
% Annotation oben
\foreach[evaluate={\K=int(\k+1)}] \k in {1,...,\N}{
\pgfmathsetmacro\text{\k==3 || \k==5 ? "" : "s"}
\path[] (C\k) -- (C\K) node[midway, above]{$\text$};
}
\foreach \k/\text in {\Ns/{n+1},\N/{n},\Np/{n-1}}{
\node[above] at (C\k) {$P_{\text}$};
}
\foreach \k in {1,2,3}{
\node[above] at (C\k) {$P_{\k}$};
}
\fi%============================
% Allgemeine Zeichnung
\ifnum\allgm=1%=====================
\pgfmathtruncatemacro{\Na}{\N-2}
\pgfmathtruncatemacro{\Nb}{3}
\fill[black!1] ([shift={(0,-2em)}]A\Na) rectangle ([shift={(-0.5*\s cm,2em)}]C\Nb) node[midway, text=black, font=\Huge]{$\boldsymbol\dots$};
\fi%=====================
\end{tikzpicture}
</math>](/matheplanet/nuke/html/latexrender/pictures/916d2e44b3976307221250d805ab0805.png "\def\allgm{1}
\def\bew{1}
\pagecolor{black!1}
% Gegebene Größen
\pgfmathsetmacro{\N}{7}
\pgfmathsetmacro{\s}{2}
\colorlet{unten}{orange!66}
\colorlet{oben}{yellow!66}
\pgfmathsetmacro{\N}{\bew==1 ? 7 : \N}
\pgfmathsetmacro{\Ns}{int(\N+1)}
\pgfmathsetmacro{\Np}{int(\N-1)}
% Annotation - Kanten
\ifnum\bew=1
\tikzset{kante/.style={text=black} }
\else
\tikzset{kante/.style={text=black!1} }
\fi
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle,
Dreieck/.style={},
every label/.style={text=black!1},
]
\foreach[count=\nI] \n in {1,...,\N}{
\begin{scope}[xshift=-\nI*\s cm]
\coordinate[label=below:$A\n$] (A\n) at (0,0);
\coordinate[label=below:$B\n$] (B\n) at (\s,0);
\coordinate[label=$C\n$] (C\n) at (60:\s);
% unten füllen
\draw[Dreieck, name path global=dreieck\n] (A\n) -- (B\n) -- (C\n) --cycle;
% Annotation unten
\ifnum\bew=1
\path[] (A\n) -- (B\n) node[midway, below]{$s$};
\fi
\end{scope}
}
% Annotation rechts
\ifnum\bew=1
\path[] (B1) -- (C1) node[midway, right]{$s$};
\fi
% Gerade 1/2
\path[name path global=gerade] (C1) -- (A\N);
\foreach[count=\kNo] \k in {2,...,\N}{
\pgfmathtruncatemacro\K{\k+1}
\path[name intersections={of={dreieck\k} and gerade, name=Point}];
\ifnum\k=\N
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-2);
\coordinate[label=Y\k] (Y\k) at (Point-1);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{n}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$s$ %$y_{\k}$
};
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-1);
\coordinate[label=Y\k] (Y\k) at (Point-2);
\ifnum\k=\Np
\path[] (X\k) -- (C\k) node[kante, pos=0.7, right]{$x_{n-1}$};
\path[] (Y\k) -- (C\k) node[kante, pos=0.7, left]{$y_{n-1}$};
\else
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$y_{\k}$};
\fi
\end{pgfonlayer}
\fi
% oben füllen
\ifnum\bew=1
\begin{pgfonlayer}{background}
\fill[oben] (X\k) -- (Y\k) -- (C\k) --cycle;
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\fill[unten] (X\k) -- (B\k) -- (A\k) -- (Y\k) --cycle;
\fill[unten] (A1) -- (B1) -- (C1) --cycle;
\end{pgfonlayer}
\fi
}
% Gerade 2/2
\draw[Dreieck] (C1) -- (A\N);
% Beweis ========================
% Annotation Strahlensatzfigur
\ifnum\bew=1%==========================
\draw[densely dashed] (C1) -- (C\N) -- +(-\s,0) coordinate[label=$C\Ns$] (C\Ns) node[midway, above]{$$} -- (A\N) node[midway, left]{$s$};
%
% Annotation oben
\foreach[evaluate={\K=int(\k+1)}] \k in {1,...,\N}{
\pgfmathsetmacro\text{\k==3 || \k==5 ? \"\" : \"s\"}
\path[] (C\k) -- (C\K) node[midway, above]{$\text$};
}
\foreach \k/\text in {\Ns/{n+1},\N/{n},\Np/{n-1}}{
\node[above] at (C\k) {$P_{\text}$};
}
\foreach \k in {1,2,3}{
\node[above] at (C\k) {$P_{\k}$};
}
\fi%============================
% Allgemeine Zeichnung
\ifnum\allgm=1%=====================
\pgfmathtruncatemacro{\Na}{\N-2}
\pgfmathtruncatemacro{\Nb}{3}
\fill[black!1] ([shift={(0,-2em)}]A\Na) rectangle ([shift={(-0.5*\s cm,2em)}]C\Nb) node[midway, text=black, font=\Huge]{$\boldsymbol\dots$};
\fi%=====================
\end{tikzpicture}")
Sei $F=\dfrac12 s^2 \sin(60^\circ)$ die Fläche eines gleichseitigen Dreiecks und seien $F_1, F_2,\dots, F_n$ die Teilschnittflächen oberhalb der Geraden mit den gleichseitigen Dreiecken bei den Punkten $P_1, P_2,\dots, P_n$.
Dann entliest man aus der Strahlensatzfigur mit Scheitelpunkt $P_1$ für die Kanten $x_k$ und $y_k$ (mit $k=1,2,\dots,n$) der Teildreiecke
$\begin{array}{l l l l}
\dfrac{x_2}{s} =\dfrac{s}{n s}
&\Leftrightarrow~
x_2 = \dfrac{s}{n} \\[1em]
\dfrac{y_2}{s} =\dfrac{s}{(n-1) s}
&\Leftrightarrow~
y_2 = \dfrac{s}{n-1}
&\Rightarrow
F_2 =\dfrac12 x_2 y_2 \sin(60^\circ)
=\dfrac12 \dfrac{s}{n} \dfrac{s}{n-1} \sin(60^\circ)
&\Leftrightarrow~ F_2 =\dfrac{F}{n(n-1)} \\[1em]
%
\dfrac{x_3}{s} =\dfrac{2s}{n s}
&\Leftrightarrow~
x_3 = \dfrac{2s}{n} \\[1em]
\dfrac{y_3}{s} =\dfrac{2s}{(n-1) s}
&\Leftrightarrow~
y_3 = \dfrac{2s}{n-1}
&\Rightarrow
F_3 =\dfrac12 x_3 y_3 \sin(60^\circ)
=\dfrac12 \dfrac{2s}{n} \dfrac{2s}{n-1} \sin(60^\circ)
&\Leftrightarrow~ F_3 =\dfrac{2^2 F}{n(n-1)} \\[1em]
%
\hphantom{\dfrac{x_2}{s}}\vdots & \hphantom{\Leftrightarrow~}\vdots
& \hphantom{\Rightarrow~}\vdots & \hphantom{\Leftrightarrow~}\vdots \\[1em]
%
\dfrac{x_n}{s} =\dfrac{(n-1) s}{n s}
&\Leftrightarrow~
x_n = \dfrac{(n-1) s}{n} \\[1em]
\dfrac{y_n}{s} =\dfrac{n s}{n s} =1
&\Leftrightarrow~
y_n = s =\dfrac{(n-1) s}{(n-1)}
&\Rightarrow
F_n =\dfrac12 x_n y_n \sin(60^\circ)
=\dfrac12 \dfrac{(n-1)s}{n} \dfrac{(n-1)s}{n-1} \sin(60^\circ)
&\Leftrightarrow~ F_n =\dfrac{(n-1)^2 F}{n(n-1)} \\[1em]
\end{array}$
Damit ergibt sich für die Summe $F_O$ der oberen Teilschnittflächen
$\begin{array}{l l}
F_O=F_1+F_2+\dots+F_n
&=\dfrac{F}{n(n-1)}\left(1^2+2^2 +\dots+ (n-1)^2 \right) \\[1em]
&=\dfrac{F}{n(n-1)} \dfrac{(n -1) n (2n -1)}{6} %\\[1em]
%&
= \underline{\underline{ \dfrac{2n-1}{6} F }}
\end{array}$
da $\displaystyle\boxed{
\sum\limits_{k=1}^{N} k^2 =\dfrac{N(N+1)(2N+1)}{6} }
$
Beweis.
Berechnung der Summe $\sum\limits_{k=1}^{N} k^2$ entspricht der Berechung des Volumens $V$ der folgenden Blockpyramide (im
Beispiel $N=5$), die sich aus Würfeln der Kantenlänge $1$ zusammensetzt.
