Solutions and comments.

79.

Let x_{0}, x_{1}, x_{2} be three positive real numbers.
A sequence { x_{n} } is defined, for n ³ 0 by
x_{n+3} = 
x_{n+2} + x_{n+1} + 1 x_{n}

. 

Determine all such sequences whose entries consist solely
of positive integers.
Solution 1. Let the first three terms of the sequence be
x, y, z. Then it can be readily checked that the sequence
must have period 8, and that the entries cycle through the
following:
x, y, z, 
y+z+1 x

, 
x + y + z + 1 + xz xy

, 


(x+y+1)(y+z+1) xyz

, 
x + y + z + 1 + xz yz

, 
x+y+1 z

. 

For all of these entries to be positive integers, it is necessary
that y + z + 1 be divisible by x, x + y + 1 be divisible by
z and (x+1)(z+1) be divisible by y. In particular,
x
£ y + z + 1 and z
£ x + y + 1.
Without loss of generality, we can assume that the smallest
entry in the sequence is y. Then,
and
whence

 
= y^{2} + yz + y £ y^{2} + x + y + 1 + y 
 


Hence
x(y^{2}  1) £ (y + 1)^{2} 

so that either y = 1 or
y £ x £ 
y+1 y1

= 1 + 
2 y1

. 

The latter yields (y
 1)
^{2} £ 2 so that y
£ 2.
Suppose that y = 1. Then x £ y + z + 1 = z + 2 £ y + x + 3 = x + 4.
Since x divides x and y + z + 1, it must divide their difference,
which cannot exceed 4. Hence x = y + z + 1 or
x must be one of 1, 2, 3, 4. Similarly, z = x + y + 1 or
z must be one of 1, 2, 3, 4.
If x = y + z + 1 = z + 2, then x + y + 1 = z + 4 and so
z divides 4. We get the periods
(6, 1, 4, 1, 6, 2, 9, 2) . 

The case z = x + y + 1 yields essentially the same periods.
Otherwise, 1
£ x, z
£ 4 and we find the additional
possible period
For each period, x
_{0} can start anywhere.
Suppose that y = 2. Then x £ y + z + 1 £ z + 3 £ x + 6,
so that x must divide some number not exceeding 6. Similarly,
z cannot exceed 6. If x = 2,
then x + y + 1 = 5 and so z = 5; this yields the period
(2, 2, 5, 4, 5, 2, 2, 1) already noted. If x = 3, then z must be
2, 3 or 6, and we obtain (3, 2, 3, 2, 3, 2, 3, 2); the possibilities
z = 2, 6 do not work. If x = 4, z = 7, which does not work.
If x = 5, then z = 2, 4 and we get a period already noted.
If x = 6, then z = 3, which does not work.
Hence there are five possible cycles and the sequence can begin
at any term in the cycle:
Solution 2. [O. Ivrii] We show that the sequence
contains a term that does not exceed 2. Suppose that
none of x_{0}, x_{1} and x_{2} is less than 3.
Let k = max(x_{1}, x_{2}). Then,
noting that k is an integer and that k ³ 3, we deduce that
x_{3} = 
x_{1} + x_{2} + 1 x_{0}

