Solutions.

304.

Prove that, for any complex numbers
z and w,
( z + w ) 
ê ê


z
z 

+ 
w
w 


ê ê

£ 2 z + w  . 

Solution 1.


 
 
= 
ê ê

z + w + 
z w
w 

+ 
w z
z 


ê ê


 
 
£ z + w + 
1
z w 

 

z

zw + 

w

zw  
 
 
= z + w + 
z w 
z w 

 

z

+ 

w

 = 2 z + w  . 

Solution 2. Let z = ae^{i a} and
w = be^{i b}, with a and b real and positive.
Then the left side is equal
to

(a + b)(e^{i a} + e^{i b})  

= ae^{i a} + ae^{i b} + be^{i a}+ be^{i b}  
 
 
£ ae^{i a} + be^{i b} + ae^{i b} + be^{i a}  . 

Observe that


= (ae^{i a} + be^{i b})(ae^{i a} + be^{i b})  
 
 
= a^{2} + b^{2} + ab[e^{i (a b)} + e^{i (b a)}] 
 
 
= (ae^{i b} + be^{i a})(ae^{i b} + be^{ i a})  

from which we find that the left side does not exceed
ae^{i a} + be^{i b} + ae^{i b} + be^{i a}  = 2 ae^{i a} + be^{i b}  = 2 z + w  . 

Solution 3. Let z = ae^{i a} and w = be^{i b},
where a and b are positive reals.
Then the inequality is equivalent to

ê ê


1
2

(e^{ia} + e^{ib}) 
ê ê

£ le^{ia} + (1  l)e^{i b}  

where
l = a/(a+b). But this simply says that the midpoint
of the segment joining e
^{ia} and e
^{ib} on the
unit circle in the Argand diagram is at least as close to the
origin as another point on the segment.
Solution 4. [G. Goldstein] Observe that, for each m Î C,

ê ê


mz
mz 

+ 
mw
mw 


ê ê

= 
ê ê


z
z 

+ 
w
w 


ê ê

, 

m[ z + w ] = mz + mw  , 

and
mz + w  = mz + mw  . 

So the inequality is equivalent to
( t + 1) 
ê ê


t
t 

+ 1 
ê ê

£ 2 t + 1  

for t
Î C. (Take
m = 1/w and t = z/w.)
Let t = r(cosq+ i sinq). Then the inequality becomes
(r + 1)  Ö

(cosq+ 1)^{2} + sin^{2} q

£ 2  Ö

(r cosq+ 1)^{2} + r^{2} sin^{2} q

= 2  Ö

r^{2} + 2rcosq+ 1

. 

Now,


 
 
= 2r^{2}(1  cosq) + 4r(cosq 1) + 2(1  cosq) 
 
 
= 2(r  1)^{2}(1  cosq) ³ 0 , 

from which the inequality follows.
Solution 5. [R. Mong] Consider complex numbers as
vectors in the plane. q = (z /w )w
is a vector of magnitude z in the direction w and
p = (w /z )z is a vector of magnitude
w in the direction z. A reflection about the angle bisector
of vectors z and w interchanges p and w, q and z.
Hence p + q  = w + z . Therefore


 
 
= z + q + p + w  £ z + w + p + q  
 
 


305.

