# Question 5

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• Nov 6th 2006, 07:47 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 \$\displaystyle A,\ B,\ C,\ O\$, show that the three angles between the bisectors of \$\displaystyle \angle AOB\$, \$\displaystyle \angle BOC\$ and \$\displaystyle \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, 05:01 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 \$\displaystyle A,\ B,\ C,\ O\$, show that the three angles between the
bisectors
of \$\displaystyle \angle AOB\$, \$\displaystyle \angle BOC\$ and \$\displaystyle \angle COA\$ are all acute, right or obtuse.

RonL

Perhaps I'm obtuse. What three angles?

-Dan
• Nov 7th 2006, 06:44 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 \$\displaystyle \angle AOB\$, \$\displaystyle \angle BOC\$, and \$\displaystyle \angle COA\$ respectivly.

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

RonL
• Nov 12th 2006, 08:23 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 \$\displaystyle A,\ B,\ C,\ O\$, show that the three angles between the bisectors of \$\displaystyle \angle AOB\$, \$\displaystyle \angle BOC\$ and \$\displaystyle \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 \$\displaystyle \bold {u, v, w}\$ along \$\displaystyle OA, OB\$ and \$\displaystyle OC\$ respectively. Then
the bisectors are collinear with \$\displaystyle \bold{u+v, v+w}\$ and \$\displaystyle \bold{u+w}\$. Also:

\$\displaystyle
(\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
\$

\$\displaystyle
(\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
\$

\$\displaystyle
(\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, 02:55 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, 07:39 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