Problems  Mathematical Olympiads' Correspondence Program  19971998
Problem Set 1
 1.

(a) Let m be any positive integer greater than
2, such that x^{2} º 1 (mod m) whenever the greatest
common divisor of x and m is equal to 1. An example is
m = 12. Suppose that n is a positive integer for which
n + 1 is a multiple of m. Prove that the sum of all of the
divisors of n is divisible by m.

(b) Does the result in (a) hold when m = 2?


(c) Find all possible values of m that satisfy the
condition in (a).


(Note: (a) and (c) are posed as Problem 2229 in
CM+MM, April, 1997. Part (c) was submitted by the
proposer without a solution.
View solution
 2.

(a) Prove that, for each pair (m, n) of integers
with 1 £ m £ n,

n å
i = 1

i(i1)(i2)¼(im+1) = 
(n+1)n(n1)¼(nm+1) m+1

. 



(b) Suppose that 1 £ r £ n and consider all
subsets of r elements of the set { 1, 2, 3, ¼, n }.
The elements of this subset are arranged in ascending order
of magnitude. For 1 £ i £ r, let t_{i} denote the
ith
smallest element in the subset, and let T(n, r, i) denote the
arithmetic mean of the elements t_{i}. Prove that
T(n, r, i) = i 
æ ç
è


n+1 r+1


ö ÷
ø

. 

(Note: Part (b) is Problem 2239 in CM+MM, May, 1997.)
View solution
 3.

Let 0 < a < b. Prove that, for any positive integer
n,

b+a 2

£ 
n æ ú
Ö

b^{n+1}  a^{n+1} (ba)(n+1)


£ 
n æ ú
Ö


. 

View solution
 4.

How many distinct acute angles a are there for
which cosacos2acos4a = 1/8?
(
Note: This is Problem 2249 in CM+MM, May, 1997.)
View solution
 5.

A convex closed figure lies inside a given circle. The figure
is seen from every point of the circumference of the circle at right
angles (that is, the two rays drawn from the point and supporting the
convex figure are perpendicular). Prove that the centre of the circle
is a centre of symmetry of the figure.
View solution
 6.

The nonisosceles right triangle ABC has ÐCAB = 90^{°}. The inscribed circle with centre T touches the sides
AB and AC at U and V respectively. The tangent through A
of the circumscribed circle meets UV produced in S. Prove
that


(a) ST BC;


(b) d_{1}  d_{2}  = r, where r is the radius of
the inscribed circle and d_{1} and d_{2} are the respective distances
from S to AC and AB.
(
Note: This is Problem 2235 in CM+MM.)
View solution
Solutions to Problem Set 1
1.
 1.

(a) First solution. Let n + 1 be a multiple
of m. Then gcd(m, n) = 1. We observe that n cannot
be a square. Suppose, if possible, that n = r^{2}. Then
gcd(r, m) = 1. Hence r^{2} º 1 (mod m).
But r^{2} + 1 º 0 (mod m) by hypothesis, so that
2 is a multiple of m, a contradiction.


As a result, if d is a divisor of n, then n/d is a
distinct divisor of n. Suppose d n (read ``d
divides n''). Since
m divides n + 1, therefore gcd(m, n) = gcd(d, m) = 1, so that
d^{2} = 1 + bm for some integer b. Also n + 1 = cm for some
integer c. Hence
d + 
n d

= 
d^{2} + n d

= 
1 + bm + cm  1 d

= 
(b + c)m d

. 

Since gcd(d, m) = 1 and d + n/d is an integer, d divides b + c
and so d + n/d
º 0 (mod m).


Hence

å
d n

d = 
å
 { (d + n/d) : d n, d < Ön } º 0 (mod m) 

as desired.


(a) Second solution. Suppose that n + 1 º 0
(mod m). As in the first solution, it can be established that
n is not a perfect square. Let x be any positive divisor of
n and suppose that xy = n; x and y are distinct. Since
gcd (x, m) = 1, x^{2} º 1 (mod m), so that
y = x^{2} y º xn º x (mod m) 

whence x + y is a multiple of m. Thus, the divisors of n comes
in pairs, each of which has sum divisible by m, and the result
follows.


(a) Third solution. [M. Boase]
As in the second solution, if
xy = n, then x^{2} º y^{2} º 1 (mod m) so that
0 º x^{2}  y^{2} º (x  y)(x + y) (mod m). 

For any divisor r of m, we have that
x(x  y) º x^{2}  xy º 1 (mod r) 

from which it follows that the greatest common divisor of m and
x  y is 1. Therefore, m must divide x + y and the solution
can be completed as before.
 1.

(b) When m = 2, the result does not hold. The hypothesis is
true. However, the conclusion fails when n = 9 since 9 + 1 is a
multiple of 2, but 1 + 3 + 9 = 13 is odd.
 1.

(c) First solution. By inspection, we find that
m = 1, 2, 3, 4, 6, 8, 12, 24 all satisfy the condition in (a).


Suppose that m is odd. Then gcd(2, m) = 1 Þ2^{2} = 4 º 1 (mod m) Þ m = 1, 3.


Suppose that m is not divisible by 3. Then gcd(3, m) = 1Þ 9 = 3^{2} º 1 (mod m) Þm = 1, 2, 4, 8. Hence any further values of m not listed in the
above must be even multiples of 3, that is, multiples of 6.


Suppose that m ³ 30. Then, since 25 = 5^{2} ¹ 1
(mod m), m must be a multiple of 5.


It remains to show that in fact m cannot be a multiple of
5. We observe that there are infinitely many primes congruent to
2 or 3 modulo 5. [To see this, let q_{1}, ¼, q_{s} be the s
smallest odd primes of this form and let Q = 5q_{1} ¼q_{s} + 2.
Then Q is odd. Also, Q cannot be a product only of primes
congruent to ±1 modulo 5, for then Q itself would be
congruent to ±1. Hence Q has an odd prime factor congruent to
±2 modulo 5, which must be distinct from q_{1}, ¼, q_{s}.
Hence, no matter how many primes we have of the desired form, we can
always find one more.] If possible, let m be a multiple of 5 with
the stated property and let q be a prime exceeding m congruent to
±2 modulo 5. Then gcd(q, m) = 1 Þ q^{2} º 1
(mod m) Þ q^{2} º 1 (mod 5) Þ
q \not º ±2 (mod 5), yielding a contradiction. Thus, we have
given a complete collection of suitable numbers m.


