Argentine National Olympiad, 2015

Let $R$ be the clockwise $90^{\circ}$ rotation about $O$. Apparently $R$ takes $A$ to $B$ and also $R(E') = F'$ for each rotated position $OE'F'$ of the initial right isosceles triangle $OEF$. Hence $R$ takes line $AE'$ to line $BF'$. The angle between a line and its image under any rotation equals the angle of rotation, hence $AE'$ and $BF'$ 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 $AE'$ and $BF'$ for some rotated position $OE'F'$ of $OEF$. A necessary condition is that line $AP$ contains a point at distance $1$ from the origin, point $E'$. Equivalently $AP$ must have a common point with the circle $\beta$ centered at $(0, 0)$ and of radius $1$.

Let $AT$ and $AT'$ be the tangents from $A$ to $\beta$, with $T$ in quadrant 2, $T'$ in quadrant 3. Then each admissible line $AP$ intersects the interior of $T\widehat{A}T'$ or coincides with one of $AT$ and $AT'$. Let $AT \cap \alpha = P_0$, $AT' \cap \alpha = P'_0$. Then all admissible positions of $P$ are contained in the closer minor arc $\widehat{P_0P'_0} = \gamma$ 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 $T\widehat{O}U = 90^{\circ}$ 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 $E' = T$, $F' = U$. (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 $P_0H$ 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 $AP_0H$ yields $y_0 = P_0H = \frac{1}{2}AP_0$. Triangle $ABP_0$ is right at $P_0$ with $\angle BAP_0 = 15^{\circ}$. One expression for $\cos 15^{\circ}$ is $\cos 15^{\circ} = \frac{1}{4}(\sqrt{2} + \sqrt{6})$. Replacing in $y_0 = \frac{1}{2}AP_0 = \frac{1}{2}AB \cos 15^{\circ}$ 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 $AP_0 = a$, $BP_0 = b$, $a > b$. The altitude from $P_0$ to $AB$ in triangle $ABP_0$ equals $h = \frac{AB}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$, by (*). Hence $ab = AB \cdot h = 2\text{área}(ABP_0) = 2\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 2$. Since

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