Maximum Triangle Area within an Ellipse

Geometry Level 3

Given the ellipse:

x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

where a > 0 a > 0 and b > 0 b > 0 , three points ( x 1 , y 1 ) (x_1, y_1) , ( x 2 , y 2 ) (x_2, y_2) , and ( x 3 , y 3 ) (x_3, y_3) are chosen to lie on it. What is the maximum area of the triangle formed by these three points?

( 2 5 / 5 ) a b (2 \sqrt{5} / 5) a b ( 3 3 / 4 ) a b (3 \sqrt{3} / 4) a b ( a 2 + b 2 ) / 2 (a^2 + b^2 ) / 2 a b / 2 a b / 2

Hosam Hajjir by Hosam Hajjir , 56, Canada

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2 solutions

Michael Mendrin
Jan 28, 2016

The area of an equilateral triangle in an unit circle is 3 3 4 \frac{3\sqrt{3}}{4} , which has the maximum area of any such circumscribed triangle. Scale both independently in orthogonal directions by a a and b b to get the answer for an ellipse of semi-axes a , b a, b . Orientation of the equilateral triangle does not matter.

4 Helpful 1 Interesting 1 Brilliant 0 Confused

Excellent argument. Thank you.

Hosam Hajjir - 5 years, 4 months ago

Why does this work? I mean I understand the intuitive reasoning, and I assume this is equivalent to the use of change of variable when integrating in calculus, the thing is, that when you do it, you usually use the jacobian as a "scale factor"

Cristian Alvarez Bran - 3 years, 11 months ago
Rohit Ner
Jan 28, 2016

As suggested by @Michael Mendrin , let ( a , 0 ) , ( a cos ⁡ θ , b sin ⁡ θ ) (a,0),(a\cos\theta,b\sin\theta) and ( a cos ⁡ θ , − b sin ⁡ θ ) (a\cos\theta,-b\sin\theta) be the vertices of the triangle lying on the ellipse.
Δ ( θ ) = 1 2 ∣ ∣ a 0 1 a cos ⁡ θ b sin ⁡ θ 1 a cos ⁡ θ − b sin ⁡ θ 1 ∣ ∣ = a b sin ⁡ θ ( cos ⁡ θ − 1 ) Δ ′ ( θ ) = a b [ sin ⁡ θ ( − sin ⁡ θ ) + cos ⁡ θ ( cos ⁡ θ − 1 ) ] = a b ( cos ⁡ 2 θ − sin ⁡ 2 θ − cos ⁡ θ ) = a b ( cos ⁡ 2 θ − cos ⁡ θ ) = 0 ...(maximum value) cos ⁡ 2 θ = cos ⁡ θ θ = 2 π 3 , 0̸ Δ ( 2 π 3 ) = a b ∣ ( 3 2 ) ( 1 2 − 1 ) ∣ = 3 3 4 a b \begin{aligned}\Delta(\theta)&=\frac{1}{2}\left|\left| \begin{matrix} a & 0 & 1 \\ a\cos { \theta } & b\sin { \theta } & 1 \\ a\cos { \theta } & -b\sin { \theta } & 1 \end{matrix} \right| \right|\\&=ab\sin\theta(\cos\theta-1)\\{\Delta}^{'}(\theta)&=ab\left[ \sin { \theta } \left( -\sin { \theta } \right) +\cos { \theta } \left( \cos { \theta }-1 \right) \right] \\&=ab\left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } -\cos { \theta } \right) \\&=ab\left( \cos { 2\theta } -\cos { \theta } \right)=0 \text{...(maximum value)}\\\cos{2\theta}&=\cos\theta \\\theta&=\frac{2\pi}{3},\not{0}\\ \Delta\left(\frac{2\pi}{3}\right)&=ab\left|\left( \frac { \sqrt { 3 } }{ 2 } \right) \left( \frac { 1 }{ 2 } -1 \right)\right| \\&\huge\color{#3D99F6}{=\boxed{\frac{3\sqrt{3}}{4}ab}}\end{aligned}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

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