Suppose that u and v are positive integer divisors of the
positive integer n and that uv < n. Is it necessarily so that
the greatest common divisor of n/u and n/v exceeds 1?
Solution 1. Let n = ur = vs. Then uv < n
Þv < r, u < s, so that n
^{2} = uvrs
Þ rs > n.
Let the greatest common divisor of r and s be g and the
least common multiple of r and s be m. Then
m
£ n < rs = gm, so that g > 1.
Solution 2. Let g = gcd (u, v), u = gs and v = gt.
Then gst £ g^{2}st < n so that st < n/g. Now s and t
are a coprime pair of integers, each of which divides n/g.
Therefore, n/g = dst for some d > 1. Therefore
n/u = n/(gs) = dt and n/v = n/(gt) = ds, so that
n/u and n/v are divisible by d, and so their greatest
common divisor exceeds 1.
Solution 3. uv < n Þnuv < n^{2} Þ n < (n/u)(n/v).
Suppose, if possible, that n/u and n/v have greatest common
divisor 1. Then the least common multiple of n/u and n/v
must equal (n/u)(n/v). But n is a common multiple of
n/u and n/v, so that (n/u)(n/v) £ n, a contradiction.
Hence the greatest common divisor of n/u and n/v exceeds 1.
Solution 4. Let P be the set of prime divisors of n,
and for each p Î P. Let a(p) be the largest integer
k for which p^{k} divides n. Since u and v are divisors
of n, the only prime divisors of either u or v must belong to
P. Suppose that b(p) is the largest value of the integer
k for which p^{k} divides uv.
If b(p) ³ a(p) for each p Î P, then n would
divide uv, contradicting uv < n. (Note that b(p) > a(p) may occur for some p.) Hence there is a
prime q Î P for which b(q) < a(q). Then
q^{a(q)} is not a divisor of either u or v, so that
q divides both n/u and n/v. Thus, the greatest common divisor
of n/u and n/v exceeds 1.
Solution 5. [D. Shirokoff] If n/u and n/v be coprime, then
there are integers x and y for which
(n/u)x + (n/v) = 1, whence n(xv + yu) = uv. Since n and
uv are positive, then so is the integer xv + yu. But uv < nÞ 0 < xv + yu < 1, an impossibility. Hence the
greatest common divisor of n/u and n/v exceeds 1.

306.

The circumferences of three circles of radius r
meet in a common point O. They meet also, pairwise, in the
points P, Q and R. Determine the maximum and minimum values
of the circumradius of triangle PQR.
Answer. The circumradius always has the value r.
Solution 1. [M. Lipnowski] ÐQPO = ÐQRO, since
OQ is a common chord of two congruent circles, and so subtends
equal angles at the respective circumferences. (Why are
angle QPO and QRO not supplementary?) Similarly, ÐOPR = ÐOQR. Let P¢ be the reflected image of P in the
line QR so that triangle P¢QR and PQR are congruent. Then


= ÐQPR + ÐQOR = ÐQPO + ÐRPO + ÐQOR 
 
 
= ÐQRO + ÐOQR + ÐQOR = 180^{°} . 

Hence P
¢ lies on the circle through OQR, and this circle
has radius r. Hence the circumradius of PQR equals the circumradius
of P
¢QR, namely r.
Solution 2. [P. Shi; A. Wice] Let U, V, W be the centres of
the circle. Then OVPW is a rhombus, so that OP and VW
intersect at right angles. Let H, J, K be the respective
intersections of the pairs (OP, VW), (OQ, UW), (OP, UV). Then
H (respectively J, K) is the midpoint of OP and VW
(respectively OQ and UW, OP and UV). Triangle PQR is
carried by a dilation with centre O and factor ^{1}/_{2} onto
HJK. Also, HJK is similar with factor ^{1}/_{2} to triangle
UVW (determined by the midlines of the latter triangle). Hence
triangles PQR and UVW are congruent. But the circumcircle of
triangle UVW has centre O and radius r, so the circumradius of
triangle PQR is also r.
Solution 3. [G. Zheng] Let U, V, W be the respective
centres of the circumcircles of OQR, ORP, OPQ. Place
O at the centre of coordinates so that
for some
a,
b,
g. Since OVPW is a
rhombus,
P ~ (r (cosb+ cosg), r (sinb+ sing)) . 

Similarly, Q
~ ( r (cos
a+ cos
g),r (sin
a+ sin
g), so that
PQ  = r  Ö

(cosb cosa)^{2}+ (sinb sina)^{2}

= UV  . 

Similarly,
PR
 =
UW
 and
QR
 =
VW
. Thus, triangles PQR and UVW are
congruent. Since O is the circumcentre of triangle UVW, the
circumradius of triangle PQR equals the circumradius of triangle
UVW which equals r.
Solution 4. Let U, V, W be the respective
centres of the circles QOR, ROP, POQ. Suppose that
ÐOVR = 2b; then ÐOPR = b. Suppose that
ÐOWQ = 2g; then ÐOPQ = g. Hence
ÐQPR = b+ g. Let r be the circumradius
of triangle PQR. Then QR  = 2rsin(b+ g).
Consider triangle QUR. The reflection in the axis OQ takes
W to U so that ÐQUO = ÐQWO = 2g. Similarly,
ÐRUO = 2g, whence ÐQUR = 2(b+ g).
Thus triangle QUR is isosceles with QU  = QR  = r
and apex angle QUR equal to 2 (b+ g). Hence
QR  = 2rsin(b+ g). It follows that
r = r.
Comment. This problem was the basis of the logo for the
40th International Mathematical Olympiad held in 1999 in Romania.

