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Nombre de usuarioheureka
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Preguntas 11
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 #1
avatar+20151 
+2

Another one of my Triangle Questions

In \(\Delta ABC\), line segments are drawn parallel to each of the sides dividing the triangle into six regions.
The areas of three regions are shown in the figure. What is the area of \(\Delta ABC\)?

 

\(\begin{array}{l} \text{Let Area $A =[ABC] $} \\\\ \text{Let Area $A_4=[E'EP] =a^2 = 4, ~a=2 $}\\ \text{Let Area $A_5=[D'DP] =b^2 = 9, ~b=3 $}\\ \text{Let Area $A_6=[F'FP] =c^2 = 16 ~c=4$} \\\\ \text{Let Area $A_1=[E'AD] $}\\ \text{Let Area $A_2=[D'BF] $}\\ \text{Let Area $A_3=[F'CE] $} \end{array}\)

 

\(\text{The triangles are similar $\\( \triangle ABC \sim \triangle E'EP \sim \triangle D'DP \sim \triangle F'FP \sim \triangle E'AD \sim \triangle D'BF \sim \triangle F'CE )$.} \)

 

The key theorem we apply here is that the ratio of the areas of 2 similar triangles is
the ratio of a pair of corresponding sides squared.

 

\(\begin{array}{|rcll|} \hline \dfrac{A_4}{A} &=& \left(\dfrac{E'P}{CB}\right)^2 \\ \dfrac{A_5}{A} &=& \left(\dfrac{PD}{CB}\right)^2 \\ \hline \dfrac{A_4}{A}\cdot \dfrac{A}{A_5} &=& \dfrac{E'P^2}{CB^2} \cdot \dfrac{CB^2}{PD^2} \\ \dfrac{A_4}{A_5} &=& \dfrac{E'P^2}{PD^2} \\ \dfrac{a^2}{b^2} &=& \dfrac{E'P^2}{PD^2} \\ \dfrac{a}{b} &=& \dfrac{E'P}{PD} \\ \boxed{\dfrac{E'P}{PD}=\dfrac{a}{b} } \\ \hline \end{array} \)

 

\(\begin{array}{|rclcrl|} \hline \dfrac{E'P}{PD} &=& \dfrac{a}{b} && E'P &=& \dfrac{a}{a+b}E'D \\ && && E'D &=& \dfrac{a+b}{a} E'P \\ \dfrac{E'D}{PD} &=& \dfrac{a+b}{a} \cdot \dfrac{E'P}{PD} \\ \dfrac{E'D}{PD} &=& \dfrac{a+b}{a} \cdot \dfrac{a}{b} \\ \boxed{\dfrac{E'D}{PD} = \dfrac{a+b}{b} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dfrac{A_1}{A_5} &=& \left( \dfrac{E'D}{PD} \right)^2 \\ A_1 &=& A_5\left( \dfrac{E'D}{PD} \right)^2 \\ A_1 &=& b^2\left( \dfrac{a+b}{b} \right)^2 \\ \mathbf{A_1} & \mathbf{=} & \mathbf{(a+b)^2} \quad & | \quad A_1 = (2+3)^2=5^2=25 \\ \hline \end{array} \)

 

analogous

\(\begin{array}{|rclcrl|} \hline \dfrac{D'P}{PF} &=& \dfrac{b}{c} && D'P &=& \dfrac{b}{b+c}D'F \\ && && D'F &=& \dfrac{b+c}{b}D'P \\ \dfrac{D'F}{PF} &=& \dfrac{b+c}{b} \cdot \dfrac{D'P}{PF} \\ \dfrac{D'F}{PF} &=& \dfrac{b+c}{b} \cdot \dfrac{b}{c} \\ \boxed{\dfrac{D'F}{PF} = \dfrac{b+c}{c} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{A_2}{A_6} &=& \left( \dfrac{D'F}{PF} \right)^2 \\ A_2 &=& A_6\left( \dfrac{D'F}{PF} \right)^2 \\ A_2 &=& c^2\left( \dfrac{b+c}{c} \right)^2 \\ \mathbf{A_2} & \mathbf{=} & \mathbf{(b+c)^2} \quad & | \quad A_2 = (3+4)^2=7^2=49 \\ \hline \end{array}\)

