| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| 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 |
Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with incenter I. BI and CI meet AC and AB at D and E
respectively. P is the foot of the perpendicular from I to DE, and IP
meets BC at Q. Suppose that IQ = 2 IP. Find angle A.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1962.1 | Answer without derivation | FLOYD::YODER | MFY | Tue Apr 18 1995 17:17 | 3 |
The answer must be pi/3, because the conditions hold when ABC is equilateral. Since nobody would propose an ill-posed problem in Crux Mathematicorum, a fuller demonstration is superfluous. :-) | |||||
| 1962.2 | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Tue Apr 18 1995 19:09 | 4 | |
But you haven't demonstrated that the conditions hold when ABC
is equilateral.
Dan
| |||||
| 1962.3 | Demonstration of .1 | FLOYD::YODER | MFY | Wed Apr 19 1995 11:25 | 7 |
Sorry, I thought it wasn't worth the effort. But here goes: For equilateral ABC, I is the center, and D and E are in the center of their sides. P is situated so that it lies on AI (note DE || BC), and so Q is in the center of BC. So IQ is equal to the inradius; angle EIP is pi/3, and EPI is a right angle, so IP is EI (also the inradius) times sin(pi/3), or 1/2 of the inradius. Thus IQ = 2 IP. | |||||