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Mathematicorum
Crux
CRUX
MATHEMATICORUM
Vol. 12, No. 9
November 1986
Published by the Canadian Mathematical Society/
Publie par la Societe Mathematique du Canada
The support of the University of Calgary Department
Statistics is gratefully acknowledged.
*
of
Mathematics
and
t
*
CRUX MATHEMATICORUM is a problem-solving journal at the senior secondary
and university undergraduate levels for those who practise or teach mathematics.
Its purpose is primarily educational, but it serves also those who
read it for professional, cultural, or recreational reasons.
It is published monthly (except July and August). The yearly subscription rate for ten issues is $22,50 for members of the Canadian Mathematical
Society and $25 for nonmembers. Back issues: $2,75 each. Bound volumes with
index: Vols. 1 & 2 (combined) and each of Vols. 3-10:
$20.
All prices
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Leo Sauve, Frederick G.B. Maskell.
Kditor: G.W. Sands, Department of Mathematics and Statistics, University
of Calgary, 2500 University Drive N.W., Calgary, Alberta, Canada, T2N 1N4.
Managing Editor: Dr. Kenneth S. Williams, Canadian Mathematical Society,
577 King Edward Avenue, Ottawa, Ontario, Canada, KIN 6N5•
ISSN 0705 - 0348.
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*
t
Return Postage Guaranteed.
t
CONTENTS
The Olympiad Corner:
Problems:
Solutions:
79
.
M.S. Klamkin
1181-1190
1010, 1048-1053
229
241
.
243
~ 228 -
- 229 -
THE OLYM>IAD CORNER: 79
M.S. KLAMKIN
All
communications
Department
of
about
Mathematics,
this
column
University
should
be
of Alberta,
sent
to
Edmonton,
M.S.
Klamkin,
Alberta,
Canada,
T6G 2G1.
This month I give three sets of problems.
The first set is a number of
solutions appear at
"Quickies" proposed by me and whose
the end of this corner. Mathematical Quickies were
initiated by C.W. Trigg in 1950 as then editor of the Problems and Questions
Department of Mathematics
Magazine*
These are problems which can be solved
laboriously, but with proper insight and Imowledge can be disposed of quickly.
Knowing a problem is a Quickie is a decided hint toward solving it.
Q-l. Determine a
two parameter
integral
(x,y9z,w)
solution
of the
Diophantine equation
w2 = 2(x4 + yA + z 4 ) .
Q-2. Determine a
three parameter
integral solution
(x,y,z,w) of the
Diophantine equation
(xyz
+ yzw + zwx * wxy)2
= (yz
- xw)(xz
- yw)(xy
Q-3. Determine the number of real solutions (x,y,z,w)
-
zw).
of the simultaneous
set of equations
x
5
+ x = 2y2z, ys + y = 2z2w, zs •¥ z ~ 2w2x, ws + w = 2x2y,
Q-4. For af b positive determine the maximum value of
y = (a - x)(x
+ Jx 2 +
b2)
over all real x (no calculus please!),
Q-5. It is known that if A, B9 C are the angles of a triangle, then
cot A/2 + cot B/2 + cot C/2 > 373"
and
cot A + cot B + cot C > J5.
Prove that in fact
3(cot A + cot B + cot C) > cot A/2 + cot B/2 + cot C/2.
Q~8. Determine the minimum value of
= cot(?T - A)/4 -f cot (IT - B)/4 -f cot(n - C)/4
. sin A + sin B + sin C
~—
where A, B, C are the angles of a triangle.
- 230 -
Q-7. Prove that the diagonals of a quadrilateral are perpendicular if and
only if the sum of the squares of one pair of opposite sides
equals
the sum of the squares of the other pair of opposite sides.
Q-8, If a, b,
{abc - az(b
c and a - x, b - yf
- y)
- bx(c
- z)
e - z are non-negative, prove that
- x)}z
- cy(a
> 4xyz(a
- x)(b
- y)(c
-
z).
Q-9. By factoring
(a2
+ b2 + t2)(a + b + c)(5 + c - a)(c + a - 6)(a + 5 -
or otherwisej show that a triangle is acute, right, or
a
2
+
2
b
+
c
2
is >
f
=
2
f
Here a9 bf
or < 8J? .
8a2b2c2
) -
obtuse,
according
as
c are the sides and R is the
circumradius of the triangle,
Q-10. Find
the
shortest
distance
between
two
non-intersecting
face
diagonals lying in adjacent faces of a unit cube«
*
The
next
two sets of problems are by courtesy of Walther Jan^s, and as
usua] we invite elegant solutions for them*
17th Austrian Mathematical Olympiad 1986
1st dayf June 4, 1986 (time allowed 4 1/2 hours)
_1.
Show that a square can be inscribed in any regular n-gon®
2*
For s9t
€ M 9 let
M = {(x,y)
| 1 < x < s, 1 < y < t9 xfy
be a given set of points in a plane•
vertices
belong
to
e m]
Determine the number of rhombuses
whose
M and whose diagonals are parallel to the x,y coordinate
axes*
_3.
Determine the set of all values of x 0 ,x x € IR such that the
sequence
defined by
x
X
n+1
1
x
= — —
,«— ,
3x . - 2x
n-1
n
li
> 1
contains infinitely many natural numbers,
2nd day (4 1/2 hours)
4_*
Determine
the
with base 10
maximum value of n such that there exists a number N
representation
t/1t/2...c/
having different digits and
such that n! |tf. Furthermore, for this maximum n9 determine all
numbers N*
the
possible
- 231 -
Show that for every convex n-gon (n > 4 ) , the arithmetic mean of the
5.
lengths of all the
lengths of all the diagonals.
6.
n
For
a
sides is
less than
the arithmetic mean of the
given positive integer, determine all functions F: M —• R
such that F(x
+ y) = F(xy
~~ n) for all x,y
€ IN with xy > n.
*
9th Austrian-Polish Mathematical Competition 1986 (Warsaw)
1st day ? June 25 f 1986 (time allowed 4 1/2 hours)
AtA2A3
_l.
is a given non-right-angled triangle*
Three circles Cit
C29
C 3 are mutually tangent and they pass through the vertices such that
A2$A3
€ Ct;
A3$At
€ C 2 ; and Ai9A2
AtA2A3
such that the triangle determined by the centers of the
e r3.
Determine the possible angles for
three
circles
is similar to the given triangle,
2*
For 12 > 1, let
13-1
be
a
a P(l)
polynomial
>
3,
having
n
distinct
0
negative
real
roots.
Prove
that
that
there
2
2n aQ*
If every point in space is either
blue
or
reds
show
exists at least one unit square whose number of blue vertices is
0,
1, or 4,
2nd day (4 1/2 hours)
4.
Determine all triplets (x,y,z) of natural numbers such that
z+1
z+1
o 100
x
- y
= 2
5*
Determine
all
quadruples
(x,y,u,v)
of real numbers satisfying the
simultaneous equations
x2 + y 2 + u2 + v 2 = 4,
xii + yv = -xv - yu,
xyu + yov + uvx
4- vxy = - 2 ,
xyov = - 1 .
69
Determine the extreme values of the ratio of the circumradius to the
inradius of a tetrahedron whose circumcenter and incenter coincide*
- 232 -
3rd day (4 1/2 hours) - Team Competition
Show that nl
7.
is divisible by k whenever n and k are natural
z
such that n
8.
> 41 > 0 and i has no prime factor > n.
m x n matrix of distinct real numbers is given.
An
numbers
The elements of
each row are rearranged (in the same row) such that the elements are
increasing from left
to
right.
Next
the
elements
of
each
column
are
rearranged (in the same column) such that the elements are increasing from top
to
bottom.
Show that now the elements in each row are still increasing from
left to right*
Determine all continuous and monotone functions Fl IR —* IR such
9*
2
F(l) = 1 and F(F(x))
= (F(x))
that
for all x e R 0
t
I
now
give
solutions
to
some previous corner problems as s*\?ll as the
"Quickies" at the beginning of this corner•
5.
