Suppose that $f:[0,1]\rightarrow R$ is a continuously differentiable function such that $f(0)=f(1)=0$ and $f(a)=\sqrt3$ for some $a\in (0,1)$.Prove that there exist two tangents to the graph of $f$ that form an equilateral triangle with an appropriate segment of the $x-$ axis.
My Attempt
I need to show that there exist two points on the graph of the function where values of derivatives are $-\sqrt3$ and $\sqrt3$. I feel that somehow Intermediate value theorem should work but am not sure how.
Let $a'$ be the maximum of $f$. So $f(a')\ge f(a) = \sqrt 3$. Thus $a'\in (0,1)$. Also $f'(a') = 0$.
By the mean value theorem, there is $b\in (0,a') $ so that
$$ f'(b) = \frac{f(a') - f(0)}{a'-0}> \sqrt 3/a' >\sqrt 3$$
similarly, there is $c\in (a', 1)$ so that
$$ f'(c) = \frac{f(1)- f(a')}{1-a} < -\sqrt 3/(1-a') < -\sqrt 3.$$
Since $f$ is continuously differentiable, $f'$ is continuous. Since $f'(a') = 0$, there is $a_1 \in (b, a')$ and $a_2 \in (a', c)$ so that
$$ f'(a_1) = \sqrt 3, \ \ \ f'(a_2) = -\sqrt 3.$$
Remark we used only the intermediate value property of $f'$, which holds true even without assuming that $f$ is continuously differentiable.
