Orthocentre, circumcentre, midpoint of a side and foot of altitude form a rectangle A rectangle $HOME$ has sides $HO=11$ and $OM=5$. A triangle $ABC$ has $H$ as intersection of altitudes, $O$ as the circumcentre, $M$ as the midpoint of $BC$ and $E$ as the foot of altitude from $A$. Find the length of $BC$.
I need some hints to start off with the problem. Also, a diagram would help. Thanks.
 A: Here is a diagram.

The altitudes are obvious. $M$ is the midpoint of $BC$, $G$ is the midpoint of $AB$, and $I$ is the midpoint of $AC$.
Here is a hint to solve your problem. We construct that diagram in a different way, starting with the rectangle (whose details we know).

Construct rectangle $HOME$ with $HO=EM=11$, $HE=MO=5$. Construct lines $\overleftrightarrow{HE}$ and $\overleftrightarrow{ME}$.
Construct arbitrary point $B$ on ray $\overrightarrow{ME}$. We need $M$ to be the midpoint of $\overline{BC}$, so construct circle $c$ with center $M$ and radius $BM$ and point $C$ the intersection of circle $c$ with ray $\overrightarrow{EM}$.
Point $O$ needs to be the circumcenter of $\triangle ABC$ and thus equidistant from points $A$ and $C$. Therefore construct circle $d$ with center $O$ and radius $CO$ and point $A$ the intersection of circle $d$ with line $\overleftrightarrow{HE}$. Construct $\triangle ABC$. We are guaranteed that point $O$ is its circumcenter.
But we also want point $H$ to be its orthocenter (intersection of the altitudes). Construct point $F$ as the intersection of lines $\overleftrightarrow{CH}$ and $\overleftrightarrow{AB}$. If lines $\overleftrightarrow{CH}$ and $\overleftrightarrow{AB}$ are perpendicular at $F$, $H$ is indeed the intersection of the altitudes. 
We want to find the position of point $B$ that makes this so.  Therefore define $r=BE$. You can now find the values of $CM$, $CE$, $OC$ and $OA$, $AH$, and $AE$. (You asked for only a hint, so I'll leave that to you.) Those desired lines are perpendicular iff triangles $\triangle BEA$ and $\triangle HEC$ are similar, so set the proportion
$$\frac{CE}{HE}=\frac{AE}{BE}$$
and solve for $r$. (That involves solving a factorable fourth-degree polynomial equation.) From that you easily find the value of $BC$.
Ask if you need the details or the final answer.