£ 
2k+1 3

< k 

so x
_{3} £ k
 1 and
x_{4} = 
x_{2} + x_{3} + 1 x_{1}

£ 
2k 3

£ k  1 

so that max(x
_{3}, x
_{4}) = k
 1. If k
 1
³ 3, we
repeat the process to get a strictly lower bound on the next two terms.
Eventually, we obtain two consecutive terms whose maximum is
less than 3. In fact, we can deduce that there is an entry
equal to either 1 or 2 arbitrarily far out in the sequence.
Suppose, from some point on in the sequence, there is no term
equal to 1. Then there are three consecutive terms a, 2, b.
The previous term is (a+3)/b and the following term is
(b+3)/a, so that a divides b + 3 and b divides a + 3.
Since a  3 £ b £ a + 3, b divides two numbers that
differ by at most 6; similarly with a. Hence, neither
a nor b exceeds 6. Testing out possibilities leads to
(a, b) = (6, 3), (5, 2), (3, 3).
Finally, we suppose that the sequence has three consecutive terms
a, 1, b and by similar arguments are led to the sequences
obtained in the first solution.
Comment. C. Lau established that x_{n} x_{n+4} = x_{n+2} x_{n+6} and thereby obtained the periodicity.
R. Barrington Leigh showed that, if all terms of the sequence
were at least equal to 2, then the sequence { y_{n} } defined
by y_{n} = max(x_{n}, x_{n+1}) satisfies y_{n+1} £ y_{n}.
Since the same recursion defines the sequence ``going backwards'',
we also have y_{n1} £ y_{n}, for all n. Hence
{ y_{n} } is a constant sequence, and so { x_{n} } is
either constant or has period 2. It is straightforward to
rule out the constant possibility. If the periodic segment
of the sequence is (a, b), then b = (a + b + 1)/a or
(a1)(b1) = 2 and we are led to the segment (2, 3).
Otherwise, there is a 1 in the sequence and we can conclude
as before.

80.

Prove that, for each positive integer n, the
series
converges to twice an odd integer not less than (n+1)!.
Solution 1. Since the series consists of nonnegative
terms, we can establish its convergence by eventually showing
that it is dominated term by term by a geometric series with
common ratio less than 1. Noting that n
£ k log
_{k}(3/2)
for k sufficiently large, we find that for large k,
k
^{n} < (3/2)
^{k} and the kth term of the series is
dominated by (3/4)
^{k}. Thus the sum of the series is
defined for each nonnegative integer n.
For nonnegative integers n, let
S_{n} = 
¥ å
k=1


k^{n} 2^{k}

. 

Then S
_{0} = 1 and

= 
¥ å
k=1


k^{n} 2^{k}

 
¥ å
k=1


k^{n} 2^{k+1}


 
= 
¥ å
k=1


k^{n} 2^{k}

 
¥ å
k=1


(k1)^{n} 2^{k}


 
= 
¥ å
k = 1


k^{n}  (k  1)^{n} 2^{k}


 
= 
¥ å
k=1


é ê
ë


æ ç
è

n
1

ö ÷
ø


k^{n1} 2^{k}

 
æ ç
è

n
2

ö ÷
ø


k^{n2} 2^{k}

+ 
æ ç
è

n
3

ö ÷
ø


k^{n3} 2^{k}

¼+ (1)^{n1} 
1 2^{k}


ù ú
û




whence
S_{n} = 2 
é ê
ë


æ ç
è

n
1

ö ÷
ø

S_{n1}  
æ ç
è

n
2

ö ÷
ø

S_{n2} + 
æ ç
è

n
3

ö ÷
ø

S_{n3}  ¼+ (1)^{n1} 
ù ú
û

. 

An induction argument establishes that S
_{n} is twice an odd integer.
Observe that S_{0} = 1, S_{1} = 2, S_{2} = 6 and S_{3} = 26.
We prove by induction that, for each n ³ 0,
from which the desired result will follow. Suppose that
we have established this for n = m
 1. Now
S_{m+1} = 2 
é ê
ë


æ ç
è

m+1
1

ö ÷
ø

S_{m}  
æ ç
è

m+1
2

ö ÷
ø

S_{m1} + 
æ ç
è

m+1
3

ö ÷
ø

S_{m2}  
æ ç
è

m+1
4

ö ÷
ø

S_{m3} + ¼ 
ù ú
û

. 

