## Wednesday, June 1, 2011

### Problem 611: Altitude of a Triangle, Perpendicular, Collinear Points, Parallel

Geometry Problem
Click the figure below to see the complete problem 611.

#### 2 comments:

1. Analytic Proof:
Let D = (o,o), A = (a,0), C = (c,0), B = (0,b)
The equations of AB, DEM are respectively
bx + ay = ab and ax-by = 0
Solving, we have
E = [ab²/(a² + b²) , a²b/(a² + b²)]
Similarly,
F = [b²c/ (b² + c²), bc²/ (b² + c²)]
It can be verified that
the equation of EF is
b (a + c)x – (b² - ac)y = abc
Hence
G = [0, abc/ (ac - b²)] and
so the equation of CG is
abx + (ac - b²)y =abc
Solving it with the equation
ax-by = 0 of DE
we obtain (eliminating x)
y-coord of M = b = y-coord of B
M and B are at the same height over AC
Hence BM is parallel to AC
Similarly BH is parallel to AC
Thus M, B, H are collinear

2. http://img840.imageshack.us/img840/7563/problem611.png

Draw circle diameter BD ( see attached sketch)
From B draw a line parallel to AC . This line cut DE and DF at M and H
AH cut BD at G and MC cut BD at G’ .
By the properties of right triangles we have BH.DC=BD^2= BM.AD
So BM/DC=BH/AD
∆MBG similar to ∆ CDG => BG’/DG’= BM/DC
∆HBG similar to ∆ADG => BG/DG= BH/AD
From above expressions we have BG’/DG’= BG/DG => G coincide to G’
M, B and H are collinear and BH // AC