Solution
by Maureen P. Cox & Albert
White, St. Bonaventure, NY.
First,
we look for a, b and c such that x4
+ 2x3 + x2 -2x
-1 + (ax + b)2 = (x2+
x+ c)2. Expanding both sides and equating like terms we have a2
= 2c, ab ö 1= c,
b2- 1 = c2. Since
a2b2 = 2c(c2
+ 1) and ab = 1+ c we have (1 + c)2 =
2c(c2+1), that is, 2c3
- c -1 = 0. A solution to this
is c = 1 and therefore
a = b
= …2
Adding
(…2x
+ …2)2
to both sides of x4 + 2x3 + x2
-2x -1 = 0 we have (x2+
x+ 1)2 = (…2x
+…2)2
This yields x2+
x+ 1 = …2x
+ …2
and x2+ x+ 1 = -(…2x + …2). Using the quadratic
formula we have the four solutions:
Also
solved by: Michael Brozinsky, Central Islip, NY; Paul M. Harms, North Newton,
KS; Armend Shabani, Prishtina, Kosova
Also
solved by: William T. Bailey, Blacksburg, VA; Michael BrozinÐsky, Central
Islip, NY; Maureen P. Cox and Albert White, St. Bonaventure, NY; Paul M.
Harms, North Newton, KS; N.J. Kuenzi, Oshkosh,WI; Armend Shabani, Prishtina,
Kosova; Brian D. Beasley,
Clinton,
SC
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4694:
When we progress from the
natural numbers N to the integers Z,
and from integers to the rational numbers Q, and from the rational numÐbers
to the real numbers R, and from the
real numbers to complex numbers C, we
ãgainä something, and we ãloseä something. E.g., when we progress from
the real numbers to the complex numbers, we lose the notion of ãorÐder.ä
Make a list of what we gain and what we lose in each step of the chain:
N¨Z¨Q¨R¨C
Solution
by Editor
Some
things are obvious: E.g., in each new set we gain the ability to solve certain
equations;
x2-
2 = 0 can be solved precisely in R but
not in Q; x2 +1 = 0 can be solved in C
but not even be approximated in R, etc.
But other notions are less obvious: E.g., In moving from N
to Z we lose the notion that
every non-empty subset of N has a
least element; but we gain the notion that every element having in Z
has a unique successor and a unique predecessor; in going from Z
to Q we lose the notion of every element having a unique successor, but we
gain the notion of denseness; we also gain an example of a field. In moving
from Q to R we lose the notion of
countability and commensurability of segment lengths, but we gain the notions
of p-adic field extensions and of completeness. And in moving from R
to C we lose the notion of order
but we gain the notion of the modulus of a number, the theory of analytic
functions of a complex variable and the basis for the theory quaternions.
4695:
Let
P be a point on the circumcircle of DABC.
Let L
be the reflection
of
P through side AB,
let M be the reflection of P through
side BC, and let
N
be the reflection of P
through AC. Prove that the points L, M
and N are
collinear.
Solution
by Editor
Let
Lâ, Nâand Mâ
be the perpendicular distances from P
to the sides AB, BC and AC respectively.
It suffices to show that the points Lâ,
Nâ and Mâ are collinear.
Angles PLB, PMC, and PNA are right angles. We will show that ÜPNM =
ÜPNL.
Construct
lines PA, PB and PC.
Quadrilateral PNBL is cyclic, so ÜPNL
= ÜPBL. It
is given that quadrilateral ABCP is
cyclic, so ÜPBL =
ÜPAC. In
right triangle PAM, angles PAC and MPA are
complementary. Quadrilateral ANMP is
cyclic so: ÜANM
= 90¡
+
ÜPNM =
180¡
- ÜMPA ÜPNM =
90¡-
ÜMPA =
ÜPAC. So
ÜPNM =
ÜPNL since
each is equal to ÜPBL.
4696:
Suppose a camel has to transport 3000 bananas to a market 1000km away. But the
camel cannot carry more than 1000 bananas at any given time, and for each
kilometer that it walks (or fraction thereof), it eats a banana (or the
corresponding fraction of it). Find the maximal number of bananas that can
reach the market.
Solution
by N.J. Kuenzi, Oshkosh, WI.
Since
the camel cannot carry more than 1000 bananas at any given time the camel must
travel from the origin O
to some point A along the route, drop off a supply of bananas at A, and
then return to O.
If x is the distance from O
to A in kilometers, then the supply of bananas deposited at A after one trip
will be 1000 - 2x. The preceding process is then repeated a second time
accumulating 2000 - 4x bananas at point A with 1000 left at O.
The camel then leaves O
for the last time with the remaining 1000 bananas and travels to point A. The
number of bananas now at point A is 3000 - 5x. In order to minimize that part
of the route which must traveled 5 times, point A should be chosen so that
1000 < 3000 - 5x <2000.
Next
leave point A with the camel carrying about half of the bananas, travel to
point B, drop off a supply of bananas at B, and return to point A. Load the
remaining bananas at A on the camel and travel to B. If y is the disÐtance
from A to B in kilometers, then the number of bananas now at point B is (3000
- 5x) - 3y. Choose B so that (3000 - 5x) - 3y £
1000. The distance from B to the Market is 1000 - (x +
y). So the number of bananas the camel can get to the Market is: (3000 -
5x - 3y) - (1000 - (x
+ y)) = 2000 - 4x - 2y = 2000 - (4x
+ 2y). Thus to get the largest number of bananas to the market, 4x +
2y should be as small as possible subject to the constraints 200 £
x <400, y > 0 and 3000 - 5x - 3y £
1000. Using the graphical techÐnique to solve this linear programming problem
yields the optimal solution when x = 200 and y = 333⅓.
Thus using the strategy outlined above the maximum number of bananas that the
camel can deliver to the market is: 2000 ö 800 - 666⅔
= 533⅓.
4697:
What is the millionth term of the sequence 1, 2, 2, 3, 3,3, 4, 4, 4, 4· in
which each positive integer n occurs in blocks of n terms?
Solution
by Charles McCracken, Dayton, OH.
The
last 1 occurs as term 1, the last 2 occurs as term 3, the last 3 occurs as
term 6, the last 4 occurs as term 10, the last 5 occurs as term 15, the last 6
occurs as term 21 and so on. In general, the last n occurs as term n(n + 1)/2,
the nth triangular
number. The 1414th triangular
number is 1000405. This marks the end of a sequence of 14141414âs.
Therefore, the millionth term is 1414.
Also
solved by: Paul M. Harms, North Newton, KS; William T.
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