For each positive integer r,

æ ç
è

m+1
2r1

ö ÷
ø

S_{m2r+2} 

 
æ ç
è

m+1
2r

ö ÷
ø

S_{m  2r + 1} 
 
³ 
é ê
ë


æ ç
è

m+1
2r1

ö ÷
ø

(m  2r + 3) 
æ ç
è

m+1
2r

ö ÷
ø


ù ú
û

S_{m2r+1} 
 
= 
æ ç
è

m+1
2r1

ö ÷
ø

[(m2r+3)  
æ ç
è


m2r+2 2r


ö ÷
ø


ù ú
û

S_{m  2r + 1} ³ 0 . 


When r = 1, we get inside the square
brackets the quantity
while when r > 1, we get
(m  2r + 3)  
æ ç
è


m2r+2 2r


ö ÷
ø

> (m  2r + 3)  (m  2r + 2) = 1 . 

Hence

³ 2 
é ê
ë


æ ç
è

m+1
1

ö ÷
ø

S_{m} 
æ ç
è

m+1
2

ö ÷
ø

S_{m1} 
ù ú
û


 
³ 2 
é ê
ë

(m+1)S_{m}  
m(m+1) 2

· 
1 m+1

S_{m} 
ù ú
û


 
= 2 
é ê
ë

m + 1  
m 2


ù ú
û

s_{m} = (m + 2)S_{m} . 


Solution 2. Define S_{n} as in the foregoing solution.
Then, for n ³ 1,

 
= 
1 2

+ 
1 2


¥ å
k=1


(k+1)^{n} 2^{k}


 
= 
1 2

+ 
1 2


¥ å
k=1


k^{n} + 
æ ç
è

n
1

ö ÷
ø

k^{n1} +¼+ 
æ ç
è

n
n1

ö ÷
ø

k + 1 
2^{k}


 
= 
1 2

+ 
1 2


é ê
ë

S_{n} + 
æ ç
è

n
1

ö ÷
ø

S_{n1} + ¼+ 
æ ç
è

n
n1

ö ÷
ø

S_{1} + 1 
ù ú
û




whence
S_{n} = 
æ ç
è

n
1

ö ÷
ø

S_{n1} + 
æ ç
è

n
2

ö ÷
ø

S_{n2} +¼+ 
æ ç
è

n
n1

ö ÷
ø

S_{1} + 2 . 

It is easily checked that S
_{k} º 2 (mod 4)
for k = 0, 1. As an induction hypothesis, suppose
this holds for 1
£ k
£ n
1. Then,
modulo 4, the right side is congruent to
2[ 
n å
k=0


æ ç
è

n
k

ö ÷
ø

 2] + 2 = 2(2^{n}  2) + 2 = 2^{n+1}  2 , 

and the desired result follows.
For n ³ 1,

= 

æ ç
è

n+1
1

ö ÷
ø

S_{n} + 
æ ç
è

n+1
2

ö ÷
ø

S_{n1}+ ¼+ 
æ ç
è

n+1
n

ö ÷
ø

S_{1} + 2 
S_{n}


 
= (n+1) + 

æ ç
è

n+1
2

ö ÷
ø

S_{n1} + 
æ ç
è

n+1
3

ö ÷
ø

S_{n2} + ¼+ (n+1)S_{1} + 2 

æ ç
è

n
1

ö ÷
ø

S_{n1} + 
æ ç
è

n
2

ö ÷
ø

S_{n2} + ¼+ nS_{1} + 2 


 


since each term in the numerator of the latter fraction
exceeds each corresponding term in the denominator.
Solution 3. [of the first part using an idea of P. Gyrya]
Let f(x) be a differentiable function
and let D be the differentiation operator. Define the
operator L by
Suppose that f(x) = (1
 x)
^{1} =
å_{k=0}^{¥} x
^{k}.
Then, it is standard that L
^{n}(f)(x) has a power series
expansion obtained by termbyterm differentiation that
converges absolutely for
x
 < 1. By induction,
it can be shown that the series given in the problem is,
for each nonnegative integer n, L
^{n}(f)(1/2).
It is straightforward to verify that
L((1  x)^{1}) = x(1  x)^{2} 

L^{2}((1  x)^{1}) = x(1 + x)(1  x)^{3} 

L^{3}((1  x)^{1}) = x(1 + 4x + x^{2})(1  x)^{4} 

L^{4}((1  x)^{1}) = x(1 + 11x + 11x^{2} + x^{3})(1  x)^{5} . 

