Take some point V and draw two rays from it. Choose some other point W in between those two rays. Then, construct a line that touches both rays and passes through W.
Now, this line forms a closed triangle together with the two rays. The point W divides this line into two segments (x1, x2). What is the ratio of these two segments such that the area of the enclosed triangle is minimal?
Does this minimal area even exist?
Let the intersections of the line with
the rays be X and Y such that
x = |WX|
y = |WY|
Let w = |WV| and
a = angle WVX
b = angle WVY
t = angle VWX
WOLOG assume a >= b. Then
w*sin(a)
x = ---------- (1)
sin(t+a)
w*sin(b)
y = ---------- (2)
sin(t-b)
A = Area = (1/2)*w*x*sin(t)
+ (1/2)*w*y*sin(180-t)
= (1/2)*w*(x+y)*sin(t) (3)
Differentiating (1)-(3),
dx -w*sin(a)*cos(t+a)
---- = -------------------- (1')
dt sin(t+a)^2
dy -w*sin(b)*cos(t-b)
---- = -------------------- (2')
dt sin(t-b)^2
dA
0 = ---- = (1/2)*w*[(x+y)*cos(t)
dt dx dy
+ (---- + ----)*sin(t)] (3')
dt dt
Combining (1),(2), and (1')-(3') we get
x = y
Using this and (1) and (2) we get
t = 90 if a = b
tan(t) = 2*tan(b) if a = 90
2*tan(a)*tan(b)
tan(t) = ----------------- if a <> 90
tan(a)-tan(b)
The line XY is therefore contructible with
straightedge and compass.
Edited on April 24, 2010, 6:13 pm
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Posted by Bractals
on 2010-03-30 23:17:18 |
Solution
