Inellipse

Geometry Level 5

Let E E be the ellipse of maximum area that can be contained in a triangle of side lengths 1 , 3 , 2 1,\sqrt3,2 .

Find the distance between the foci of the ellipse E E .


Try this

2 3 7 4 \dfrac{2}{3}\sqrt[4]{7} 1 3 7 \dfrac{1}{3}\sqrt{7} 2 3 6 4 \dfrac{2}{3}\sqrt[4]{6} 1 3 6 \dfrac{1}{3}\sqrt{6}

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4 solutions

Digvijay Singh
Jun 9, 2019

The Steiner inellipse has the maximum area of any inellipse of a triangle.

According to Marden's theorem , If the vertices of a triangle z 1 , z 2 , z 3 z_1,z_2,z_3 are the complex roots of a cubic polynomial f ( x ) f(x) , then the foci of the Steiner inellipse are the roots of the derivative f ( x ) f'(x) .

In this case letting z 1 = 0 , z 2 = 1 , z 3 = 3 i z_1=0, z_2=1, z_3=\sqrt{3}i

We have f ( x ) = x ( x 1 ) ( x 3 i ) f(x)=x(x-1)(x-\sqrt{3}i)

Letting f ( x ) = 0 f'(x)=0 , we get 3 x 2 2 ( 1 + 3 i ) x + 3 i = 0 3x^2-2(1+\sqrt{3}i)x+\sqrt{3}i=0

Now, the distance between the foci of the ellipse is equal to the absolute value of the difference between the roots of the above equation.

Which is 2 3 2 3 i = 2 3 7 4 \left|\dfrac{2}{3}\sqrt{-2-\sqrt{3}i}\right|=\boxed{\dfrac{2}{3}\sqrt[4]{7}}


  • Click here for the proof that the Steiner inellipse has the greatest area of all inellipses of a triangle.

  • Click here for the proof of Marden's theorem.

Sir, can you please post a solution for this: https://brilliant.org/problems/confusing-question-no-way-out/

Jake Tricole - 2 years ago
David Vreken
Jun 9, 2019

A Steiner inellipse has the maximum area of any inellipse of a triangle, and for a triangle with sides a a , b b , and c c it has semi-major and semi-minor axes 1 6 a 2 + b 2 + c 2 ± 2 Z \frac{1}{6}\sqrt{a^2 + b^2 + c^2 \pm 2Z} where Z = a 4 + b 4 + c 4 a 2 b 2 b 2 c 2 c 2 a 2 Z = \sqrt{a^4 + b^4 + c^4 - a^2b^2 - b^2c^2 - c^2a^2} . Therefore, for a triangle with sides a = 1 a = 1 , b = 3 b = \sqrt{3} , and c = 2 c = 2 , the semi-major and semi-minor axes a e a_e and b e b_e are 1 6 8 ± 2 7 \frac{1}{6}\sqrt{8 \pm 2\sqrt{7}} .

The distance d d between the foci of an ellipse is twice its linear eccentricity c e c_e , and c e = a e 2 b e 2 c_e = \sqrt{a_e^2 - b_e^2} . Therefore, d = 2 c e d = 2c_e = = 2 a e 2 b e 2 2\sqrt{a_e^2 - b_e^2} = = 2 ( 1 6 8 + 2 7 ) 2 ( 1 6 8 2 7 ) 2 2\sqrt{(\frac{1}{6}\sqrt{8 + 2\sqrt{7}})^2 - (\frac{1}{6}\sqrt{8 - 2\sqrt{7}})^2} = = 2 3 7 4 \boxed{\frac{2}{3}\sqrt[4]{7}} .

Sir, can you please post a solution for this: https://brilliant.org/problems/confusing-question-no-way-out/

Jake Tricole - 2 years ago

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Yes, I can.

David Vreken - 2 years ago

Since solution does not compile in this space, I have to give it below.
The given formulas are from link below.
link text

T h i s i s a S t e i n e r E l l i p s e o f Δ w i t h s i d e s s a y a , b , c . T h e n E = 2 3 Z , w h e r e Z i s , Z 2 = a 4 + b 4 + c 4 a 2 b 2 b 2 c 2 c 2 a 2 = 1 4 + ( 3 ) 4 + 2 4 1 2 2 2 ( 3 ) 2 ( 1 2 + 2 2 ) = 26 19 = 7. E = 2 / 3 Z = 2 / 3 7 4 . This ~is~a~Steiner~Ellipse~of~\Delta~with~sides~say~a, ~b,~c.\\ Then~E=\dfrac 2 3*\sqrt{Z}, ~where~Z~is,\\ Z^2=a^4+b^4+c^4-a^2*b^2-b^2*c^2-c^2*a^2\\ =1^4+(\sqrt3)^4+2^4-1^2*2^2-(\sqrt3)^2(1^2+2^2)\\ =26-19=7.\\ \therefore~E=2/3*\sqrt{Z}=\Large \color{#D61F06}{2/3*\sqrt[4] 7.}

Niranjan Khanderia - 2 years ago
Vijay Simha
Jun 10, 2019

What is the area of this Steiner inellipse ?

It is pi/6

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