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# Kepler Triangle

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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

#2
+18948
+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 }

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

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

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

heureka  Jun 12, 2015
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#1
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(I realize that golden rectangle and kepler triangle are different,Thank you CPhill!)

fiora  Jun 12, 2015
#2
+18948
+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 }

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

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

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

heureka  Jun 12, 2015
#3
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Yep....correct....nice work, fiora and heureka.....!!!!!

CPhill  Jun 12, 2015
#4
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Was it a test Chris ?

Melody  Jun 12, 2015
#5
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LOL!!!!!......nah.......it's just something I thought up that might make a nice problem.......

CPhill  Jun 12, 2015
#6
+91780
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Did you already know the answer and how to get there ?

Melody  Jun 12, 2015
#7
+82757
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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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