Hosam Hajjir vs Maria Kozlowska

Geometry Level 4

Twin ellipses have semiminor axes of length 1 and semimajor axes of length 2. In one ellipse, the semiminor axis is horizontal; in the other, the semimajor axis is horizontal. The twin ellipses are tangent and the horizontal axes (semiminor for one, semimajor for the other) lie on the same line. The twin ellipses lie on opposite sides of their common tangent line.

What is the minimum diameter of the circle that encloses both ellipses?

Provide your answer to 3 decimals places.


The answer is 6.058.

W Rose by W Rose , 63, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Mar 17, 2019

Let the center of the ellipse with semiminor axis horizontal be the origin of the x y xy -plane and the radius of the circle be r r . Then the equations of the circle and the ellipse are:

{ ( x + 5 − r ) 2 + y 2 = r 2 . . . ( 1 ) 4 x 2 + y 2 = 4 . . . ( 2 ) \begin{cases} (x+5-r)^2 + y^2 = r^2 & ...(1) \\ 4x^2 + y^2 = 4 & ...(2) \end{cases}

Differentiate both equations w.r.t x x :

{ 2 ( x + 5 − r ) + 2 y d y d x = 0 . . . ( 1 a ) 8 x + 2 y d y d x = 0 . . . ( 2 a ) \begin{cases} 2(x+5-r) + 2y\dfrac {dy}{dx} = 0 & ...(1a) \\ 8x + 2y\dfrac {dy}{dx} = 0 & ...(2a) \end{cases}

Since the tangential points of the ellipse and circle have the same gradients, the points satisfy the above system of equations.

From ( 2 a ) − ( 1 a ) : 6 x − 10 + 2 r = 0 ⟹ r = 5 − 3 x (2a)-(1a): \ 6x - 10 + 2r = 0 \implies r = 5 - 3x . Substituting this in

⟹ ( 1 ) : ( x + 5 − 5 + 3 x ) 2 + y 2 = ( 5 − 3 x ) 2 16 x 2 + y 2 = 9 x 2 − 30 x + 25 7 x 2 + 30 x − 25 + y 2 = 0 . . . ( 3 ) \begin{aligned} \implies (1): \quad (x+5-5+3x)^2 + y^2 & = (5-3x)^2 \\ 16x^2 + y^2 & = 9x^2 - 30x + 25 \\ 7x^2 + 30x - 25 + y^2 & = 0 & ...(3) \end{aligned}

⟹ ( 3 ) − ( 2 ) 3 x 2 + 30 x − 21 = 0 x 2 + 10 x − 7 = 0 \begin{aligned} \implies (3)-(2) \quad 3x^2+ 30x - 21 & = 0 \\ x^2 + 10x - 7 & = 0 \end{aligned}

⟹ x = − 10 + 1 0 2 + 28 2 = 4 2 − 5 Since x > 0 r = 5 − 3 ( 4 2 − 5 ) = 20 − 12 2 2 r = 40 − 24 2 ≈ 6.059 \begin{aligned} \implies x & = \frac {-10+\sqrt {10^2 + 28}}2 = 4\sqrt 2 - 5 & \small \color{#3D99F6} \text{Since }x > 0 \\ r & = 5 - 3(4\sqrt 2 - 5) = 20 - 12\sqrt 2 \\ 2r & = 40 - 24\sqrt 2 \approx \boxed{6.059} \end{aligned}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Inventive solution using calculus in lieu of trigonometry.

W Rose - 2 years, 3 months ago

Log in to reply

Not really inventive (I mean out of nothing), if y y and d y d x \dfrac {dy}{dx} are not the same for the ellipse and circle the above system of equations does not work. It is just presentation.

Chew-Seong Cheong - 2 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...