(c) Second solution. [D. Arthur] Suppose that
m = ab satisfies the condition of part (a), where the greatest
common divisor of a and b is 1. Let
gcd (x, a) = 1. Since a and b are coprime, there exists a
number t such that at º 1  x (mod b), so that
z = x + at and b are coprime. Hence, the greatest common divisor
of z and ab equals 1, so that z^{2} º 1 (mod ab), whence
x^{2} º z^{2} º 1 (mod a). Thus a (and also b) satisfies
the condition of part (a).


When m is odd and exceeds 3, then gcd (2, m) = 1, but
2^{2} = 4 \not º 1 (mod m), so m does not satisfy the
condition. When m = 2^{k} for k ³ 4, then gcd (3, m) = 1,
but 3^{2} = 9 \not º 1 (mod m). It follows from the first
paragraph that if m satisfies the condition, it cannot be
divisible by a power of 2 exceeding 8 nor by an odd number exceeding
3. This leaves the possibilities 1, 2, 3, 4, 6, 8, 12, 24, all of
which satisfy the condition.
2.
 2.

(a) First solution.
i(i1)(i2) ¼(im+1) = [([(i+1)(im)])/(m+1)]i(i1)(i2) ¼(im+1)
= 
(i+1)i(i1) ¼(im+1)  i(i1)(i2)¼(im+1)(im) m+1



so that


n å
i = 0

i(i1)(i2)¼(im+1) 

= 
n+1 å
i = 2


i(i1)¼(im) m+1

 
n å
i = 1


i(i1)¼(im) m+1


 
= 
(n+1)n(n1)¼(nm+1) m+1

 0 
 
= 
(n+1)n(n1)¼(nm+1) m+1

. 

 



(a) Second solution. [K. Yeats]
Let n = m + k. Then

n å
i = 1

i(i1)(i2)¼(im+1) = m! + 
(m+1)! 1!

+ ¼+ 
n! (nm)!



= 
1 (m+1)k!


é ê
ë

(m+1)!k! + 
(m+1)!k!(m+1) 1!

+ 
(m+2)!k!(m+1) 2!

+ ¼+ n!(m+1) 
ù ú
û



= 
(m+1)! (m+1)k!


é ê
ë

k! + 
k! 1!

(m+1) + 
k! 2!

(m+2)(m+1) + ¼+ n(n1)¼(m+2)(m+1) 
ù ú
û

. 

The quantity in square brackets has the form (with q = 0)

k! q!

+ 
k! (q+1)!

(m+1) + 
k! (q+2)!

(m+q+2)(m+1) + 
k! (q+3)!

(m+q+3)(m+q+2)(m+1) + ¼ 

+ 
k! k!

n(n1)¼(m+q+2)(m+1) 

= (m+q+2) 
é ê
ë


k! (q+1)!

+ 
k! (q+2)!

(m+1) + 
k! (q+3)!

(m+q+3)(m+1)+ ¼+ 
k! k!

n ¼(m+q+3)(m+1) 
ù ú
û

. 

Applying this repeatedly with q = 0, 1, 2,
¼, k1
leads to the expression for the left sum in the problem of

(m+k+1)! (m+1)k!


é ê
ë


k! k!


ù ú
û

= 
(n+1)! (m+1)(nm)!

= 
(n+1)n(n1) ¼(nm+1) m+1

. 

[A variant, due to D. Nicholson, uses an induction on
r to prove that, for m
£ r
£ n,

r å
i = m

i(i1)¼(im+1) = 
(r+1)! (rm)!(m+1)

.] 



(a) Third solution. For 1 £ i £ m1,
i(i1)¼(im+1) = 0. For m £ i £ n,
i(i1)¼(im+1) = m!(i  m). Also,

(n+1)n ¼(nm+1) m+1

= m! 
æ ç
è

n+1
m+1

ö ÷
ø



so the statement is equivalent to

n å
m


æ ç
è

i
m

ö ÷
ø

= 
æ ç
è

n+1
m+1

ö ÷
ø

. 

This is clear for n = m. Suppose it holds for n = k
³ m. Then

k+1 å
i = m


æ ç
è

i
m

ö ÷
ø

= 
æ ç
è

k+1
m+1

ö ÷
ø

+ 
æ ç
è

k+1
m

ö ÷
ø

= 
æ ç
è

k+2
m+1

ö ÷
ø



and the result follows by induction.


(a) Fourth solution. Use induction on n. If n = 1,
then m = 1 and both sides of the equation are equal to 1. Suppose
that the result holds for n = k and 1 £ m £ k. Then, for
1 £ m £ k,


k+1 å
i = 1

i(i1)¼(im+1) 

= 
(k+1)k(k1)¼(km+1) m+1

+ (k+1)k(k1) ¼(km+2) 
 
= 
(k+1)k(k1)¼(km+2) m+1

[(km+1)+ (m+1)] 
 
= 
(k+2)(k+1)k(k1)¼(km+2) m+1



 

as desired. When m = n = k+1, all terms on the left have k+1 terms
and so they vanish except for the one corresponding to i = k+1.
This one is equal to (k+1)! and so to the right side.
 2.

(b) First solution.
For 1 £ i £ r £ n, let
S(n, r, i) be the sum of the elements t_{i} where (t_{1}, t_{2}, ¼, t_{r}) runs over rtples with 1 £ t_{1} < t_{2} < ¼ < t_{r} £ n. Then S(n, r, i) = (n  r)T(n, r, i).
For 1 £ k £ n,
1 £ i £ r, the number of ordered rtples
(t_{1}, t_{2}, ¼, t_{r}) with t_{i} = k is
((k1)  (i1))((nk)  (ri)) where (0  0) = 1
and (a  b) = 0 when b > a. Hence

æ ç
è

n
r

ö ÷
ø

= 
n å
k = 1


æ ç
è

k1
i1

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

. 