Competition
[1981: 43]
Britain,
Luxembourg,
Ten gamblers
Each
in
in
Mersch,
The Netherlands,
started playing
turn threw five dice*
Luxembourg
and
each with
that
the same
after
payment
the beginning.,
amount of money*
At each stage the gambler who had thrown paid
moment, where u is the total shown by the dice.
one after the other.
Great
Yugoslavia).
to each of his nine opponents 1/n times the amount which that
at
(Belgium,
opponent
owned
They threw and p>aid
At the tenth throw the dice showed a total
of
12,
and
it turned out that every gambler had the same sum as he had at
Determine if possible the totals shown by the dice at each
of
the other throws.
Solution
by Andy Liu,
University
of Alberta,
Edmonton,
Alberta.
Suppose fl was one of the gamblers and N was the next gambler • Before M's
play,
both
M and N had received exactly the same amounts from each previous
Suppose M3s die-roll was
7* - x
m; then M's total payment to the other gamblers amounted to — — where T was
gambler and so still had the same amount x of money.
the
total
wealth
of
gamblers other than fl.
all
the
gamblers and T - x was the wealth of all the
After M's payment, fl had
x - . _ _ _ = x(l + —) - —
m
m m
- 233 -
Now suppose Nfs die-roll was n.
while N had x(l + — ) .
Then*
in the same
manner, after N's payment M had
W 1 + b ~ &<J + b - x<* + b*1 + ^> - & 1 + ^
m
m
n
m
n
m
n
and N h a d
r - x(i + -)
1
1
1
rr
x(l + i ) - _ ^ ™ ^ ~ i ^ = x(l + i ) ( l + i ) - ~ .
a
«
a
n
n
These two amounts must be equal, since M and N had equal amounts at the end of
the game. Thus
Id
m
+
1)
=1,
n
»
that is, m = n + 1.
Since in our problem* we have 10 pleyers with the last die-roll being 12,
the previous die-roll must have been 12+1 and then the die-roll before that
12+1+1, etc. Hence the ten die-rolls are 21, 20, 19, 18, 17, 16, 15, 14, 13,
and
12 in that order*
It is to be noted that the information about the five
dice being used, though consistent with the answers, is in fact redundant.
19.
[1985: 4]
From
the
1984
All-Union
Olympiad
(proposed
by
Balotov).
Find xy + Zyz + 3zx if x, y, z are positive numbers for which
x
2
+ xy + y2/3 = 25,
Solution
by George
y2/3 + z 2 = 9,
Evagelopoulos,
z 2 + zx + x2 = 16.
and
Law student,
Athens,
Greece,
Let
A = x 2 + xy + y2/3,
B = y2/3 + z 2 ,
C = z 2 + zx + x 2 ,
D - xy + 2yz + 3zx.
Since A = B + C, we get xy = z(x + 2z>* Also,
17 = y(x + 2z) + 3zx = y(xy)/z
+ 3xz = (y2/3 + z2)3x/z = 27x/z.
Furthermore,
D = xy + 2yz + 3zx = 2z2 + zx + 2yz + 3zx = 2z(2x + y + z),
so that
2x + y + z = Z?/2z = 27x/2z2,
Next,
A-B+C=
Since x,y > 0, x/z = 875/9*
32 = x ( 2 x + y + z ) = 27x2/2z2.
Finally, D = 27x/z = 2473"*
A.B.
- 234 -
This problem appears in KVANTf
23.
From
[1985: 5]
the
November 1984, p*61 9
1984
All-Union
Olympiad
(proposed
by
Yu.
Mikheyev)•
For what integers m and n do we have
(5 + 3J2)m
= (3 + 5jZf
and
(a + bM)m
= (b + a ^ ) 1 5 ,
where a and Z> are relatively prime integers, while d is a uatural
number
not
divisible by the square of a prime?
solution
We
bj
by George
Evagelopoulos,
Law student,
Athens,
Greece.
will show that, with the assumption (implicit in the problem) that a,
and d are all greater than one f the only solution to either equation is the
obvious one m = n = 0.
Suppose that
= (5 4- avO")12
(a + bMf
and that one of z» and J? is nonzero; then both are, and in fact they both
have
Without loss of generality! w e may take m and n positive
the same sign.
and a > b*
must
Then since
1 < a + bJ3 < b + av^"?
it follows that m > n,
By expanding (a + &v£7)
and (b
+
aJ7)
by
the binomial
theorem
and
equating rational and irrational parts, it also follows that
(a - bj&f
- (b - av6T)J
(a - £>vEF)a
'5 - ajcf]
and thus
[iTOTJ
We will now show that (1) cannot hold*
u =
a - 5^
a + 57d
and
v =
Let
B -I- avfc/
Then since a > 5, b - a,/7 < 0, so
v = a7d + 5
We will use the fact that for positive p and q,
2g
q . i
= 1
TpTgFTT
p + q
~
p + g
i n c r e a s e s with p/g*
Now i f a - 5v3" > 0 ,
then
a - bvET"
u = a + bjd
and
(1!
- 235 -
a
and thus u < v.
. aJJ
On the other hand if a - hJU < 0, then
b<JI - a
Kjd + a
and
&J7
and
again
it
implies that u
24 a
follows
that
u
<
a J7
<
v.
In either case 0 < uf
v < 1 and m > n
< v $ contradicting (1).
From
[1985: 5]
the
1984
All-Union
Olympiad
(proposed
by
O.R.
Musin)•
If
72 >
3
natural numbers are written around a circle so that the
ratio of the sum of the two neighbours of any number to itself
is
a
natural
number, prove that the sum of all such ratios is
(a)
not less than 2n f
(b)
not more than 3n.
Solution
by George
Evagelopoulos,
Law student,
Athens,
Let the 72 natural numbers be denoted by ai tan,
a
l
+ a
n
2
a
A
• • . ,a
A
a
l + a3 A
Greece.
and let
n
. + a.
12-1
1
12
(a)
Since
x/y
+ y/x
> 2
for x,y > 0, It follows that
s 12 =
\a2
a
Q-i
2^
L
(b)
l]
a
1
+
S
0
[ 2
J
We will show that S
[a3
a
2]
a.,
a
> 2n.
Q-
a
l
%
3J
< 3i2 for 12 > 3 by mathematical induction.
12
For 72 = 3, we consider two cases.
(I) All the numbers a, are equal*
(li) At
least
two of at,
a29
In this case S3
a 3 are not equal*
6 < 3*3,
Suppose at
> a2 and at
> a3»
Then
at > £iJL£i so 2 > £» + a* .
Since the latter ratio equals a natural number, at
a
a
1+ » + J +
+ a.
+
= a2 + a 3 .
2 ^ + 2a,
Consequently§
- 236 -
where k = 2a 3 /a 2 and € = 2a a /a 3 are natural numbers.
Since k€
=
4*
we
can
only have the following possibilities:
* = 1, € = 4, then S 3 = 8 < 9;
1 = 2, e = 2, then S 3 = 7 < 9;
A = 4, * = 1, then S 3 = 8 < 9.
We assume now that the inequality is valid for S
S
1
and we will show that
= 2n < 3n*
Otherwise? we choose
< Zn.
n
If all the numbers a
are equal, then 5
largest of the n numbers.
the
If there is more than one such number, we take
one of them which is greater than at least one of its nearest neighbors.
Let
this number be a . Then
n
a
a
1
12-1
>
+ a1
n
so
2 >
13-1
1
Since the latter ratio equals a natural number, a = a ., 4- a., so that
^
n
73-1
1
a + a0
a 0 + a
1 + r,
13
2 = 1 + s,
n-1
where
a
0
+ a1
13-Z
a i
i?-l
1
and
a
are
natural
numbers.
-5 + a 0
13-1
2
Consequently the sequence a.?a2,•*.9a
inductive hypothesis and so S _, < 3 (12 - ! ) •
a
S
=
-1 +
n
a0
a
l
+ a
3
0
+ a
, + 3 < 3/2,
13-1
i
Finally,
- F a
i
a + a
0
13-2
13 ,
13-1
1 ^ 13
2
+ — . — _ — _ _ + — — _ — — — . -i- — — - — _
a
^
a
8...