307.

Let p be a prime and m a positive integer for which
m < p and the greatest common divisor of m and p is equal to
1. Suppose that the decimal expansion of m/p has period 2k for
some positive integer k, so that

m
p

= .ABABABAB ... = (10^{k} A + B)(10^{2k} +10^{4k} + ¼) 

where A and B are two distinct blocks of k digits. Prove that
(For example, 3/7 = 0.428571 ... and 428 + 571 = 999.)
Solution. We have that

m
p

= 
10^{k}A + B
10^{2k}  1

= 
10^{k}A + B
(10^{k}  1)(10^{k} + 1)



whence
m(10^{k}  1)(10^{k} + 1) = p(10^{k} A + B) = p(10^{k}  1)A + p(A + B) . 

Since the period of m/p is 2k, A
¹ B and p does not
divide 10
^{k}  1. Hence 10
^{k}  1 and p are coprime and
so 10
^{k}  1 must divide A + B. However, A
£ 10
^{k}  1 and
B
£ 10
^{k}  1 (since both A and B have k digits),
and equality can occur at most once. Hence
A + B < 2 ×10
^{k}  2 = 2(10
^{k}  1). It follows that
A + B = 10
^{k}  1 as desired.
Comment. This problem appeared in the College Mathematics
Journal 35 (2004), 2630. In writing up the solution, it is clearer
to set up the equation and clear fractions, so that you can argue
in terms of factors of products.

308.

Let a be a parameter. Define the sequence
{ f_{n} (x) : n = 0, 1, 2, ¼} of polynomials by
f_{n+1} (x) = x f_{n} (x) + f_{n} (ax) 

for n ³ 0.


(a) Prove that, for all n, x,
f_{n} (x) = x^{n} f_{n} (1/x) . 



(b) Determine a formula for the coefficient of x^{k}
(0 £ k £ n) in f_{n} (x).
Solution 1. The polynomial f
_{n} (x) has degree n for each
n, and we will write
f_{n} (x) = 
n å
k=0

b(n, k) x^{k} . 

Then
x^{n} f_{n} (1/x) = 
n å
k=0

b(n, k)x^{nk} = 
n å
k=0

b(n, n  k)x^{k} . 

Thus, (a) is equivalent to b(n, k) = b(n,n
 k) for
0
£ k
£ n.
When a = 1, it can be established by induction that
f_{n} (x) = (x + 1)^{n} = å_{k=0}^{n} (n  k) x^{n}.
Also, when a = 0, f_{n} (x) = x^{n} + x^{n1} + ¼+ x + 1 = (x^{n+1}  1)(x  1)^{1}. Thus, (a) holds in these cases
and b(n, k) is respectively equal to (n  k) and 1.
Suppose, henceforth, that a ¹ 1. For n ³ 0,


= 
n å
k=0

b(n, k)x^{k+1} + 
n å
k=0

a^{k} b(n, k)x^{k} 
 
 
= 
n å
k=1

b(n, k1) x^{k} + b(n, n)x^{n+1} + b(n, 0)+ 
n å
k=1

a^{k} b(n, k) x^{k} 
 
 
= b(n, 0) + 
n å
k=1

[b(n, k1) + a^{k} b(n, k)]x^{k}+ b(n, n)x^{n+1} , 

whence b(n + 1, 0) = b(n, 0) = b(1, 0) and b(n + 1, n + 1) = b(n, n) = b(1, 1) for all n
³ 1. Since f
_{1} (x) = x + 1,
b(n, 0) = b(n, n) = 1 for each n. Also
b(n + 1, k) = b(n, k1) + a^{k} b(n, k) 
 (1) 
for 1
£ k
£ n.
We conjecture what the coefficients b(n, k) are from an
examination of the first few terms of the sequence:
f_{0}(x) = 1; f_{1}(x) = 1 + x; f_{2}(x) = 1 + (a + 1)x + x^{2}; 