In general, a straightforward induction argument yields that
for each positive integer n,
L^{n}(f)(x) = x(1 + a_{n,1}x + ¼+ a_{n,n2}x^{n2} +x^{n1})(1  x)^{(n+1)} 

for some integers a
_{n,1},
¼, a
_{n,n2}. Hence
L^{n}(f)(1/2) = 2(2^{n1} + a_{n,1}2^{n2} + ¼+ a_{n,n2}2 + 1) , 

yielding the desired result.

81.

Suppose that x ³ 1 and that x = ëx û+ { x }, where ëx û
is the greatest integer not exceeding x and the
fractional part { x } satisfies 0 £ x < 1.
Define


(a) Determine the smallest number z
such that f(x) £ z for each x ³ 1.


(b) Let x_{0} ³ 1 be given, and for
n ³ 1, define x_{n} = f(x_{n1}). Prove that
lim_{n ® ¥} x_{n} exists.
81.
Solution. (a) Let x = y + z, where y =
ëx
û and
z = { x }. Then
which is less than 2 because
Ö[yz]
£ ^{1}/
_{2}(y+z)
by the arithmeticgeometric means inequality. Hence 0
£ f(x)
£ Ö2 for each value of x. Taking y = 1,
we find that

lim
x 2

f(x)^{2} = 
lim
z 1


æ ç
è

1 + 
2Öz 1 + z


ö ÷
ø

= 2 , 

whence sup{ f(x): x
³ 1 } =
Ö2.
(b) In determining the fate of { x_{n} }, note that after the
first entry, the sequence lies in the interval [1, 2). So,
without loss of generality, we may assume that 1 £ x_{0} < 2.
If x_{n} = 1, then each x_{n} = 1 and the limit is 1. For the rest,
note that f(x) simplifies to (1 + Ö[(x1)])/Öx on
(1, 2). The key point now is to observe that
there is exactly one value v between 1 and 2 for which
f(v) = v, f(x) > x when 1 < x < v and f(x) < x when
v < x < 2. Assume these facts for a moment. A derivative
check reveals that f(x) is strictly increasing on (1, 2),
so that for 1 < x < v, x < f(x) < f(v) = v, so that
the iterates { x_{n} } constitute a bounded, increasing
sequence when 1 < x_{0} < v which must have a limit.
(In fact, this limit must be a fixed point of f and so
must be v.) A similar argument shows that, if v < x_{0} < 2,
then the sequence of iterates constitute a decreasing convergent
sequence (with limit v).
It remains to show that a unique fixed point v exists.
Let x = 1 + u with u > 0. Then it can be checked that
f(x) = x if and only if 1 + 2Öu + u = 1 + 3u + 3u^{2} + u^{3}
or u^{5} + 6u^{4} + 13u^{3} + 12u^{2} + 4u  4 = 0. Since the left side
is strictly increasing in u, takes the value 4 when u = 0
and the value 32 when u = 1, the equation is satified for exactly
one value of u in (0, 1); now let v = 1 + u. The
value of V turns out to be about 1.375, (Note that
f(x) > x if and only if x < u.)

82.

(a) A regular pentagon has side length
a and diagonal length b. Prove that

b^{2} a^{2}

+ 
a^{2} b^{2}

= 3 . 



(b) A regular heptagon (polygon with seven equal
sides and seven equal angles) has diagonals of two
different lengths. Let a be the length of a side,
b be the length of a shorter diagonal and c be the
length of a longer diagonal of a regular heptagon
(so that a < b < c). Prove that:

a^{2} b^{2}

+ 
b^{2} c^{2}

+ 
c^{2} a^{2}

= 6 

and

b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}

= 5 . 