13-1
13
1
a !
13-Z
1
+ 1 + 1 + 1 + 13-1
n-1
= S
a
. satisfies the
+ a0
Z
- 237 -
Editorial
comment.
By letting a.
_ 1 + 3
Q
pi - 2) + n . (n - 1) + 1 . 12 + 2
2
12
= i for all i, we obtain
12-1
1
12
= 2 + . . . + 2 + l + ( j * + 2)
= 2(n - 2) + n + 3
= 3l2 - 1,
which is the least upper bound*
t
Solutions to the Quickies
Q-l. Letting z = x + y, we find that w = ±2(x2 4- xy -I- y 2 ) •
Q-2, Expanding out the left and right sides, we obtain
2 x2y2z2
+ 2xyzw
2 xy = 2 x 2 y 2 z 2 - xyzw 2 x 2
or
xyzw(x + y + z - f w ) 2 = 0 .
Thus all solutions are given by xyzw = O o r x r + y + z-l-w = Os
Q-3. Multiplying the equations together, we obtain
+ x 4 ) - 16x 2 y 2 z 2 w 2 } = 0,
xyzw{Jl(l
Since 1 + x4
> 2x 2 with equality if
and
only
if
x
=
±1,
etc.,
all
the
solutions are given by
(x,y,z,w)
provided
= (0,0,0,0) or (±1,±1,±1,±1)
x and z have the same sign and the same for y and w.
Thus there are
five real solutions*
Q-4» Letting x = b sinh B,
sinh B + &Jsinh20 + 1)
2y = 2(a - b sinh B)(b
= (2a + 5(e
@
= 5 2 + be9(2a
~ e 0 ))&e @
- &e0).
Thus y takes on its maximum value for 5e
2
=
^raax ""
b
= a,
Finally,
2
+ a
2""~™°™
and is taken on for
_ b(a/b
x _ — ^
- 6/a) _ a 2 - b-z
- - 2 a
'
Q-5. Let a, 5, c be the sides of the triangle, s the semiperimeter, and r
the inradius.
Since
- 233 -
cot A/2 =
etc.
and
1
2 cot A = cot A/2
etc.
the inequality is equivalent to
a
r
^-HM
+
| „ (s - b
Is - c
P
r
a ,s - b
_
.
4-
a
s - c
4 —
or
M-
a ,s - 5 ,s - c
4 ___—™—. 4
,
s - 6
^
s - c
or
s = 3s - (a 4 5 + c) > 3r 2
s - a
s~b
<1]
s - c
Since it is known that
r2s
= (s - a)(s - 6)(s - c ) ,
(1) is equivalent successively to
s2
s
2
> 3[(s - 6)(s - c) + (s - c)(s - a) 4 (s - a)(s - 6)],
> 3[3s2 - 2s(a
4 I? 4 c) 4 be 4 ca 4 a&],
(a 4 5 4 c ) 2 > 3(6c 4 ca 4 afc),
and finally
(b - c ) 2 + (c - a)2
4 (a - 6 ) 2 > 0,
which is true, with equality if and only if the triangle is equilateral.
Q-So For any triangle ABC$ there is a
non-obtuse
triangle
A'B'C
with
angles
A# = (u - A)/2,
B' =
(IT
- B)/2,
C" = (u - C)/2.
So equivalently, we want the minimum of
S
for
non-obtuse
, _ cot A 7 2 + cot 5 7 2 + cot C / 2
=
sin W + sinTET 4 sin IT7""
triangles,-
It will turn out that we can drop the primes in
this expression and consider all triangles? since the minimum will occur
only
for equilateral triangles•
Since the area F of a triangle ABC is given by
F = ft2{sin 2A 4 sin 2B 4 sin 2C}/2 = r2{cot A/2 4 cot B/2 + cot C/2}
where
R
and r are the circumradius and inradiusf respectively, S' will equal
(tf/r)2/2.
equilateral
It is well known that (R/r)
triangles.
.
2 and is taken on only for
mm
Finally! the minimum values of S and S' are then both
- 239 -
equal to 2,
Q~7, This problem appeared in the 1912 Botvos Competition,
are
given
in
the
Hungarian
Problem
Book
II,
Library, MAA, Washington, D.C., 1963, pp*50-54 where it is
figure
is
coplanar*
Two solutions
New Mathematical
assumed
that
the
We give a vectorial solution which is simpler and which
does not assume the figure is coplanar*
Denoting the vertices of the quadrilateral by vectors A, B f C, D
from
a
common origin, the4 result follows immediately from the identity
(A - B ) 2 + (B - C ) 2 - (A - B ) 2 - (D - C ) 2 = 2(A - C)•(B - D ) .
Q-8» The
result
follows
immediately
from
the
equivalent
inequality
since this inequality implies
[xyz
+ (a - x)(b
- y)(c
- z ) ] 2 > ixyz(a
- x)(b
- y)(c
- z)
and the left side reduces to
[abc - az(b - y) - bx(c - z) - cy(a - x)] 2 *
The given inequality is related to a result of J*F, Rigby,
"Inequalities
concerning the areas obtained when one
Math.
triangle is inscribed in another",
Mag. 45 (1972) 113-116.
Let a, bf
c
denote the sides of a triangle AM? with
an inscribed triangle DEF where BD = xf
CE = vi AF = z, and let A, A , zL , A ,
J
a b c
and A~ denote the areas of ABC, AFE, BFD,
CDE? and DEF, respectively.
a-x
Then
(a -
so dividing the given inequality by (abc)2
zl
2L
y)y
gives
A
A
A^ A
> 4 -£. A JE.
-
J
^1
4
or
zlid > 4z! zL zl
0
abc
with
equality if and only if the three cevians AD, BE,
Thus
A® + Zt^Zt
0
0 a
+ 21*21. 4- A%A
05
> 421 21, A
0 c -
abc
and CF are concurrent,
- 240 -
find it. follows quickly that
a
0 -
b
c
with equality if and only if the three cevians are medians*
Q-9, Letting S = a2 4- b2 + c 2 , we have
(d2 -f b2 + c2)(a
4- & 4- c)(b
4- c - a)(c
-ha-
b)(a
8a2b2c2
4- 6 - c) -
= S[2(k 2 c 2 4- c 2 d 2 4- a 2 5 2 ) - (<2^4- fc« 4- c 4 )] - 8a 2 & 2 c 2
= S'[4(&2c2 4- c2a2
4- d2£>2) - S 2 ] - 8a 2 5 2 c 2
= 4$(5 2 c 2 4- c 2 a 2 4- a 2 6 2 ) - S 3 - 8a 2 5 2 c 2
= S'3 - 2 S 2 U 2 + 5 2 4- c 2 ) 4- 4S(5 2 c 2 4- c2a2
4- d 2 5 2 ) - 8a 2 5 2 c 2
= (S - 2a2)(S - 252)(S - 2r a )
= (52 4- c 2 - a2)(c2
•¥ a2 ~ b2)(a2
+ 52 - c2),
(1)
Also, since a6c = 4/?F and
(d 4- b 4- c)(5 4- c - a)(c 4- a - b) (a 4- b - c) = 16F2
where # and F are the circumradius and area, respectively,
(1)
is
also
equal
to
16F2(a2
4- b 2
of
the
triangle,,
4- c 2 - 8i?2) • Finally, since (1) is
positivef 0, or negative according to whether the triangle is acute, right, or
obtuse, the desired result follows,
Q-1Q» In the following figure we have two
adjacent
unit
cubes
face to face and we will determine the shortest distance
EB and FC.
h
between
D
Since* FJ\\EB, h is the distance from B to the plane of FJ and FC.
of tetrahedron BFCJ = i = (|.) -Area(FCJ).
*
touching
*
The volume V
Since Area(FCJ) = -g-, h = - ^
*
- 241 -
P R O B L E M S
Tiollem
proposals
and solulloas
should
le sent to the editor, whose
uild^vess appeals on the giant pafe o§ this Issue.
Tioposuls
should, wA>enewei
possllle,
le accompanied ty, a solution}
legeiences,
and othei InslqAts which
aie llfcelf to le of help to the edltoi.
