Problem of the Week # 276 - Aug 15, 2017

[Ackbach]

Here is this week's POTW:

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Can an arc of a parabola inside a circle of radius 1 have a length greater than 4?

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[Ackbach]

Re: Problem Of The Week # 276 - Aug 15, 2017

This was Problem A-6 in the 2001 William Lowell Putnam Mathematical Competition.

Congratulations to MarkFL for his correct solution, which follows:

Spoiler

I would consider the parabola:

\(\displaystyle y=ax^2-1\) where $0<a$

And given its symmetry about the $x$-axis, I will only consider the right half of the parabola. And so we need to show that this arc-length can be greater than 2 units.

So, the objective function here will be the arc-length of the parabola:

\(\displaystyle f(a,b)=\int_0^b\sqrt{1+4a^2x^2}\,dx=\frac{2ab\sqrt{(2ab)^2+1}+\arsinh(2ab)}{4a}\)

Subject to the constraint:

\(\displaystyle g(a,b)=b^2+\left(ab^2-1\right)^2-1=a^2b^4-(2a-1)b^2=b^2\left(a^2b^2-2a+1\right)=0\)

The objective function is zero for $b=0$ thus, we need only consider the constraint:

\(\displaystyle g(a,b)=a^2b^2-2a+1=0\)

This implies:

\(\displaystyle ab=\sqrt{2a-1}\)

And so the objective function can be written in terms of $a$ alone:

\(\displaystyle f(a)=\frac{2\sqrt{2a-1}\sqrt{(2\sqrt{2a-1})^2+1}+\arsinh(2\sqrt{2a-1})}{4a}=\frac{2\sqrt{(8a-3)(2a-1)}+\arsinh(2\sqrt{2a-1})}{4a}\)

Differentiating, equating the result to zero, and using a numeric method to solve, results in the critical value:

\(\displaystyle a\approx94.0913\)

We then find:

\(\displaystyle f(94.0913)\approx2.00133514884\)

This shows that there are values of $a$ for which the objective function is greater than 2, which means that there are parabolic arcs within a circle of radius 1 whose length is greater than 4.