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Jules
science forum beginner
Joined: 30 Jul 2005
Posts: 9
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 Posted: Sun May 21, 2006 11:23 pm Post subject:
Geometry Problem
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Hi All:
Solving a geometry problem I run into some difficulties.
There is a triangle ABC with its incircle, O as a center. The points of
tangency with AC and BC are M and N respectively. BO intersects MN at P.
My text book starts solving the problem which is larger, saying that the
trick to solve it is realizing that angle APB is 90 degrees.
I had to run over complex trigonometric substitutions to show that this
angle is indeed 90 degrees. Is there any other simpler way to demonstrate
it?
Thanks in advance. |
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BSK
science forum beginner
Joined: 16 Jan 2006
Posts: 14
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| Posted: Mon May 22, 2006 11:05 am Post subject:
Re: Geometry Problem
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| Quote: | Hi All:
Solving a geometry problem I run into some
difficulties.
There is a triangle ABC with its incircle, O as a
center. The points of
tangency with AC and BC are M and N respectively. BO
intersects MN at P.
My text book starts solving the problem which is
larger, saying that the
trick to solve it is realizing that angle APB is 90
degrees.
I had to run over complex trigonometric substitutions
to show that this
angle is indeed 90 degrees. Is there any other
simpler way to demonstrate
it?
Thanks in advance.
|
(angle MNO) = (angle ACB)/2
(angle NPB) = Pi/2 - (angle ACB)/2 - (angle CBA)/2 =
= (angle BAC)/2 = (angle MAO)
(angle MAO) + (angle MPO) = Pi
Therefore the points A, M, P and O are cyclic (?) and
(angle APO) = (angle AMO) = Pi/2 |
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Jules
science forum beginner
Joined: 30 Jul 2005
Posts: 9
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| Posted: Mon May 22, 2006 3:12 pm Post subject:
Re: Geometry Problem
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Thanks Pal, it was an elegant demonstration.
"BSK" <sasa@ict.sc.ru> wrote in message
news:15062294.1148295967623.JavaMail.jakarta@nitrogen.mathforum.org...
| Quote: | Hi All:
Solving a geometry problem I run into some
difficulties.
There is a triangle ABC with its incircle, O as a
center. The points of
tangency with AC and BC are M and N respectively. BO
intersects MN at P.
My text book starts solving the problem which is
larger, saying that the
trick to solve it is realizing that angle APB is 90
degrees.
I had to run over complex trigonometric substitutions
to show that this
angle is indeed 90 degrees. Is there any other
simpler way to demonstrate
it?
Thanks in advance.
(angle MNO) = (angle ACB)/2
(angle NPB) = Pi/2 - (angle ACB)/2 - (angle CBA)/2 =
= (angle BAC)/2 = (angle MAO)
(angle MAO) + (angle MPO) = Pi
Therefore the points A, M, P and O are cyclic (?) and
(angle APO) = (angle AMO) = Pi/2 |
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