Official solution: Let $E$ be the point where the boy turned westward, reaching the shore at $D$. Let the ray $DE$ cut $AC$ at $F$ and the shore again at $G$. Then $EF = AE = x$ (because $AEF$ is an equilateral triangle) and $ FG = DE = y$. From $AE \cdot EB = DE \cdot EG$ 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 $y = 2y_1$. The above equation can be rewritten as \[(x + y_1 - 43)^2 + (2y_1)^2 = (43 - y_1)^2.\] Since $y_1 < 43$, we have $\gcd(2y_1, 43-y_1) = 1$, and thus $(|x+y_1-43|, 2y_1, 43-y_1)$ is a primitive Pythagorean triple. Consequently there exist integers $a > b > 0$ such that $y_1 = ab$ and $43 - y_1 = a^2 + b^2$. We obtain that $a^2 + b^2 + ab = 43$, 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 $x(344 - 3x) = (2y + x)^2$ is a square, and since $x$ is even, we have $(x, 344 - 3x) = 2$ or $4$. Consequently $x, 344 - 3x$ are either both squares or both two times squares. The rest is easy.

Amir Hossein wrote: Namely, we have that $x(344 - 3x) = (2y + x)^2$ is a square, and since $x$ is even, we have $(x, 344 - 3x) = 2$ or $4$. Not quite. If $d=\gcd(x,344-3x)$, then $d\mid 344=8\cdot 43$. Since always $2\mid d$ ($x$ is even), we cannot also have $43\mid d$, since then $2\cdot 43 = 86 \mid x < 86$; but we can have $d=2$, $d=4$, or $d=8$. Correspondingly, if $x=dz$, we need have $z,172-3z$ perfect squares (with $z$ odd), or $z,86-3z$ perfect squares (with $z$ odd), or $z,43-3z$ perfect squares. Casework yields $z=1$ (for the first case, so $x=2$), and $z=9$ (for the third case, so $x=72$), as only possibilities. There are other ways, also. Since both $x$ and $y$ must be even, denoting $d=\gcd(x/2,y/2)$, $x=2da$, $y=2db$, with $\gcd(a,b)=1$, the relation obtained writes $a(43-da) = db(a+b)$. This implies $a\mid d$, so $d=ea$, and we can write $43-ea^2 = eb(a+b)$. So $e\mid 43$, and since we cannot have $e=43$ since then $x\geq 86$, we need have $e=1$, thus $43 = a^2+ab+b^2$. Working with the discriminant, it is immediate that the only solutions are $\{a,b\}=\{1, 6\}$. Are you sure the official solution is that first presented, working with Pythagorean triples, and hugely complicating things?