This week's problem went unsolved by the Bradley students. Ray
Kremer submitted a partial solution.
A correct solution was submitted by Philippe Fondanaiche who also
correctly solved the more challenging problems. Sorry, Philippe, the
dollar must go to a Bradley student!
Refer to the diagram in the statement of the problem.
Let P = (x,y), O denote the origin, and
R = (x,0). The triangles TOQ and PRQ are
similar, giving us
or
|
T =
|
sy
s -
x
|
=
|
sy(s + x)
s2 - x2
|
.
|
|
From the right triangle POR we get x2 +
y2 = s2 which after substituting
and simplifying gives
|
T =
|
|
|
xn(Ö(x2
+ x2n))
(x + Ö(x2 +
x2n))
x2n
|
=
|
1 + x2n - 2 + Ö(1 + x2n - 2)
xn - 2
|
|
|
Now look at cases. If n < 2,
then 2 - n is positive, so T = ( 1
+ x2n - 2 + Ö(1 +
x2n - 2) )
x2 - n with the first term going to 2 and the
second term going to 0, as x ® 0.
So T ® 0 in this case. If n
> 2, the term
|
|
1 + Ö(1 +
x2n - 2)
x2n
- 2
|
® ¥
|
|
since the numerator approaches 2, as x goes to 0, while the denominator
gets arbitrarily small. On the other hand
x2n - 2
xn - 2
|
= xn ®
0 as x ® 0.
|
|
So in this case T ®
¥. Finally, if n = 2, then
T = 1 + x2 + Ö(1 +
x2) ® 2.
Finally we have the answer.
|
T ®
|
| 0, |
if n < 2 |
| 2, |
if n = 2 |
| ¥ |
if n > 2 |
|
This proves, I guess, that 2 is between 0 and ¥.
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Page last updated 23 April 1998.