82.
Solution 1. (a) Let ABCDE be the regular
pentagon, and let triangle ABC be rotated about C
so that B falls on D and A falls on E. Then
ADE is a straight angle and triangle CAE is similar
to triangle BAC. Therefore

a+b b

= 
b a

Þ 
b a

 
a b

= 1Þ 
b^{2} a^{2}

+ 
a^{2} b^{2}

 2 = 1 

so that b
^{2}/a
^{2} + a
^{2}/b
^{2} = 3, as desired.
(b) Let A, B, C, D, E be consecutive vertices of
the regular heptagon. Let AB, AC
and AD have respective lengths a, b, c, and let
ÐBAC = q. Then q = p/7, the length
of BC, of CD and of
DE is a, the length of AE is c,
ÐCAD = ÐDAE = q, since the angles
are subtended by equal chords of the circumcircle of the heptagon,
ÐADC = 2q, ÐADE = ÐAED = 3q
and ÐACD = 4q.
Triangles ABC and ACD can be glued together along BC and
DC (with C on C) to form a triangle similar to
DABC, whence
Triangles ACD and ADE can be glued together along CD and
ED (with D on D) to form a triangle similar to
DABC, whence
Equation (2) can be rewritten as
^{1}/
_{b} =
^{1}/
_{a} ^{1}/
_{c}. whence
Substituting this into (1) yields
which simplifies to
a^{3}  a^{2}c  2ac^{2} + c^{3} = 0 . 
 (3) 
Note also from (1) that b
^{2} = a
^{2} + ac.

a^{2} b^{2}

+ 
b^{2} c^{2}

+ 
c^{2} a^{2}

 6 

= 
a^{4} c^{2} + b^{4} a^{2} + c^{4} b^{2}  6a^{2}b^{2}c^{2} a^{2}b^{2}c^{2}


 
= 
a^{4}c^{2} + (a^{4} + 2a^{3}c + a^{2}c^{2})a^{2} + c^{4}(a^{2} + ac) 6a^{2}c^{2}(a^{2} + ac) a^{2} b^{2} c^{2}


 
= 
a^{6} + 2a^{5}c  4a^{4}c^{2}  6a^{3}c^{3} + a^{2}c^{4} + ac^{5} a^{2}b^{2}c^{2}


 
= 
a(a^{2} + 3ac + c^{2})(a^{3}  a^{2}c  2ac^{2} + c^{3}) a^{2}b^{2}c^{2}

= 0 `. 



b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}

 5 

= 
(a^{4} + 2a^{3}c + a^{2}c^{2})c^{2} + a^{2}c^{4} + a^{4}(a^{2} + ac) 5a^{2}c^{2}(a^{2} + ac) a^{2}b^{2}c^{2}


 
= 
a^{6} + a^{5}c  4a^{4}c^{2}  3a^{3}c^{3} + 2a^{2}c^{4} a^{2}b^{2}c^{2}


 
= 
a^{2}(a + 2c)(a^{3}  a^{2}c  2ac^{2} + c^{3}) a^{2} b^{2} c^{2}

= 0 . 


Solution 2. (b) [R. Barrington Leigh] Let the heptagon be
ABCDEFG; let AD and BG intersect at P,
and BF and CG intersect at Q. Observe that
PD  = GE  = b,
AP  = c  b, GP  = DE  = a,
BP  = b  a, GQ  = AB  = a,
CQ  = c  a. From similarity of
triangles, we obtain the following:

a c

= 
cb a

Þ 
a c

 
c a

+ 
b a

= 0 (DAPG ~ DADE) 


ca a

= 
c b

Þ 
c a

 
c b

= 1 (DQBC ~ DCEG) 


cb a

= 
ba b

Þ 
c a

 
b a

+ 
a b

= 1 (DAPG ~ DDPB) 


ba a

= 
b c

Þ 
b a

 
b c

= 1 (DABP ~ DADB) . 