Replacing n by n+1 and r by r+1 yields a reading

æ ç
è

n+1
r+1

ö ÷
ø

= 
n+1 å
k = 1


æ ç
è

k1
i1

ö ÷
ø


æ ç
è

n+1k
r(i1)

ö ÷
ø

for 1 £ i £ r+1 . 

Replacing i1 by i yields

æ ç
è

n+1
r+1

ö ÷
ø

= 
n+1 å
k = 1


æ ç
è

k1
i

ö ÷
ø


æ ç
è

n+1k
ri

ö ÷
ø

for 0 £ i £ r . 

When 1
£ i
£ r, the first term of the sum is 0, so that

æ ç
è

n+1
r+1

ö ÷
ø

= 
n+1 å
k = 2


æ ç
è

k1
i

ö ÷
ø


æ ç
è

n(k1)
ri

ö ÷
ø

= 
n å
k = 1


æ ç
è

k
i

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

. 

Thus
S(n,r,i) = 
n å
k = 1

k 
æ ç
è

k1
i1

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

= i 
n å
k = 1


æ ç
è

k
i

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

= i 
æ ç
è

n+1
r+1

ö ÷
ø



so
 2.

(b) Second solution. When r = 1, we have that
T(n, 1, 1) = 
1 + 2 + ¼+ n n

= 
n+1 2

. 

When r = 2, the subsets are {1, 2 }, { 1, 3 },
¼,{ 1, n} , { 2, 3 }, { 2, 4 },
¼{ 2, n },
¼, { n1 , n }, so that


= 
1 ×(n1) + 2 ×(n2) + ¼+ (n1) ×1


 
= 
[(n1) + (n2) + ¼+ 1] + [(n2) + (n3) + ¼+ 1]+ ¼+ 1


 
= 
n1 å
j = 1

[1 + 2 + ¼+ (nj)] 
n(n1)/2

= 
n1 å
j = 1

(nj+1)(nj)/2 
n(n1)/2


 
= 
(1/6)(n+1)n(n1) (1/2)n(n1)

= 
n+1 3

, 

 

and


= 
(n1)×n + (n2)×(n1) + ¼+ 1 ×2


 
= 
(n+1)n(n1)/3 n(n1)/2

= 2 
æ ç
è


n+1 3


ö ÷
ø

. 

 

Thus, the result holds for n = 1, 2 and all i, r with 1
£ i
£ r
£ n, and for all n and 1
£ i
£ r
£ 2. Suppose as
an induction hypothesis, we have established the result up to n1
and all appropriate r and i, and for n and 1
£ i
£ r1.
The relement subsets of { 1, 2,
¼, n } have
((n1)  r) instances without n and ((n1)  (r1))
instances with n.


Let 1 £ i £ r1. Then


= 

æ ç
è

n1
r

ö ÷
ø

T(n1, r, i) + 
æ ç
è

n1
r1

ö ÷
ø

T(n1, r1, i) 


 
= 
i[ 
æ ç
è

n1
r

ö ÷
ø


n r+1

+ 
æ ç
è

n1
r1

ö ÷
ø


n r

] 

= 
i[ 
æ ç
è

n
r+1

ö ÷
ø

+ 
æ ç
è

n
r

ö ÷
ø

] 


 

 

Also


= 

æ ç
è

n1
r

ö ÷
ø

T(n1, r, r) + 
æ ç
è

n1
r1

ö ÷
ø

n 


 
= 

æ ç
è

n1
r

ö ÷
ø


rn r+1

+ 
æ ç
è

n1
r1

ö ÷
ø


rn r


= 
r[ 
æ ç
è

n
r+1

ö ÷
ø

+ 
æ ç
è

n
r

ö ÷
ø

] 

= r 
æ ç
è


n+1 r+1


ö ÷
ø

. 

 



(b) Third solution.
For 1 £ i £ r £ n, let
S(n, r, i) be the sum of the elements t_{i} where (t_{1}, t_{2}, ¼, t_{r}) runs over rtples with 1 £ t_{1} < t_{2} < ¼ < t_{r} £ n. Then S(n, r, i) = (n  r)T(n, r, i).
We observe first that
S(n, r, i) = S(n1, r1, i) + S(n2, r1, i) + ¼+ S(r1, r1, i) 

for 1
£ i
£ r1. This is true, since, for each j with
1
£ j
£ nr+1, S(nj, r1, i) adds the t
_{i}
over all rtples for which t
_{r} = n  j + 1.


Now S(n, 1, 1) = 1 + 2 + ¼+ n = ^{1}/_{2}(n+1)n
and S(n, 2, 1) = ^{1}/_{2}n(n1) + ¼+ 1 = [1/3!](n+1)n(n1). As an induction hypothesis, suppose that
S(n, r1, 1) = [1/r!](n+1)n(n1)¼(nr+2). Then


= 
n1 å
k = r1

S(k, r1, 1) 
 
= 
1 r!


n1 å
k = r1

(k+1)k(k1) ¼(kr+2) = 
1 r!


n å
k = 1

k(k1) ¼(kr+1) 
 
= 
1 (r+1)!

(n+1)n(n1) ¼(nr+1) = 
æ ç
è


n+1 r+1


ö ÷
ø


n! r!(nr)!

= 
æ ç
è


n+1 r+1


ö ÷
ø


æ ç
è

n
r

ö ÷
ø

. 

 

Thus, for each r with 1
£ r
£ n,
S(n, r, 1) = (n  r)(n+1)/(r+1) so that
T(n, r, 1) = (n+1)/(r+1).


