Solutions and Comments
Write your solutions to be read. You may have to rework
and reedit your solution to make the argument clear,
close gaps in the reasoning, define symbols used and
present ideas in logical order. Always check your work.
It is too easy to miscopy or misrepresent a symbol
and to make a computational mistake. In a competition,
your goal is to maximize the number of points you receive.
In the following solutions, [A ¼Z] represents the
area of the figure A ¼Z, dist (A, BC) is the
perpendicular distance from point A to line BC, and
dist (a, b) is the perpendicular distance between
parallel lines a and b.
 13.

Suppose that x_{1}, x_{2}, ¼,x_{n} are nonnegative real numbers for which
x_{1} + x_{2} + ¼+ x_{n} < ^{1}/_{2}.
Prove that
(1  x_{1}) (1  x_{2}) ¼(1  x_{n}) > 
1 2

, 

Comments. Some of you tried a ``moving variables'' argument,
or an ``extreme case'' argument, i.e., if we make
x_{i} as large as possible, we reduce the product, so that
if it works for (x_{1}, x_{2}, ¼, x_{n}) = (^{1}/_{2},0, 0, ¼, 0), then we are in business. Some felt that
if it works for the extreme case with each x_{i} = 1/2n, then
we are ok. Unless you back this up with solid argument and
detailed analysis that relates the general case to the
extreme case, then it is worthless. ``Moving variable''
arguments are always risky and best avoided.
Both approaches often muddy the situation rather than clarify
it.
Solution 1. If 0 < u, v < 1, then
(1  u)(1  v) = 1  (u + v) + uv > 1  (u + v) . 

It can be shown by induction that
(1  x_{1})(1  x_{2}) ¼(1  x_{k}) > 1  (x_{1} + x_{2} + ¼+ x_{k}) 

for 2
£ k
£ n, whence
(1  x_{1}) ¼(1  x_{n}) > 1  (x_{1} + ¼+ x_{n}) > 
1 2

. 

Comments. You should carry out the necessary induction
argument.
Note that the problem asks for > rather than ³ .
Solution 2. [R. Barrington Leigh] Let
t_{k} = 
å
 { x_{i1}x_{12} ¼x_{ik}: 1 £ i_{1} < i_{2} < ¼ < i_{k} £ n } 

for 1
£ k
£ n. Then
(1  x_{1})(1  x_{2}) ¼(1  x_{n}) = 1  t_{1} + t_{2}  t_{3} + t_{4}  ¼+ (1)^{n} t_{n} . 

For 1
£ k
£ n1, we have that


= 
å
 { x_{i1} x_{i2} ¼x_{ik}: 1 £ i_{1} < i_{2} < ¼ < i_{k} £ n } 
 
> 
å
 { x_{i1} x_{i2} ¼x_{ik}(x_{ik + 1} + x_{ik + 2} + ¼+ x_{n}): 1 £ i_{1} < i_{2} < ¼ < i_{k} £ n } = t_{k+1} 

 

so that
(1  x_{1})(1  x_{2}) ¼(1  x_{n}) = (1  t_{1}) + (t_{2}  t_{3}) + ¼ > (1  t_{1}) = 1  (x_{1} + ¼+ x_{n}) > 
1 2

. 

 14.

Given a convex quadrilateral, is it
always possible to determine a point in its interior
such that the four line segments joining the point
to the midpoints of the sides divide the
quadrilateral into four regions of equal area?
If such a point exists, is it unique?
Note. Let ABCD be the quadrilateral and let
P, Q, R, S be the respective midpoints of AB,
BC, CD, DA. Recall that PS
 QR and
PQ
 SR (prove it!).
Solution 1. Observe that
[APS] = ^{1}/_{4}[ABD] and [CRQ] = ^{1}/_{4}[CDB], so that
[APS] + [CRQ] = ^{1}/_{4}[ABCD] (why?).
Since
[ABCD] = 
1 2

BD ·(dist (A, BD)+ dist (C, BD)) 

[APS] = 
1 4

BD ·dist (A, PS) 

[CRQ] = 
1 4

BD ·dist (C, QR) 

we have that


+ dist (C, QR) = 
1 2

[ dist (A, BD)+ dist (C, BD) 
 

 

Hence it is possible to draw a line m parallel to
and between PS and QR for which
dist (m, QR) = dist (A, PS) and dist (m, PS)
= dist (C, QR). For any point M on m, we have
that [MQR] = [APS] and [MSP] = [CRQ], so that
[MQCR] = [APMS] =
^{1}/
_{4}[ABCD].
In a similar way, we can draw a line n parallel to
and between PQ and SR for which dist (n, PQ) =
dist (D, SR) and dist (n, SR) = dist (B, PQ),
so that, for any point N on n,
[NPQ] = [DSR], [NRS] = [BRP] and
[NPBQ] = [DSNR] = ^{1}/_{4}[ABCD].
The lines m and n are not parallel, and, so, intersect
in a unique point O for which
[OQCR] = [APOS] = [OPBQ] = [DSOR] = 
1 4

[ABCD] . 