An a^te^il^h
(*) agtei
a numlei
Indicates a piollem sulmltted without a solution.
Vnltylfial pioltems
me paitlculaily,
sought.
%ut ethei
lateiesttng,
pioltems may also le acceptable pnowlded thef
aie not too well known mid
legeiences aie g,lven as to theln provenance.
Gidlnailly, t l§ the oilplnatoi og
a piollem can le located. 6t should not le sulmltted Ip somelodp else without
his o% her peimlsalon.
To facilitate
the It consideration,
'four solutions,
typewritten
or
neatly,
handwritten,
on signed,
separate
sheets, should piegeially, le mailed to the
edit01 legoie fane 1, 1987, although solutions received agtei that
date
will
also le considered until the time when a solution 6<^ punished.
t
*
1181.
Proposed
by
D.S.
Belgrade,
Belgrade,
t
Mitrinovic
and J.E.
Yugoslavia.
Pecaric,
University
of
(Dedicated to Leo Sauve.)
Let x, y, z be real numbers such that
xyz(x
+ y + z) > 0,
and let af b, c be the sides, m , m^ . w the medians
a
triangle•
(a)
(b)
b
and F
the area
of a
c
Prove that
jyz-a2 + zxb2
+ xyc2 j > AFJxyz(x
2
2
+ y + z)
m
jyzffl
-f- zxn?^ + xym 1 > 3FvSyzTx~T y°'™4" z)
1
1182.
a
D
c* "*
.
Proposed by Peter Andrews and Edward T.H. Wang, Wilfrid
University,
Waterloo, Ontario.
Let a^ ^a^, > « » fa
Laurier
(Dedicated to Leo Sauve*)
denote positive reals where n > 2. Prove that
" < tan-»£i- + tan-i£2. + .. . + tan-* Jl < ( " " 1 } "
Z a2
a3
at ~
2
and for each inequality determine when equality holds*
1183.
Proposed by Roger Izard,
Let
Dallas,
Texas.
ABCD be a convex quadrilateral and let points Et G lie on
BD and F, H lie on AC such that AE, BF, CG, DH bisect angles
CDA respectively.
parallelogram.
Suppose
DAB, ABC,
BCD,
that AE = CG and BF = DH, Prove that ABCD is a
- 242 -
Proposer! by J.T. Groenman, Arnhem, The Netherlands.
1184.
Let ABC be a nonequilateral triangle and let 0, J, H, F denote
the
circumcenter?
incenter*
orthocenter*
and the center of the nine-point
Can either of the triangles OIF or IFH be equilateral?
circle, respectively.
t
Proposed
Austria.
1185.
Determine
by
Walther
Janous,
(Jrsulinengywnasium,
Innsbruck,
the set M of all integers k > 2 such that there
exists a positive real number u
satisfying [u ] = n (mod k) for all natural
A
A
numbers n$ where [x] denotes the greatest integer < x.
(Problem A-5 in the 1983 Putnam Competition is equivalent to showing that
2 € M. )
1186.
Proposed
by Svetoslav
Bilchev,
Velikova,
Mathematical gymnasium, Russe,
If a, £>, «: are the sides
Technical
University
and Emilia
Bulgaria.
of a triangle
aind s,
R,
r the
Equipment
Corp.,
semiperimeter, circumradius* and inradius? respectively? prove that
2(6 + c - <*)vST > 4r(4fl + r) ^R
+ F
URs"
where the sum is cyclic over a, 6, c.
Proposed
1187.
by Stanley
Rabmowitz,
Digital
Nashua, New Hampshire.
Find
2
+2
a polynomial
with
integer
coefficients
that
has
as a root,
Proposed by Dan Sokolowsky, Williamsburg,
1188.
Given
a circle
Virginia.
A' and distinct points Af B in the plane of ft*,
o
construct a chord PQ offtsuch that B lies on the line PQ and ZPAO = 90 .
t
1189.
Proposed by Kee-wai Lau, HOIK/ Kong.
Find the surface area and volume of the solid
intersection
formed
by
of eight unit spheres whose centres are located at the vertices
of a unit cube.
t
1190.
the
Proposed by Richard I. Hess, Rancho Palos Verdest
Prove or disprove that
California.
- 243 -
n-1
lim
2
- 2n
7
*=0
S O L U T I O N S
A'# piaC-teai l& evei peimanefitty,
cto-^ett.
The e<&Lt&% mltl atmai/A, &e p-tea^e<$,
to cofibLdet tfoi p-u&tLcatLon a&w §,otutLoa& oi new- LnA,LgAt& on p-nt^t psio&temb.
1010?
[1985:
17;
1986:
Ursulinengymnasium,
Innsbruck,
by
Walther
i7 an integer, contains infinitely
determine all such k.
many
Janous,
Austria.
there integers k t 1 such that the sequence {3n2k
Are
squares?
+ ink2
If the answer
+ i3}3,
is yes3,
(The case k = 1 is dealt with in Crux 873 ,[1984: 335].)
Comment by Richard
1.
Proposed
113]
K. Guy,
University
of Calgary,
Calgary,
Alberta.
show that k must be of the form p^ *. .p €2 or 3D.,,,p .€2 where ro > 0
l
r
l
m
*i '©+1
and each p . is a prime s 1 mod 6; that is? the condition conjectured by Hess
We
[1986: 115] Is at least necessary for k to be a solution of the problem*
(In
the comment following this one?
not
however,
G. Walsh
shows
that
It is
sufficient.)
If we put k = kxk\
where ki
z
2
2
= 3n k
is squarefree? and write
+ 3nk2
= 3n2ktk2
then
2
2
k1k i\z t
+ k3
+ 3nklk*
that ktk2\z
and it follows
+ *}*| ,
since kt is squarefree*
(1)
Putting
z = ir1A"2z1, (1) becomes
klklzl
= Sn 2 !^ 2 -f 3n*2*2 + AfAf
JtjL^f = 3n2 + taJ^A* + Af**.
Hence l i j3n
first 3iki9
2
?
f
so (as in Kierstead s argument) either kt\n
(2)
or 3 |A"4 •
Assuming
put n = ^ii^i • Then (2) becomes
Axzf = Zk\n\
+ 3iflii2i + Af*«
zf = S l ^ f + 3Jt1Jt|/?1 + ktk*
so we can put zt
= k1z29
and then
SO
4* t z| = 3(2/2! + JSr|)2 + *«.
(3)
- 244 -
3)1!, we put kt
If
= 3k3$
and with similar calculations (2) can be written in
the form
4Jr3zf = (2nt
+ 3 1 2 ) 2 + 3k*»
(4)
Thus we need to show that if (3) (resp, (4)) has infinitely many solutions for
a given kt
(resp, k3) then kk
(resp, k3)
is a product of primes all s 1 mod
6
(and k3 t 1).
we show that 2fii in (3) and that 2fjt3 in (4), oince kt
First
thus both kt
and
Ar3
= 3i 3 and
are squarefreef we need to show that both
3(2i2s + I f ) 2 + k*
and
(2nt
+ 3 £ 2 ) 2 + 3Jr^
have even powers of 2 in their prime factorizations.
this
for
expressions
2
the form 3X
of
2
X
if
and
Y
is
enough
to
show
But this is obvious if X and F
+ r .
have opposite parityf and follows by induction if X
Finally,
It
2
2
are both odd, then 3,Y + Y
and
Y
are
both
even.
= 4 mod 8, so 3,Y + Y2 is
2
divisible by 4 but not 8.
Next we show that k3 £ 1 in (4). Otherwise, (4) becomes
4z| = (2nt
+ 3 1 2 ) 2 + 3*J,
or
3*5 = (2z2
+ 2nt
+ 3*|)(2z2 - 2nt - 3i 2 )
which obviously has only a finite number of solutions for each k2«
It is clear since kt
p
=
2
mod
is square free that 3fjr3, Thus we suppose that
3 is a divisor of ^
this is impossible.
Since kt
some
in (3) or Jf3 in (4), and it remains to show
and k3 are squarefree, this means that either
3(21^ + k2)2
+ k\
or
(2nt
-f 3ir|)2 -4- 3*J
has an odd power of p in its prime factorization.
