Similar Triangles and Centroids

Lemma: $\triangle M_1OM_2\sim \triangle AOB\sim \triangle BOC$

Proof: let the midpoints of $AB$ and $BC$ be $P_1$ and $P_2$ , respectively. Note that $\triangle P_1OP_2$ and $\triangle M_1OM_2$ are related by a homothety with center $O$ and scale $\dfrac{3}{2}$ , so they are similar. Thus it remains to prove that $P_1OP_2\sim AOB$ .

Note that $P_1O$ and $P_2O$ are the medians from $O$ of $\triangle AOB$ and $\triangle BOC$ respectively. Thus $\dfrac{P_1O}{P_2O} = \dfrac{AO}{BO}$ .

Also, note that $\angle P_1OP_2 = \angle P_1OB + \angle BOP_2 = \angle P_1OB+\angle AOP_1=\angle AOB$ .

Thus, $\triangle P_1OP_2 \sim \triangle AOB$ and we proved our lemma.

---

Since $\triangle M_1OM_2$ is right, then that implies that $\triangle AOB$ and $\triangle BOC$ are also right.

Since $M_1M_2=AB$ , it means $\triangle M_1OM_2\cong \triangle AOB$ .

Now there are two cases: $\angle AOB=90^{\circ}$ or $\angle OAB = 90^{\circ}$ , because clearly $\angle ABO\ne 90^{\circ}$ .

Case 1: $\angle AOB=90^{\circ}$

Let $M_1O=AO=x$ . Note that $P_1O=\dfrac{3}{2}x$ . Since $P_1A=P_1B=P_1O$ , then $AB=3x$ . Finally, by Pythagorean Theorem, $BO = 2x\sqrt{2}$ .

This gives $\cos^2 \angle ABO = \left(\dfrac{2\sqrt{2}}{3}\right)^2=\dfrac{8}{9}$

Case 2: $\angle OAB = 90^{\circ}$

Let $M_1O=AO=x$ . Note that $P_1O=\dfrac{3}{2}x$ . By Pythagorean Theorem, $P_1A = \dfrac{\sqrt{5}}{2}x$ so $AB = x\sqrt{5}$ . By Pythagorean Theorem, $BO = x\sqrt{6}$ .

This gives $\cos^2 \angle ABO = \left(\dfrac{\sqrt{5}}{\sqrt{6}}\right)^2=\dfrac{5}{6}$

Finally, $\dfrac{8}{9}+\dfrac{5}{6}=\dfrac{31}{18}$ so our answer is $31+18=\boxed{49}$ .