A problem by Sudeep Salgia

Level 2

Consider a standard ellipse:
x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
where a a and b b are the lengths of semi-major and semi-minor axes respectively.

Now, consider a circle concentric with the ellipse and the radius equal to the length of the semi-major axis of the ellipse, i.e. ,
x 2 + y 2 = a 2 x^2 + y^2 = a^2

A set of complimentary points are defined on these two conics, P P and P ′ P' with a parameter, θ \theta . Point P P is lies on the ellipse and P ′ P' on the circle.

The points are defined as :
P ( θ ) = ( a cos ⁡ θ , b sin ⁡ θ ) P(\theta) = (a\cos \theta , b\sin \theta)
P ′ ( θ ) = ( a cos ⁡ θ , a sin ⁡ θ ) P'(\theta) = (a\cos \theta , a\sin \theta)

Given, the pairs of complimentary points namely ( A , A ′ A , A' ) , ( B , B ′ B , B' ) and ( C , C ′ C , C' ) with parameters α \alpha , β \beta and γ \gamma , i.e., the points on the ellipse are A ( α ) A(\alpha) , B ( β ) B(\beta) and C ( γ ) C(\gamma) .

Find the ratio of the area of the triangle A B C ABC to that of the triangle A ′ B ′ C ′ A'B'C' .

Details and Assumptions:
a = 5 a = 5
b = 3 b = 3


The answer is 0.6.

Sudeep Salgia by Sudeep Salgia , 24, India

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3 solutions

Sudeep Salgia
Feb 3, 2014

One of the ways to calculate the area of a triangle when the coordinates of the vertices, namely ( x 1 , y 1 ) , ( x 2 , y 2 ) a n d ( x 3 , y 3 ) (x_{1}, y_{1}) , (x_{2}, y_{2}) and (x_{3}, y_{3}) , are given is,

A r e a = 1 2 ∣ ( 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ) ∣ Area = \frac{1}{2} |\begin{pmatrix} 1 & x_{1} & y_{1} \\ 1 & x_{2} & y_{2}\\ 1 & x_{3} & y_{3} \end{pmatrix}|
(That is the absolute value of a determinant and not that of a matrix)

Using the above, we get,

A r e a ( Δ A B C ) = 1 2 ∣ ( 1 a cos ⁡ α b sin ⁡ α 1 a cos ⁡ β b sin ⁡ β 1 a cos ⁡ γ b sin ⁡ γ ) ∣ = a b 2 ∣ ( 1 cos ⁡ α sin ⁡ α 1 cos ⁡ β sin ⁡ β 1 cos ⁡ γ sin ⁡ γ ) ∣ Area(\Delta ABC) = \frac{1}{2} |\begin{pmatrix} 1 & a\cos \alpha & b\sin \alpha \\ 1 & a\cos \beta & b\sin \beta\\ 1 & a\cos \gamma & b\sin \gamma \end{pmatrix}| = \frac{ab}{2} |\begin{pmatrix} 1 & \cos \alpha & \sin \alpha \\ 1 & \cos \beta & \sin \beta\\ 1 & \cos \gamma & \sin \gamma \end{pmatrix}|

A r e a ( Δ A ′ B ′ C ′ ) = 1 2 ∣ ( 1 a cos ⁡ α a sin ⁡ α 1 a cos ⁡ β a sin ⁡ β 1 a cos ⁡ γ a sin ⁡ γ ) ∣ = a 2 2 ∣ ( 1 cos ⁡ α sin ⁡ α 1 cos ⁡ β sin ⁡ β 1 cos ⁡ γ sin ⁡ γ ) ∣ Area(\Delta A'B'C') = \frac{1}{2} |\begin{pmatrix} 1 & a\cos \alpha & a\sin \alpha \\ 1 & a\cos \beta & a\sin \beta\\ 1 & a\cos \gamma & a\sin \gamma \end{pmatrix}| = \frac{a^{2}}{2} |\begin{pmatrix} 1 & \cos \alpha & \sin \alpha \\ 1 & \cos \beta & \sin \beta\\ 1 & \cos \gamma & \sin \gamma \end{pmatrix}|

Clearly, on taking the ratio,we get,

A r e a ( Δ A B C ) A r e a ( Δ A ′ B ′ C ′ ) \frac{Area(\Delta ABC)}{Area(\Delta A'B'C')} = a b a 2 \frac{ab}{a^{2}} = b a \frac{b}{a}

Plug in the values to get the answer.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Of course a tricky quick way to find the answer is assuming three points (0,5), (0,-5) and (5,0) on the circle for which the corresponding points on ellipse will be (0,3), (0,-3) and (5,0). The areas of both these triangles can be found very simply using area of triangle 1/2bh...