![<math>
\pgfmathsetlengthmacro{\u}{0.567cm}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
y={(\u,0cm)},
x={({0.3*\u}, {-0.45*\u})},
z={(0cm,\u)},
every label/.style={text=black!1},
]
% Ansicht
\def\vgl{0}
% Anzahl Stufen
\pgfmathsetmacro{\N}{5}
% Einstellungen
\def\vglvol{1}
\def\abschlusspyramide{1}
\def\vglwuerfel{1}
\def\cosy{0}
\colorlet{wuerfel}{lightgray}
\colorlet{Prisma}{green!77!black}
\colorlet{prisma}{Prisma!44}
\colorlet{pyramide}{red!55}
\ifnum\vgl=0
\def\abschlusspyramide{0}
\def\vglvol{0}
\def\vglwuerfel{1}
\else
\def\abschlusspyramide{1}
\def\vglwuerfel{0}
\fi
%\pgfmathsetmacro\k{4}
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
% Blockpyramide
\foreach[count=\K from 0] \k in {\N,...,1}{%==========================
\pgfmathsetmacro{\kp}{\k-1}
\begin{scope}[shift={(0,0,\K)}, local bounding box=blockpyramide]
\begin{pgfonlayer}{background}
\coordinate[label=left:$A\k$] (A\k) at (0,0,0);
\coordinate[label=left:$B\k$] (B\k) at (\k,0,0);
\coordinate[label=right:$C\k$] (C\k) at (\k,\k,0);
\coordinate[label=right:$D\k$] (D\k) at (0,\k,0);
\coordinate[label=left:$A\k"$] (A\k1) at (0,0,1);
\coordinate[label=left:$B\k"$] (B\k1) at (\k,0,1);
\coordinate[label=right:$C\k"$] (C\k1) at (\k,\k,1);
\coordinate[label=right:$D\k"$] (D\k1) at (0,\k,1);
\end{pgfonlayer}
% Grund- und Deckfläche
\draw[] (A\k1) -- (B\k1) -- (C\k1) -- (D\k1) --cycle;
\draw[] (A\k) -- (B\k) -- (C\k) ;
\foreach \P in {A\k,B\k,C\k} \draw[] (\P) -- (\P1);
% Füllung der Würfelflächen 1/3
\foreach \x in {0,...,\kp}{
\foreach \y in {0,...,\kp}{
\fill[wuerfel] ([shift={(\x,\y,0)}]A\k1) -- ++(1,0,0) -- ++(0,1,0) -- ++(-1,0,0) --cycle;
}}
\foreach \x in {0,...,\kp}{
% Füllung der Würfelflächen 2/3
\fill[wuerfel] ([shift={(\x,0,0)}]A\k) -- ++(1,0,0) -- ++(0,0,1) -- ++(-1,0,0) --cycle;
}
\foreach \y in {0,...,\kp}{
% Füllung der Würfelflächen 3/3
\fill[wuerfel] ([shift={(0,\y,0)}]B\k) -- ++(0,1,0) -- ++(0,0,1) -- ++(0,-1,0) --cycle;
% Hilfslinien 2/2
\draw[] ([shift={(0,\y,0)}]B\k) -- ++(0,0,1) -- ++(-\k,0,0);
}
% Hilfslinien 2/2
\foreach \x in {0,...,\k}{
\draw[] ([shift={(\x,0,0)}]A\k) -- ++(0,0,1) -- ++(0,\k,0);
}
\ifnum\vgl=1%==============================
% Kantenpyramiden rechts
\draw[Prisma] (C\k) -- ++(0,1,0) coordinate(P) -- ++(-\k,0,0) coordinate(Q) -- (D\k1) coordinate(R) -- (C\k1) coordinate(S)
--(P);
\fill[prisma] (P) -- (Q) -- (R) -- (S) --cycle;
% Eckpyramiden
\draw[red] (C\k) -- ++(0,1,0) coordinate(P) -- ++(1,0,0) coordinate(Q) -- ++(0,-1,0) coordinate(R) --cycle;
\path[draw=red, fill=pyramide] (P) -- (Q) -- (C\k1) --cycle;
\path[draw=red, fill=pyramide] (Q) -- (R) -- (C\k1) --cycle;
% Kantensäulen vorne
\draw[Prisma] (C\k) -- ++(1,0,0) coordinate(P) -- ++(0,-\k,0) coordinate(Q) -- ++(-1,0,0) coordinate(R) -- ++(0,0,1)
coordinate(S) -- (C\k1) coordinate(T);
\fill[prisma] (P) -- (Q) -- (S) -- (T) --cycle;
\fill[prisma] (Q) -- (R) -- (B\k1) --cycle;
\draw[Prisma] (P) -- (Q);
\draw[Prisma] (Q) -- (S);
\draw[Prisma] (P) -- (T);
\fi%==============================
\end{scope}
}%==========================
% Abschlußpyramide oben
\ifnum\abschlusspyramide=1
\path[] (A11) -- +(0,0,1) coordinate(S);
\path[draw=red, fill=pyramide] (S) -- (B11) -- (C11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A11) -- (B11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C11) -- (D11) --cycle;
\fi
\ifnum\cosy=1
\begin{scope}[-latex, shift={(0,3,4)}]
\foreach \P/\s/\Pos in {(1,0,0)/x/below, (0,1,0)/y/above, (0,0,1)/z/right}
\draw[] (0,0,0) -- \P node[\Pos, pos=0.9,inner sep=2pt]{$\s$};
\end{scope}
\fi
% Vergleichskörper
\ifnum\vglvol=1%===========================
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\draw[Prisma] (0,0,0) coordinate(P) -- (2,0,0) coordinate(Q) node[near start, below]{1} -- ++(0,3,0) coordinate(R) node
[midway, below]{$k$} -- ++(-2,0,0) coordinate(S) -- cycle;
\draw[Prisma] (0,0,2) coordinate(Ps) -- (0,3,2) coordinate(Ss);
\fill[prisma] (Q) -- (R) -- (Ss) -- (Ps) --cycle;
\fill[prisma] (P) -- (Q) -- (Ps) --cycle;
\draw[Prisma] (Ps) -- (P) node[midway, left]{1} (Ps) -- (Q);
\draw[Prisma] (Ss) -- (R);
\node[anchor=west] at ($(R)!0.5!(Ss)$) {$
V_\text{KS}(k) =G h = \frac12 \cdot 1 \cdot 1 \cdot k
= \underline{ \frac12 k =V_\text{KS}(k) }$};
\end{scope}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-0,-5.75)$)}]
\coordinate(S) at (0,0,2);
\coordinate(A) at (0,0,0);
\coordinate(B) at (2,0,0);
\coordinate(C) at (2,2,0);
\coordinate(D) at (0,2,0);
\draw[red] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\path[draw=red, fill=pyramide] (S) -- (B) -- (C) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A) node[red, midway, left]{1} -- (B) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C) -- (D) --cycle;
\node[anchor=west, yshift=-2cm] at ($(R)!0.5!(Ss)$) {$
V_\text{EPy0}=\frac13 G h = \frac13 \cdot 1 \cdot 1
= \underline{ \frac13 =V_\text{EPy0} }$};
\end{scope}
\fi%===========================
\ifnum\vglwuerfel=1%===========================
\pgfmathsetmacro\e{1.3}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\coordinate(A) at (0,0,0);
\coordinate(B) at (\e,0,0);
\coordinate(C) at (\e,\e,0);
\coordinate(D) at (0,\e,0);
\coordinate(A1) at (0,0,\e);
\coordinate(B1) at (\e,0,\e);
\coordinate(C1) at (\e,\e,\e);
\coordinate(D1) at (0,\e,\e);
\draw[] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\fill[wuerfel] (A) -- (B) -- (B1) -- (A1) --cycle;
\fill[wuerfel] (B) -- (C) -- (C1) -- (B1) --cycle;
\fill[wuerfel] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\draw[] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\path[] (A1) -- (A) node[midway, left]{1};
\foreach \P in {A,B,C} \draw[] (\P) -- (\P1);
\node[anchor=west] at ($(C)!0.5!(C1)$) {$
V_\text{W} =a^3 = 1^3
= \underline{ 1 =V_\text{W} }$};
\end{scope}
\fi%===========================
\end{tikzpicture}
</math>](/matheplanet/nuke/html/latexrender/pictures/5a95c8812105d58983fd64d9f068f9ca.png "\pgfmathsetlengthmacro{\u}{0.567cm}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
y={(\u,0cm)},
x={({0.3*\u}, {-0.45*\u})},
z={(0cm,\u)},
every label/.style={text=black!1},
]
% Ansicht
\def\vgl{0}
% Anzahl Stufen
\pgfmathsetmacro{\N}{5}
% Einstellungen
\def\vglvol{1}
\def\abschlusspyramide{1}
\def\vglwuerfel{1}
\def\cosy{0}
\colorlet{wuerfel}{lightgray}
\colorlet{Prisma}{green!77!black}
\colorlet{prisma}{Prisma!44}
\colorlet{pyramide}{red!55}