2
the form 3X
+ Y
2
can have this property.
We show that no integer
This is clear if p divides just one
of X and Yf and follows by induction if p divides both of X and F,
that plY.
Then
3A'2 + Y2 E 0 mod p
may be written
9X2Y~2 + 3 E 0 mod p,
or
(3XY-1)2
of
= -3 mod p.
So suppose
- 245 -
However it is known that -3 is a quadratic non-residue of such primes p.
This
contradiction completes the proof*
11. Comment by Gary Walsh,
student,
University
of
Calgary,
Calgary,
Alberta.
was conjectured by Hess [1986: 115] that for any number k of the
it
2
form p<.««*p £
there
are
or 3D. • * .p B*£29
infinitely
integer z.
many
We give a
where m > 0 and each p. is , prime E 1 mod
integers n such that (n *¥ k)3
constructive
proof
that
there
6,
- n3 = z 2 for some
are
infinitely
many
counterexamples to this conjecture.
use the following facts, where m and n denote squarefree odd integers
We
greater than 1 (for the first, see D.T\ Walker, The equation mXz - nY2
Amer.
Math.
Monthly
=
±1,
74 (1967) 504-513; the second follows from the first by a
simple induction):
(i) if a positive solution (X9Y)
smallest positive solution (AfB)t
a
to mX2 - nY2
= 1 exists, then there
is
and all positive solutions (A,,B,) are
A
&
obtained from
A, JH + B.Jfi = (Ajfi + Bjn)
A
2#+l
K
with ir = 0,1,2, • • • .
is the smallest positive solution to mX2 - nY2
(ii) If (A,B)
B
and
are
of opposite parityf and if A (resp« B) is even, then A
B ) is even for every kf
and B
K
p.
E
1 mod
6,
the
A
(resp*
(resp. A ) is odd for every i.
K
K
As well, we need the result proved in [1986: 113], that for k =
each
= 1, then
problem
of
p^«»*p
with
finding infinitely many n such that
(n + & ) 3 - n* is a square is equivalent to the solvability of the equation
Uaz
- 3& 2 = 1
for some integers a, 5»
Suppose that x, y, A" are positive integers with k squarefree and k
and
suppose 12x
solvable.
(XfY)
We
= <y,2x).
z
2
2
+ 1 = ky .
have
that
We claim that the equation Aka
the
equation
2
kX
-
2
3Y
=
1
>
1,
2
- 3i> = 1 is not
is
solvable
by
Thus
y^F + 2x75" = A.JE + B.V5" = (A^F + BV5")2i+1
for
kX
z
some
- 3Y
2
integer
= 1.
/,
where
(AfB)
is
the
smallest
positive
solution to
Since B. = 2x, B . is even and A . is odd for every /.
Now
if
- 246 -
a, b are integers such that 41a2 - 3b2 = 1, then (2a,3) = (A ,B ) for some €t
contradiction•
We now have that any squarefree integer k > 1 for which 12x2 + 1 = ky2 is
solvable is not a solution to the original problem. Next we
integer k
show that any
of this form is a product of primes each congruent to 1 mod 6*
If
pj k, then
12x2 + 1 E 0 mod p
so clearly p ^ 2 or 3. Also
(6x)2 = -3 mod p
so p as 2 mod 3 since -3 is a quadratic nonresidue modulo such primes.
Thus
p = 1 mod 6*
integer k > 1 for which 12x2 + 1 = ky2 holds is a
Thus any squarefree
counterexample of Messrs conjecture; for instance,
x = 1, k = 13,
x = 3, * = 109,
x = 4, 1 = 193,
x = 5, k = 301,
and so on. For # = 1,2,... put 12n2
+
1 = k V2
where each k
nJn
j-
> 1 is
n
.
squarefree. It remains to show that the set K = {k |T? = 1,2,...} is infinite.
We suppose for a contradiction that K is finite, say \K\ = s. For each k € Kf
the equation
1 = ky2 can be written ky2 - 3(2n)2
12/22 +
positive solutions are all of the form (yt2n)
= 1, and thus its
= (A.,B.) where
A.JE + B.V3" = (A^f 4- B7S")2i+1
and (A,B) is the smallest positive solution. Since
A.^JF
+ B.^JS
2+1
2+1
= (A^F + B ^ " ) 2 i + 3
= {A.JF + B.^")(AvT + Bv^")2
= {A„(1A2 + 3B2) + 6ABB.}JF + {B.(kA2
+ 3B2) + 21ABAi}75",
we have
= B.(iA2 + 3B2) + 2MBA i ,
B
and in particular B . „. > 4B.. Thus \B.
7+I
B.9B.
i
> s*
j -
I
f
1
- B.I
/8
Now consider the equations
12s2 + 1 = k y2
> 3s whenever
i
*
j
and
- 247 -
12(s + l) 2 + 1 = k ^ y *
12<2S)2
Then from \K\ = s, two of {k ,k
o
with i £ j
t
1,
ST X
+
1 = * 2s y| s .
. . . ,k0
} must be equal, say k
=k
1
S TJ
bS
ST
=k
and so
*yL,-
-
3
^
s
+
*>J2 = x
and
*y*+J But 2(s + i), 2(s + j)
3 2
[ <s
+
i)i
2
= !•
> 2s > s and
|2(s + i) - 2(s + j)j = 2|i - j| < 2s < 3s,
which is a contradiction.
*
1048.
*
Proposed
[1985: 147]
*
by
J.T.
Groenman,
Arnhem,
The
Netherlands.
In base ten, 361 = 192. Find at least three other bases in
which 361 is a perfect square.
Solution
It
by Hayo Ah1 burg, Benidorm,
Alicante,
Spain.
is well known that Euler [1] solved the following problem:
find all
triangular numbers N = _ - _ _ which are also pentagonal numbers N =
_ _ _ .
This leads to
1 ± JTZF2~~T-JW~TT
i -—
E~
the expression R under the square root should be a square to make £ an
Y - 1
integer. With k = —-r^— we obtain the Fermat-Pell equation
where
/? = 3x2 - 2 = y2.
Euler gives recursion formulas, in our notation
x
x
o0 = JYo
° = 1»
. 1 = 2x
y , ^n+l
y .1 = 3xn + J2y
n+1
n +^n
n ,
Following methods developed by Lagrange [2], we can also give x and y in
closed form:
x
13
perhaps by few people?
T$T^ (2 +
^
+ l,
(l)
- 248 -
[H4
(2 + 73")
•]•
where
n
=
0,1,2,.«.
and
[a] is the greatest integer < a.
The triangular-
pentagonal numbers N are now found from
3e2 - e
P H
M =
with
k =
x -1
_ i ± y
that is,
"2T
"ff"
The first values are
1?
0
i
2
3
4
5
6
7
X
1
3
11
41
153
571
2131
7953
y
1
5
19
71
265
989
3691
13775
N
0
1
15
210
2926
40755
567645
7906276
Now? 1 hope the patient reader who has stayed with me thus far will be as
enchanted as I am to discover cross-connections
problems.
In
Crux
between
seemingly
unrelated
1048 we are looking for those positive integers b (base)
and y which are solutions of
3b2 + 6£ + 1 = y 2 .
Substituting b = x - 1, this becomes
3x2 - 2 = y2»
Voila!
b = 2, 10, 40, 152, 570, 2130, 7952, etc.
(The base b in closed
form
i s, according to (1),
b
=
[T73^ ( 2
+ JS)n
]
f n= x 2 3
'''-
)
References:
[1]
Leonhard Euler, Vollstandige Anleitung zur Algebra? St„ Petersburg, 1770,
Part II, Section 2, Chapter 6, %91, Problem V.
[2]
Joseph
Louis
Lagrange,
Additions
to the French translation of Euler?s
Algebra, Lyon, 1774, SVII, #75.
Also
Rruxelles,
solved
by SAM BAETHGE,
Belgique;
Antonio,
Texas;
C.
