1. ## Question 5

ImPerfectHacker has requested that I set a problem this week so here it is:

Problem of the Week 5

Given any four distinct points $A,\ B,\ C,\ O$, show that the three angles between the bisectors of $\angle AOB$, $\angle BOC$ and $\angle COA$ are all acute, right or obtuse.

RonL

(Clarification: the bisectors are of the non-reflex angles that correspond to the specified points)

2. Originally Posted by CaptainBlack
InPerfectHacker has requested that I set a problem this week so here it is:

Problem of the Week 5

Given any four distinct points $A,\ B,\ C,\ O$, show that the three angles between the
bisectors
of $\angle AOB$, $\angle BOC$ and $\angle COA$ are all acute, right or obtuse.

RonL
Perhaps I'm obtuse. What three angles?

-Dan

3. Originally Posted by topsquark
Perhaps I'm obtuse. What three angles?

-Dan
There are three lines of interest and they are the bisectors of $\angle AOB$, $\angle BOC$, and $\angle COA$ respectivly.

See the diagram where $B_1,\ B_2,\ B_3$ are the three bisectors.

RonL

4. Originally Posted by CaptainBlack
ImPerfectHacker has requested that I set a problem this week so here it is:

Problem of the Week 5

Given any four distinct points $A,\ B,\ C,\ O$, show that the three angles between the bisectors of $\angle AOB$, $\angle BOC$ and $\angle COA$ are all acute, right or obtuse.

RonL

(Clarification: the bisectors are of the non-reflex angles that correspond to the specified points)
Introduce unit vectors $\bold {u, v, w}$ along $OA, OB$ and $OC$ respectively. Then
the bisectors are collinear with $\bold{u+v, v+w}$ and $\bold{u+w}$. Also:

$
(\bold{u}+\bold v).(\bold v+\bold w) = \bold v.\bold v + \bold u.\bold v + \bold u.\bold w + \bold v.\bold w = 1 +\bold u.\bold v + \bold u.\bold w + \bold v.\bold w
$

$
(\bold u+\bold v).(\bold u+\bold w) = \bold u.\bold u + \bold v.\bold u + \bold u.\bold w + \bold v.\bold w = 1 + \bold v.\bold u + \bold u.\bold w + \bold v.\bold w
$

$
(\bold v+\bold w).(\bold u+\bold w) = \bold w.\bold w + \bold v.\bold u + \bold w.\bold u + \bold v.\bold w = 1 + \bold v.\bold u + \bold w.\bold u + \bold v.\bold w
$

So these three dot products are equal, so in particular are all positive,
zero or negative. Which implies that the angles between the bisectors
are all acute, right or obtuse.

RonL

5. A couple of questions before I try to solve it: (The previous solution used concepts I am unfamiliar with so the problem isn't ruined for me )

Can I assume A,B,C, and O are noncollinier?
Am I trying to prove that all of the angles have the same quality of acute, obtuse, and/or right, or only that none of them are reflexive or lines?
Also, how much geometry is required at mininum to find the proof? I've only taken 1 year.

6. Originally Posted by The Pondermatic
A couple of questions before I try to solve it: (The previous solution used concepts I am unfamiliar with so the problem isn't ruined for me )

Can I assume A,B,C, and O are noncollinier?
You can assume so, I see an exeptional and/or ambiguous cases
if the can be colinear. What I actualy want is the O does not lie on
any of the segmants AB, BC, AC.

Am I trying to prove that all of the angles have the same quality of acute, obtuse, and/or right, or only that none of them are reflexive or lines?
Also, how much geometry is required at mininum to find the proof? I've only taken 1 year.
As I don't know a synthetic proof, I can't say.

RonL