\ifnum\vgl=0
\def\abschlusspyramide{0}
\def\vglvol{0}
\def\vglwuerfel{1}
\else
\def\abschlusspyramide{1}
\def\vglwuerfel{0}
\fi
%\pgfmathsetmacro\k{4}
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
% Blockpyramide
\foreach[count=\K from 0] \k in {\N,...,1}{%==========================
\pgfmathsetmacro{\kp}{\k-1}
\begin{scope}[shift={(0,0,\K)}, local bounding box=blockpyramide]
\begin{pgfonlayer}{background}
\coordinate[label=left:$A\k$] (A\k) at (0,0,0);
\coordinate[label=left:$B\k$] (B\k) at (\k,0,0);
\coordinate[label=right:$C\k$] (C\k) at (\k,\k,0);
\coordinate[label=right:$D\k$] (D\k) at (0,\k,0);
\coordinate[label=left:$A\k\"$] (A\k1) at (0,0,1);
\coordinate[label=left:$B\k\"$] (B\k1) at (\k,0,1);
\coordinate[label=right:$C\k\"$] (C\k1) at (\k,\k,1);
\coordinate[label=right:$D\k\"$] (D\k1) at (0,\k,1);
\end{pgfonlayer}
% Grund- und Deckfläche
\draw[] (A\k1) -- (B\k1) -- (C\k1) -- (D\k1) --cycle;
\draw[] (A\k) -- (B\k) -- (C\k) ;
\foreach \P in {A\k,B\k,C\k} \draw[] (\P) -- (\P1);
% Füllung der Würfelflächen 1/3
\foreach \x in {0,...,\kp}{
\foreach \y in {0,...,\kp}{
\fill[wuerfel] ([shift={(\x,\y,0)}]A\k1) -- ++(1,0,0) -- ++(0,1,0) -- ++(-1,0,0) --cycle;
}}
\foreach \x in {0,...,\kp}{
% Füllung der Würfelflächen 2/3
\fill[wuerfel] ([shift={(\x,0,0)}]A\k) -- ++(1,0,0) -- ++(0,0,1) -- ++(-1,0,0) --cycle;
}
\foreach \y in {0,...,\kp}{
% Füllung der Würfelflächen 3/3
\fill[wuerfel] ([shift={(0,\y,0)}]B\k) -- ++(0,1,0) -- ++(0,0,1) -- ++(0,-1,0) --cycle;
% Hilfslinien 2/2
\draw[] ([shift={(0,\y,0)}]B\k) -- ++(0,0,1) -- ++(-\k,0,0);
}
% Hilfslinien 2/2
\foreach \x in {0,...,\k}{
\draw[] ([shift={(\x,0,0)}]A\k) -- ++(0,0,1) -- ++(0,\k,0);
}
\ifnum\vgl=1%==============================
% Kantenpyramiden rechts
\draw[Prisma] (C\k) -- ++(0,1,0) coordinate(P) -- ++(-\k,0,0) coordinate(Q) -- (D\k1) coordinate(R) -- (C\k1) coordinate(S)
--(P);
\fill[prisma] (P) -- (Q) -- (R) -- (S) --cycle;
% Eckpyramiden
\draw[red] (C\k) -- ++(0,1,0) coordinate(P) -- ++(1,0,0) coordinate(Q) -- ++(0,-1,0) coordinate(R) --cycle;
\path[draw=red, fill=pyramide] (P) -- (Q) -- (C\k1) --cycle;
\path[draw=red, fill=pyramide] (Q) -- (R) -- (C\k1) --cycle;
% Kantensäulen vorne
\draw[Prisma] (C\k) -- ++(1,0,0) coordinate(P) -- ++(0,-\k,0) coordinate(Q) -- ++(-1,0,0) coordinate(R) -- ++(0,0,1)
coordinate(S) -- (C\k1) coordinate(T);
\fill[prisma] (P) -- (Q) -- (S) -- (T) --cycle;
\fill[prisma] (Q) -- (R) -- (B\k1) --cycle;
\draw[Prisma] (P) -- (Q);
\draw[Prisma] (Q) -- (S);
\draw[Prisma] (P) -- (T);
\fi%==============================
\end{scope}
}%==========================
% Abschlußpyramide oben
\ifnum\abschlusspyramide=1
\path[] (A11) -- +(0,0,1) coordinate(S);
\path[draw=red, fill=pyramide] (S) -- (B11) -- (C11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A11) -- (B11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C11) -- (D11) --cycle;
\fi
\ifnum\cosy=1
\begin{scope}[-latex, shift={(0,3,4)}]
\foreach \P/\s/\Pos in {(1,0,0)/x/below, (0,1,0)/y/above, (0,0,1)/z/right}
\draw[] (0,0,0) -- \P node[\Pos, pos=0.9,inner sep=2pt]{$\s$};
\end{scope}
\fi
% Vergleichskörper
\ifnum\vglvol=1%===========================
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\draw[Prisma] (0,0,0) coordinate(P) -- (2,0,0) coordinate(Q) node[near start, below]{1} -- ++(0,3,0) coordinate(R) node
[midway, below]{$k$} -- ++(-2,0,0) coordinate(S) -- cycle;
\draw[Prisma] (0,0,2) coordinate(Ps) -- (0,3,2) coordinate(Ss);
\fill[prisma] (Q) -- (R) -- (Ss) -- (Ps) --cycle;
\fill[prisma] (P) -- (Q) -- (Ps) --cycle;
\draw[Prisma] (Ps) -- (P) node[midway, left]{1} (Ps) -- (Q);
\draw[Prisma] (Ss) -- (R);
\node[anchor=west] at ($(R)!0.5!(Ss)$) {$
V_\text{KS}(k) =G h = \frac12 \cdot 1 \cdot 1 \cdot k
= \underline{ \frac12 k =V_\text{KS}(k) }$};
\end{scope}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-0,-5.75)$)}]
\coordinate(S) at (0,0,2);
\coordinate(A) at (0,0,0);
\coordinate(B) at (2,0,0);
\coordinate(C) at (2,2,0);
\coordinate(D) at (0,2,0);
\draw[red] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\path[draw=red, fill=pyramide] (S) -- (B) -- (C) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A) node[red, midway, left]{1} -- (B) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C) -- (D) --cycle;
\node[anchor=west, yshift=-2cm] at ($(R)!0.5!(Ss)$) {$
V_\text{EPy0}=\frac13 G h = \frac13 \cdot 1 \cdot 1
= \underline{ \frac13 =V_\text{EPy0} }$};
\end{scope}
\fi%===========================
\ifnum\vglwuerfel=1%===========================
\pgfmathsetmacro\e{1.3}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\coordinate(A) at (0,0,0);
\coordinate(B) at (\e,0,0);
\coordinate(C) at (\e,\e,0);
\coordinate(D) at (0,\e,0);
\coordinate(A1) at (0,0,\e);
\coordinate(B1) at (\e,0,\e);
\coordinate(C1) at (\e,\e,\e);
\coordinate(D1) at (0,\e,\e);
\draw[] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\fill[wuerfel] (A) -- (B) -- (B1) -- (A1) --cycle;
\fill[wuerfel] (B) -- (C) -- (C1) -- (B1) --cycle;
\fill[wuerfel] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\draw[] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\path[] (A1) -- (A) node[midway, left]{1};
\foreach \P in {A,B,C} \draw[] (\P) -- (\P1);
\node[anchor=west] at ($(C)!0.5!(C1)$) {$
V_\text{W} =a^3 = 1^3
= \underline{ 1 =V_\text{W} }$};
\end{scope}
\fi%===========================
\end{tikzpicture}")
Mit der Ergänzung von Eckpyramiden und Kantensäulen ergibt sich eine Vergleichspyramide:
![<math>
\pgfmathsetlengthmacro{\u}{0.567cm}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
y={(\u,0cm)},
x={({0.3*\u}, {-0.45*\u})},
z={(0cm,\u)},
every label/.style={text=black!1},
]
% Ansicht
\def\vgl{1}
% Anzahl Stufen
\pgfmathsetmacro{\N}{5}
% Einstellungen
\def\vglvol{1}
\def\abschlusspyramide{1}
\def\vglwuerfel{1}
\def\cosy{0}
\colorlet{wuerfel}{lightgray}
\colorlet{Prisma}{green!77!black}
\colorlet{prisma}{Prisma!44}
\colorlet{pyramide}{red!55}
\ifnum\vgl=0
\def\abschlusspyramide{0}
\def\vglvol{0}
\def\vglwuerfel{1}
\else
\def\abschlusspyramide{1}
\def\vglwuerfel{0}
\fi
%\pgfmathsetmacro\k{4}
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
% Blockpyramide
\foreach[count=\K from 0] \k in {\N,...,1}{%==========================
\pgfmathsetmacro{\kp}{\k-1}
\begin{scope}[shift={(0,0,\K)}, local bounding box=blockpyramide]
\begin{pgfonlayer}{background}
