Elliptical Sangaku

Geometry Level 5

For some ellipses it is possible to inscribe seven circles, such that each circle is tangential to the ellipse at precisely two points, and such that successive circles are tangential to each other. It is not necessary for the seven circles to be symmetrically arranged within the ellipse.

An ellipse can be inscribed in this way with seven mutually tangent circles precisely when its eccentricity e e is greater than a critical value α \alpha . Give as answer the value of 1 0 6 α \lfloor10^6\alpha\rfloor .


The answer is 974927.

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1 solution

Mark Hennings
Oct 23, 2020

By symmetry, we are looking for circles inside the ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 that are tangent to the ellipse at two points, and whose centres lie on the x x -axis. If we consider the point ( u , 0 ) (u,0) , where u < a |u| < a , then the square of the distance from ( u , 0 ) (u,0) to the point ( a cos t , b sin t ) (a\cos t,b\sin t) on the ellipse is D ( t ) = ( a cos t u ) 2 + b 2 sin 2 t D(t) \; = \; (a\cos t - u)^2 + b^2\sin^2t and hence D ( t ) = 2 a sin t ( a cos t u ) + 2 b 2 sin t cos t = 2 sin t ( a u ( a 2 b 2 ) cos t ) = 2 a sin t ( u a e 2 cos t ) D'(t) \; = \; -2a\sin t(a\cos t - u) + 2b^2\sin t \cos t \; = \; 2\sin t\big(au - (a^2-b^2)\cos t\big) \; = \; 2a\sin t\big(u - ae^2\cos t\big) If u a e 2 |u| \ge ae^2 , there are only two turning points of D D (at t = 0 t=0 and t = π t=\pi ) and hence any circle centred at ( u , 0 ) (u,0) can only have one point of tangency. We must have u < a e 2 |u| < ae^2 .

Define the angle 0 < φ < 1 2 π 0 < \varphi < \tfrac12\pi such that cos φ = e \cos\varphi = e .

For any u < a e 2 |u| < ae^2 we can find a value w w such that u = a e sin w u = ae \sin w ; since sin w < e = cos φ |\sin w| < e = \cos\varphi we deduce that w < 1 2 π φ |w| < \tfrac12\pi - \varphi . The minimum distance from ( u , 0 ) (u,0) to the ellipse occurs at the point ( a cos θ , b sin θ ) (a\cos\theta,b\sin\theta) where cos θ = u a e 2 = e 1 sin w \cos\theta = \frac{u}{ae^2} = e^{-1}\sin w , and this minimum distance is therefore r 2 r^2 , where r 2 = ( a e 1 sin w a e sin w ) 2 + b 2 sin 2 θ = a 2 e 2 ( 1 e 2 ) 2 sin 2 w + b 2 ( 1 e 2 sin 2 w ) = b 2 e 2 ( 1 e 2 ) sin 2 w + b 2 ( 1 e 2 sin 2 w ) = b 2 cos 2 w \begin{aligned} r^2 & = \; \big(ae^{-1}\sin w - ae\sin w\big)^2 + b^2\sin^2\theta \; = \; a^2e^{-2}(1 - e^2)^2\sin^2w + b^2\big(1 - e^{-2}\sin^2w\big) \\ & = \; b^2e^{-2}(1 - e^2)\sin^2w + b^2\big(1 - e^{-2}\sin^2w\big) \; = \; b^2\cos^2w \end{aligned} Thus we can draw a circle with centre at the point ( u , 0 ) (u,0) and radius r r , where u = a e sin w u = ae\sin w and r = b cos w r = b\cos w , which is tangential to the ellipse at precisely two points for any w < 1 2 π φ |w| < \tfrac12\pi - \varphi .

