9. From a point $L$ two straight roads depart, forming an acute angle $\alpha$. Along one of the two roads there are two street lamps, positioned at $P$ and $Q$, such that $L P=40 \mathrm{~m}$ and $L Q=90 \mathrm{~m}$. Eva is at $E$ on the other road, and sees the two street lamps under an angle $P \widehat{E} Q$. How far from $L$ is Eva, if $P \widehat{E} Q$ has the maximum possible amplitude?
(A) $40 \mathrm{~m}$
(B) $60 \mathrm{~m}$
(C) $65 \mathrm{~m}$
(D) $90 \mathrm{~m}$
(E) the distance depends on $\alpha$.
9. The answer is (B). Let $f$ be the road where Eva is located and $s$ the one where the street lamps are. Consider the circle passing through $P, Q, E$, which exists because $P, Q$ belong to $s$ while $E$ does not belong to it, otherwise the angle $P \hat{E} Q$ would be 0, and suppose for contradiction that it is not tangent to $f$. Then there exists a point $E^{\prime}$ of further intersection of this circle with $f$ and let $E^{\prime \prime}$ be a point on the arc $E E^{\prime}$ not containing $P$ and $Q$, we have $P \hat{E} Q = P \hat{E}^{\prime \prime} Q$ since they subtend the same arc $P Q$; let $D$ be the intersection of $f$ and $P E^{\prime \prime}$, then $P \hat{D} Q > P \hat{E}^{\prime \prime} Q = P \hat{E} Q$ because it is an exterior angle of the triangle $P D E^{\prime \prime}$ not adjacent to the angle at $E^{\prime \prime}$. Therefore, the point $E$ could not have given the maximum angle, and thus the circle through $P, Q, E$ must necessarily be tangent to $f$ at $E$. But then, by the tangent-secant theorem, $L E^{2} = L P \cdot L Q$, so $L E = \sqrt{40 \cdot 90} = 60$.

(Kuzmin)