FESTRAETS-HAMOIR,
RICHARD I. HESS, Rancho Palos Vercles, California; MLTHER
JANGUS,
Ursulinengywnasinm,
Cuyahoga
Falls,
Ohio;
San
J.A,
Innsbruck,
Austria;
MCCALWM, Medicine
Hat,
FRIEND
Alberta;
H.
KIERSTEAD JR. ,
STEWART METCHETTE,
- 249 -
Culver
City,
Ohio;
California;
BOB PRIELIPP,
RABINOWITZ,
Federal
University
Digital
Republic
proposer.
LEROY F. MEYERS, The Ohio State
Equipment
of
There
of
Wisconsin,
Corp.,
Germany;
was one partial
solution
*
Oshkosh,
Nashua,
KENNETH M.
New Hampshire;
WILKE,
Proposed
[1985: 148]
ABC
Let
A > B > C and A'
and
> B' > C*.
Topeka,
J.
STANLEY
SUCK,
Kansas;
and
Essen,
the
received.
t
by Jack
Garfunkel,
Flushing,
N.Y.
A^B'C* be two nonequilateral triangles such that
Prove that
r
r
where s, r and s'9
Columbus,
Wisconsin;
t
1049,
A'B'C
University,
r* are the semiperimeter and inradius of triangles ABC
and
, respectively.
Comment by U.S.
Since
Klamkin,
University
of Alberta,
Edmonton,
Alberta.
the perimeter of a triangle circumscribed about a given circle can
o
be unbounded if either two angles of the triangle are close
close
to
o
0 ,
the
result
stated
is clearly f
le,
to
For
90
a
or
else
specific
counterexample, just let
A = 120°,
f
A
= 89°,
Then A - C > A' - C , but
s
^ A
— = C O t -rr
while
r
B = 30°,
B* = 89°,
+ COt
TJ
C = 30°,
C
= 2°
+ COt rr ** 8
Z
— T = C O t yr— + C O t rr— 4" C O t yj— ?« 59 .
r
Z
Z
Z
One can also find counterexamples involving only acute triangles, for example
A = 89°, B = 86,5°,
C = 4.5°,
A' = 88°, B' = 88°, C" = 4°.
Nevertheless? since cot(x/2) is convex in (0,ir), we can obtain a positive
result by the Majorization Inequality (e.g. see p.30 of E*F. Beckenbach and R,
Bellman, Inequalities,
Springer-Verlag, Heidelberg, 1965).
A > B > C,
A' > B' > C ,
and also that
A > A'
A 4- B > A' + B' ;
and
then
- = Z cot A/2 > 2 cot A V 2 = ^
r
-
r* .
Suppose that
- 250 -
*
*
*
[1985: 148] Proposed by Stanley
1050.
Corp.,
In
Descartes.
Digital
Equipment
Nashua, New Hampshii-e.
the
Show
Rabinowitz,
plane,
how
to
you are given the curve known as the folium of
construct
the
asymptote
to
this
curve
using
straightedge and compasses only.
Solution
by J.A.
McCailum, Medicine
Hat,
Alberta.
We assume that the folium is in general position*
Then the only definite
point in the plane is the point where the folium intersects itself.
C*
C
With
as centre and any convenient radius draw a circle.
Call this
This circle
will cut the two tails of the folium in
points P and Q.
(It may also cut the
loop of the folium but we ignore these
points if they occur,)
Join PQ and find
Join DC and produce it
Its midpoint D«
to cut the loop of the folium at A.
then the apex of the folium.
A is
Trisect AC9
and produce AC to E so that CE = ^AC.
Finally, at E erect a line perpendicular
to ECAf and extend this line Indefinitely
far in both directions.
construction
works
This is the required asymptote to the
is
The
because of the known symmetry of the folium and the known
fact that the apex of the folium is three times as far from
asymptote
folium.
(easy
to
prove
point
3
C
as
the
+ y 3 = 3axy of the
from the usual equation x
folium)•
Also solved
Ohio
State
University,
assumed that
solutions
by ELWN ADAMS, Gainesville,
Columbus,
the coordinate
assuming
as
Ohio;
Florida;
and the proposer.
axes were given.
given
not only
LEROY F.
Two other
the axes,
but also
MEYERS, The
Adams,
readers
however,
submitted
the equation
of
the
folium.
The proposer
wonders if asymptotes
to other
y = tan x or y = log x, can be constructed
*
1051.
familiar
curves,
with straightedge
*
[1985: 187]
Proposed
and
for
example
compasses.
*
by
George
Tsintsif&s,
Thessaioniki
,
- 251 -
Greece.
Let a? b9 c be the side lengths of a triangle of area K,
and
let uf v9 w be positive real numbers. Prove that
ua< +_vb^__^J^_>
y + w
w + u
u + v
BK2>
When does equality occur? Some interesting triangle inequalities may result
if we assign specific values to u, vf w. Find a few.
I.
Generalization
by D.S. Mitrinovic
Bel grade, Bel grade,
Yugoslavia.
and J.B,
Pe6aricf
University
of
Let r., , • • •, r be real numbers such that
1
n
n
Rk - 2 r > 0,
* = 1,. ..,73,
i=l
and let x 1 t .. M x
be positive numbers. We will prove that
n
r.
."1
1=1
73
1
xz >
/?. i - I? -
r
1
2 X
1=1 J
(1)
J
i=l
Indeed,
13
r.
i=l
i
2
73
1=1
2
(.1=1
1
r . - R.
"
x
IT
r .
2
1=1
J
i
_
1
73
X2
1=1 "
n
2 ff .
l
1=1
73
2
-
x.
n
2 X2
J
i =l
73
F^T
2
x
-
2 x2
i=l *
where we used Cauchy*s inequality in the last step. Equality is valid in (1)
if and only if
*T
Putting 7? = 3, rt
becomes
R
= w , r 2 = vf r3 = w and xl = a2, x 2 = b2, x3 = c2,
(1)
- 252 -
_,,
-r — r —
V 4- W
T
W 4- U
J ^ L _ > i(a 2 + 5 2 + c 2 ) 2 - (a4 4- 5 4 4- c 4 )
U 4- V ~ 2
= ^[2a 2 5 2 + 25 2 c 2 4- 2c 2 a 2 - (a4 4- 5 4 4- c 4 )]
1
•(a 4- 6 + c)(a
4- 6 - c)(6 4- c - a)(c 4- a - 6)
with equality if and only if
y 4- w
Generalization
II.
by
w4-o
M.S.
u 4- y ®
Klamkin,
University
of Alberta,
Edmonton,
Alberta,
More generally? we show that for 0 < n < 1,
4/2
4/2
ua
vb
U + W
¥4-11
.2/2
4/2
0
3 4&
wc
U 4- V
(2)
75"
The proposed inequality corresponds to the case n = 1,
Letting S be the left side of (2), we have
o r o
2[S
fU 4- V 4- W
, 4J3, ,4/3,
4/2*,
s
4- (a
+ 5
4- c
) ]
[(V
> (a
by Cauchy*s inequality.
, U 4- V 4- W , 4 / 2 , U 4- V 4- W 4/21
44- _ U
— _4- —
V— - c
W 4-, U — - 6
4/2
,4/2
4/2
4- (w 4- if) 4- (u 4- y ) ]
\V + W
W + U
U 4- V
F 4- W
4- w)
4-5
4/2
a
4- c
)
(3
Equality holds in (3) if and only if
2/?
a
V-fW
2/?
2/2
W4-H
1/4-F
(3) is equivalent to
2S > (a
4-5
4-c
) - 2(a
4-5
+ c )
,
4/2
^ ,4/2 ^
4/2,
2n
2n
2/2
2J2
2/2.2/2
0h
A 0
A 0
25
c
4- 2c a
4- 2a 5
(a
+5
4- c )
, /2 , , n 8 n w , i ) , //
/? w // , //
, 12 w n , , //
n.
= (a 4 - 5 4 - c ) ( 5 4 - c - a ) ( c 4- a - 5 ) ( a 4- 5 - c ) .
Since af
b9 c are the edges of a triangle, {a ,5 ,c }
also
form
a
t Ax
(4)
triangle
whose area we denote by F . Then (4) becomes
72
S > 8F 2
72
by
?