f_{3}(x) = 1 + (a^{2} + a + 1)x + (a^{2} + a + 1)x^{2} + x^{3}; 

f_{4}(x) = 1 + (a^{3} + a^{2} + a + 1)x + (a^{4} + a^{3} + 2a^{2} + a + 1)x^{2}+ (a^{3} + a^{2} + a + 1)x^{3} + x^{4}; 

f_{5}(x) = (1 + x^{5}) + (a^{4} + a^{3} + a^{2} + a + 1)(x + x^{4})+ (a^{6} + a^{5} + 2a^{4} + 2a^{3} + 2a^{2} + a + 1)(x^{2} + x^{3}) . 

We make the empirical observation that
b(n + 1, k) = a^{n+1k}b(n, k1) + b(n, k) 
 (2) 
which, with (1), yields
(a^{n+1k}  1)b(n, k1) = (a^{k}  1)b(n, k) 

so that
b(n+1, k) = 
é ë


a^{k}  1
a^{n+1k}  1

+ a^{k} 
ù û

b(n, k) = 
é ë


a^{n+1}  1
a^{n+1k}  1


ù û

b(n, k) 

for n
³ k. This leads to the conjecture that
b(n, k) = 
æ è


(a^{n}  1)(a^{n1}  1) ¼(a^{k+1}  1)
(a^{nk}  1)(a^{nk1}  1) ¼(a  1)


ö ø

b(k, k) 
 (3) 
where b(k, k) = 1.
We establish this conjecture. Let c(n, k) be the right side of
(3) for 1 £ k £ n1 and c(n, n) = 1. Then
c(n, 0) = b(n, 0) = c(n, n) = b(n, n) = 1 for each n.
In particular, c(n, k) = b(n, k) when n = 1.
We show that
c(n + 1, k) = c(n, k1) + a^{k} c(n, k) 

for 1
£ k
£ n, which will, through an induction argument,
imply that b(n, k) = c(n, k) for 0
£ k
£ n. The right
side is equal to

æ è


a^{n}  1
a^{nk}  1


ö ø

¼ 
æ è


a^{k+1}  1
a  1


ö ø


é ë


a^{k}  1
a^{nk+1}  1

+ a^{k} 
ù û

= 
(a^{n+1}  1)(a^{n}  1) ¼(a^{k+1}  1)
(a^{n+1k}  1)(a^{nk}  1) ¼(a  1)

= c(n+1, k) 

as desired. Thus, we now have a formula for b(n, k) as required
in (b).
Finally, (a) can be established in a straightforward way, either
from the formula (3) or using the pair of recursions (1) and (2).
Solution 2. (a) Observe that f_{0} (x) = 1, f_{1} (x) = x + 1
and f_{1} (x)  f_{0} (x) = x = a^{0} x f_{0} (x/a). Assume as an
induction hypothesis that f_{k} (x) = x^{k} f(1/x) and
f_{k} (x)  f_{k1} (x) = a^{k1} x f_{k1}(x/a) 

for 0
£ k
£ n. This holds for k = 1.
Then


= x[f_{n}(x)  f_{(n1)}(x)] + [f_{n}(ax)  f_{n1}(ax)] 
 
 
= a^{n1}x^{2} f_{n1}(x/a) + a^{n1}axf_{n1}(x) 
 
 
= a^{n}x [f_{n1}(x) + (x/a)f_{n1}(x/a) = a^{n} x f_{n}(x/a) , 

whence


= f_{n} (x) + a^{n} x f_{n}(x/a) = f_{n}(x) + a^{n} x(x/a)^{n}f_{n}(a/x) 
 
 
= x^{n} f_{n}(1/x) + x^{n+1} f_{n} (a/x) = x^{n+1} [(1/x)f_{n}(1/x) + f_{n}(a/x)] = x^{n+1}f_{n+1}(1/x) . 

The desired result follows.
Comment. Because of the appearance of the factor a  1
in denominators, you should dispose of the case a = 1 separately.
Failure to do so on a competition would likely cost a mark.

