# Question 5

• Nov 6th 2006, 08:47 PM
CaptainBlack
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)
• Nov 7th 2006, 06:01 AM
topsquark
Quote:

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
• Nov 7th 2006, 07:44 AM
CaptainBlack
Quote:

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
• Nov 12th 2006, 09:23 PM
CaptainBlack
Quote:

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
• Nov 20th 2006, 03:55 PM
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?
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.
• Nov 20th 2006, 08:39 PM
CaptainBlack
Quote:

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.

Quote:

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