Heron s formula.
We now obtain (2) by applying Oppenheimfs generalization
of the Finsler-Hadwiger inequality! i.e..
- 253 -
_£>
4F
' ^
AF
JZ
JJ
0 < n < 1
with equality (for n & 1) if and only if a =
Inequalities
Piihlications
involving
elements
de la Faculte
5
=
c.
(See
A*
Oppenheim,
of triangles, quadrilaterals or tetrahedra,
d'Electrotechnique
de
1'Universite
a
Belgrade,
No.461 - No.497, (1974) 257-263.)
Note
American
also
that the case n = y gives the proposer's problem E3150 in the
Mathematical
Monthly
(May 1986, p.400).
It is of interest to compare the given inequality with the not so
easily
proven, and incomparable, inequality
(See
M.S. Klamkin,
ub2c2
vc2a2
wa2b2
V + W
W + U
Tl * V -
Two
non-negative
ft2
forms, El em,
quadratic
der Math.,
28
(1973) 141-146.)
3
Finally, note that if we put a = 5 = c, we then have Kz = y^ a 4 ,
and
so
the given inequality reduces to the known inequality
u
v
w
> 3
+
+
V + W
W4-U
for any positive real numbers ti, v9 w.
III. Further
special
cases
by
Innsbruck,
£! + V ~ "2"
Walther
Janous,
Ursulinengymnasium,
Austria.
(i) Putting u - v = w into the given inequality, we obtain
a 4 + 5 4 + c 4 > 16IT2
which i s ( [ 1 ] , 4 , 1 0 ) .
(ii) Putting u = a , y = 6 , w = c , we obtain
2s - a + 2F^~5" * 2S~=~F - 8l4'2>
where s is the semiperimeter.
(iii) Putting u = a" 4 , v = 6"4, w = c"4, we obtain
ft«c<
5
4
-!» c
+
4
c<a<
c
4
+ a
+
4
a<ft«
a
4
+ 6
^
4
which, because of
54c4
5 4 + c4
&2c2
x /t
, e x.c. ,
Z
is stronger than ([1], 4.12).
(iv) Putting u = a"2, v = 6~2, w = c"2, we obtain
- 254 -
1
a2b2c2
1
> 8K''
a2 * h2
which, using 4K7? = abc where R is the circumradius, can be written
1
1
1
2R2
+ b2
Because of
<
b2
* c2
etc. ,
w'
this is stronger than the first inequality of ([l] f 5*24),
(v) Putting u = b + c, v = c + a , w = a + 6 , we obtain
'2s - a' 4
(2s - b' A
'2s - c" c«
a +
b +
> 8ff2
2s 4- c_
[2s + a
[2s + j
b
(vi)
Since
4^2
-
2i}2
= Z>aA? = c 2 A 2 ,
b
c
i r e t h e a l t i t u d e s , we o b t a i n
a
a
a
b
c
u
v +
Putting u - h2,
a
v
h2
5
a2
W
2
V
4
A
a
w = h2,
c
" w *
U
A»
b
we obtain
C
+
°
V
2,
h* c
> 2,
h2 + i?2
c
a
A? + i32
6
_.*! + _JL_.il >
i? 2
4- /> 2
a
b
which is ([1], 6,7).
Next we apply Klamkin's duality
3a 35 3c 3/T
I (a ,b ,c ,m ,m, ,m ,K) > 0 •=» I
a
fc
c
> 0
"
where m t m, * m are the medians.
a
b
c
(This follows
immediately
from
the
fact
the medians of a triangle with area K are themselves sides of a triangle
3a 36 3c
,
3K
with medians j — , -j—, -*— and area —-. See Klamkin's solution to Aufgabe 677 9
that
ff.^m. <fcr Mat/?, 28 (1973) 130.) Tsintsifas 1 inequality then reads
— — ~
ID- +
v + w
a
> "A2,
r - 2
1
m? + -^—m«
u -f v
w + tz h
We may now use (5) to obtain more or Jess interesting duals of (i) - (v)
v =ro."4f w = m
instance, putting n - m~
4
into (5) yields
M /flM
4
b c
D
4
c
c
4
Z
c a
Q
> 77 A ?
—
a
a
which we show is stronger than ([!]* 8.6)* i.e.
6
(5)
For
- 255 -
m2 + mf +ffl2> 37SK.
a
ft
c ~
SqIJHr i ri£, Lh i s beec >mes
(M* + < -I- «/") + 2(m2mf + i » > 2 + ra2wi2) > 27tf2f
a
A>
c
a b
be
c a and t h u s we h a v e t o show t h a t
m*m*
o c
6
< 2 /s^ + 2 2 izifffl
'6 c
+ 2ZT
ffl,
But this follows because of
6 c
a!
1
< 2/»>2
5 c
4
-I- i?i
5
etc.
c
find
< + m4
b
c
m*m*
6 c
etc.
m£ +
b
Reference:
[1]
0. Bottema et al, Geometric
InequalHies,
Wolters-Noordhoff f
Groningenf
1969.
Also
solved
by C. FESTRAETS-HAMOIR, Bruxelles,
Arnhem, The Netherlands;
Middle town,
Pennsylvania;
author unknown, sent
Bulgaria,
VEDULA
to
who received
and
this
N.
the
MURTY,
proposer.
office
by
it from the editor
Pennsylvania
Svetoslav
(2),
From a
[1985: 187]
Trinity
State
Bilchev
University,
solution,
of
Russe,
Sosinskij.
using the same method.
by other solvers,
(i) by Klamkin and Murty, and (iv) by Murty and the
1052.
J.
of KVANT, A.B.
cases were noticed
J .T. GROENMAN,
There was a further
Janous also obtained Klamkin's generalization
A couple of Janous' special
Belgique;
in
particular
proposer.
College,
Cambridge, examination
p&fter dated June 5, 1901.
Prove that
1
2
_
3
l -3 -5
Solution
2
1
2
3
1
3 -5 -7
2
by George Szekeres,
2
3
5 -7 *9
_
2
University
= I1^_
9
?L
IT"
e
2s
2
of New South
Wales,
Kensington
AustraIia.
How
would
a
Trinity
student of 1901 evaluate this sum?
He may choose
- 256 -
from a number of different methods,
Partial
I.
fractions.
First,
1
2 3
(n-2) n (n±2)'
J_ +
1
32n
ISn
1
3
1
64
n-2
n+2
1
128 (n-2)
(//+2)2J
2
(not very hard to determine the coefficients9 the answer in a way gives
away
the expected form of the partial fraction decomposition).
From here the required sum is
Y
(
^
n=3
;/ odd
-i)<"+1>/2
JL - _i
(ji-2)2n3(n+2)2
32
L+ L
1
32
_i
_1
16
16
1
3"
Hi -1) - 1\
4
3] 32" [
BT
1
1
1
32" + TB" + 3ZT + TBrgr
+
+
l-i_
33
1
F
+
I_-
53
1
T2"8~
T E J 1 3- + 5-
1
TB"
_ 1
1
' S" TB"
1
3
1
+X
3"
TB"
1-1-+I
The answer follows now from
>-W
and
i - i _3
3
The
+
i
53
32
first one is trivial from the arctan series, the second is Euler's.
method is perhaps the most
remembers
the second
straightforward
but assumes
that
This
the student
sum. Of course he can still bluff his way through the
problem by working backwards from the given answer*
TT.
Theory of
lesidues.
This is no doubt the most direct method (my own favourite) and best
suiled
for examination
conditions
memorize any filing.