 

analogous

\(\begin{array}{|rclcrl|} \hline \dfrac{F'P}{PE} &=& \dfrac{c}{a} && F'P &=& \dfrac{c}{a+c}F'E \\ && && F'E &=& \dfrac{a+c}{c}F'P \\ \dfrac{F'E}{PE} &=& \dfrac{a+c}{c} \cdot \dfrac{F'P}{PE} \\ \dfrac{F'E}{PE} &=& \dfrac{a+c}{c} \cdot \dfrac{c}{a} \\ \boxed{\dfrac{F'E}{PE} = \dfrac{a+c}{a} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{A_3}{A_4} &=& \left( \dfrac{F'E}{PE} \right)^2 \\ A_3 &=& A_4\left( \dfrac{F'E}{PE} \right)^2 \\ A_3 &=& a^2\left( \dfrac{a+c}{a} \right)^2 \\ \mathbf{A_3} & \mathbf{=} & \mathbf{(a+c)^2} \quad & | \quad A_3 = (2+4)^2=6^2=36 \\ \hline \end{array} \)


\(\mathbf{A=\ ?}\)

\(\begin{array}{|rcll|} \hline A_1+A_2+A_3 &=& \mathbf{A}+ A_4+A_5 + A_6 \\ 25+49+36 &=& A+ 4+9+16 \\ 110 &=& A + 29 \\ A &=& 110 - 29 \\ \mathbf{A} & \mathbf{=} & \mathbf{81} \\ \hline \end{array}\)

 

laugh

heureka 16-nov-2018
 #1
avatar+20151 
+4

Wie kann ich die Formel:
\(\displaystyle \large{q^1 + q^2 + q^3 +\ldots+ q^n = \dfrac{q^{n+1} - q}{q-1} }\)
durch völlstandige Induktion beweisen ?
für alle \( n \in \mathbb{N}\):

 

\(\bf{\text{Beweise mit vollständiger Induktion:}} \\ \displaystyle q^1 + q^2 + q^3 +\ldots+ q^n = \dfrac{q^{n+1} - q}{q-1} \)

 

\(\text{Induktionsanfang:}\)

\(\begin{array}{|lll|} \hline n=1 & \text{linke Seite:} & q^1 \\ & & =q \\\\ & \text{rechte Seite:} & \dfrac{q^{1+1} - q} {q-1} \\ & &= \dfrac{q^{2} - q} {q-1} \\ & &= \dfrac{q\cdot q - q} {q-1} \\ & &= \dfrac{q\cdot( q - 1)} {q-1} \\ & &= q\cdot \left( \dfrac{q - 1} {q-1} \right) \\ & &= q \\ \hline \end{array}\)

 

\(\text{Für $\mathbf{n=1}$ sind beide Seiten gleich, und die Aussage ist wahr!}\)

 

\(\text{Die Induktionsannahme (I.A.) lautet:}\)

\(\begin{array}{|rcll|} \hline \displaystyle q^1 + q^2 + q^3 +\ldots+ q^n = \dfrac{q^{n+1} - q}{q-1} \\ \hline \end{array}\)

 

\(\text{Der Induktionsschluss von $\mathbf{n}$ nach $\mathbf{n+1}$:}\)

\(\begin{array}{|rcll|} \hline \displaystyle q^1 + q^2 + q^3 +\ldots+ q^n + q^{n+1} &=& \dfrac{q^{(n+1)+1} - q}{q-1} \\ \hline \end{array}\)

 

\(\bf{\text{linke Seite:}}\)

\(\begin{array}{|llrcll|} \hline &\mathbf{ q^1 + q^2 + q^3 +\ldots+ q^n + q^{n+1} }\\\\ \overset{I.A.}{=} & \dfrac{q^{n+1} - q}{q-1} + q^{n+1} \\\\ = & \dfrac{q^{n+1} - q}{q-1} + q^{n+1}\cdot \left(\dfrac{q-1}{q-1}\right) \\\\ = & \dfrac{q^{n+1} - q+ q^{n+1}(q-1)}{q-1} \\\\ = & \dfrac{q^{n+1} - q+ q^{n+2}-q^{n+1} }{q-1} \\\\ = & \dfrac{ - q+ q^{n+2}}{q-1} \\\\ \mathbf{=} & \mathbf{\dfrac{ q^{n+2}- q }{q-1} }\\ \hline \end{array} \)

 

\(\bf{\text{rechte Seite:}}\)

\(\begin{array}{|ll|} \hline & \mathbf{\dfrac{q^{(n+1)+1} - q}{q-1} } \\\\ \mathbf{=}& \mathbf{\dfrac{ q^{n+2}- q }{q-1} }\\ \hline \end{array}\)

 

laugh

heureka 08-nov-2018
 #2
avatar+20151 
+4

Vollständige Induktion!

