A geometry problem by Ilham Saiful Fauzi

Using Vieta's formula, $\small{\tan p+\tan q=-a,\tan p\tan q=2016}$ .

$\small{\tan 2(p+q)=\dfrac{\tan p+\tan q}{1-\tan p \tan q}=\dfrac a{2015}}$

$\small{\sin 2(p+q)=\dfrac{2\tan (p+q)}{1+\tan^2 (p+q)}=\dfrac{\frac{2a}{2015}}{(1+\tan^2(p+q)}}$

$\small{\cos 2(p+q)=\dfrac{1-\tan^2(p+q)}{1+\tan^2(p+q)}=\dfrac{\frac{1}{2015}\left(2015-\frac{a^2}{2015}\right)}{1+\tan^2(p+q)}}$ Write the given expression as: $\mathfrak {F}=a\sin(2(p+q))+4030\underbrace{\cos^{2}(p+q)}_{\dfrac{1+\cos 2(p+q)}{2}}$

Substitute values of $\sin 2(p+q)$ and $\cos 2(p+q)$ found previously so that : $\mathfrak F=\boxed{4030}$

Using Vieta's formula, $\small{\tan p+\tan q=-a,\tan p\tan q=2016}$ .

$\small{\tan 2(p+q)=\dfrac{\tan p+\tan q}{1-\tan p \tan q}=\dfrac a{2015}}$

This gives us $2015 \tan(p+q) = a$

$a\sin(2(p+q))+4030\cos^{2}(p+q) = 2015 \tan(p+q)sin(2(p+q))+4030\cos^{2}(p+q)$

But since $\tan(p+q) = \frac{\sin(p+q)}{\cos(p+q)}$ and $\sin(2(p+q)) = 2\sin(p+q)\cos(p+q)$ ,

Therefore $2015 \tan(p+q)sin(2(p+q))+4030\cos^{2}(p+q) = 4030 (\frac{\sin(p+q)}{\cos(p+q)})( \sin(p+q)\cos(p+q))+4030\cos^{2}(p+q)$

$= 4030(\sin^2(p+q)+\cos^2(p+q)) = 4030$ (Since $\cos^2 \theta + \sin^2 \theta = 1$ for all $\theta$ )

Therefore our answer is $\boxed{4030}$