Note that the required sum is
as it does
not require the student to
- 257 -
(-D
}
n-i
("D
(2fi-l)2(2/3+l)3(2n+3)2
73-1
2
L
,
12=1
(2l3-l)
(213-1
(2/H-l)3(2l2+3)2
(-D
(2/2~l)2(2/2+l)3(2/2-f3):
n=-2
I
7
(-D
D=-oo
1
1
7*
S"
12-1
(-D-1
1
(2/7-1 )2(2//-fl)3(2«-f-3):
n-1
2"
(-D-
(-1)2(1)3(3)2
(-3)2(-l)3(l)2
where
(~D
J3=-oo
13-1
(2i3-l)2(2ri-l-l)3(2i3+3)2
To evaluate S, we consider the function
IT
f(z)
2
(z - l) (z + l) 2 z 3 cos nz
Let iV be a positive integer and let C be
having
corners
(±N9±N)*
the
square
in
the
complex
plane
From the fact that jcos H Z | > 1 for z on C we see
that
lim [ f(z)dz = 0,
Hence the sum of all the residues of f(z)
is zero.
On the
other
handf
f(z)
has simple poles at z = k + ~r (& and integer), poles of order 2 at z = ±1, and
a pole of order 3 at z = 0,
lim
z-tk+y
TT
z - k
2
y
2
The residue of f(z)
at & + ^ is
lim
71
3
(z - l ) ( z + l ) z c o s xrz
3
z-^ir+rr
hrt-rhy
128w
2
3
(2k - I ) ( 2 * + I ) (21 + 3 )
(-1)
2
&-1
k
1
c o s rrz
lim
2
-128
(2k - 1 ) ( 2 1 + 1 ) 3 ( 2 * + 3 ) 2
The residue of f(z) at z = 1 is
Z
z-#*+£ - u s i n TFZ
- 258 -
H
lim
JZ
z^l
the residue of f(z)
IT
(z + l)2z3cos nz
at z = -1 is similarly
d
u
lim
Uz
(Z - l) 2 Z 3 COS HZ
z-r-1
and the residue of f(z) at z = 0 is
Id*
lim :
z-»0
2 dz2\ (z - 1)2(Z 4- l)2COS HZ
J: = —
2
+ 2lt
Thus
128S -f- u + u + _
+ 2ir = 0,
S =
and so the required sum is
1
c x 1 - 1
7^
II1»
Fourier
This
+
«
**
^ - g- ST " "512" #
series.
is
more
elaborate
than the first two, but gives quite a bit
more (definitely not recommended to the run-of-the-mill student).
Let S be as before and define
F(
sin(2/3 4- l)x
„=l
n=-oo
(1)
(2D - l)2(2n + l)3(2n 4- 3 ) 2
then clearly
- = •*[?] •
The following two Fourier expansions are well known
e.g.
Knopp,
501,
#15-8):
Unendliche
}
Reihen,
(any
standard
386-388, or Apostol, Mathematical
S^~Tssii n ( 2 n
- l)x =
textbook,
Analysis,
(2)
n=-oo
i
J2=-o©
cos(2/? - l ) x =
(2n
-
1)
:
IT*
7fX
(3)
- 259 -
both valid for 0 < x < IT. There are corresponding formulae for
(n 9 2TI ) but we need not bother about them.
First, F'"(x)
= ~G(x)9
the
interval
Here we go.
where
00
cos(2i? + l ) x
G?(x)
(2n -
l)2(2n
+ 3)2
Then
oo
(2n + l ) 2 c o s ( 2 n
<r- x) = -
+ l)x
2
(2n - l ) ( 2 n + 3 ) a
^
/l=-oo
=
+
- i 5 fcr=-r
sTr-Trf003*2"
+ 1)x
n=-oo
oo
1
oo
X
eos(2n
+ l ) x + cos(2i7 - 3 ) x
= -T2
1
i2n
.
/ 0
7 c
L
V
)
os(2x)
_ 1)2
I V
— " ^ 2 Tsr^TTTsr+^r
COs(2l2 - 1)X
1 „y .
.
1— _ L _ - v H(x)
Z
(2n ~ I ) 2
L
72 = - o o
where
Hfv>
H{x)
-
\
"
2
cos(2ri + l ) x
12n - I ) ( 2 n + 31
77=-oo
and we used the identity
o o
A - £
A + B
c o s A 4- c o s B = 2 c o s — ^
cos -—«—
Thus
oo
H'/V\
tf
m
{2n
- _
^
"
Z
+
D(sin(2n
+ l)x
(2n - l)(JJn + 3)
n=-«o
oo
1
~
V
? L
1
In - 1
+
1
' sin(2n + l)x
iin + 3
n~-oo
1
V
1 I
/} = - »
cos(2j2 + l ) x
sin(2n + l ) x + sin(2n - 3)x
In - 1
//8.
(4)
- 260 -
oo
v \
-cos(2x) >
/0
sin(2n - l)x
2n - t
J2 = -oo
= - -n- COS
by
(2) and
the
2x,
identity
A + B
A - B
sin A + sin B = 2 sin — n — - cos — J J —
Integrating? and observing that H\^\
= 0, we get
ff(x) = ~ x
sin
2x,
hence from (3) and (4)
flT2
1
TTXl
IT
G"(x) = - -^ oos(2x)« U- - •£-- + g- sin 2x
u
IT2
nx
•g sin 2x - ^- cos 2x + -j— cos 2x*
Using
,
x sin 2x cos 2x
0
x cos 2x dx =
^—
+ — j — —
and
~ _,
-x cos 2x . sin 2x
x sin 2x ax =
+ —*
n
Z
4
and noting G'(0) = 0, G|£ = 0 we get
G'(x)
= - yyr sin 2x + —^ sin 2x
and
7T
7TX
IT
G(x) = -^ cos 2x - j7T cos 2x + _
sin 2x«
Thus
F"'(x) = - Tp^ cos 2x -I- jrr cos 2x - m sin 2x.
Finally,
three times and observing that F#/(0) = 0, F'LJ
integrating
F(0) = 0, we obtain successively
«#//
7T 2
v
.
0
, UX
r/
F' (x) = - _: sin 2x + ^
^r/
i
l* 2
FMx) = ^
o
cos 2x -
.
W X
M
0
s
IT
0
7T
sin 2x 4- •— cos 2x - *«• ,
o
.
3lT
cos 2x + m
o
TTX ^ IT 2
sin 2x - ^
+^
,
and
JT
F(x) = ^
Thus
2
TT V
sin 2x - J 2 ? sin 2x -
IT
M
TTX
cos 2x - ^
7T
V
TF
+ E5~ + CT
- 261 -
7X3
IT
vfa]
IT 3
H
IT3
IT
as required.
Admittedly the Fourier argument is heavier than the other two, but as a
bonus, formula (5) evaluates the series (1) for all 0 < x < n«
Also
solved
by
C.
KIERSTEAD JR., Cuyahoga
Southwold,
Suffolk,
Townsville,
Austral
FESTRAFFS-EAMOIR,
Falls,
Ohio;
England;
and
KEE-WAI
BASIL
Bruxelles,
Ik^lgique;
LAU,
Kong;
C.
Hong
R.C.
LYNESS,
RENNIE, James Cook
University,
la.
%
%
1053.
FRIEND H.
[1985: 187]
Proposed
Corp., Nashua,
Exhibit
%
by Stanley
New
Rabinowitz,
Digital
Equipment
Hampshire.
a bisection
between
the points in the plane and the
lines in the plane.
Solution
by Leroy
F
Meyers,
The Ohio State
University,
Columbus,
Ohio.
Every nonvertical line in the (Euclidean) plane has a unique equation of
the
form y = mx + 5, where m and b are arbitrary real numbers, and conversely
every pair (m,b) of real numbers determines a nonvertical line with
y = mx + b.
Since
the vertical
$
haven t
lines
correspondence, we must modify the correspondence.
line with
equation
x
been
Thus,
= b we associate the point (09b);
integer
and the point
(mfb)
included
to each
in the
vertical
to each nonvertical
line with equation y = mx + b we associate the point (m +
nonnegative
equation
1,5) if m is a
if HI is otherwise.
This is the
required one-to-one correspondence.
Also
solved
by J.
PARMENTER, Memorlal
There
omit
University
was one incorrect
projective
plane
SUCK, Essen,
(nn
Federal
of
solution.
the
surface,
Republic
Newfound1 andf
Two readers
an easier
of
St.
Germany;
J
John s,
Newfound1
*
M M.
and.
submitted
solutions
for
the
problem).
I flatly
decided
to
these.
*
and
*