Zeigen Sie mit vollständiger Induktion
\(\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1}\cdot k=\dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right]\)
für alle \(n \in \mathbb{N}\):

 

\(\bf{\text{Beweise mit vollständiger Induktion:}}\)

\(\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1}\cdot k=\dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right]\)

 

\(\text{Induktionsanfang:}\)

\(\begin{array}{|lll|} \hline n=1 & \text{linke Seite:} & (-1)^{1-1}\cdot 1 \\ & &= 1 \\\\ & \text{rechte Seite:} & \dfrac{1}{4}[~1+(-1)^{1-1}\cdot (2\cdot 1+1)~] \\ & &= \dfrac{1}{4}[~1+1\cdot 3~] \\ & &= \dfrac{1}{4}\cdot(4) \\ & &= 1 \\ \hline \end{array} \)

 

\(\text{Für $\mathbf{n=1}$ sind beide Seiten gleich, und die Aussage ist wahr!}\)

 

\(\text{Die Induktionsannahme (I.A.) lautet:}\)

\(\begin{array}{|rcll|} \hline \displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1}\cdot k &=& \dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right] \\ \hline \end{array}\)

 

\(\text{Der Induktionsschluss von $\mathbf{n}$ nach $\mathbf{n+1}$:}\)

\(\begin{array}{|rcll|} \hline \displaystyle \sum\limits_{k=1}^{n+1} (-1)^{k-1}\cdot k &=& \dfrac{1}{4}\left[~1+(-1)^{(n+1)-1}\cdot (2(n+1)+1)~\right] \\ \hline \end{array}\)

 

\(\bf{\text{linke Seite:}}\)

\(\begin{array}{|llrcll|} \hline &\mathbf{ \sum\limits_{k=1}^{n+1} (-1)^{k-1}\cdot k }\\\\ = & \sum\limits_{k}^{n} (-1)^{k-1}\cdot k + (-1)^{(n+1)-1}(n+1) \\\\ = & \sum\limits_{k}^{n} (-1)^{k-1}\cdot k + (-1)^{n}(n+1) \\\\ \overset{I.A.}{=} & \dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right] + (-1)^{n}(n+1) \\\\ = & \dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right] + (-1)^{n}(n+1) \cdot \dfrac{4}{4} \\\\ = & \dfrac{1}{4}\left[~1+\dfrac{(-1)^{n}}{-1}\cdot (2n+1)+ 4\cdot(-1)^{n}(n+1) ~\right] \\\\ = & \dfrac{1}{4}\left[~1- (-1)^{n}\cdot (2n+1)+ 4\cdot(-1)^{n}(n+1) ~\right] \\\\ = & \dfrac{1}{4}\left\{~1 +(-1)^n\cdot[~ 4(n+1)-(2n+1)~] ~\right\} \\\\ = & \dfrac{1}{4}\left[~1 +(-1)^n\cdot (4n+4-2n-1)~\right] \\\\ \mathbf{=} & \mathbf{\dfrac{1}{4}\left[~1 +(-1)^n\cdot(~ 2n+3) ~\right] }\\ \hline \end{array}\)

 

\(\bf{\text{rechte Seite:}} \)

\(\begin{array}{|ll|} \hline & \mathbf{\dfrac{1}{4}\left[~1+(-1)^{(n+1)-1}\cdot (2(n+1)+1)~\right] } \\\\ =& \dfrac{1}{4}\left[~1+(-1)^{n}\cdot (2n+2+1)~\right] \\\\ \mathbf{=}& \mathbf{\dfrac{1}{4}\left[~1+(-1)^{n}\cdot (2n+3)~\right]} \\ \hline \end{array}\)

 

laugh

heureka 25-oct-2018