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| Title: | Mathematics at DEC |
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| Moderator: | RUSURE::EDP |
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| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
1680.0. "Olympiad Corner No. 131, 13" by RUSURE::EDP (Always mount a scratch monkey.) Mon Oct 26 1992 09:38
For an acute triangle ABC, M is the midpoint of the segment BC, P is a
point on the segment AM such that PM=BM, H is the foot of the
perpendicular line from P to BC, Q is the point of intersection of the
segment AB and the line passing through H that is perpendicular to PB,
and finally R is the point of intersection of the segment AC and the
line passing through H that is perpendicular to PC. Show that the
circumcircle of triangle QHR is tangent to the side BC at the point H.
| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1680.1 | | 3D::ROTH | Geometry is the real life! | Mon Oct 26 1992 13:05 | 5 |
| >> -< Olympiad Corner No. 131, 13 >-
That sounds like the kind of problem Stan would come up with.
- Jim
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| 1680.2 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Mon Oct 26 1992 14:34 | 3 |
| Can you post a diagram? :-)
Dan
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