The problem below was posed by Hadamard:
Given $\angle AOB$, and its bisector $\angle AOM$, show that for any point $C \neq O$, $\angle COM$
- is one half the difference of $\angle COA$ and $\angle COB$ if $C$ is inside $\angle AOB$
- is the
supplement of half the difference if $C$ is inside angle $\angle A'OB'$ which is vertical
to $\angle AOB$- and is one half the sum of $\angle COA$ and $\angle COB$ if $C$ is anywhere else.
Note: $A', B', M'$ are the reflection of $A,B,M$ through $O$.
I request verification, critique, and suggestions on my solution below:
Notation: $\angle$ denotes angle measure, which is always between $0$ and $\pi$ inclusive.
Solution:
We solve this by letting point $C$ start at a simple location, and examine how the angles vary as $C$ moves. Let $C$ start coincident with $B$. Then $\angle COB = 0$ and $\angle COM = \frac {\angle COA} 2 = \frac {\angle COA + \angle COB} 2$.
Now, let $C$ move in a circular arc in the direction away from $A$ (that is, away from the interior of $\angle AOB$). For every angular step $s$ that $C$ takes on this arc, we have $$\Delta \angle COA = \Delta \angle COM = \Delta \angle COB = s$$ so the identity $$\angle COM = \frac {\angle COA + \angle COB} 2$$ still holds. This remains the case until $C$ crosses $A', M', B'$, or $A$.
When $C$ reaches $A'$, we have
$$\begin{align*}
\angle COA &= \pi \\
\angle COM &= \frac {\pi + \angle COB} 2 \\
&= \pi – \frac {\angle COA – \angle COB} 2 \\
&= \pi – \frac {|\angle COA – \angle COB|} 2 \quad \text{(since $\angle COA > COB$)} \\
\end{align*}$$ so $\angle COM$ is supplementary to half the difference. As $C$ continues to step on its arc, we have $\Delta \angle COA = -s$ (since $\angle COA$ is now shrinking) and $\Delta \angle COB = \Delta \angle COM = s$, so $\angle COM = \pi – \frac {|\angle COA – \angle COB|} 2$ continues to hold.
This remains the case as $C$ reaches $M'$, at which point $\angle COA = \angle COB$, so $\angle COM = \pi – \frac {|\angle COA – \angle COB|} 2 = \pi – \frac {\angle COB – \angle COA} 2$. As $C$ continues to step, we have $\Delta \angle COA = -s = \Delta \angle COM$ and $\Delta \angle COB = s$, so this continues to hold.
Finally, when $C$ reaches $B'$, we have $\angle COB = \pi, \angle COM = \pi – \frac {\angle COB – \angle COA} 2 = \frac {\angle COA + \pi} 2 = \frac {\angle COA + \angle COB} 2$. And, as $C$ continues to step, $\Delta \angle COA = \Delta \angle COM = \Delta \angle COB = -s$, so this continues to hold until $C$ reaches $A$. At this point, it is easy to verify that as $C$ enters the interior, $\angle COM = \frac {|\angle COA – \angle COB|} 2$.
The null angle case of $B = A$ follows the above as well, whereas the straight angle case of $B = A'$ does not have a well-defined bisector and must be excluded, completing the proof.
Discussion:
I first tried to solve this using directed angle measure, in which I believe the identity $2\measuredangle COM = \measuredangle COA + \measuredangle COB$ holds true regardless of where $C$ is. I expected this to greatly simplify things, but, despite my best effort, I failed to map this to absolute (i.e. nonnegative) angle measures. I'd be grateful if anyone could explain how to use directed angles here or explain why they won't work.
Instead, the approach I used above was inspired by Hadamard's technique of viewing geometry problems as dynamic rather than static: find a configuration where the result is easy, and explore how the result changes as things move.
Questions:
- Is my proof correct?
- Is the approach good? Is there a simpler or better approach?
- How about the exposition? While it's quite long, the number of cases prevented a shorter exposition. Can it be improved (or shortened)? Is it clear?
- What are your thoughts on the approach I tried but dropped of using directed angle measure?
Best Answer
We take given lines as given or fixed inclined at $2 \beta$ to each other and radius vector at C makes variable angle $\theta$ in the modern anticlockwise direction convention/sense which we consistently use.
The same sign convention applies to angular velocities $ \dfrac{d\theta}{dt}= \omega$ as positive in dynamics.
Then the first case
$$ \frac{(\beta-\theta)-(\beta+\theta)}{2}= - \theta$$
meaning that we need to rotate in the clockwise direction when starting at A to reach B.
In the second case we rotate radius vector at C by $\pi$ so we can add this rotation algebraically:
$$ \frac{(\beta-\theta)-(\beta+\theta)}{2}=\pi - \theta$$
The third case makes no sense by adopting same consistent convention.