Let n ³ 2. Suppose that for 1 £ k £ n1 and
1 £ i £ r £ k, it has been established that S(k, r, i) = iS(k, r, 1). Then for 1 £ i £ r £ n,


= S(n1, r1, i) + S(n  2, r1, i) +¼+ S(r  1, r 1, i) 
 
= i[S(n1, r1, 1) + S(n2, r1, 1) + ¼+ S(r1, r1, 1) = iS(n, r, 1) . 

 

Dividing by (n  r) yields
T(n, r, i) = iT(n, r, 1) = i 
æ ç
è


n+1 r+1


ö ÷
ø

. 



Comments. (1) There is a oneone correspondence
(t_{1}, t_{2}, ¼, t_{r}) « (n+1  t_{r}, n + 1  t_{r1} , ¼, n + 1  t_{r}) 

of the set of suitable rtples to itself, it follows that


= 
æ ç
è

n
r

ö ÷
ø

(n+1) S(n, r, 1) = 
æ ç
è

n
r

ö ÷
ø

(n+1) 
é ê
ë

1  
1 r+1


ù ú
û


 
= 
r(n+1) r+1


æ ç
è

n
r

ö ÷
ø

= rS(n, r, 1) 

 

from which T(n, r, r) = r(n+1)/(r+1) = rT(n, r, 1).


(2) To illustrate another method for getting and
using the recursion, we prove first that
T(n, r, 2) = 2T(n, r, 1) for 2 £ r £ n. Consider the
case r = 2. For 1 £ t_{1} < t_{2} £ n,
(t_{1}, t_{2}) « (t_{2}  t_{1}, t_{2}) defines a oneone
correspondence between suitable pairs. Since t_{2} = t_{1} + (t_{2} t_{1}), it follows from this correspondence that S(n, 2, 2) = 2S(n, 2, 1). Dividing by (n  2) yields
T(n, 2, 2) = 2T(n, 2, 1).


Suppose that r ³ 2. For each positive integer j
with 1 £ j £ n  r + 1, we define a oneone correspondence
between rtples (t_{1}, t_{2}, ¼, t_{r}) with
1 £ t_{1} < t_{2} < ¼ < t_{r} £ n and t_{3}  t_{2} = j and
(r1)tples (s_{1}, s_{2}, s_{3}, ¼, s_{r}) = (t_{1}, t_{2}, t_{4}  j, ¼, t_{r}  j) with
1 £ s_{1} = t_{1} < s_{2} = t_{2} < s_{3} = t_{4}  j < ¼ < s_{r} = t_{r}  j £ n  j. The sum of the elements t_{2} over all
rtples with t_{3}  t_{2} = j is equal to the sum of t_{2} over
all the (r  1)tples. Hence
S(n, r, 2) = S(n  1, r  1, 2) + S(n  2, r  1, 2) + ¼+ S(r  1, r  1, 2) . 



More generally,
for 1 £ j £ n  r  1, there is a oneone
correspondence between rtples (t_{1}, t_{2}, ¼, t_{r}) with
t_{i+1}  t_{i} = j and (r  1)tples (s_{1}, s_{2}, ¼, s_{r1}) = (t_{1}, ¼, t_{i}, t_{i+2}  j, ¼, t_{r}  j) with
1 £ s_{1} = t_{1} < ¼ < s_{i} = t_{i} < s_{i+1} = t_{i+2}  j < ¼ < s_{r1} = t_{r}  j £ n  j. We now use induction on
r. We have that


= S(n1, r1, i) + S(n  2, r1, i) +¼+ S(r  1, r 1, i) . 

 



(b) Fourth solution. [Y. Shen] We establish that

i+(nr) å
k = i


æ ç
è

k
i

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

= 
æ ç
è

n+1
r+1

ö ÷
ø

. 

Consider the (r+1)element sets where t
_{i+1} = k+1
and t
_{r+1} £ n + 1. We must
have i
£ k
£ n  (ri) and there are (k  i)((nk)  (ri)) ways of selecting t
_{1},
¼, t
_{i} and
t
_{i+2},
¼, t
_{r+1}. The desired equation follows from
a counting argument over all possibilities for t
_{i+1}.


In a similar way, we note that t_{i} = k for
((k1)  (i1))((nk)  (ri)) sets { t_{1}, ¼, t_{r} }
chosen from {1, ¼, n }, where 1 £ k £ nr+1. Observe
that

æ ç
è

k1
i1

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

= 
i k


æ ç
è

k
i

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø

. 

Then


= 
nr+1 å
k = i

k 
æ ç
è

k1
i1

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø



 
= 
i 
nr+i å
k = i


æ ç
è

k
i

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø



 

 



(b) Fifth solution. [Christopher So]
Note that

nr+i å
k = i


æ ç
è

k
i

ö ÷
ø


æ ç
è

nk
ri

ö ÷
ø



is the coefficient of x
^{i} y
^{ri} in the polynomial

nr+i å
k = i

(1 + x)^{k} (1 + y)^{nk} 

or in


n+1 å
k = 0

(1 + x)^{k} (1 + y)^{nk} 

= 
(1 + y)^{n+1}  (1 + x)^{n+1} y  x


 
= 
n+1 å
j = 0


æ ç
è

n+1
j

ö ÷
ø

(y^{j}  x^{j}) 
yx

. 

 

Now the only summand which involves terms of degree r corresponds to
j = r+1, so that the coefficient of x
^{i} y
^{r1} in the sum is
the coefficient in the single term

æ ç
è

n+1
r+1

ö ÷
ø


y^{r+1}  x^{r+1} yx



namely, ((n+1)  (r+1)). We can now complete the argument as in
the fourth solution.


