+0  
 
+9
367
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avatar+78744 

What is the area of the largest Kepler Triangle that can be inscribed in the circle whose equation is x^2 + y^2  = 9   ????

 

{A Kepler Triangle is a right triangle whose sides are in the ratio of 1 : √Phi : Phi....where Phi = [1 + √5] / 2 }

 

 

CPhill  Jun 11, 2015

Best Answer 

 #2
avatar+18715 
+10

What is the area of the largest Kepler Triangle that can be inscribed in the circle whose equation is x^2 + y^2  = 9   ????

{A Kepler Triangle is a right triangle whose sides are in the ratio of 1 : √Phi : Phi....where Phi = [1 + √5] / 2 }

 

$$\boxed{a:b:c = 1:\sqrt{\varphi}:\varphi}$$

 

$$r_{circle} = \sqrt{9} = 3\qquad c = 2\cdot r_{circle} \\\\
\small{\text{$
\begin{array}{rcl}
\sqrt{ \varphi } & \cdot & \dfrac{ \varphi }{ \sqrt{ \varphi } } = \varphi \\\\
b & \cdot & \dfrac { \varphi } { \sqrt{ \varphi } } = c \\\\
\textcolor[rgb]{1,0,0}{b} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi }}\\\\
\hline
&\\
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
1 & \cdot & \dfrac{ \varphi }{ 1 } = \varphi \\\\
a & \cdot & \dfrac{ \varphi }{ 1 } = c \\\\
\textcolor[rgb]{1,0,0}{a} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\dfrac{ c } { \varphi } }\\\\
\hline
&\\
\end{array}
$}}\\\\$$

 

$$\small{\text{$
\begin{array}{rcl}
A &=& \dfrac{a\cdot b}{2} \\\\
A &=& \dfrac{\dfrac{ c } { \varphi }\cdot c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi } }{2} \\\\
A&=& \dfrac{c^2}{\varphi^2} \cdot \dfrac{ \sqrt{ \varphi } } { 2}\\\\
A &=& 2\cdot r^2 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\
A &=& 18 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\
A &=& 8.74562889162 \\\\
\end{array}
$}}$$

 

heureka  Jun 12, 2015
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7+0 Answers

 #1
avatar+519 
+10

(I realize that golden rectangle and kepler triangle are different,Thank you CPhill!)

fiora  Jun 12, 2015
 #2
avatar+18715 
+10
Best Answer

What is the area of the largest Kepler Triangle that can be inscribed in the circle whose equation is x^2 + y^2  = 9   ????

{A Kepler Triangle is a right triangle whose sides are in the ratio of 1 : √Phi : Phi....where Phi = [1 + √5] / 2 }

 

$$\boxed{a:b:c = 1:\sqrt{\varphi}:\varphi}$$

 

$$r_{circle} = \sqrt{9} = 3\qquad c = 2\cdot r_{circle} \\\\
\small{\text{$
\begin{array}{rcl}
\sqrt{ \varphi } & \cdot & \dfrac{ \varphi }{ \sqrt{ \varphi } } = \varphi \\\\
b & \cdot & \dfrac { \varphi } { \sqrt{ \varphi } } = c \\\\
\textcolor[rgb]{1,0,0}{b} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi }}\\\\
\hline
&\\
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
1 & \cdot & \dfrac{ \varphi }{ 1 } = \varphi \\\\
a & \cdot & \dfrac{ \varphi }{ 1 } = c \\\\
\textcolor[rgb]{1,0,0}{a} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\dfrac{ c } { \varphi } }\\\\
\hline
&\\
\end{array}
$}}\\\\$$

 

$$\small{\text{$
\begin{array}{rcl}
A &=& \dfrac{a\cdot b}{2} \\\\
A &=& \dfrac{\dfrac{ c } { \varphi }\cdot c \cdot \dfrac{ \sqrt{ \varphi } } { \varphi } }{2} \\\\
A&=& \dfrac{c^2}{\varphi^2} \cdot \dfrac{ \sqrt{ \varphi } } { 2}\\\\
A &=& 2\cdot r^2 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\
A &=& 18 \cdot \dfrac{ \sqrt{ \varphi } } { \varphi^2 } \\\\
A &=& 8.74562889162 \\\\
\end{array}
$}}$$

 

heureka  Jun 12, 2015
 #3
avatar+78744 
0

Yep....correct....nice work, fiora and heureka.....!!!!!

 

 

CPhill  Jun 12, 2015
 #4
avatar+91045 
+5

Was it a test Chris ?

Melody  Jun 12, 2015
 #5
avatar+78744 
0

LOL!!!!!......nah.......it's just something I thought up that might make a nice problem.......

 

 

CPhill  Jun 12, 2015
 #6
avatar+91045 
0

Did you already know the answer and how to get there ?

Melody  Jun 12, 2015
 #7
avatar+78744 
0

Yep......I used fiora's  approach  [more or less ].......but.....I like the way heureka did it, too.....

 

 

CPhill  Jun 12, 2015

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