3. (USA 3) Let $A, B$, and $C$ be three points on the edge of a circular chord such that $B$ is due west of $C$ and $A B C$ is an equilateral triangle whose side is 86 meters long. A boy swam from $A$ directly toward $B$. After covering a distance of $x$ meters, he turned and swam westward, reaching the shore after covering a distance of $y$ meters. If $x$ and $y$ are both positive integers, determine $y$.

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3. Let $E$ be the point where the boy turned westward, reaching the shore at $D$. Let the ray $D E$ cut $A C$ at $F$ and the shore again at $G$. Then $E F=$ $A E=x$ (because $A E F$ is an equilateral triangle) and $F G=D E=y$. From $A E \cdot E B=D E \cdot E G$ we obtain $x(86-x)=y(x+y)$. If $x$ is odd, then $x(86-x)$ is odd, while $y(x+y)$ is even. Hence $x$ is even, and so $y$ must also be even. Let $y=2 y_{1}$. The above equation can be rewritten as  $$ \left(x+y_{1}-43\right)^{2}+\left(2 y_{1}\right)^{2}=\left(43-y_{1}\right)^{2} \text {. } $$  Since $y_{1}b>0$ such that $y_{1}=a b$ and $43-y_{1}=a^{2}+b^{2}$. We obtain that $a^{2}+b^{2}+a b=43$, which has the unique solution $a=6, b=1$. Hence $y=12$ and $x=2$ or $x=72$. Remark. The Diophantine equation $x(86-x)=y(x+y)$ can be also solved directly. Namely, we have that $x(344-3 x)=(2 y+x)^{2}$ is a square, and since $x$ is even, we have $(x, 344-3 x)=2$ or 4 . Consequently $x, 344-3 x$ are either both squares or both two times squares. The rest is easy.