(b) Comment. Let r and n be fixed values and
consider i to be variable. The (n  r) rterm sets
contain altogether r (n  r) numbers, each number occurring
equally often: ^{r}/_{n}(n  r) times. The sum of all the
elements in the set is
S(n, r, 1) + S(n, r, 2) + ¼+ S(n, r, r) = 
r n


æ ç
è

n
r

ö ÷
ø

(1 + 2 + ¼+ n) = 
r(n+1) 2


æ ç
è

n
r

ö ÷
ø



where S(n, r, i) is the sum of the elements t
_{i} over the
(n  r) subsets. The ordered relement subsets
(t
_{1}, t
_{2},
¼, t
_{r}) can be mapped oneone to themselves by
(t_{1}, t_{2}, ¼, t_{r}) « (n+1  t_{r}, n+1  t_{r1}, ¼, n + 1  t_{1}) . 

From this, we see that, for 1
£ r,
S(n, r, r+1i) = 
æ ç
è

n
r

ö ÷
ø

(n+1)  S(n, r, i) 

so that
S(n, r, 1) + S(n, r, r) = S(n, r, 2) + S(n, r, r1) = ¼ = S(n, r, i) + S(n, r, r + 1  i) = ¼ = 
æ ç
è

n
r

ö ÷
ø

(n+1) . 

This is not enough to imply that S(n, r, i) is an arithmetic
progression in i, but along with this fact would give a quick
solution to the problem.
3.
 3.

First solution. Dividing the inequality through
by (b+a)/2 yields the equivalent inequality
1 £ 
n æ ú
Ö

b¢^{n+1}  a¢^{n+1} (b¢ a¢)(n+1)


£ 
n æ ú
Ö




with a
¢ = (2a)/(b+a) and b
¢ = (2b)/(b+a). Note that
(a
¢+ b
¢)/2 = 1, and we can write
b
¢ = 1 + u and a
¢ = 1  u with 0 < u < 1. The central term
becomes the nth root of


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



= 
2[(n+1)u + 
æ ç
è

n+1
3

ö ÷
ø

u^{3} + 
æ ç
è

n+1
5

ö ÷
ø

u^{5} + ¼] 
2(n+1)u


 
= 1 + 
1 3


æ ç
è

n
2

ö ÷
ø

u^{2} + 
1 5


æ ç
è

n
4

ö ÷
ø

u^{4} +¼ 

 

which clearly exceeds 1 and gives the left inequality. The right
term becomes the nth root of

1 2

[(1 + u)^{n} + (1  u)^{n} ] = 1 + 
æ ç
è

n
2

ö ÷
ø

u^{2}+ 
æ ç
è

n
4

ö ÷
ø

u^{4} + ¼ 

and the right inequality is clearly true.
 3.

Second solution. The inequality

n æ ú
Ö

b^{n+1}  a^{n+1} (ba)(n+1)


£ 
n æ ú
Ö




is equivalent to
0 £ (n+1)(a^{n} + b^{n})  
2(b^{n+1}  a^{n+1}) ba

. 

The right side is equal to


 2(b^{n} + b^{n1}a + b^{n2}a^{2} ¼+ b^{2} a^{n2} + ba^{n1} + a^{n}) 
 
= (a^{n}  b^{n}) + (a^{n}  b^{n1}a) + (a^{n}  b^{n2}a) + ¼+ (a^{n}  b a^{n1}) + (a^{n}  a^{n}) 
 
+ (b^{n}  b^{n}) +(b^{n}  b^{n1}a) + ¼+ (b^{n}  ba^{n1}) + (b^{n}  a^{n}) 
 
= (a^{n}  b^{n}) + a(a^{n1}  b^{n1}) + a^{2}(a^{n2}  b^{n2}) +¼+ a^{n1}(a  b) + 0 
 
+ 0 + b^{n1}(b  a) + ¼+ b(a^{n1}  a^{n1}) + (b^{n}  a^{n}) 
 
= 0 + (ba)(b^{n1}  a^{n1}) + (b^{2}  a^{2})(b^{n2}  a^{n2})+ ¼+ (b^{n1}  a^{n1})(b  a) 
 

 



The left inequality

b+a 2

£ 
n æ ú
Ö

b^{n+1}  a^{n+1} (ba)(n+1)




is equivalent to

æ ç
è


b+a 2


ö ÷
ø

n

£ 
b^{n+1}  a^{n+1} (ba)(n+1)

. 

Let v =
^{1}/
_{2}(ba) so that b + a = 2(a + v). Then


b^{n+1}  a^{n+1} (ba)(n+1)

 
æ ç
è


b+a 2


ö ÷
ø

n



= 
(a+2v)^{n+1}  a^{n+1} (ba)(n+1)

 (a + v)^{n} 
 
= 
1 2v(n+1)


æ ç
è


n+1 å
k = 1


æ ç
è

n+1
k

ö ÷
ø

a^{n+1k} (2v)^{k} 
ö ÷
ø

 (a + v)^{n} 
 
= 
1 n+1


æ ç
è


n+1 å
k = 1


æ ç
è

n+1
k

ö ÷
ø

a^{n(k1)} (2v)^{k1} 
ö ÷
ø

 (a + v)^{n} 
 
= 
1 n+1


æ ç
è


n å
k = 0


æ ç
è

n+1
k+1

ö ÷
ø

a^{nk} (2v)^{k} 
ö ÷
ø

 
n å
k = 0


æ ç
è

n
k

ö ÷
ø

a^{nk} v^{k} 
 
= 
æ ç
è


n å
k = 0


1 k+1


æ ç
è

n
k

ö ÷
ø

a^{nk} (2v)^{k}  
n å
k = 0


æ ç
è

n
k

ö ÷
ø

a^{nk} v^{k} 
ö ÷
ø


 
= 
n å
k = 0


æ ç
è


2^{k} k+1

 1 
ö ÷
ø


æ ç
è

n
k

ö ÷
ø

a^{nk} v^{k} ³ 0 , 

 

since 2
^{k} = (1 + 1)
^{k} = 1 + k + (k  2) +
¼ ³ 1 + k
with equality if and only if k = 0 or 1. The result follows.
 3.

Third solution. [D. Nicholson]
(partial) Let n ³ 2i + 1. Then
(b^{ni}a^{i} + b^{i} a^{ni})  (b^{ni1}a^{i+1} + b^{i+1}a^{ni1}) = (b  a)a^{i} b^{i} (b^{n2i1}  a^{n2i1}) ³ 0 . 