309.

Let ABCD be a convex quadrilateral for which all
sides and diagonals have rational length and AC and BD intersect
at P. Prove that AP, BP, CP, DP all have rational
length.
Solution 1. Because of the symmetry, it is enough to show
that the length of AP is rational. The rationality of the
lengths of the remaining segments can be shown similarly.
Coordinatize the situation by taking
A
~ (0, 0), B
~ (p, q), C
~ (c, 0),
D
~ (r, s) and P
~ (u, 0).
Then, equating slopes, we find that
so that
whence
u = r
 [(sr
 ps)/(s
 q)] = [(ps
 qr)/(s
 q)].
Note that AB ^{2} = p^{2} + q^{2},
AC ^{2} = c^{2}, BC ^{2} = (p^{2}  2pc + c^{2})+ q^{2}, CD ^{2} = (c^{2}  2cr + r^{2}) + s^{2}
and AD ^{2} = r^{2} + s^{2}, we have that
2rc = AC^{2} + AD^{2}  CD^{2} 

so that, since c is rational, r is rational. Hence s
^{2} is rational.
Similarly
2pc = AC^{2} + AB^{2}  BC^{2} . 

Thus, p is rational, so that q
^{2} is rational.
2qs = q^{2} + s^{2}  (q  s)^{2} = q^{2} + s^{2}  [(p  r)^{2} + (q  s)^{2}] + p^{2}  2pr + r^{2} 

is rational, so that both qs and q/s = (qs)/s
^{2} are rational.
Hence
is rational.
Solution 2. By the cosine law, the cosines of all of the angles
of the triangle ACD, BCD, ABC and ABD are rational.
Now
and

CP
BC

= 
sinÐPBC
sinÐBPC

. 

Since
ÐAPB +
ÐBPC = 180
^{°}, therefore
sin
ÐAPB = sin
ÐBPC and


= 
AB sinÐABP
BC sinÐPBC

= 
AB sinÐABP sinÐPBC
BC sin^{2} ÐPBC


 
 
= 
AB (cosÐABP cosÐPBC  cos(ÐABP + ÐPBC))
BC (1  cos^{2} ÐPBC)


 
 
= 
AB (cosÐABD cosÐDBC  cosÐABC)
BC (1  cos^{2} ÐDBC)



is rational. Also AP + CP is rational, so that
(AP/CP)(AP + CP) = ((AP/CP) + 1)AP is rational. Hence
AP is rational.

310.

(a) Suppose that n is a positive integer. Prove that
(x + y)^{n} = 
n å
k=0


æ è

n
k

ö ø

x (x + k)^{k1} (y  k)^{nk} . 



(b) Prove that
(x + y)^{n} = 
n å
k=0


æ è

n
k

ö ø

x (x  kz)^{k1} (y + kz)^{nk} . 

Comments. (a) and (b) are equivalent. To obtain (b) from
(a), replace x by
x/z and y by
y/z. On the other hand,
the substitution z =
1 yields (a) from (b).
The establishment of the identities involves the recognition of
a certain sum which arise in the theory of finite differences.
Let f(x) be a function of x and define the following operators
that take functions to functions:
Ef(x) = f(x + 1) = (I + D)f(x) 

Df(x) = f(x + 1)  f(x) = (E  I)f(x) . 

For any operator P, P
^{n}f(x) is defined recursively by
P
^{0} f(x) = f(x) and P
^{k+1}f(x) = P(P
^{k1}) f(x)), for
k
³ 1. Thus E
^{k}f(x) = f(x + k) and
D^{2} f(x) = Df(x + 1)  Df(x) = f(x + 2)  2f(x + 1) + f(x) = (E^{2}  2E + I)f(x) = (E  I)^{2}f(x) . 

