36. Buried Treasures. A certain pirate decided to hide his treasures on the shore of an uninhabited island. Nearby were two boulders $A$ and $B$, and further from the shore grew three coconut palms $C_{1}$, $C_{2}, C_{3}$. Standing at $C_{1}$, the pirate laid out a segment $C_{1} A_{1}$, equal and perpendicular to the segment $C_{1} A$, directing it from the line $C_{1} A$ in the direction opposite to the one where the triangle $A C_{1} B$ was. Similarly, he laid out a segment $C_{1} B_{1}$, equal and perpendicular to the segment $C_{1} B$ and also directed away from the triangle $A C_{1} B$. Then he marked $P_{1}$, the intersection point of $A B_{1}$ and $B A_{1}$. Standing sequentially at $C_{2}$ and $C_{3}$, he marked points $P_{2}$ and $P_{3}$ in a similar manner and, finally, buried his treasures at the center of the circle circumscribed around the triangle $P_{1} P_{2} P_{3}$.

Returning to the island several years later, the pirate found that after a strong hurricane, no trace of the coconut palms remained. How can he find his hidden treasures?
36. Since triangles $A B_{1} C_{1}$ and $A_{1} B C_{1}$ are equal to each other, angle $C_{1} A B_{1}$ is equal to angle $C_{1} A_{1} B$. Further, angle $A P_{1} A_{1}$ is equal to angle

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$A C_{1} A_{1}$, which in turn is equal to $90^{\circ}$. Therefore, angle $A P_{1} B=90^{\circ}$. This means that $P_{1}$ and similarly $P_{2}, P_{3}$ lie on a circle whose diameter coincides with $A B$. Therefore, the treasure is buried at the midpoint $T$ of segment $A B$.