Hence, for 0
£ j
£ ^{1}/
_{2}(n+1),
b^{n} + a^{n} ³ b^{n1}a + ab^{n1} ³ ¼ ³ b^{nj}a^{j} +b^{j} a^{nj} . 

When n = 2k+1,
b^{n} + b^{n1}a + ¼+ ba^{n1} + a^{n} = 
k å
i = 0

(b^{ni}a^{i} + b^{i} a^{ni}) £ (k+1)(b^{n} + a^{n}) = 
n+1 2

(b^{n} + a^{n}) 

and when n = 2k, we use the ArithmeticGeometric Means Inequality
to obtain b
^{k} a
^{k} £ ^{1}/
_{2}(a
^{2k} + b
^{2k}), so that
b^{n} + b^{n1}a + ¼+ ba^{n1} + a^{n} = 
k1 å
i = 0

(b^{ni} a^{i} + b^{i} a^{ni}) + b^{k} a^{k} £ k(b^{n} + a^{n}) + 
b^{n} + a^{n} 2

= 
n+1 2

(b^{n} + a^{n}) . 

Hence

b^{n+1}  a^{n+1} (ba)(n+1)

£ 
b^{n} + a^{n} 2

. 

 3.

Fourth solution. [Y. Shen] Let 1 £ k £ n and
1 £ i £ k. Then
(b^{k+1} + a^{k+1})  (b^{i} a^{k+1i} + a^{i} b^{k+1i}) = (b^{i}  a^{i})(b^{k+1i}  a^{k+1i}) ³ 0 . 

Hence
k(b^{k+1} + a^{k+1}) ³ 
k å
i = 1

(b^{i} a^{k+1i} + a^{i} b^{k+1i}) = 2 
k å
i = 1

b^{i} a^{k+1i} . 

This is equivalent to

(2k+2) 
k+1 å
i = 0

b^{i} a^{k+1i} 

= (2k + 2)(b^{k+1} + a^{k+1}) + (2k + 2) 
k å
i = 1

b^{i} a^{k+1i} 
 
³ (k+2) (b^{k+1} + a^{k+1}) + (2k + 4) 
k å
i = 1

b^{i} a^{k+1i} 
 
= (k + 2) (b^{k+1} + 2 
k å
i = 1

b^{i} a^{k+1i} + a^{k+1}) 
 
= (k + 2) (b + a) 
k å
i = 0

b^{i} a^{ki} 

 

which in turn is equivalent to

k+1 å
i = 0

b^{i} a^{k+1i} 
k+2

³ 
(b+a)( 
k å
i = 0

b^{i} a^{ki}) 
2(k+1)

. 

We establish by induction that

æ ç
è


b+a 2


ö ÷
ø

n

£ 
1 n+1


n å
i = 0

b^{i} a^{ni} 

which will yield the left inequality. This holds for n = 1.
Suppose that it holds for n = k. Then


= 
æ ç
è


b+a 2


ö ÷
ø

· 
æ ç
è


b+a 2


ö ÷
ø

k


 
£ 
æ ç
è


b+a 2


ö ÷
ø

· 
æ ç
è


1 k+1


ö ÷
ø


k å
i = 0

b^{i} a^{ki} £ 
1 k+2


k+1 å
i = 0

b^{i} a^{ki} . 

 



As above, we have, for k = n1,
(n1)(b^{n} + a^{n}) ³ 2 
n1 å
i = 1

b^{i} a^{ni} 

so that
(n+1)(b^{n} + a^{n}) ³ 2 
n å
i = 0

b^{i} a^{ni} = 2 
æ ç
è


b^{n+1}  a^{n+1} ba


ö ÷
ø



from which the right inequality follows.
 3.

Comment. The inequality

b+a 2

£ 
n æ ú
Ö

b^{n+1}  a^{n+1} (ba)(n+1)




is equivalent to
0 £ 
2^{n}(b^{n+1}  a^{n+1}) b  a

 (n+1)(b+a)^{n} . 

When n = 1, the right side is equal to 0. When n = 2, it is equal
to
4(b^{2} + ba + a^{2})  3(b + a)^{2} = (b  a)^{2} > 0 . 

When n = 3, we have
8(b^{3} + b^{2}a + ba^{2} + a^{3})  4(b + a)^{3} = 4b^{3}  4b^{2}a  4ba^{2} + 4a^{3} = 4(b^{2}  a^{2})(b  a) = 4(b + a)(b  a)^{2} > 0 . 

When n = 4, 5 and 6, the right side is, respectively,
(11b^{2} + 18ab + 11a^{2})(b  a)^{2} 

(26b^{3} + 54b^{2}a + 54ba^{2} + 26a^{3})(b  a)^{2} 

(57b^{4} + 136b^{3}a + 174b^{2}a^{2} + 136ba^{3} + 57a^{4})(b  a)^{2} . 

There is a pattern here; can anyone express it in a general way
that will yield the result, or at least show that the right side
is the product of (b  a)^{2} and a polynomial with positive
coefficients?
4.
 4.

First solution.
8 sinacosacos2acos4a = 4 sin2 acos2acos4a = 2 sin4acos4a = sin8a . 

Hence the given equation is equivalent to sin8
a = sin
a
with sin
a ¹ 0. Hence, we must have
a+ 8
a =
p, 8
a = 2
p+
a or 8
a = 3
p
a, since
a lies strictly between 0 and
p/2. These lead respectively
to
a =
p/9 (20
^{°}),
a = 2
p/ 7 and
a =
p/3 (60
^{°}). Thus, there are three acute angles
a for
which the equation is satisfied.


Comment. Alternatively, note that
0 = sin8a sina = 2sin 
7a 2

cos 
9a 2



whence 7
a = 2
p or 9
a =
p or 3
p.


