+0  
 
+11
739
7
avatar+88871 

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+20006 
+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
 #1
avatar+519 
+10

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

fiora  Jun 12, 2015
 #2
avatar+20006 
+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+88871 
0

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

 

 

CPhill  Jun 12, 2015
 #4
avatar+93305 
+5

Was it a test Chris ?

Melody  Jun 12, 2015
 #5
avatar+88871 
0

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

 

 

CPhill  Jun 12, 2015
 #6
avatar+93305 
0

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

Melody  Jun 12, 2015
 #7
avatar+88871 
0

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

 

 

CPhill  Jun 12, 2015

19 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.