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Suppose a parabola defined by $y=\lambda x^2$ in the $xy$-plane partitions the unit disk (centered at the origin) into two parts.

Which $\lambda>1$, if any, satisfy that the area of the lower part is exactly $\lambda$ times larger than the area of the upper part?

Of course, it is necessary that the upper area be $\frac{\pi}{\lambda+1}$ and the lower area be $\frac{\pi\lambda}{\lambda+1}$.

So far I've tried computing $\lambda$ directly, i.e. $$\int_I\left[\sqrt{1-x^2}-\lambda x^2\right]\text{d}x=\dfrac{\pi}{\lambda+1},$$ where $I$ is the interval whose endpoints are the abscissae of the points where the parabola and semicircle intersect. I couldn't derive an explicit solution for $\lambda$, but graphing seems to show that no $\lambda>1$ satisfies the equality (and hence no parabola yields the desired partition).

Is this the right conclusion? Any easier or more clever proofs that demonstrate this?

Corellian
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  • It seems that a value of $\lambda$ satisfies the equality (if I made no mistake : see my answer below, to be checked). In order to check your result, would you mind show your calculus and what functions you graphed. – JJacquelin Oct 02 '16 at 10:45
  • @JJacquelin With a different approach I arrived at the same result, so I suppose we can trust it – Hagen von Eitzen Oct 02 '16 at 11:13
  • @JJacquelin Ah, thanks for clarifying the result! I came to the same equality as the one written in your answer, albeit left in terms of $\lambda$ (messy, messy). Guess I made some error when I tried to input both sides into my TI-83 to inspect for intersection. – Corellian Oct 04 '16 at 05:32

2 Answers2

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The abscisse du point d'intersection cercle parabole is $x_m$ $$\sqrt{1-x_m^2}=\lambda x_m^2\quad\to\quad \lambda^2x^4+x^2-1=0$$ $$x_m= \sqrt{\frac{-1+\sqrt{1+4\lambda^2}}{2\lambda^2}}$$ The condition to satisfy is : $$\int_{-x_m}^{x_m}\left(\sqrt{1-x^2}-\lambda x^2\right)dx = \frac{\pi}{\lambda+1}=-\frac{2}{3}\lambda x_m^3+x_m\sqrt{1-x_m^2}+\sin^{-1}(x_m)$$ $$\frac{\pi}{\frac{\sqrt{1-x_m^2}}{x_m^2} +1}=-\frac{2}{3}\frac{\sqrt{1-x_m^2}}{x_m^2} x_m^3+x_m\sqrt{1-x_m^2}+\sin^{-1}(x_m)$$

One cannot express the solution on a closed form. But numerical calculus provides a solution : $$x_m\simeq 0.4688388857...$$ $$\lambda=\frac{\sqrt{1-x_m^2}}{x_m^2}\simeq 4.018398...$$

JJacquelin
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The parabola intersects the circle in the first quadrant in a point $(\cos\phi,\sin \phi)$ with $0<\phi<\frac\pi2$. The upper area consists of a wedge of area $\frac \pi2-\phi$ and two pieces cut off by the parabola from triangles, each of which is (as was already known to Euclid) $\frac13$ of said triangles. In other words, the upper part has area $$A_+=\frac\pi2-\phi+\frac13\sin\phi\cos\phi. $$ Using $\sin\phi=\lambda\cos^2\phi$, we arrive at the condition $$\frac\pi2-\phi-\frac13\sin\phi\cos\phi =\frac\pi{\lambda+1}=\frac{\pi\cos^2\phi}{\sin\phi+\cos^2\phi}$$ For $\phi=0$, the left hnd side is too small; for $\phi=\frac\pi2$ we have equality (but that corresponds to $\lambda=\infty$). By plotting the graph, we spot that there is exactly one more solution, and numerically solving the equality we find $\phi\approx 1.0828205509$ and hence $$\lambda=\frac{\sin\phi}{\cos^2\phi}\approx 4.01839816749552.$$

Hagen von Eitzen
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  • Excellent answer, thank you! Interesting approach using polar coordinates and no calculus. I initially understood all except the bit about the triangles, since I wasn't familiar with that little geometric theorem on parabolic segments. Euclid was quite brilliant. – Corellian Oct 04 '16 at 05:00