Second solution. Let x = cosa. Then
cos2a = 2x^{2}  1 and cos4a = 2(2x^{2}  1)^{2}  1 = 8x^{4}  8x^{2} + 1, so that
cosacos2acos4a = x(2x^{2}  1)(8x^{4}  8x^{2} + 1) . 

Let f(x) = x(2x
^{2}  1)(8x
^{4}  8x
^{2} + 1)  1/8. Using the fact that
a =
p/3 corresponding to x = 1/2 is an obvious solution
of the equation f(x) = 0, we find that


= 128x^{7}  192x^{5} + 80x^{3}  8x  1 
 
= (2x  1)(64x^{6} + 32x^{5}  80x^{4}  40x^{3} + 20x^{2} + 10x + 1) . 

 

Consider the factor


= 64x^{6} + 32x^{5}  80x^{4}  40x^{3} + 20x^{2} + 10x + 1 
 
= (2x + 1)(32x^{5}  40x^{3} + 10x) + 1 . 

 

By Descartes' Rule of Signs, g(x) = 0 has at most two positive
roots (there are two changes of signs as one reads along the coefficients
from left to right). Now g(1) = 7, g(
^{1}/
_{2}) = 3 and
g(
^{3}/
_{4}) = (
^{5}/
_{2})([(177)/32]) + 1 < 0.
Hence g(x) = 0 has precisely two roots, one in each of the intervals
(1/2, 3/4) and (3/4, 1). These along with x = 1/2, constitute
the three roots of f(x) = 0. Since all roots lie in (0, 1),
they are cosines of acute angles
a for which
cos
acos2
acos4
a = 1/8.


Comment. g(x) = (8x^{3} + 4x^{2}  4x  1)(8x^{3}  6x  1).
By Descartes' Rule of Signs, each factor has one positive root, and
since each is negative for x = 0 and positive for x = 1, that
root must lie between 0 and 1. We need to dispose of the possibility
that both factors have the same root. If c is such a common root,
then 8c^{3} + 8c^{2}  4c  1 = 8c^{3}  6c  1 = 0 Þ8c^{2} + 2c = 0 which yields a contradiction.


Third solution. [A. Birka] An obvious solution is
a = p/3. Observe that the product cosacos2acos4a vanishes exactly when a = p/8 and a = p/4. On the interval [0, p/8],
cosa, cos2a, cos4a are all nonnegative and
decreasing, so the product decreases from 1 to 0. Hence, by the
Intermediate Value Theorem and the continuity of the functions,
cosacos2acos4a = 1/8 has precisely
one solution in this interval. On the interval (p/8, p/4),
cosa and cos2a are positive and cos4a
is negative, so there are no solutions. Similarly, on the
interval (3p/8, p/2), cosa and cos4a are
positive and cos2a is negative, so there are no solutions.
Hence any remaining solutions must lie in (p/4, 3p/8),
where cosa > 0, cos2a < 0 and cos4a < 0.


We need a preliminary result. Suppose that u(a) and
v(a) are nonnegative twice differentiable functions on an open
interval (a, b) such that u(a) is increasing, v(a) is
decreasing and both functions are concave (i.e., have negative
second derivative). Then u(a) v(a) is also concave. This
follows immediately from (uv)¢¢ = u¢¢v + 2u¢v¢+ uv¢¢.


Apply this to the interval ([( p)/4], [(p)/3])
and u(a) = cos2a, v(a) = cosa to obtain
that u(a) v(a) is concave. Observe that u(a)v(a) is nonnegative and


= [cosa(cos2a)]¢ = (cosa 2cos^{3} a)¢ 
 
= sina(6cos^{2} a 1) ³ 
1 2

sina(6 cos^{2} 
p 3

 1) = 
1 4

sina > 0 

 

so u(
a) v(
a) is increasing.


On the interval ([(p)/4], [(p)/3]),
cos4a is nonnegative, decreasing and concave, so we can
apply the result again with u(a) = cosacos2a
and v(a) = cos4a to obtain that
f(a) º cosacos2acos4a is nonnegative
and concave on ([(p)/4], [(p)/3]). It can be checked
that f([(p)/4]) = 0, f([(p)/3]) = ^{1}/_{8} and
f¢([(p)/3]) = 3 Ö3/8 < 0, so an analysis of the graph
reveals exactly one solution of f(a) = 1/8 for
p/4 < a < p/3.


Finally, we consider p/ 3 < a < 3p/ 8. If
g(a) º cos2acos4a = 2 cos^{3} 2a cos2a, then g¢(a) = 2sin2a[1  6cos^{2} 2 a] < 0, so g(a) is positive and decreasing on ([(p)/3],[(3p)/8]). Since g(p/ 3) = 1/4, it follows that
f(a) < (cosa) /4 < 1/8 for p/3 < a < 3p/8.
The result follows.


Comment: It is interesting that the three solutions,
given in increasing order of complexity, with the second requiring
some sophisticated algebraic machinery and the last a reasonably
sophisticated analysis, are also increasingly loath to provide
information. The first solution is not only the easiest, but in
fact allows us to explicitly state the values of a.
In the second solution, we at least have the tools to make some
kind of approximation to the solutions, although the one
corresponding to the factor (2x  1) can be given exactly.
The analytic theory (intermediate value theorem) in the third
solution just tells us that some interval has a solution without
specifying very exactly where.
5.
 5.

First solution.
Let the figure be denoted by F and the circle
by C, and let r be the central reflection through
the centre of the circle. Suppose that m is any line of support
for F and that it intersects the circle in P and
Q. Then there are lines p and q through P and Q respectively,
perpendicular to m, which support F. Let p meet
the circle in P and R, and q meet it in Q and S; let t
be the line RS. Since PQRS is concyclic with adjacent right
angles, it is a rectangle, and t is a line of support of
F. Since PS and RQ are both diameters of
C, it follows that
S = r(P), R = r(Q) and t = r(m).


Hence, every line of support of F is carried
by r into a line of support of F. We note that
F must be on the same side of its line of support
as the centre of the circle.


