AoPS - Problem

Source: 2015 Argentina OMA L3 p3

Tags: geometry, analytic geometry, rotation, coordinate geometry, geometric transformation

parmenides51

20.06.2020 00:09

Consider the points $O = (0,0), A = (- 2,0)$ and $B = (0,2)$ in the coordinate plane. Let $E$ and $F$ be the midpoints of $OA$ and $OB$ respectively. We rotate the triangle $OEF$ with a center in $O$ clockwise until we obtain the triangle $OE'F'$ and, for each rotated position, let $P = (x, y)$ be the intersection of the lines $AE'$ and $BF'$. Find the maximum possible value of the $y$-coordinate of $P$.

Points $E(-1,0),F(0,1)$. We rotate clockwise $F'(\sin \theta,\cos \theta)$ and $E'(-\cos \theta,\sin \theta)$. The line $AE'\ :\ y=\frac{\sin \theta}{-\cos \theta +2}(x+2)$. The line $BF'\ :\ y=2+\frac{\cos \theta-2}{\sin \theta}(x)$. The lines cut in the point $P(\frac{2\sin \theta(\sin \theta+\cos \theta-2}{4\cos \theta-5},\frac{2\sin \theta(\cos \theta-\sin \theta-2}{4\cos \theta-5})$. $y_{P}=f(\theta)$ has a maximum if $f'(\theta)=0$, if $(2\cos \theta-1)(2\cos^{2}\theta-2\sin \theta\cos \theta+4\sin \theta-4\cos \theta+3)=0$, so $\theta=\frac{\pi}{3}$ and the maximum of $y_{P}=\frac{\sqrt{3}+1}{2}$.