\coordinate[label=left:$A\k$] (A\k) at (0,0,0);
\coordinate[label=left:$B\k$] (B\k) at (\k,0,0);
\coordinate[label=right:$C\k$] (C\k) at (\k,\k,0);
\coordinate[label=right:$D\k$] (D\k) at (0,\k,0);
\coordinate[label=left:$A\k"$] (A\k1) at (0,0,1);
\coordinate[label=left:$B\k"$] (B\k1) at (\k,0,1);
\coordinate[label=right:$C\k"$] (C\k1) at (\k,\k,1);
\coordinate[label=right:$D\k"$] (D\k1) at (0,\k,1);
\end{pgfonlayer}
% Grund- und Deckfläche
\draw[] (A\k1) -- (B\k1) -- (C\k1) -- (D\k1) --cycle;
\draw[] (A\k) -- (B\k) -- (C\k) ;
\foreach \P in {A\k,B\k,C\k} \draw[] (\P) -- (\P1);
% Füllung der Würfelflächen 1/3
\foreach \x in {0,...,\kp}{
\foreach \y in {0,...,\kp}{
\fill[wuerfel] ([shift={(\x,\y,0)}]A\k1) -- ++(1,0,0) -- ++(0,1,0) -- ++(-1,0,0) --cycle;
}}
\foreach \x in {0,...,\kp}{
% Füllung der Würfelflächen 2/3
\fill[wuerfel] ([shift={(\x,0,0)}]A\k) -- ++(1,0,0) -- ++(0,0,1) -- ++(-1,0,0) --cycle;
}
\foreach \y in {0,...,\kp}{
% Füllung der Würfelflächen 3/3
\fill[wuerfel] ([shift={(0,\y,0)}]B\k) -- ++(0,1,0) -- ++(0,0,1) -- ++(0,-1,0) --cycle;
% Hilfslinien 2/2
\draw[] ([shift={(0,\y,0)}]B\k) -- ++(0,0,1) -- ++(-\k,0,0);
}
% Hilfslinien 2/2
\foreach \x in {0,...,\k}{
\draw[] ([shift={(\x,0,0)}]A\k) -- ++(0,0,1) -- ++(0,\k,0);
}
\ifnum\vgl=1%==============================
% Kantenpyramiden rechts
\draw[Prisma] (C\k) -- ++(0,1,0) coordinate(P) -- ++(-\k,0,0) coordinate(Q) -- (D\k1) coordinate(R) -- (C\k1) coordinate(S) --(P);
\fill[prisma] (P) -- (Q) -- (R) -- (S) --cycle;
% Eckpyramiden
\draw[red] (C\k) -- ++(0,1,0) coordinate(P) -- ++(1,0,0) coordinate(Q) -- ++(0,-1,0) coordinate(R) --cycle;
\path[draw=red, fill=pyramide] (P) -- (Q) -- (C\k1) --cycle;
\path[draw=red, fill=pyramide] (Q) -- (R) -- (C\k1) --cycle;
% Kantensäulen vorne
\draw[Prisma] (C\k) -- ++(1,0,0) coordinate(P) -- ++(0,-\k,0) coordinate(Q) -- ++(-1,0,0) coordinate(R) -- ++(0,0,1) coordinate(S) -- (C\k1) coordinate(T);
\fill[prisma] (P) -- (Q) -- (S) -- (T) --cycle;
\fill[prisma] (Q) -- (R) -- (B\k1) --cycle;
\draw[Prisma] (P) -- (Q);
\draw[Prisma] (Q) -- (S);
\draw[Prisma] (P) -- (T);
\fi%==============================
\end{scope}
}%==========================
% Abschlußpyramide oben
\ifnum\abschlusspyramide=1
\path[] (A11) -- +(0,0,1) coordinate(S);
\path[draw=red, fill=pyramide] (S) -- (B11) -- (C11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A11) -- (B11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C11) -- (D11) --cycle;
\fi
\ifnum\cosy=1
\begin{scope}[-latex, shift={(0,3,4)}]
\foreach \P/\s/\Pos in {(1,0,0)/x/below, (0,1,0)/y/above, (0,0,1)/z/right}
\draw[] (0,0,0) -- \P node[\Pos, pos=0.9,inner sep=2pt]{$\s$};
\end{scope}
\fi
% Vergleichskörper
\ifnum\vglvol=1%===========================
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\draw[Prisma] (0,0,0) coordinate(P) -- (2,0,0) coordinate(Q) node[near start, below]{1} -- ++(0,3,0) coordinate(R) node[midway, below]{$k$} -- ++(-2,0,0) coordinate(S) -- cycle;
\draw[Prisma] (0,0,2) coordinate(Ps) -- (0,3,2) coordinate(Ss);
\fill[prisma] (Q) -- (R) -- (Ss) -- (Ps) --cycle;
\fill[prisma] (P) -- (Q) -- (Ps) --cycle;
\draw[Prisma] (Ps) -- (P) node[midway, left]{1} (Ps) -- (Q);
\draw[Prisma] (Ss) -- (R);
\node[anchor=west] at ($(R)!0.5!(Ss)$) {$
V_\text{KS}(k) =G h = \frac12 \cdot 1 \cdot 1 \cdot k
= \underline{ \frac12 k =V_\text{KS}(k) }$};
\end{scope}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-0,-5.75)$)}]
\coordinate(S) at (0,0,2);
\coordinate(A) at (0,0,0);
\coordinate(B) at (2,0,0);
\coordinate(C) at (2,2,0);
\coordinate(D) at (0,2,0);
\draw[red] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\path[draw=red, fill=pyramide] (S) -- (B) -- (C) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A) node[red, midway, left]{1} -- (B) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C) -- (D) --cycle;
\node[anchor=west, yshift=-2cm] at ($(R)!0.5!(Ss)$) {$
V_\text{EPy0}=\frac13 G h = \frac13 \cdot 1 \cdot 1
= \underline{ \frac13 =V_\text{EPy0} }$};
\end{scope}
\fi%===========================
\ifnum\vglwuerfel=1%===========================
\pgfmathsetmacro\e{1.3}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\coordinate(A) at (0,0,0);
\coordinate(B) at (\e,0,0);
\coordinate(C) at (\e,\e,0);
\coordinate(D) at (0,\e,0);
\coordinate(A1) at (0,0,\e);
\coordinate(B1) at (\e,0,\e);
\coordinate(C1) at (\e,\e,\e);
\coordinate(D1) at (0,\e,\e);
\draw[] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\fill[wuerfel] (A) -- (B) -- (B1) -- (A1) --cycle;
\fill[wuerfel] (B) -- (C) -- (C1) -- (B1) --cycle;
\fill[wuerfel] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\draw[] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\path[] (A1) -- (A) node[midway, left]{1};
\foreach \P in {A,B,C} \draw[] (\P) -- (\P1);
\node[anchor=west] at ($(C)!0.5!(C1)$) {$
V_\text{W} =a^3 = 1^3
= \underline{ 1 =V_\text{W} }$};
\end{scope}
\fi%===========================
\end{tikzpicture}
</math>](/matheplanet/nuke/html/latexrender/pictures/a8ca4dc54c03728eb0d0221ff68bc426.png "\pgfmathsetlengthmacro{\u}{0.567cm}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
y={(\u,0cm)},
x={({0.3*\u}, {-0.45*\u})},
z={(0cm,\u)},
every label/.style={text=black!1},
]
% Ansicht
\def\vgl{1}
% Anzahl Stufen
\pgfmathsetmacro{\N}{5}
% Einstellungen
\def\vglvol{1}
\def\abschlusspyramide{1}
\def\vglwuerfel{1}
\def\cosy{0}
\colorlet{wuerfel}{lightgray}
\colorlet{Prisma}{green!77!black}
\colorlet{prisma}{Prisma!44}
\colorlet{pyramide}{red!55}
\ifnum\vgl=0
\def\abschlusspyramide{0}
\def\vglvol{0}
\def\vglwuerfel{1}
\else
\def\abschlusspyramide{1}
\def\vglwuerfel{0}
\fi
%\pgfmathsetmacro\k{4}
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
% Blockpyramide
\foreach[count=\K from 0] \k in {\N,...,1}{%==========================
\pgfmathsetmacro{\kp}{\k-1}
\begin{scope}[shift={(0,0,\K)}, local bounding box=blockpyramide]
\begin{pgfonlayer}{background}
\coordinate[label=left:$A\k$] (A\k) at (0,0,0);
\coordinate[label=left:$B\k$] (B\k) at (\k,0,0);
\coordinate[label=right:$C\k$] (C\k) at (\k,\k,0);
\coordinate[label=right:$D\k$] (D\k) at (0,\k,0);
\coordinate[label=left:$A\k\"$] (A\k1) at (0,0,1);
\coordinate[label=left:$B\k\"$] (B\k1) at (\k,0,1);
\coordinate[label=right:$C\k\"$] (C\k1) at (\k,\k,1);
\coordinate[label=right:$D\k\"$] (D\k1) at (0,\k,1);
\end{pgfonlayer}
% Grund- und Deckfläche