Consider two of these circles, given by values 1 2 π + φ < w 1 < w 2 < 1 2 π φ -\tfrac12\pi + \varphi < w_1 < w_2 < \tfrac12\pi - \varphi . Then u 2 u 1 = a e ( sin w 2 sin w 1 ) = 2 a cos φ cos 1 2 ( w 1 + w 2 ) sin 1 2 ( w 2 w 1 ) r 1 + r 2 = b ( cos w 1 + cos w 2 ) = 2 a sin φ cos 1 2 ( w 1 + w 2 ) cos 1 2 ( w 2 w 1 ) \begin{aligned} u_2 - u_1 & = \; ae(\sin w_2 - \sin w_1) \; = \; 2a\cos\varphi \cos\tfrac12(w_1+w_2)\,\sin\tfrac12(w_2-w_1) \\ r_1 + r_2 & = \; b(\cos w_1 + \cos w_2) \; = \; 2a\sin\varphi \cos\tfrac12(w_1+w_2)\cos\tfrac12(w_2-w_1) \end{aligned} and hence these two circles are tangential to each other provided that cos φ sin 1 2 ( w 2 w 1 ) = sin φ cos 1 2 ( w 2 w 1 ) \cos\varphi \sin\tfrac12(w_2-w_1) = \sin\varphi \cos\tfrac12(w_2-w_1) , and hence so that tan 1 2 ( w 2 w 1 ) = tan φ \tan\tfrac12(w_2-w_1) = \tan\varphi , and hence provided that w 2 w 1 = 2 φ w_2 - w_1 = 2\varphi .

Thus we can have seven mutually tangent circles inscribed in the ellipse provided we can choose w j = w 0 + 2 j φ w_j = w_0 + 2j\varphi for 0 j 6 0 \le j \le 6 such that 1 2 π + φ < w j < 1 2 π φ -\tfrac12\pi + \varphi < w_j < \tfrac12\pi - \varphi for all 0 j 6 0 \le j \le 6 . This means that 12 φ = w 6 w 0 12\varphi = w_6 - w_0 must be less than π 2 φ \pi - 2\varphi , and hence that φ < 1 14 π \varphi < \tfrac{1}{14}\pi . Thus we must have e = cos φ > cos 1 14 π e = \cos\varphi > \cos\tfrac{1}{14}\pi , so we deduce that α = cos 1 14 π \alpha = \cos\tfrac{1}{14}\pi , and hence 1 0 6 α = 974927 \lfloor 10^6\alpha \rfloor = \boxed{974927} .

Sorry if this is something obvious, but I'm having a hard time understanding why u a e 2 |u| \geq ae^{2} would imply that D D has only one turning point.

N. Aadhaar Murty - 7 months, 3 weeks ago

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In that case the only turning points are at t = 0 t=0 and t = π t=\pi . Either you have a circle that is internally tangent at the end of a semimajor axis, or else you have a circle which is externally tangent at the end of the other semimajor axis.

Mark Hennings - 7 months, 3 weeks ago

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Thanks, now I get the whole process. This is a great exercise. Also, maybe this can be generalized for any no of circles?

N. Aadhaar Murty - 7 months, 3 weeks ago

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@N. Aadhaar Murty Yes. You need e > cos π 2 n e > \cos \tfrac{\pi}{2n} for n n circles.

Mark Hennings - 7 months, 3 weeks ago

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@Mark Hennings Oh yeah, should've seen that. Neat result indeed :)

N. Aadhaar Murty - 7 months, 3 weeks ago

@Mark Hennings Sorry, I want to discuss the newest problem I posted. I accidentally entered the pendulum's maximum position as the answer, when I meant to enter the amplitude, approximately 0.1 0.1 rad.

The problem is re-uploaded. I seem to be getting the same answer for the bonus too, π / 2 \pi/2 . However, this is only true when the initial position of the pendulum is at π / 2 \pi/2 , and the pendulum furiously oscillated between 0 0 and π \pi when the initial condition is 0 0 .

Your maths seems to be the same as mine, thanks for helping.

Krishna Karthik - 5 months, 3 weeks ago

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