Adding these equations in pairs yields

b a

+ 
a c

 
c b

= 1Þ 
b^{2} a^{2}

+ 
a^{2} c^{2}

+ 
c^{2} b^{2}

+ 2 
æ ç
è


b c

 
c a

 
a b


ö ÷
ø

= 1 

and

c a

+ 
a b

 
b c

= 2Þ 
c^{2} a^{2}

+ 
a^{2} b^{2}

+ 
b^{2} c^{2}

+ 2 
æ ç
è


c b

 
b a

 
a c


ö ÷
ø

= 4 . 

The desired result follows from these equations.
Solution 3. (b) [of the second result by J. Chui]
Let the heptagon be ABCDEFG and q = p/7.
Using the Law of Cosines in the indicated
triangles ACD and ABC, we obtain the following:
cos2q = 
a^{2} + c^{2}  b^{2} 2ac

= 
1 2


æ ç
è


a c

+ 
c a

 
b^{2} ac


ö ÷
ø



cos5q = 
2a^{2}  b^{2} 2a^{2}

= 1  
1 2


æ ç
è


b a


ö ÷
ø

2



from which, since cos2
q =
cos5
q,
1 + 
1 2


æ ç
è


b a


ö ÷
ø

2

= 
1 2


æ ç
è


a c

+ 
c a

 
b^{2} ac


ö ÷
ø



or

b^{2} a^{2}

= 2 + 
a c

+ 
c a

 
b^{2} ac

. 
 (1) 
Examining triangles ABC and ADE, we find that
cosq = b/2a and cosq = (2c^{2}  a^{2})/(2c^{2}) = 1  (a^{2}/2c^{2}), so that
Examining triangles ADE and ACF, we find that
cos3
q = a/2c and cos3
q = (2b
^{2}  c
^{2})/(2b
^{2}),
so that
Adding equations (1), (2), (3) yields

b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}

= 6 + 
c^{2}  bc  b^{2} ac

. 

By Ptolemy's Theorem, the sum of the products of pairs of
opposite sides of a concylic quadrilaterial is equal to the
product of the diagonals. Applying this to the
quadrilaterals ABDE and ABCD, respectively, yields
c
^{2} = a
^{2} + bc and b
^{2} = ac + a
^{2}, whence
c
^{2}  bc
 b
^{2} = a
^{2} + bc
 bc
 ac
 a
^{2} =
ac and
we find that

b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}

= 6  1 = 5 . 

Solution 4. [of the second result by X. Jin]
By considering isosceles triangles
with sidebase pairs (a, b), (c, a) and (b, c), we find
that b^{2} = 2a^{2}(1  cos5q), a^{2} = 2c^{2}(1  cosq),
c^{2} = 2b^{2}(1  cos3q), where q = p/7. Then

b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}

= 2[3  (cosq+ cos3q+ cos5q)] . 

Now,
sinq(costheta + cos3q+ cos5q) 

= 
1 2

[ sin2q+ (sin4q sin2q)+ (sin6q sin4q)] 
 


so that cos
q+ cos3
q+ cos5
q = 1/2.
Hence b
^{2}/a
^{2} + c
^{2}/b
^{2} + a
^{2}/c
^{2} = 2(5/2) = 5.
Solution 5. (b) There is no loss of generality in assuming that the
vertices of the heptagon are placed at the seventh roots of unity on
the unit circle in the complex plane. Then z = cos(2p/7) + isin(2p/7) be the fundamental seventh root
of unity. Then z^{7} = 1,
1 + z+ z^{2} + ¼+ z^{6} = 0
and (z, z^{6}), (z^{2}, z^{5}), (z^{3},z^{4}) are pairs of complex conjugates. We have that
a = z 1  = z^{6}  1  

b = z^{2}  1  = z^{9}  1  

c = z^{3}  1  = z^{4}  1  . 

