b must be higher than 7^.5 and lower than 5

Let $D,E,F$ be midpoints of $BC,CA,AB$, and $H$ is the orthocenter of $\triangle ABC$. Then the feet of perpendiculars from $H$ to the sides of $\triangle DEF$ are collinear, so $H$ lies on the circumcircle of $\triangle DEF$. But note that the midpoint of $AH$, $H_a$, is the orthocenter of $\triangle AEF$, and lies on the circumcircle of $DEF$. Similarly define $H_b,H_c$, then $H,H_a,H_b,H_c$ lie on a circle thus $H,A,B,C$ lie on a circle. Then now it is easy to show that $\triangle ABC$ must be right. Since $\triangle ABC$ is right, $AC=5\text{ or }\sqrt7$.