Suppose that X Î F. Let Y = r(X). Suppose,
if possible that Y \not Î F. Then there must be a disc
containing Y that does not intersect F, so we can find
a line m of support for F such that F is on one side
and Y is strictly on the other side of m. Let n = r(m).
Then n is a line of support for F which has X = r(Y)
on one side and O = r(O) on the
other. But this is not possible. Hence
Y Î F and so r(F) Í F.
Now r°r is the identity mapping,
so F = r(r(F)) Í r(F).
It follows that F = r(F) and the result follows.


Second solution. Let P be any point on the circle
C. There are two perpendicular lines of support from
P meeting the circle in Q and S. As in the first solution,
we see that P is one vertex of a rectangle PQRS each of
whose sides supports F. Let G be the intersection
of all the rectangles as P ranges over the circumference of the
circle C. Since each rectangle has central symmetry
about the centre of C, the same is true of G.
It is clear that F Í G. It remains to
show that G Í F. Suppose a point X
in G does not
belong to F. Then there is a line r of support to
F for which X and F are on opposite sides. This
line of support intersects C at the endpoints of a chord
which must be a side of a supporting rectangle for F.
The point X lies outside this rectangle, and so must lie outside
of G. The result follows.


Third solution. [D. Arthur] If the result is false, then
there is a line through the centre of the circle such that OP > OQ,
where P is where the line meets the boundary of the figure on one
side and Q is where it meets the boundary on the other. Let m be
the line of support of the figure through Q. Then, as shown in
Solution 1, its reflection t in the centre of the circle is also
a line of support. But then P and O lie on opposite sides of
t and we obtain a contradiction.
6.
 6.

(a) First solution.
Wolog, suppose that the situation is as diagrammed.
ÐBAC = ÐAUT = ÐAVT = 90^{°},
so that AUVT is a rectangle with AU = AV and UT = VT.
Hence AUTV is a square with diagonals AT and UV which
rightbisect each other at W.
Since SW rightbisects AT, by reflection in
the line SW, we see that DASU º DUST, and
so ÐUTS = ÐUAS.


Let M be the midpoint of BC. Then M is the circumcentre
of DABC, so that MA = MC and ÐMCA = ÐMAC.
Since AS is tangent to the circumcircle of DABC,
AS ^AM. Hence
ÐUTS = ÐUAS = ÐSAM  ÐBAM = 90^{°} ÐBAM = ÐMAC = ÐMCA . 

Now UT
^AB implies that UT
 AC. Since
ÐUTS =
ÐACB, it follows that ST
 BC.


(a) Second solution. Wolog, suppose that S is
on the opposite side of AB to C.


BT, being a part of the diameter produced of the inscribed
circle, is a line of reflection that takes the circle to itself and
takes the tangent BA to BC. Hence ÐUBT = ^{1}/_{2}ÐABC. Let a = ÐABT. By the tangentchord
theorem applied to the circumscribed circle, ÐXAC = ÐABC = 2a, so that ÐSAU = 90^{°}  2a.


Consider triangles SAU and STU. Since AUTV is a square
(see the first solution), AU = UT and ÐAUV = ÐTUV = 45^{°} so ÐSUA = ÐSUT = 135^{°}. Also
SU is common. Hence DSAU º DSTU, so
ÐSTU = ÐSAU = 90^{°}  2a. Therefore,
ÐSTB = ÐUTB  ÐSTU = (90^{°}  a) (90^{°}  2a) = a = ÐTBC 

from which it results that ST
 BC.


(a) Third solution. As before
DAUS º DTUS, so ÐSAU = ÐSTU.
Since UT  AC, ÐSTU = ÐSYA. Also, by the
tangentchord theorem, ÐSAB = ÐACB. Hence
ÐSYA = ÐSTU = ÐSAB = ÐACB, so
ST  BC.


(a) ³Fourth solution. In the Cartesian plane, let
A ~ (0, 0), B ~ (0, b), C ~ (c, 0).
The centre of the circumscribed circle is at M ~ (c/2, b/2). Since the slope of AM is b/c, the equation of
the tangent to the circumscribed circle through A is
y = (c/b)x. Let r be the radius of the inscribed circle.
Since AU = AV, the equation of the line UV is y = x  r.
The abscissa of S is the solution of x  r = (cx)/b, so
S ~ ([br/(bc)], [cr/(bc)]). Since T ~ (r, r), the slope of ST is b/c and the result follows.
 6.

(b) First solution. [¼] denotes area.
Wolog, suppose that d_{1} > d_{2}, as diagrammed.


Let r be the inradius of DABC. Then
[AVU] = ^{1}/_{2}r^{2}, [AVS] = ^{1}/_{2}rd_{1} and
[AUS] = ^{1}/_{2}rd_{2}. From [AVU] = [AVS]  [AUS], it
follows that r^{2} = rd_{1}  rd_{2}, whence r = d_{1}  d_{2}.


(b) Second solution. [F. Crnogorac]
Suppose that the situation is as diagrammed.
Let P and Q be the respective feet of the perpendiculars
from S to AC and AB. Since ÐPVS = 45^{°}
and ÐSPV = 90^{°}, DPSV is isosceles and
so PS = PV = PA + AV = SQ + AV, i.e., d_{1} = d_{2} + r.


(b) Third solution. Using the coordinates of the
fourth solution of (a), we find that
d_{1} = 
ê ê
ê


cr bc


ê ê
ê

and d_{2} = 
ê ê
ê


br bc


ê ê
ê



whence
d
_{2}  d
_{1}  = r as desired.


(b) Fourth solution. [M. Boase]
Wolog, assume that the configuration
is as diagrammed.


Since ÐSUB = ÐAUV = 45^{°}, SU is parallel to the external bisector of ÐA.
This bisector is the locus of points equidistant from AB and
CA produced. Wolog, let PS meet this bisector in W, as in
the diagram. Then PW = PA so that PS  PA = PS  PW = SW = AU
and thus d_{1}  d_{2} = r.