We have an
operational calculus in which we can treat
polynomials in I, E and
D as satisfying the regular rules
of algebra. In particular
E^{n} f(x) = (I + D)^{n} f(x) = 
å
 
æ è

n
k

ö ø

D^{k} f(x) 

and
D^{n} f(x) = (E  I)^{n} f(x) = 
n å
k = 0

(1)^{nk} 
æ è

n
k

ö ø

E^{k} f(x) = 
n å
k = 0

(1)^{nk} 
æ è

n
k

ö ø

f(x + k) , 

for each positive integer n, facts than can be verified directly
by unpacking the operational notation.
Now let f(x) be a polynomial of degree d ³ 0. If f(x) is
constant (d = 0), then Df(x) = 0. If d ³ 1, then
Df(x) is a polynomial of degree d  1. It follows that
D^{d} f(x) is constant, and D^{n} f(x) = 0 whenever
n > d. This yields the identity

n å
k=0

(1)^{k} 
æ è

n
k

ö ø

f(x + k) = 0 

for all x whenever f(x) is a polynomial of degree strictly less
than n.
Solution 1. [G. Zheng]


x(x + k)^{k1} (y  k)^{nk} = 
n å
k=0


æ è

n
k

ö ø

x(x + k)^{k1} [(x + y)  (x + k)]^{nk} 
 
 
= 
n å
k=0


nk å
j=0


æ è

n
k

ö ø

x(x + k)^{k1} 
æ è

nk
j

ö ø

(x + y)^{j} (1)^{nkj}(x + k)^{nkj} 
 
 
= 
å
0 £ k £ n  j £ n

(1)^{nkj} 
æ è

n
k

ö ø


æ è

nk
j

ö ø

x (x + k)^{nj1}(x + y)^{j} 
 
 
= 
n å
j=0


nj å
k = 0

(1)^{nkj} 
æ è

n
j

ö ø


æ è

nj
k

ö ø

x(x + k)^{nj1} (x + y)^{j} 
 
 
= 
n å
j=0

(1)^{nj} 
æ è

n
j

ö ø

(x + y)^{j} 
nj å
k=0

(1)^{k} 
æ è

nj
k

ö ø

x(x + k)^{nj1} 
 
 
= (x + y)^{n} x(x + 0)^{1} + x 
n1 å
j=1

(1)^{nj} 
æ è

n
j

ö ø

(x + y)^{j} 
nj å
k=0

(1)^{k} 
æ è

nj
k

ö ø

(x + k)^{nj1} . 

Let m = n  j so that 1 £ m £ n. Then


nj å
k=0

(1)^{k} 
æ è

nj
k

ö ø

(x + k)^{nj1} 

= 
m å
k=0

(1)^{k} 
æ è

m
k

ö ø

(x + k)^{m1} 
 
 
= 
m å
k=0

(1)^{k} 
æ è

m
k

ö ø


m1 å
l=0


æ è

m1
l

ö ø

x^{ml}k^{l} 
 
 
= 
m1 å
l=0


æ è

m1
l

ö ø

x^{ml} 
m å
k=0

(1)^{k} 
æ è

m
k

ö ø

k^{l} = 0 . 

The desired result now follows.
Solution 2. [M. Lipnowski] We prove that

n å
k=0


æ è

n
k

ö ø

x (x  kz)^{k1} (y + kz)^{nk} = (x + y)^{n} 

by induction. When n = 1, this becomes
1 ·x(x)^{1}y + 1 ·x(x  z)^{0}(y+z)^{0} = y + x = x + y . 

Assume that for n
³ 2,

n1 å
k=0


æ è

n1
k

ö ø

x(x  kz)^{k1}(y + zk)^{nk1} = (x + y)^{n1} . 

Let f(y) = (x + y)
^{n} and g(y) =
å_{k=0}^{n} (n  k)x(x
 kz)
^{k1} (y + kz)
^{nk}. We can establish that f(y) = g(y)
for all y by showing that f
¢(y) = g
¢(y) for all y
(equality of the derivatives with respect to y) and
f(
x) = g(
x) (equality when y is replaced by
x).
That f¢(y) = g¢(y) is a consequence of the induction hypothesis and
the identity (n  k)(n  k) = n ((n1)  k). Also


= 
n å
k=0


æ è

n
k

ö ø

x (x  kz)^{k1}(x + kz)^{nk} 
 
 
= x 
n å
k=0

(1)^{nk} 
æ è

n
k

ö ø

(x  kz)^{n1} = 0 , 

by appealing to the finite differences result. The desired result now
follows.