\draw[] (A\k1) -- (B\k1) -- (C\k1) -- (D\k1) --cycle;
\draw[] (A\k) -- (B\k) -- (C\k) ;
\foreach \P in {A\k,B\k,C\k} \draw[] (\P) -- (\P1);
% Füllung der Würfelflächen 1/3
\foreach \x in {0,...,\kp}{
\foreach \y in {0,...,\kp}{
\fill[wuerfel] ([shift={(\x,\y,0)}]A\k1) -- ++(1,0,0) -- ++(0,1,0) -- ++(-1,0,0) --cycle;
}}
\foreach \x in {0,...,\kp}{
% Füllung der Würfelflächen 2/3
\fill[wuerfel] ([shift={(\x,0,0)}]A\k) -- ++(1,0,0) -- ++(0,0,1) -- ++(-1,0,0) --cycle;
}
\foreach \y in {0,...,\kp}{
% Füllung der Würfelflächen 3/3
\fill[wuerfel] ([shift={(0,\y,0)}]B\k) -- ++(0,1,0) -- ++(0,0,1) -- ++(0,-1,0) --cycle;
% Hilfslinien 2/2
\draw[] ([shift={(0,\y,0)}]B\k) -- ++(0,0,1) -- ++(-\k,0,0);
}
% Hilfslinien 2/2
\foreach \x in {0,...,\k}{
\draw[] ([shift={(\x,0,0)}]A\k) -- ++(0,0,1) -- ++(0,\k,0);
}
\ifnum\vgl=1%==============================
% Kantenpyramiden rechts
\draw[Prisma] (C\k) -- ++(0,1,0) coordinate(P) -- ++(-\k,0,0) coordinate(Q) -- (D\k1) coordinate(R) -- (C\k1) coordinate(S) --(P);
\fill[prisma] (P) -- (Q) -- (R) -- (S) --cycle;
% Eckpyramiden
\draw[red] (C\k) -- ++(0,1,0) coordinate(P) -- ++(1,0,0) coordinate(Q) -- ++(0,-1,0) coordinate(R) --cycle;
\path[draw=red, fill=pyramide] (P) -- (Q) -- (C\k1) --cycle;
\path[draw=red, fill=pyramide] (Q) -- (R) -- (C\k1) --cycle;
% Kantensäulen vorne
\draw[Prisma] (C\k) -- ++(1,0,0) coordinate(P) -- ++(0,-\k,0) coordinate(Q) -- ++(-1,0,0) coordinate(R) -- ++(0,0,1) coordinate(S) -- (C\k1) coordinate(T);
\fill[prisma] (P) -- (Q) -- (S) -- (T) --cycle;
\fill[prisma] (Q) -- (R) -- (B\k1) --cycle;
\draw[Prisma] (P) -- (Q);
\draw[Prisma] (Q) -- (S);
\draw[Prisma] (P) -- (T);
\fi%==============================
\end{scope}
}%==========================
% Abschlußpyramide oben
\ifnum\abschlusspyramide=1
\path[] (A11) -- +(0,0,1) coordinate(S);
\path[draw=red, fill=pyramide] (S) -- (B11) -- (C11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A11) -- (B11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C11) -- (D11) --cycle;
\fi
\ifnum\cosy=1
\begin{scope}[-latex, shift={(0,3,4)}]
\foreach \P/\s/\Pos in {(1,0,0)/x/below, (0,1,0)/y/above, (0,0,1)/z/right}
\draw[] (0,0,0) -- \P node[\Pos, pos=0.9,inner sep=2pt]{$\s$};
\end{scope}
\fi
% Vergleichskörper
\ifnum\vglvol=1%===========================
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\draw[Prisma] (0,0,0) coordinate(P) -- (2,0,0) coordinate(Q) node[near start, below]{1} -- ++(0,3,0) coordinate(R) node[midway, below]{$k$} -- ++(-2,0,0) coordinate(S) -- cycle;
\draw[Prisma] (0,0,2) coordinate(Ps) -- (0,3,2) coordinate(Ss);
\fill[prisma] (Q) -- (R) -- (Ss) -- (Ps) --cycle;
\fill[prisma] (P) -- (Q) -- (Ps) --cycle;
\draw[Prisma] (Ps) -- (P) node[midway, left]{1} (Ps) -- (Q);
\draw[Prisma] (Ss) -- (R);
\node[anchor=west] at ($(R)!0.5!(Ss)$) {$
V_\text{KS}(k) =G h = \frac12 \cdot 1 \cdot 1 \cdot k
= \underline{ \frac12 k =V_\text{KS}(k) }$};
\end{scope}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-0,-5.75)$)}]
\coordinate(S) at (0,0,2);
\coordinate(A) at (0,0,0);
\coordinate(B) at (2,0,0);
\coordinate(C) at (2,2,0);
\coordinate(D) at (0,2,0);
\draw[red] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\path[draw=red, fill=pyramide] (S) -- (B) -- (C) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A) node[red, midway, left]{1} -- (B) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C) -- (D) --cycle;
\node[anchor=west, yshift=-2cm] at ($(R)!0.5!(Ss)$) {$
V_\text{EPy0}=\frac13 G h = \frac13 \cdot 1 \cdot 1
= \underline{ \frac13 =V_\text{EPy0} }$};
\end{scope}
\fi%===========================
\ifnum\vglwuerfel=1%===========================
\pgfmathsetmacro\e{1.3}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\coordinate(A) at (0,0,0);
\coordinate(B) at (\e,0,0);
\coordinate(C) at (\e,\e,0);
\coordinate(D) at (0,\e,0);
\coordinate(A1) at (0,0,\e);
\coordinate(B1) at (\e,0,\e);
\coordinate(C1) at (\e,\e,\e);
\coordinate(D1) at (0,\e,\e);
\draw[] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\fill[wuerfel] (A) -- (B) -- (B1) -- (A1) --cycle;
\fill[wuerfel] (B) -- (C) -- (C1) -- (B1) --cycle;
\fill[wuerfel] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\draw[] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\path[] (A1) -- (A) node[midway, left]{1};
\foreach \P in {A,B,C} \draw[] (\P) -- (\P1);
\node[anchor=west] at ($(C)!0.5!(C1)$) {$
V_\text{W} =a^3 = 1^3
= \underline{ 1 =V_\text{W} }$};
\end{scope}
\fi%===========================
\end{tikzpicture}")
Das Volumen der Vergleichspyramide berechnet sich zu
$V_\text{Py} =\frac13 G h =\frac13 \cdot (N+1)^2 \cdot (N+1)
=\dfrac{(N+1)^3}{3}.$
Das Gesamtvolumen aller Kantensäulen berechnet sich zu
$V_\text{KS}
=2\sum\limits_{k=1}^{N} V_\text{KS}(k)
=2\sum\limits_{k=1}^{N} \frac12 k
=\sum\limits_{k=1}^{N} k
=\dfrac{N(N+1)}{2}
$
da $\boxed{ \sum\limits_{k=1}^{N} k=\dfrac{N(N+1)}{2} }$
Beweis.
$\begin{array}{l l l}
\hphantom{=+}\sum\limits_{k=1}^{N} k &= 1 +2 +3 +\dots +N \\
\hphantom{=}+\sum\limits_{k=1}^{N} k
&= N +(N-1) +(N-2) +\dots +1 \\ \hline
=2\sum\limits_{k=1}^{N} k &= (N+1) + (N+1) +\dots +(N+1) = N\cdot(N+1) \hspace{11mm}\square
%=2\sum\limits_{k=1}^{N} k &= N\cdot(N+1) \hspace{11mm}\square
\end{array}$
Das Gesamtvolumen aller Eckpyramiden berechnet sich zu
$V_\text{EPy}
=(N+1) \cdot V_\text{EPy0}
=(N+1) \cdot \dfrac13
=\dfrac{N+1}{3}.$
Damit ergibt sich für das Volumen der Blockpyramide
$\begin{array}{l l}
\sum\limits_{k=1}^{N} k^2 =V
=V_\text{Py} -V_\text{KS} -V_\text{EPy}
&=\dfrac{(N+1)^3}{3} -\dfrac{N(N+1)}{2} -\dfrac{N+1}{3} \\[1em]
&=(N+1)\cdot \dfrac{2(N+1)^2 -3N -2}{6} \\[1em]
&=(N+1)\cdot \dfrac{2(N^2+2N+1) -3N -2}{6} \\[1em]
&=(N+1)\cdot \dfrac{2N^2 +N}{6} \\[1em]
&=(N+1)\cdot \dfrac{N(2N +1)}{6} \hspace{11mm}\square
\end{array}$
$\begin{array}{l l l}
\Leftrightarrow~
\displaystyle\sum\limits_{k=1}^{n-1} k^2
=\left( \sum\limits_{k=1}^{n} k^2 \right) -n^2
&=\dfrac{n(n+1)(2n+1)}{6} -n^2 %\\
&=\dfrac{n(n+1)(2n+1)-6n^2}{6} \\
&=n\dfrac{(n+1)(2n+1)-6n}{6} %\\
&=n\dfrac{2n^2+3n+1-6n}{6} \\
&=n\dfrac{2n^2-3n+1}{6} %\\
&=n\dfrac{2n^2-n -2n+1}{6} \\[0.75em]
&=n\dfrac{n(2n-1)-(2n-1)}{6} % \\
&=\dfrac{(n -1) n (2n -1)}{6}
\end{array}$
Die gesuchte Summe der unteren Teilschnittflächen $F_U$ ergibt sich entsprechend zu
$\begin{array}{l l l}
F_U=n F -F_O
&=nF -\dfrac{2n-1}{6}F %\\[1em]
&= \underline{\underline{ \dfrac{4n+1}{6} F }}.
