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Let $O$ be the orthocenter (intersection of heights) of the triangle $ABC$. If $\overline{OC}$ equals $\overline{AB}$, find the angle $\angle$ACB.

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Position the circumcenter $P$ of the triangle at the origin, and let the vectors from the $P$ to $A$, $B$, and $C$ be $\vec{A}$, $\vec{B}$, and $\vec{C}$. Then the orthocenter is at $\vec{A}+\vec{B}+\vec{C}$. (Proof: the vector from $A$ to this point is $(\vec{A}+\vec{B}+\vec{C})-\vec{A} = \vec{B}+\vec{C}$. The vector coinciding with the side opposite vertex $A$ is $\vec{B}-\vec{C}$. Now $(\vec{B}+\vec{C})\cdot(\vec{B}-\vec{C}) = |\vec{B}|^2 - |\vec{C}|^2 = R^2-R^2 = 0$, where $R$ is the circumradius. So the line through $A$ and the head of $\vec{A}+\vec{B}+\vec{C}$ is the altitude to $BC$. Similarly for the other three altitudes.)

Now the vector coinciding with $OC$ is $\vec{O}-\vec{C}=\vec{A}+\vec{B}$. Thus $|OC|=|AB|$ if and only if $$|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$$ if and only if $$\vec{A}\cdot\vec{A} + \vec{B}\cdot\vec{B} + 2\vec{A}\cdot\vec{B} = \vec{A}\cdot\vec{A} + \vec{B}\cdot\vec{B} - 2\vec{A}\cdot\vec{B}$$ if and only if $$4\vec{A}\cdot\vec{B} = 0$$ if and only if $$\angle APB = \pi /2$$ if and only if $$\boxed{\angle ACB = \pi/4 = 45^\circ}.$$

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Let point $P$ on $AC$ be the foot of the perpendicular $BO$, and note that $\angle OCA$ is the complement of $A$. Then, $$\begin{eqnarray} |AC|&=&|AP|+|PC|\\ &=&|AB|\cos A+|OC|\cos\angle OCA \\ &=&|AB| \cos A+|OC| \sin A \\ &=&|AB|(\cos A+\sin A) \end{eqnarray}$$ Conveniently scaling to unit circumdiameter ---so that $|AC| = \sin B$, $|AB| = \sin C$, and $|BC| = \sin A$ (which we may assume is non-zero)--- we have

$$\begin{eqnarray} \sin B &=& \sin C \; (\cos A+\sin A) \\ \implies\sin(A+C) &=& \cos A \sin C + \sin A \sin C \\ \implies\sin A \cos C + \cos A \sin C &=& \cos A \sin C + \sin A \sin C \\ \implies\sin A \cos C &=& \sin A \sin C \\ \implies\cos C &=& \sin C \\ \implies C &=& \pi/4 \end{eqnarray}$$

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Assuming there is an answer, then it is $45^\circ$ or $\pi/4$, as a symmetric right-angled triangle (half a square), where $C$ is not the right angle, satisfies this: the orthocentre is at the right angle.

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