11. Let $A B C$ be an equilateral triangle with unit side length, and let $P$ be a point on the side opposite the line $A B$ with respect to point $C$, such that the angle $\widehat{A P B}$ measures $60^{\circ}$. Suppose that the angle bisector of $\overline{A P B}$ intersects the segments $A B$ and $A C$ at points $X$ and $Y$, respectively. What is the minimum possible value for the area of triangle $A X Y$?
(A) $\frac{1}{3 \sqrt{3}}$
(B) $\frac{1}{2 \sqrt{3}}$
(C) $\frac{\sqrt{3}}{12}$
(D) $\frac{\sqrt{3}}{8}$
(E) None of the above.
11. The answer is $\mathbf{( A )}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_04503179b337e2a9d877g-10.jpg?height=654&width=962&top_left_y=747&top_left_x=593)

Let $O$ be the center of the triangle. The point $P$ lies on the arc of the circumcircle of $A O B$ outside the triangle by hypothesis. The bisector of the angle $A \hat{P} B$ passes through the midpoint of the arc $A B$ opposite to the one on which it lies, which is precisely $O$.

![](https://cdn.mathpix.com/cropped/2024_04_17_04503179b337e2a9d877g-10.jpg?height=554&width=643&top_left_y=1648&top_left_x=752)

As the lines passing through $O$ and intersecting the segments $A B$ and $A C$ at $X$ and $Y$ vary, the one for which the area of triangle $A X Y$ is minimized is the one parallel to $B C$. Let $X^{\prime}$ and $Y^{\prime}$ be the intersections corresponding to this latter line, and assume, by symmetry, that $B XO X^{\prime}=O Y^{\prime}>O Y$.

The minimum possible value for the area of triangle $A X Y$ is thus

$$
\frac{4}{9} \cdot[A B C]=\frac{4}{9} \cdot \frac{\sqrt{3}}{4}=\frac{1}{3 \sqrt{3}}
$$

Alternatively. Suppose we fix a Cartesian coordinate system with origin at $O$. The points $X$ and $Y$ depend linearly on the slope of the line through $O$, so the area of $A X Y$ depends quadratically on this parameter: hence the minimum can only be at the vertex of the parabola, which by symmetry must correspond to the line parallel to $B C$, or at the endpoints, one of which corresponds to the altitude from $B$. In the first case, the area is $4 / 9$ that of $A B C$, in the other it is half.