\end{array}$
Für die Verhältnisse der gesamten Schnittflächen zur Gesamtfläche aller Dreiecke erhält man
$\dfrac{F_U}{nF}
=\dfrac{\dfrac{4n+1}{6} F}{nF}
=\ \dfrac{2}{3} + \dfrac{1}{n}
$ bzw. $\lim\limits_{n\to\infty}\dfrac{F_U}{nF}
=\dfrac{2}{3}$
und $\dfrac{F_O}{nF}
=\dfrac{\dfrac{2n-1}{6} F}{nF}
=\ \dfrac{1}{3} - \dfrac{1}{n}
$ bzw. $\lim\limits_{n\to\infty}\dfrac{F_O}{nF}
=\dfrac{1}{3}.$
![<math>
\def\allgm{0}
\def\bew{0}
\pagecolor{black!1}
% Gegebene Größen
\pgfmathsetmacro{\N}{7}
\pgfmathsetmacro{\s}{2}
\colorlet{unten}{orange!66}
\colorlet{oben}{yellow!66}
\pgfmathsetmacro{\N}{\bew==1 ? 7 : \N}
\pgfmathsetmacro{\Ns}{int(\N+1)}
\pgfmathsetmacro{\Np}{int(\N-1)}
% Annotation - Kanten
\ifnum\bew=1
\tikzset{kante/.style={text=black} }
\else
\tikzset{kante/.style={text=black!1} }
\fi
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle,
Dreieck/.style={},
every label/.style={text=black!1},
]
\foreach[count=\nI] \n in {1,...,\N}{
\begin{scope}[xshift=-\nI*\s cm]
\coordinate[label=below:$A\n$] (A\n) at (0,0);
\coordinate[label=below:$B\n$] (B\n) at (\s,0);
\coordinate[label=$C\n$] (C\n) at (60:\s);
% unten füllen
\draw[Dreieck, name path global=dreieck\n] (A\n) -- (B\n) -- (C\n) --cycle;
% Annotation unten
\ifnum\bew=1
\path[] (A\n) -- (B\n) node[midway, below]{$s$};
\fi
\end{scope}
}
% Annotation rechts
\ifnum\bew=1
\path[] (B1) -- (C1) node[midway, right]{$s$};
\fi
% Gerade 1/2
\path[name path global=gerade] (C1) -- (A\N);
\foreach[count=\kNo] \k in {2,...,\N}{
\pgfmathtruncatemacro\K{\k+1}
\path[name intersections={of={dreieck\k} and gerade, name=Point}];
\ifnum\k=\N
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-2);
\coordinate[label=Y\k] (Y\k) at (Point-1);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{n}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$s$ %$y_{\k}$
};
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-1);
\coordinate[label=Y\k] (Y\k) at (Point-2);
\ifnum\k=\Np
\path[] (X\k) -- (C\k) node[kante, pos=0.7, right]{$x_{n-1}$};
\path[] (Y\k) -- (C\k) node[kante, pos=0.7, left]{$y_{n-1}$};
\else
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$y_{\k}$};
\fi
\end{pgfonlayer}
\fi
% oben füllen
\ifnum\bew=1
\begin{pgfonlayer}{background}
\fill[oben] (X\k) -- (Y\k) -- (C\k) --cycle;
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\fill[unten] (X\k) -- (B\k) -- (A\k) -- (Y\k) --cycle;
\fill[unten] (A1) -- (B1) -- (C1) --cycle;
\end{pgfonlayer}
\fi
}
% Gerade 2/2
\draw[Dreieck] (C1) -- (A\N);
% Beweis ========================
% Annotation Strahlensatzfigur
\ifnum\bew=1%==========================
\draw[densely dashed] (C1) -- (C\N) -- +(-\s,0) coordinate[label=$C\Ns$] (C\Ns) node[midway, above]{$$} -- (A\N) node[midway, left]{$s$};
%
% Annotation oben
\foreach[evaluate={\K=int(\k+1)}] \k in {1,...,\N}{
\pgfmathsetmacro\text{\k==3 || \k==5 ? "" : "s"}
\path[] (C\k) -- (C\K) node[midway, above]{$\text$};
}
\foreach \k/\text in {\Ns/{n+1},\N/{n},\Np/{n-1}}{
\node[above] at (C\k) {$P_{\text}$};
}
\foreach \k in {1,2,3}{
\node[above] at (C\k) {$P_{\k}$};
}
\fi%============================
% Allgemeine Zeichnung
\ifnum\allgm=1%=====================
\pgfmathtruncatemacro{\Na}{\N-2}
\pgfmathtruncatemacro{\Nb}{3}
\fill[black!1] ([shift={(0,-2em)}]A\Na) rectangle ([shift={(-0.5*\s cm,2em)}]C\Nb) node[midway, text=black, font=\Huge]{$\boldsymbol\dots$};
\fi%=====================
\end{tikzpicture}
</math>](/matheplanet/nuke/html/latexrender/pictures/3bd639088f74cf835fcd64e5725fdbe9.png "\def\allgm{0}
\def\bew{0}
\pagecolor{black!1}
% Gegebene Größen
\pgfmathsetmacro{\N}{7}
\pgfmathsetmacro{\s}{2}
\colorlet{unten}{orange!66}
\colorlet{oben}{yellow!66}
\pgfmathsetmacro{\N}{\bew==1 ? 7 : \N}
\pgfmathsetmacro{\Ns}{int(\N+1)}
\pgfmathsetmacro{\Np}{int(\N-1)}
% Annotation - Kanten
\ifnum\bew=1
\tikzset{kante/.style={text=black} }
\else
\tikzset{kante/.style={text=black!1} }
\fi
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle,
Dreieck/.style={},
every label/.style={text=black!1},
]
\foreach[count=\nI] \n in {1,...,\N}{
\begin{scope}[xshift=-\nI*\s cm]
\coordinate[label=below:$A\n$] (A\n) at (0,0);
\coordinate[label=below:$B\n$] (B\n) at (\s,0);
\coordinate[label=$C\n$] (C\n) at (60:\s);
% unten füllen
\draw[Dreieck, name path global=dreieck\n] (A\n) -- (B\n) -- (C\n) --cycle;
% Annotation unten
\ifnum\bew=1
\path[] (A\n) -- (B\n) node[midway, below]{$s$};
\fi
\end{scope}
}
% Annotation rechts
\ifnum\bew=1
\path[] (B1) -- (C1) node[midway, right]{$s$};
\fi
% Gerade 1/2
\path[name path global=gerade] (C1) -- (A\N);
\foreach[count=\kNo] \k in {2,...,\N}{
\pgfmathtruncatemacro\K{\k+1}
\path[name intersections={of={dreieck\k} and gerade, name=Point}];
\ifnum\k=\N
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-2);
\coordinate[label=Y\k] (Y\k) at (Point-1);
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{n}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$s$ %$y_{\k}$
};
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\coordinate[label=X\k] (X\k) at (Point-1);
\coordinate[label=Y\k] (Y\k) at (Point-2);
\ifnum\k=\Np
\path[] (X\k) -- (C\k) node[kante, pos=0.7, right]{$x_{n-1}$};
\path[] (Y\k) -- (C\k) node[kante, pos=0.7, left]{$y_{n-1}$};
\else
\path[] (X\k) -- (C\k) node[kante, midway, right]{$x_{\k}$};
\path[] (Y\k) -- (C\k) node[kante, midway, left]{$y_{\k}$};
\fi
\end{pgfonlayer}
\fi
% oben füllen
\ifnum\bew=1
\begin{pgfonlayer}{background}
\fill[oben] (X\k) -- (Y\k) -- (C\k) --cycle;
\end{pgfonlayer}
\else
\begin{pgfonlayer}{background}
\fill[unten] (X\k) -- (B\k) -- (A\k) -- (Y\k) --cycle;
\fill[unten] (A1) -- (B1) -- (C1) --cycle;
\end{pgfonlayer}
\fi
}
% Gerade 2/2
\draw[Dreieck] (C1) -- (A\N);
% Beweis ========================
% Annotation Strahlensatzfigur
\ifnum\bew=1%==========================
\draw[densely dashed] (C1) -- (C\N) -- +(-\s,0) coordinate[label=$C\Ns$] (C\Ns) node[midway, above]{$$} -- (A\N) node[midway, left]{$s$};
%
% Annotation oben
\foreach[evaluate={\K=int(\k+1)}] \k in {1,...,\N}{
\pgfmathsetmacro\text{\k==3 || \k==5 ? \"\" : \"s\"}
\path[] (C\k) -- (C\K) node[midway, above]{$\text$};
}
\foreach \k/\text in {\Ns/{n+1},\N/{n},\Np/{n-1}}{
\node[above] at (C\k) {$P_{\text}$};
}
\foreach \k in {1,2,3}{
\node[above] at (C\k) {$P_{\k}$};
}
\fi%============================
% Allgemeine Zeichnung
\ifnum\allgm=1%=====================
\pgfmathtruncatemacro{\Na}{\N-2}
\pgfmathtruncatemacro{\Nb}{3}
\fill[black!1] ([shift={(0,-2em)}]A\Na) rectangle ([shift={(-0.5*\s cm,2em)}]C\Nb) node[midway, text=black, font=\Huge]{$\boldsymbol\dots$};
\fi%=====================
\end{tikzpicture}")
