Inspired by Keen Mun

Geometry Level 5

There are 1 bigger circle and 2 same size circles inscribed in an ellipse. The two red points are the foci of the ellipse. Find the ratio of areas A : B A:B .

Give your answer to 3 significant figures.


The answer is 0.04289.

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Dexter Woo Teng Koon by Dexter Woo Teng Koon , 20, Malaysia

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2 solutions

In normal notation, let a be semimajor axis, b semiminor axis and distance between focii as 2c. R A a n d R B t h e r a d i i . W e k n o w t h a t a 2 = b 2 + c 2 . B u t i n t h i s p r o b l e m R A = b = c , ⟹ a 2 = 2 R A 2 = 2 c 2 . ⟹ a = 2 ∗ R A . W e a l s o s a w c = R A . . . . . . ∗ . N o w h e r e w e h a v e 2 R B = a − c , ⟹ R B = 1 2 ∗ ( a − c ) = . . . . . ∗ 1 2 ∗ ( 2 R A − R A ) = 1 2 ( 2 − 1 ) R A . S o t h e r a t i o s o f a r e a s = π ∗ R B 2 π ∗ R A 2 = { 1 2 ( 2 − 1 ) R A ) } 2 R A 2 = 1 4 ∗ ( 3 − 2 2 ) . T h e r a t i o s o f a r e a s = 0.0428932 \text{In normal notation, let a be semimajor axis, b semiminor axis and distance between focii as 2c.}\\ R_A\ \ and\ \ R_B\ the\ radii.\\ We\ know\ that \ \ a^2\ =\ b^2\ +\ c^2.\ \ \ But\ in\ this\ problem\ R_A=b=c,\ \ \implies\ a^2\ =\ 2R_A^2=2c^2.\\ \implies\ \ a=\sqrt2*R_A.\ \ \ We\ also\ saw\ \ \ c=R_A.\ \ .....\color{#3D99F6}{*}.\\ Now\ here\ we\ have\ \ 2R_B= a - c,\ \ \implies R_B=\frac 1 2*(a - c)=.....\color{#3D99F6}{*} \frac 1 2*(\sqrt2{R_A} - R_A)=\frac 1 2(\sqrt2 - 1)R_A.\\ So\ the\ ratios\ of\ areas\ =\dfrac{\pi*R_B^2} {\pi*R_A^2}=\dfrac{\{\frac 1 2(\sqrt2 - 1)R_A)\}^2} {R_A^2}=\frac 1 4 *(3 - 2\sqrt2).\\ The\ ratios\ of\ areas= \color{#D61F06}{0.0428932}

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