Yathish Dhavala - 7 years, 4 months ago

Just a note: The circle in the question is called director circle.

Shivam Gautam - 7 years, 4 months ago

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The circle is called auxiliary circle. Director circle is the one having radius equal to a 2 + b 2 \sqrt{a^2 +b^2}

Sudeep Salgia - 7 years, 4 months ago

it is called auxilliary circle and not director circle

Priyesh Pandey - 7 years, 2 months ago
Vaibhav Nayak
Feb 15, 2014

For a triangle with vertices ( 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}) ; area is given by:

A = 1 2 [ x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 2 ) + x 3 ( y 1 − y 2 ] A=\frac{1}{2}[x_{1}(y_{2} - y_{3})+x_{2}(y_{3} - y_{2})+x_{3}(y_{1} - y_{2}]

So area of triangle A B C ABC , say A 1 A_{1} will be:

A 1 = 1 2 [ a cos ⁡ α ( b sin ⁡ β − b sin ⁡ γ ) + a cos ⁡ β ( b sin ⁡ γ − b sin ⁡ α ) + a cos ⁡ γ ( b sin ⁡ α − b sin ⁡ β ) ] A_{1}=\frac{1}{2}[a\cos \alpha(b\sin \beta - b\sin \gamma)+a\cos \beta(b\sin \gamma - b\sin \alpha)+a\cos \gamma(b\sin \alpha - b\sin \beta)]

∴ A 1 = 1 2 [ a b ( cos ⁡ α sin ⁡ β − cos ⁡ α sin ⁡ γ ) + a b ( cos ⁡ β sin ⁡ γ − cos ⁡ β sin ⁡ α ) + a b ( cos ⁡ γ sin ⁡ α − cos ⁡ γ sin ⁡ β ) ] \therefore A_{1}=\frac{1}{2}[ab(\cos \alpha\sin \beta - \cos \alpha\sin \gamma)+ab(\cos \beta\sin \gamma - \cos \beta\sin \alpha)+ab(\cos \gamma\sin \alpha - \cos \gamma\sin \beta)]

∴ A 1 = a b 2 [ ( cos ⁡ α sin ⁡ β − cos ⁡ β sin ⁡ α ) + ( cos ⁡ β sin ⁡ γ − cos ⁡ γ sin ⁡ β ) + ( cos ⁡ γ sin ⁡ α − cos ⁡ α sin ⁡ γ ) ] \therefore A_{1}=\frac{ab}{2}[(\cos \alpha\sin \beta - \cos \beta\sin \alpha)+(\cos \beta\sin \gamma - \cos \gamma\sin \beta)+(\cos \gamma\sin \alpha - \cos \alpha\sin \gamma)]

∴ A 1 = a b 2 [ ( cos ⁡ α sin ⁡ β − cos ⁡ β sin ⁡ α ) + ( cos ⁡ β sin ⁡ γ − cos ⁡ γ sin ⁡ β ) + ( cos ⁡ γ sin ⁡ α − cos ⁡ α sin ⁡ γ ) ] \therefore A_{1}=\frac{ab}{2}[(\cos \alpha\sin \beta - \cos \beta\sin \alpha)+(\cos \beta\sin \gamma - \cos \gamma\sin \beta)+(\cos \gamma\sin \alpha - \cos \alpha\sin \gamma)]

∴ A 1 = a b 2 [ cos ⁡ ( α + β ) + cos ⁡ ( β + γ ) + cos ⁡ ( γ + α ) ] \therefore A_{1}=\frac{ab}{2}[\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)]

Through similar operations for triangle A ′ B ′ C ′ ) A^{'}B^{'}C^{'}) , it turns out that its area A 2 A_{2} :

A 2 = a 2 2 [ cos ⁡ ( α + β ) + cos ⁡ ( β + γ ) + cos ⁡ ( γ + α ) ] A_{2}=\frac{a^{2}}{2}[\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)]

So, required ratio:

A 1 A 2 = a b a 2 = b a = 3 5 = 0.6 \frac{A_{1}}{A_{2}}=\frac{ab}{a_{2}}=\frac{b}{a}=\frac{3}{5}=\boxed{0.6} s q . u n i t s sq. units

1 Helpful 0 Interesting 0 Brilliant 0 Confused

And there's a wonderful solution above mine. I'm learning :)

Vaibhav Nayak - 7 years, 3 months ago
Ahmed R. Maaty
Mar 10, 2014

SA of ABC= 15

SA of A'B'C'=25

ratio= 15/25=0.6

Click Here

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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