![<math>
\pgfmathsetlengthmacro{\u}{0.567cm}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
y={(\u,0cm)},
x={({0.3*\u}, {-0.45*\u})},
z={(0cm,\u)},
every label/.style={text=black!1},
]
% Ansicht
\def\vgl{1}
% Anzahl Stufen
\pgfmathsetmacro{\N}{5}
% Einstellungen
\def\vglvol{0}
\def\abschlusspyramide{1}
\def\vglwuerfel{1}
\def\cosy{0}
\colorlet{wuerfel}{lightgray}
\colorlet{Prisma}{green!77!black}
\colorlet{prisma}{Prisma!44}
\colorlet{pyramide}{red!55}
\ifnum\vgl=0
\def\abschlusspyramide{0}
\def\vglvol{0}
\def\vglwuerfel{1}
\else
\def\abschlusspyramide{1}
\def\vglwuerfel{0}
\fi
%\pgfmathsetmacro\k{4}
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
% Blockpyramide
\foreach[count=\K from 0] \k in {\N,...,1}{%==========================
\pgfmathsetmacro{\kp}{\k-1}
\begin{scope}[shift={(0,0,\K)}, local bounding box=blockpyramide]
\begin{pgfonlayer}{background}
\coordinate[label=left:$A\k$] (A\k) at (0,0,0);
\coordinate[label=left:$B\k$] (B\k) at (\k,0,0);
\coordinate[label=right:$C\k$] (C\k) at (\k,\k,0);
\coordinate[label=right:$D\k$] (D\k) at (0,\k,0);
\coordinate[label=left:$A\k"$] (A\k1) at (0,0,1);
\coordinate[label=left:$B\k"$] (B\k1) at (\k,0,1);
\coordinate[label=right:$C\k"$] (C\k1) at (\k,\k,1);
\coordinate[label=right:$D\k"$] (D\k1) at (0,\k,1);
\end{pgfonlayer}
% Grund- und Deckfläche
\draw[] (A\k1) -- (B\k1) -- (C\k1) -- (D\k1) --cycle;
\draw[] (A\k) -- (B\k) -- (C\k) ;
\foreach \P in {A\k,B\k,C\k} \draw[] (\P) -- (\P1);
% Füllung der Würfelflächen 1/3
\foreach \x in {0,...,\kp}{
\foreach \y in {0,...,\kp}{
\fill[wuerfel] ([shift={(\x,\y,0)}]A\k1) -- ++(1,0,0) -- ++(0,1,0) -- ++(-1,0,0) --cycle;
}}
\foreach \x in {0,...,\kp}{
% Füllung der Würfelflächen 2/3
\fill[wuerfel] ([shift={(\x,0,0)}]A\k) -- ++(1,0,0) -- ++(0,0,1) -- ++(-1,0,0) --cycle;
}
\foreach \y in {0,...,\kp}{
% Füllung der Würfelflächen 3/3
\fill[wuerfel] ([shift={(0,\y,0)}]B\k) -- ++(0,1,0) -- ++(0,0,1) -- ++(0,-1,0) --cycle;
% Hilfslinien 2/2
\draw[] ([shift={(0,\y,0)}]B\k) -- ++(0,0,1) -- ++(-\k,0,0);
}
% Hilfslinien 2/2
\foreach \x in {0,...,\k}{
\draw[] ([shift={(\x,0,0)}]A\k) -- ++(0,0,1) -- ++(0,\k,0);
}
\ifnum\vgl=1%==============================
% Kantenpyramiden rechts
\draw[Prisma] (C\k) -- ++(0,1,0) coordinate(P) -- ++(-\k,0,0) coordinate(Q) -- (D\k1) coordinate(R) -- (C\k1) coordinate(S) --(P);
\fill[prisma] (P) -- (Q) -- (R) -- (S) --cycle;
% Eckpyramiden
\draw[red] (C\k) -- ++(0,1,0) coordinate(P) -- ++(1,0,0) coordinate(Q) -- ++(0,-1,0) coordinate(R) --cycle;
\path[draw=red, fill=pyramide] (P) -- (Q) -- (C\k1) --cycle;
\path[draw=red, fill=pyramide] (Q) -- (R) -- (C\k1) --cycle;
% Kantensäulen vorne
\draw[Prisma] (C\k) -- ++(1,0,0) coordinate(P) -- ++(0,-\k,0) coordinate(Q) -- ++(-1,0,0) coordinate(R) -- ++(0,0,1) coordinate(S) -- (C\k1) coordinate(T);
\fill[prisma] (P) -- (Q) -- (S) -- (T) --cycle;
\fill[prisma] (Q) -- (R) -- (B\k1) --cycle;
\draw[Prisma] (P) -- (Q);
\draw[Prisma] (Q) -- (S);
\draw[Prisma] (P) -- (T);
\fi%==============================
\end{scope}
}%==========================
% Abschlußpyramide oben
\ifnum\abschlusspyramide=1
\path[] (A11) -- +(0,0,1) coordinate(S);
\path[draw=red, fill=pyramide] (S) -- (B11) -- (C11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A11) -- (B11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C11) -- (D11) --cycle;
\fi
\ifnum\cosy=1
\begin{scope}[-latex, shift={(0,3,4)}]
\foreach \P/\s/\Pos in {(1,0,0)/x/below, (0,1,0)/y/above, (0,0,1)/z/right}
\draw[] (0,0,0) -- \P node[\Pos, pos=0.9,inner sep=2pt]{$\s$};
\end{scope}
\fi
% Vergleichskörper
\ifnum\vglvol=1%===========================
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\draw[Prisma] (0,0,0) coordinate(P) -- (2,0,0) coordinate(Q) node[near start, below]{1} -- ++(0,3,0) coordinate(R) node[midway, below]{$k$} -- ++(-2,0,0) coordinate(S) -- cycle;
\draw[Prisma] (0,0,2) coordinate(Ps) -- (0,3,2) coordinate(Ss);
\fill[prisma] (Q) -- (R) -- (Ss) -- (Ps) --cycle;
\fill[prisma] (P) -- (Q) -- (Ps) --cycle;
\draw[Prisma] (Ps) -- (P) node[midway, left]{1} (Ps) -- (Q);
\draw[Prisma] (Ss) -- (R);
\node[anchor=west] at ($(R)!0.5!(Ss)$) {$
V_\text{KS}(k) =G h = \frac12 \cdot 1 \cdot 1 \cdot k
= \underline{ \frac12 k =V_\text{KS}(k) }$};
\end{scope}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-0,-5.75)$)}]
\coordinate(S) at (0,0,2);
\coordinate(A) at (0,0,0);
\coordinate(B) at (2,0,0);
\coordinate(C) at (2,2,0);
\coordinate(D) at (0,2,0);
\draw[red] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\path[draw=red, fill=pyramide] (S) -- (B) -- (C) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A) node[red, midway, left]{1} -- (B) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C) -- (D) --cycle;
\node[anchor=west, yshift=-2cm] at ($(R)!0.5!(Ss)$) {$
V_\text{EPy0}=\frac13 G h = \frac13 \cdot 1 \cdot 1
= \underline{ \frac13 =V_\text{EPy0} }$};
\end{scope}
\fi%===========================
\ifnum\vglwuerfel=1%===========================
\pgfmathsetmacro\e{1.3}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\coordinate(A) at (0,0,0);
\coordinate(B) at (\e,0,0);
\coordinate(C) at (\e,\e,0);
\coordinate(D) at (0,\e,0);
\coordinate(A1) at (0,0,\e);
\coordinate(B1) at (\e,0,\e);
\coordinate(C1) at (\e,\e,\e);
\coordinate(D1) at (0,\e,\e);
\draw[] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\fill[wuerfel] (A) -- (B) -- (B1) -- (A1) --cycle;
\fill[wuerfel] (B) -- (C) -- (C1) -- (B1) --cycle;
\fill[wuerfel] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\draw[] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\path[] (A1) -- (A) node[midway, left]{1};
\foreach \P in {A,B,C} \draw[] (\P) -- (\P1);
\node[anchor=west] at ($(C)!0.5!(C1)$) {$
V_\text{W} =a^3 = 1^3
= \underline{ 1 =V_\text{W} }$};
\end{scope}
\fi%===========================
\end{tikzpicture}
</math>](/matheplanet/nuke/html/latexrender/pictures/c87f17ee080ac631856d61580db2515b.png "\pgfmathsetlengthmacro{\u}{0.567cm}
\begin{tikzpicture}[%scale=0.7,
font=\footnotesize,
background rectangle/.style={draw=none, fill=black!1, rounded corners}, show background rectangle,
y={(\u,0cm)},
x={({0.3*\u}, {-0.45*\u})},
z={(0cm,\u)},
every label/.style={text=black!1},
]
% Ansicht
\def\vgl{1}
% Anzahl Stufen
\pgfmathsetmacro{\N}{5}
% Einstellungen
\def\vglvol{0}
\def\abschlusspyramide{1}
\def\vglwuerfel{1}
\def\cosy{0}
\colorlet{wuerfel}{lightgray}
\colorlet{Prisma}{green!77!black}
\colorlet{prisma}{Prisma!44}
\colorlet{pyramide}{red!55}
\ifnum\vgl=0
\def\abschlusspyramide{0}
\def\vglvol{0}
\def\vglwuerfel{1}
\else
\def\abschlusspyramide{1}
\def\vglwuerfel{0}
\fi
