Argentine National Olympiad 2015

Let $R$ be the clockwise $90^{\circ }$ rotation about $O$. Apparently $R$ takes $A$ to $B$ and also $R(E^{\prime })=F^{\prime }$ for each rotated position $O{E}^{\prime }{F}^{\prime }$ of the initial right isosceles triangle $OEF$. Hence $R$ takes line $A{E}^{\prime }$ to line $B{F}^{\prime }$. The angle between a line and its image under any rotation equals the angle of rotation, hence $A{E}^{\prime }$ and $B{F}^{\prime }$ are perpendicular. In other words $\angle APB=90^{\circ }$, meaning that $P$ lies on the circle with diameter $AB$, i.e., on the circle $\alpha$ with center $(-1,1)$ and radius $\sqrt{2}$. Naturally not every point $P\in \alpha$ can be obtained as the intersection of lines $A{E}^{\prime }$ and $B{F}^{\prime }$ for some rotated position $O{E}^{\prime }{F}^{\prime }$ of $OEF$. A necessary condition is that line $AP$ contains a point at distance $1$ from the origin, point $E^{\prime }$. Equivalently $AP$ must have a common point with the circle $\beta$ centered at $(0,0)$ and of radius $1$.

Let $AT$ and $A{T}^{\prime }$ be the tangents from $A$ to $\beta$, with $T$ in quadrant 2, $T^{\prime }$ in quadrant 3. Then each admissible line $AP$ intersects the interior of $T\hat{A}{T}^{\prime }$ or coincides with one of $AT$ and $A{T}^{\prime }$. Let $AT\cap \alpha =P_0$, $A{T}^{\prime }\cap \alpha =P_0^{\prime }$. Then all admissible positions of $P$ are contained in the closer minor arc $\widehat{P_0{P}_0^{\prime }}=\gamma$ of circle $\alpha$ (the arc not containing $A$ and $B$). Note that $P_0$ is in quadrant 1. The entire $\gamma$ is under the line through $P_0$ parallel to the $x$-axis. Hence the $y$-coordinate of an admissible point $P$ does not exceed the $y$-coordinate $y_0$ of $P_0$. In fact $y_0$ is the desired maximum value because $P_0$ is admissible. Indeed let $BU$ be the tangent to $\beta$ from $B$, with $U$ in quadrant 1. Rotations preserve tangency, so, given $R(A)=B$, rotation $R$ takes tangent $AT$ to tangent $BU$. This yields $T\hat{O}U=90^{\circ }$ on the one hand, and $AT\perp BU$ on the other. The latter means that $AT$ and $BU$ intersect on $\alpha$, and since $P_0$ is defined by $AT\cap \alpha =P_0$, we find $AT\cap BU=P_0$. Hence $P_0$ is admissible, with $E^{\prime }=T$, $F^{\prime }=U$. (It follows from the computation below that $P$ is obtained through a $60^{\circ }$-clockwise rotation of $OEF$ about the origin.)

It remains to evaluate $y_0$, i.e., the length of the perpendicular $P_{0}H$ from $P_0$ to $x$-axis. Triangle $OAT$ is right at $T$ with $OA=2$, $OT=1$, therefore $\angle OAT=30^{\circ }$. Hence the right triangle $A{P}_{0}H$ yields $y_0=P_{0}H=\frac{1}{2}A{P}_0$. Triangle $AB{P}_0$ is right at $P_0$ with $\angle BA{P}_0=15^{\circ }$. One expression for $\cos 15^{\circ }$ is $\cos 15^{\circ }=\frac{1}{4}(\sqrt{2}+\sqrt{6})$. Replacing in $y_0=\frac{1}{2}A{P}_0=\frac{1}{2}AB\cos 15^{\circ }$ leads to the answer: $y_{\max }=y_0=\frac{1}{2}(1+\sqrt{3})$.

Remark. Using $\cos 15^{\circ }$ can be avoided by applying the following elementary fact: the hypotenuse of a $15^{\circ }$-$75^{\circ }$-$90^{\circ }$ triangle is $4$ times greater than its respective altitude. (*)

Let the triangle $ABC$ with $\angle C=90^{\circ }$, $\angle B=15^{\circ }$ and altitude $CH=h$. Take the midpoint $M$ of $AB$. It is known that $MA=MB=MC$, so $\angle CMH=\angle MBC+\angle MCB=30^{\circ }$. Thus triangle $MCH$ is $30^{\circ }$-$60^{\circ }$-$90^{\circ }$, hence $MC=2CH=2h$ and $AB=2MC=4h$.

Then the computation of $y_0$ can go as follows. Set $A{P}_0=a$, $B{P}_0=b$, $a>b$. The altitude from $P_0$ to $AB$ in triangle $AB{P}_0$ equals $h=\frac{AB}{4}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}$, by (*). Hence $ab=AB\cdot h=2\text{aˊrea}(AB{P}_0)=2\sqrt{2}\cdot \frac{\sqrt{2}}{2}=2$. Since

$$a^2+b^2=(2\sqrt{2}{)}^2=8,\text{ we obtain }(a\pm b{)}^2=8\pm 4.\text{ Therefore}$$ $$ a+b=2\sqrt{3}, \quad a-b=2 \text{ and so } a=1+\sqrt{3}, \quad y_0=\frac{1}{2}a=\frac{1}{2}(1+\sqrt{3})