| 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.
ABCD is a square with incircle G. A tangent l to G meets the sides AB
and AD and the diagonal AC at P, Q and R respectively. Prove that
AP AR AQ
-- + -- + -- = 1.
PB RC QD
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 2032.1 | pedestrian algebraic method | TDCIS4::ROTH | Geometry is the real life! | Mon Mar 25 1996 14:38 | 30 |
Suppose G is the unit circle.
The equation of the line l tangent to G at (c,s)
is x c + y s = 1; this line meets the other lines
x = 1, y = 1, and x = y giving very simple equations,
so you can just write down the sum as
1 - (1 - c)/s 1 - (1 - s)/c 1 - 1/(s + c)
------------- + ------------- + -------------
1 + (1 - c)/s 1 + (1 - s)/c 1 + 1/(s + c)
s + c - 1 c + s - 1 s + c - 1
--------- + --------- + ---------
s - c + 1 c - s + 1 s + c + 1
1 1 1
(s + c - 1) ( --------- + --------- + --------- )
s - c + 1 c - s + 1 c + s + 1
and stuff just collapses to
1 1
(s + c - 1) ( ----- + --------- ) = 1
c s c + s + 1
I'm more curious about how they thought of it, and if there
is a way to think of parallel resistors or barycentric coordinates
giving this kind of relation. That would be kind of neat.
- Jim
| |||||