%\pgfmathsetmacro\k{4}
% Layers
\pgfdeclarelayer{background}
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
% Blockpyramide
\foreach[count=\K from 0] \k in {\N,...,1}{%==========================
\pgfmathsetmacro{\kp}{\k-1}
\begin{scope}[shift={(0,0,\K)}, local bounding box=blockpyramide]
\begin{pgfonlayer}{background}
\coordinate[label=left:$A\k$] (A\k) at (0,0,0);
\coordinate[label=left:$B\k$] (B\k) at (\k,0,0);
\coordinate[label=right:$C\k$] (C\k) at (\k,\k,0);
\coordinate[label=right:$D\k$] (D\k) at (0,\k,0);
\coordinate[label=left:$A\k\"$] (A\k1) at (0,0,1);
\coordinate[label=left:$B\k\"$] (B\k1) at (\k,0,1);
\coordinate[label=right:$C\k\"$] (C\k1) at (\k,\k,1);
\coordinate[label=right:$D\k\"$] (D\k1) at (0,\k,1);
\end{pgfonlayer}
% Grund- und Deckfläche
\draw[] (A\k1) -- (B\k1) -- (C\k1) -- (D\k1) --cycle;
\draw[] (A\k) -- (B\k) -- (C\k) ;
\foreach \P in {A\k,B\k,C\k} \draw[] (\P) -- (\P1);
% Füllung der Würfelflächen 1/3
\foreach \x in {0,...,\kp}{
\foreach \y in {0,...,\kp}{
\fill[wuerfel] ([shift={(\x,\y,0)}]A\k1) -- ++(1,0,0) -- ++(0,1,0) -- ++(-1,0,0) --cycle;
}}
\foreach \x in {0,...,\kp}{
% Füllung der Würfelflächen 2/3
\fill[wuerfel] ([shift={(\x,0,0)}]A\k) -- ++(1,0,0) -- ++(0,0,1) -- ++(-1,0,0) --cycle;
}
\foreach \y in {0,...,\kp}{
% Füllung der Würfelflächen 3/3
\fill[wuerfel] ([shift={(0,\y,0)}]B\k) -- ++(0,1,0) -- ++(0,0,1) -- ++(0,-1,0) --cycle;
% Hilfslinien 2/2
\draw[] ([shift={(0,\y,0)}]B\k) -- ++(0,0,1) -- ++(-\k,0,0);
}
% Hilfslinien 2/2
\foreach \x in {0,...,\k}{
\draw[] ([shift={(\x,0,0)}]A\k) -- ++(0,0,1) -- ++(0,\k,0);
}
\ifnum\vgl=1%==============================
% Kantenpyramiden rechts
\draw[Prisma] (C\k) -- ++(0,1,0) coordinate(P) -- ++(-\k,0,0) coordinate(Q) -- (D\k1) coordinate(R) -- (C\k1) coordinate(S) --(P);
\fill[prisma] (P) -- (Q) -- (R) -- (S) --cycle;
% Eckpyramiden
\draw[red] (C\k) -- ++(0,1,0) coordinate(P) -- ++(1,0,0) coordinate(Q) -- ++(0,-1,0) coordinate(R) --cycle;
\path[draw=red, fill=pyramide] (P) -- (Q) -- (C\k1) --cycle;
\path[draw=red, fill=pyramide] (Q) -- (R) -- (C\k1) --cycle;
% Kantensäulen vorne
\draw[Prisma] (C\k) -- ++(1,0,0) coordinate(P) -- ++(0,-\k,0) coordinate(Q) -- ++(-1,0,0) coordinate(R) -- ++(0,0,1) coordinate(S) -- (C\k1) coordinate(T);
\fill[prisma] (P) -- (Q) -- (S) -- (T) --cycle;
\fill[prisma] (Q) -- (R) -- (B\k1) --cycle;
\draw[Prisma] (P) -- (Q);
\draw[Prisma] (Q) -- (S);
\draw[Prisma] (P) -- (T);
\fi%==============================
\end{scope}
}%==========================
% Abschlußpyramide oben
\ifnum\abschlusspyramide=1
\path[] (A11) -- +(0,0,1) coordinate(S);
\path[draw=red, fill=pyramide] (S) -- (B11) -- (C11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A11) -- (B11) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C11) -- (D11) --cycle;
\fi
\ifnum\cosy=1
\begin{scope}[-latex, shift={(0,3,4)}]
\foreach \P/\s/\Pos in {(1,0,0)/x/below, (0,1,0)/y/above, (0,0,1)/z/right}
\draw[] (0,0,0) -- \P node[\Pos, pos=0.9,inner sep=2pt]{$\s$};
\end{scope}
\fi
% Vergleichskörper
\ifnum\vglvol=1%===========================
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\draw[Prisma] (0,0,0) coordinate(P) -- (2,0,0) coordinate(Q) node[near start, below]{1} -- ++(0,3,0) coordinate(R) node[midway, below]{$k$} -- ++(-2,0,0) coordinate(S) -- cycle;
\draw[Prisma] (0,0,2) coordinate(Ps) -- (0,3,2) coordinate(Ss);
\fill[prisma] (Q) -- (R) -- (Ss) -- (Ps) --cycle;
\fill[prisma] (P) -- (Q) -- (Ps) --cycle;
\draw[Prisma] (Ps) -- (P) node[midway, left]{1} (Ps) -- (Q);
\draw[Prisma] (Ss) -- (R);
\node[anchor=west] at ($(R)!0.5!(Ss)$) {$
V_\text{KS}(k) =G h = \frac12 \cdot 1 \cdot 1 \cdot k
= \underline{ \frac12 k =V_\text{KS}(k) }$};
\end{scope}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-0,-5.75)$)}]
\coordinate(S) at (0,0,2);
\coordinate(A) at (0,0,0);
\coordinate(B) at (2,0,0);
\coordinate(C) at (2,2,0);
\coordinate(D) at (0,2,0);
\draw[red] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\path[draw=red, fill=pyramide] (S) -- (B) -- (C) --cycle;
\path[draw=red, fill=pyramide] (S) -- (A) node[red, midway, left]{1} -- (B) --cycle;
\path[draw=red, fill=pyramide] (S) -- (C) -- (D) --cycle;
\node[anchor=west, yshift=-2cm] at ($(R)!0.5!(Ss)$) {$
V_\text{EPy0}=\frac13 G h = \frac13 \cdot 1 \cdot 1
= \underline{ \frac13 =V_\text{EPy0} }$};
\end{scope}
\fi%===========================
\ifnum\vglwuerfel=1%===========================
\pgfmathsetmacro\e{1.3}
\begin{scope}[shift={($(blockpyramide.north east)+(0,\N-1,-2)$)}]
\coordinate(A) at (0,0,0);
\coordinate(B) at (\e,0,0);
\coordinate(C) at (\e,\e,0);
\coordinate(D) at (0,\e,0);
\coordinate(A1) at (0,0,\e);
\coordinate(B1) at (\e,0,\e);
\coordinate(C1) at (\e,\e,\e);
\coordinate(D1) at (0,\e,\e);
\draw[] (A) -- (B) node[near start, below]{1} -- (C) node[midway, below]{1} -- (D) --cycle;
\fill[wuerfel] (A) -- (B) -- (B1) -- (A1) --cycle;
\fill[wuerfel] (B) -- (C) -- (C1) -- (B1) --cycle;
\fill[wuerfel] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\draw[] (A1) -- (B1) -- (C1) -- (D1) --cycle;
\path[] (A1) -- (A) node[midway, left]{1};
\foreach \P in {A,B,C} \draw[] (\P) -- (\P1);
\node[anchor=west] at ($(C)!0.5!(C1)$) {$
V_\text{W} =a^3 = 1^3
= \underline{ 1 =V_\text{W} }$};
\end{scope}
\fi%===========================
\end{tikzpicture}")
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haribo
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Dabei seit: 25.10.2012
Mitteilungen: 2718
| Beitrag No.6, eingetragen 2021-01-14
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haribo
Senior
Dabei seit: 25.10.2012
Mitteilungen: 2718
| Beitrag No.7, eingetragen 2021-01-18
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nicht wirklich meine idee, aber hier ist nochmal eine geometrische 2D herleitung der formel für die summe von quadraten von 1 bis n
\ rechts die grauen aufeinander gestellten quadrate mit der höhe (n*(n+1))/2, das ist die gauss´e summenformel in die fläche mit der breite (2n+1) passt dann genau die dreifache quadrate-fläche: -die grauen rechts; -links dazu gespiegelt die in teilflächen zerlegten bunten quadrate; -und im zwischenraum die flächengleich hier weitgehend gelben; somit wäre die summenformel aller quadrate bis n² (n*(n+1))/2 * (2n+1)/3 = (n*(n+1)*(2n+1))/6 wie gehabt
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haribo
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Dabei seit: 25.10.2012
Mitteilungen: 2718
| Beitrag No.8, eingetragen 2021-01-18
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dito mit weniger schnitten
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