Official solution
Area of triangle ABC
\[A = {1 \over 2} \times BC \times AM\]
\[ = {1 \over 2} \times 2\sqrt {{r^2} - {x^2}} \times (r + x)\]
\[A = (r + x)\sqrt {{r^2} - {x^2}} \]
\[{{dA} \over {dx}} = \sqrt {{r^2} - {x^2}} - {x \over {\sqrt {{r^2} - {x^2}} }} \times (r + x) \]
\[= {{{r^2} - {x^2} - rx - {x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{{r^2} - rx - 2{x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{ - (x + r)(2x - r)} \over {\sqrt {{r^2} - {x^2}} }}\]
\[{{dA} \over {dx}} = 0 \Rightarrow x = {r \over 2}\]
Sign change of \[{{dA} \over {dx}}\] at \[x = {r \over 2}\]
\[ \Rightarrow \] A has maximum at \[x = {r \over 2}\]
\[BC = 2\sqrt {{r^2} - {x^2}} = \sqrt 3 r\],
\[AM = r + {1 \over 2}r\] = \[{3 \over 2}r\]
\[ \Rightarrow AB = AC = \sqrt 3 r\]