Sum of roots of trigonometric equation

Question

This is the hardest problem on Georgian (country) high school math exam.

Find all values for parameter $a$ for which the sum of all the roots of the equation:

$$\sin\left(\sqrt {ax-x^2}\right)=0 $$

equal to $100$.

Note that you can't use calculus and we assume only real roots!

Answer 1

We have $a>0$, and the equation reads

$$ax-x^2=k^2\pi^2.$$

By Vieta, when you add the roots in pairs, the sum is $a$.

Hence with $k\ge0$

$$(k+1)a=100$$ with $$a\ge 2k\pi$$

or

$$(k+1)a=100\ge 4(k+1)k\pi.$$

Finally,

$$a=\frac{100}{k+1}$$ with $k=0,1,2.$

Note that as $a$ is rational and $\pi$ transcendental, there is no risk of equal roots.

Answer 2

Hint: what is the sum of the roots of $\sqrt{a x - x^2} = n \pi$?

Answer 3

Suppose the real zeros of $ax-x^2=k\pi$ are $x_{k1}$ and $x_{k2}$, then from the condition we have $$\sum_{\text{all possible $k$}}(x_{k1}+x_{k2})=100.$$

Now since the zeros are real, we also need $a^2-4k^2\pi^2\geq 0$ which gives us an upper bound of the possible $k$:$\frac{a}{2\pi}$, while it is also evident to see the lower bound should be $\frac{a}{2\pi}-1$, then apply Vieta's formula,(note 0 only be counted once, so there are $N+1$ $a$'s.) And we have $$\frac{100}{\frac{a}{2\pi}+1}\leq a\leq \frac{100}{\frac{a}{2\pi}}.$$ A rough estimate gives that $22.12\leq a\leq 25.06$, however since $N$ is an integer, and $a=\frac{100}{N+1}$, so $a=25$.

Note:If $a=25$, then we have $k=0, 1, 2, 3$ and sum all up gives $25*4$ which is 100.