It follows from this that

b a

= z+ 1  
c b

= z^{2} + 1  
a c

= z^{3} + 1  , 

whence

b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}



= (z+ 1)(z^{6} + 1) + (z^{2} + 1)(z^{5} + 1)+ (z^{3} + 1)(z^{4} + 1) 
 
= 2 + z+ z^{6} + 2 + z^{2} + z^{5} + 2 + z^{3} + z^{4} 
 
= 6 + (z+ z^{2} + z^{3} + z^{4} + z^{5}+ z^{6}) = 6  1 = 5 . 


Also

a b

= z^{4} + z^{2} + 1  
b c

= z^{6} + z^{3} + 1  
c a

= z^{2} + z+ 1  , 

whence

+ 
b^{2} c^{2}

+ 
c^{2} a^{2}

= (z^{4} + z^{2} + 1)(z^{3} + z^{5} + 1)+ (z^{6} + z^{3} + 1)(z+ z^{4} + 1)+ (z^{2} + z+ 1)(z^{5} + z^{6} + 1) 
 
= (3 + 2z^{2} + z^{3} + z^{4} + 2z^{5})+ (3 + z+ 2z^{3} + 2z^{4} + z^{6})+ (3 + 2z+ z^{2} + z^{5} + 2z^{6}) 
 
= 9 + 3(z+ z^{2} + z^{3} + z^{4} + z^{5}+ z^{6}) = 9  3 = 6 . 


Solution 6. (b) Suppose that the circumradius of the
heptagon is 1. By considering isosceles triangles with base
equal to the sides or diagonals of the heptagon and apex at the
centre of the circumcircle, we see that

= 2sinq = 2sin6q = 2 sin8q 
 
 


where
q =
p/7 is half the angle subtended
at the circumcentre by each side of the heptagon.
Observe that
cos2q = 
1 2

(z+ z^{6}) cos4q = 
1 2

(z^{2} + z^{5}) cos6q = 
1 2

(z^{3} + z^{4}) 

where
z is the fundamental primitive root of unity.
We have that

b a

= 2cosq = 2cos6q 
c b

= 2cos2q 
a c

= 2cos4q 

whence

b^{2} a^{2}

+ 
c^{2} b^{2}

+ 
a^{2} c^{2}



= 4cos^{2} 6q+ 4cos^{2} 2q+ 4cos^{2} 4q 
 
= (z^{3} + z^{4})^{2} + (z+ z^{6})^{2} + (z^{2} + z^{5})^{2} 
 
= z^{6} + 2 + z+ z^{2} + 2 + z^{5}+ z^{4} + 2 + z = 6  1 = 5 . 


Also

= 
sin6q sin2q

= 4cos^{2} 2q 1 = (z+ z^{6})^{2}  1 = 1 + z^{2} + z^{5} 
 
= 
sin9q sin3q

= 4cos^{2} 3q 1 = 4cos^{2} 4q 1 = (z^{2} + z^{5})^{2}  1 = 1 + z^{4} + z^{3} 
 
= 
sin3q sinq

= 4cos^{2} 6q 1 = (z^{3} + z^{4})^{2}  1 = 1 + z^{6} + z , 


whence

a^{2} b^{2}

+ 
b^{2} c^{2}

+ 
c^{2} a^{2}



= (3 + 2z^{2} + z^{3} + z^{4} + 2z^{5})+ (3 + z+ 2z^{3} + 2z^{4} + z^{6})+ (3 + 2z+ z^{2} + z^{5} + 2z^{6}) 
 
= 9 + 3(z+ z^{2} + z^{3} + z^{4} + z^{5}+ z^{6}) = 9  3 = 6 . 



83.

