A geometry problem by Kshetrapal Dashottar

For $\Delta ABC$ let $|AB| = \sqrt{2}$ and $|BC| = \sqrt{3}$ . Also, let $M,N$ be the respective midpoints of $AB$ and $BC$ , and let $O$ be the centroid, i.e., the point where the medians intersect.

Now the centroid of any triangle divides each of the medians in a $2:1$ ratio. So letting $|CM| = x$ and $|AN| = y$ , and noting that with the medians intersecting at a right angle we have that $\Delta AOM$ and $\Delta CON$ are right triangles, we apply Pythagoras to these two triangles to find that

$\left(\dfrac{2}{3}y\right)^{2} + \left(\dfrac{1}{3}x\right)^{2} = \left(\dfrac{\sqrt{2}}{2}\right)^{2}$ and $\left(\dfrac{1}{3}y\right)^{2} + \left(\dfrac{2}{3}x\right)^{2} = \left(\dfrac{\sqrt{3}}{2}\right)^{2}$ .

Adding these equations together yields that $\dfrac{5}{9}(y^{2} + x^{2}) = \dfrac{5}{4} \Longrightarrow x^{2} + y^{2} = \dfrac{9}{4}$ .

But as $\Delta AOC$ is also a right triangle we see that

$|AC|^{2} = \left(\dfrac{2}{3}y\right)^{2} + \left(\dfrac{2}{3}x\right)^{2} = \dfrac{4}{9}(y^{2} + x^{2})$ ,

and so $|AC|^{2} = \dfrac{4}{9} \times \dfrac{9}{4} = 1 \Longrightarrow |AC| = \boxed{1}$ .