How to solve this problem:
Also, most people would use trigonometry, but is there a way to use derivative to solve this too?
[Math] Hard math contest trigonometry type problem
contest-mathgeometrytrigonometry
Related Solutions
note $$\cos{(x+y)}=\cos{[(x+y+z)-z]}=\cos{(x+y+z)}\cos{z}+\sin{(x+y+z)}\sin{z}$$ and $$\cos{(y+z)}=\cos{(x+y+z)}\cos{x}+\sin{(x+y+z)}\sin{x}$$ $$\cos{(z+x)}=\cos{(x+y+z)}\cos{y}+\sin{(x+y+z)}\sin{y}$$ add this three \begin{align*} &\cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}\\ &=(\cos{x}+\cos{y}+\cos{z})\cos{(x+y+z)}+(\sin{x}+\sin{y}+\sin{z})\sin{(x+y+z)}\\ &=p\cos^2{(x+y+z)}+p\sin^2{(x+y+z)}\\ &=p \end{align*}
The trick here is to shift the perpendicular between the two diagonal lines upwards so that two similar triangles are obtained. Then, by applying the Pythagorean theorem, we obtain $$\frac ax=\frac b{\sqrt{b^2+(c-x)^2}}$$ which may be manipulated to obtain a quadratic polynomial for $x$: $$a\sqrt{b^2+(c-x)^2}=bx$$ $$a^2(b^2+(c-x)^2)=b^2x^2$$ $$b^2+c^2-2cx+x^2=\frac{b^2}{a^2}x^2$$ $$\left(1-\frac{b^2}{a^2}\right)x^2-2cx+b^2+c^2=0$$ After solving, you need to check whether the obtained $x$ values are sensible – they have to lie within 0 and $c$.
Best Answer
$$ x=\frac{1200\sqrt{3}}{54}=38.490017\ldots $$
No trig or calculus needed, only geometry.