Let \frak C be a circle with centre
O and radius 1, and let \frak F be a closed
convex region inside \frak C. Suppose from each point
on \frak C, we can draw two rays tangent to \frak F
meeting at an angle of 60^{°}. Describe
\frak F.
Solution. Let A be an arbitrary point on the
circumference of \frak C. Draw rays AC, AB,
BD tangent to \frak F; let AC and BD intersect
at P. Since
ÐCAB =
ÐABD = 60
^{°},
ÐAPM = 60
^{°} and
DAPB is equilateral
and contains \frak F. Suppose, if possible, that
P lies strictly inside the circle. Let CE be the second ray
from C tangent to \frak F. Then
ÐACE = 60
^{°},
so CE is parallel to DB and lies strictly on the opposite
side of BD to \frak F; thus, it cannot be tangent to
\frak F and we have a contradiction. Similarly, if
P lies strictly outside the circle, the second ray from C,
CE, tangent to \frak F is parallel to and distinct from
BD. We have that \frak F is tangent to AC, BD and
CE, an impossibility since DE, within the circle, lies
between CE and DB.
Hence C = D = P and ABC is an equilateral triangle containing
\frak F with all its sides tangent to \frak F. Since
A is arbitrary, \frak F is contained within the intersection
of all such triangles, namely the circle \frak D with centre
O and radius 1/2, and every chord of the given circle tangent
to \frak F has length Ö3. If \frak F were a
proper subset of \frak D, there would be a point Q
on the circumference of \frak D outside \frak F. and
a tangent of \frak F separating \frak F from Q. This
tangent chord would intersect the interior of \frak D and so
be longer than Ö3, yielding a contradiction. Hence
\frak F must be the circle \frak D.

84.

Let ABC be an acuteangled triangle,
with a point H inside. Let U, V, W be
respectively the reflected image of H with respect
to axes BC, AC, AB. Prove that H is the
orthocentre of DABC if and only if
U, V, W lie on the circumcircle of
DABC,
Solution 1. Suppose that H is the orthocentre of
DABC. Let P, Q, R be the respective feet
of the altitudes from A, B, C. Since BC right
bisects HU,
DHBP
º DUBP and so
ÐHBP =
ÐUBP. Thus

= ÐQCB = 90^{°}  ÐQBC = 90^{°}  ÐHBP 
 
= 90^{°}  ÐUBP = ÐPUB = ÐAUB , 


so that ABUC is concyclic and U lies on the circumcircle
of
DABC. Similarly V and W lie on the circumcircle.
Now suppose that U, V, W lie on the circumcircle. Let
\frak C_{1}, \frak C_{2}, \frak C_{3} be the respective
reflections of the circumcircle about the axes BC, CA,
AB. These three circles intersect in the point H.
If H¢ is the orthocentre of the triangle, then by the
first part of the solution, the reflective image of
H¢ about the three axes lies on the circumcircle, so that
H¢ belongs to \frak C_{1}, \frak C_{2}, \frak C_{3}
and H = H¢ or else HH¢ is a common chord of the three
circles. But the latter does not hold, as the common chords
AH, BH and CH of pairs of the circles intersect only
in H.
Solution 2. Let H be the orthocentre, and
P, Q, R the pedal points as defined in the first solution.
Since ARHQ is concyclic
ÐBAC + ÐBUC = ÐBAC + ÐBHC = ÐRAQ + ÐRHQ = 180^{°} 

and so ABUC is concyclic. A similar argument holds for
V and W.
[A. Lin] Suppose that U, V, W are on the circumcircle.
>From the reflection about BC, ÐBCU = ÐBCH.
>From the reflections about BA and BC, we see that
BW = BH = BU, and so, since the equal chords BW and BU
subtend equal angles at C, ÐBCW = ÐBCU.
Hence ÐBCW = ÐBCH, with the result that
C, H, W are collinear and CW is an altitude.
Similarly, AU and BV are altitudes that contain H, and so their
point H of intersection must be the orthocentre.