Un \text{Un} -Chebyshev Magic

Geometry Level 4

In the rectangular plane, the graph uniquely presented by the cubic polynomial function y = a 3 x 3 + a 2 x 2 + a 1 x + a 0 y = a_3x^3 + a_2x^2 + a_1x + a_0 passes through the square inscribed in a unit circle centered at the origin. There exists the rotation angle θ \theta of the vertex ( 1 , 0 ) (1,0) about the origin, such that the cubic equation passes through the circle exactly at 4 4 vertices.

If θ \theta can be expressed as A B π \dfrac{A}{B}\pi radians, where A , B A,B are coprime positive integers and 0 < A < 4 B 0 < A < 4B , input the product A B AB as your answer.


The answer is 6.

Michael Huang by Michael Huang , 29, USA

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1 solution

Mark Hennings
Apr 6, 2021

By symmetry, the cubic must be odd. Then the cubic y = a x 3 + b x y = ax^3 + bx passes through the points ( ± cos ⁡ θ , ± sin ⁡ θ ) (\pm \cos\theta, \pm\sin\theta) and ( ± sin ⁡ θ , ∓ cos ⁡ θ ) (\pm \sin\theta, \mp \cos\theta) , and the gradient of the cubic at the latter two points is tan ⁡ θ \tan\theta . Thus a cos ⁡ 3 θ + b cos ⁡ θ = sin ⁡ θ a sin ⁡ 3 θ + b sin ⁡ θ = − cos ⁡ θ 3 a sin ⁡ 2 θ + b = tan ⁡ θ a\cos^3\theta + b\cos\theta \; =\; \sin\theta \hspace{1cm} a\sin^3\theta + b\sin\theta \; =\; -\cos\theta \hspace{1cm} 3a\sin^2\theta + b \; = \; \tan\theta Thus we deduce that a = 1 sin ⁡ θ cos ⁡ θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) b = − cos ⁡ 4 θ + sin ⁡ 4 θ sin ⁡ θ cos ⁡ θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) a \; = \; \frac{1}{\sin\theta \cos\theta (\cos^2\theta - \sin^2\theta)} \hspace{2cm} b \; = \; - \frac{\cos^4\theta + \sin^4\theta}{\sin\theta \cos\theta(\cos^2\theta - \sin^2\theta)} and hence 3 sin ⁡ 2 θ sin ⁡ θ cos ⁡ θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) − cos ⁡ 4 θ + sin ⁡ 4 θ sin ⁡ θ cos ⁡ θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) = tan ⁡ θ 3 sin ⁡ 2 θ − cos ⁡ 4 θ − sin ⁡ 4 θ = sin ⁡ 2 θ ( cos ⁡ 2 θ − sin ⁡ 2 θ ) 3 tan ⁡ 2 θ ( 1 + tan ⁡ 2 θ ) − 1 − tan ⁡ 4 θ = tan ⁡ 2 θ ( 1 − tan ⁡ 2 θ ) 3 tan ⁡ 4 θ + 2 tan ⁡ 2 θ − 1 = 0 ( 3 tan ⁡ 2 θ − 1 ) ( tan ⁡ 2 θ + 1 ) = 0 \begin{aligned} \frac{3\sin^2\theta}{\sin\theta \cos\theta (\cos^2\theta - \sin^2\theta)} - \frac{\cos^4\theta + \sin^4\theta}{\sin\theta \cos\theta (\cos^2\theta - \sin^2\theta)} & = \; \tan\theta \\ 3\sin^2\theta - \cos^4\theta - \sin^4\theta & = \; \sin^2\theta(\cos^2\theta - \sin^2\theta) \\ 3\tan^2\theta(1 + \tan^2\theta) - 1 - \tan^4\theta & = \; \tan^2\theta(1 - \tan^2\theta) \\ 3\tan^4\theta + 2\tan^2\theta - 1 & = \; 0 \\ (3\tan^2\theta - 1)(\tan^2\theta + 1) & = \; 0 \end{aligned} so that tan ⁡ θ = 1 3 \tan\theta = \tfrac{1}{\sqrt{3}} , and hence θ = 1 6 π \theta = \tfrac16\pi , making the answer 1 × 6 = 6 1 \times 6 = \boxed{6} .

4 Helpful 3 Interesting 4 Brilliant 0 Confused

it's just beautiful, absolutely magical

Stolz Cesàro - 1 month, 4 weeks ago

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