Note that m and n intersect inside PQRS and so inside
ABCD.
We show that O is the only point with this property.
Let L be another point inside ABCD. Then L lies in
one of the four partitioning quadrilaterals, say
[OQCR], so that [LQCR] < [OQCR] = ^{1}/_{4}[ABCD]. Hence, L will satisfy the conditions of the
problem.
Comments. A careful solution will contain the observation
that m and n are not parallel, and will intersect within
the quadrilateral. The fact that m and n have a unique
point of intersection does not, in and of itself, establish that the
requisite point cannot be found in some other way. Therefore,
uniqueness needs to be explicitly handled, either by showing that
the required point must lie on m and n or by some other
argument.
Solution 2. Analysis. Suppose such a point O exists.
Then [OPBQ] = [OCQR] = ^{1}/_{4}[ABCD], so that
[AOB] = 2[POB] = 2([OPBQ]  [OBQ]) = 2([OQCR]  [OQC]) = 2[ROC] = [DOC] . 

Since [AOB] =
^{1}/
_{2}AB ·dist (O, AB)
and [DOC] =
^{1}/
_{2} CD ·dist (O, CD), we must have that

dist (O, AB) dist (O, CD)

= 
CD AB

. 

The locus of points O with this property is a line h through
the intersection of AB and CD (or parallel to them
if they do not intersect)
which lies between them
(prove this!). Similarly, O must lie on a line k defined by

dist (O, BC) dist (O, AD)

= 
AD BC



through the intersection of BC and AD (if they intersect)
that lies between them.
These lines are not parallel and intersect within ABCD (why?).
If O exists, it must lie on the intersection of h and k.
Synthesis. Construct lines h and k to satisfy the
foregoing conditions. They must intersect in a unique point
O within ABCD. Now
[AOB] = 
1 2

AB ·dist (O, AB) = 
1 2

CD ·dist (O, CD) = [DOC] . 

Hence [AOP] = [POB] = [COR] = [DOR]. Let
a be the common value. Similarly,
[BOQ] = [COQ] = [AOS] = [DOS] =
b, say. Then
each of [APOS], [BQOP], [CROQ] and [DSOR] has the
value
a+
b and O is the desired point.
 15.

Determine all triples (x, y, z)
of real numbers for which
x(y + 1) = y (z + 1) = z(x + 1) . 

Comments. In solving a system of equations,
one begins by assuming a solution and determining
what properties it must have. Since this may
involve oneway reasoning, such properties may
not necessarily yield a solution and extraneous
solutions may arise. Thus, when you have solved
the equations, for a complete solution, you should
check that the solution is valid.
If, in your manipulations, you divide by a certain
quantity or find a quantity in the denominator of a
fraction, you should explore the possibility that the
quantity could vanish, Remember that you cannot divide
by zero. When you write up your final solution, it is
often a good idea to deal with this possibility ahead
of time and get it out of the way.
Solution 1. Suppose one of the variable, say x, is
0. Then z(x + 1) = 0, so z = 0, and y(z + 1) = 0 so
y = 0. Suppose one of the variables, say x is 1.
The x(y + 1) = 0, so y = 1 and y(z + 1) = 0, so
z = 1. Hence, if any variable assumes either of the
values 1 and 0, then all are equal. Henceforth,
we assume that none of them have either value.
>From the equations, we find that

y(z + 1) y+1

= x = 
yz + y  z z



whence
yz(z + 1) = (y + 1)yz + (y + 1)(y  z) 

or
yz(z  y) = (y + 1)(y  z) . 

Hence, either y = z or yz + y + 1 = 0.
If y = z, then z + 1 = x + 1. so that z = x and
x = y = z. Conversely, the system is satisfied when
x = y = z.
Suppose that yz + y + 1 = 0. Then x(y + 1) = z(x + 1) = y(z + 1) = 1, and so xy + x + 1 = xz + z + 1 = 0.
Suppose z = t, Then x = (t + 1)/t and
y = 1/(t + 1). Conversely, it is straightforward to check
that the system is satisfied by
(x, y, z) = 
æ ç
è

 
t + 1 t

,  
1 1 + t

,t 
ö ÷
ø



where t
¹ 0, 1. Thus, we have obtained exactly the
complete set of solutions.
Comment. The solution with x, y, z unequal may look
nonsymmetrical. Verify that, if x = s = (t + 1)/t, then
y = 1/(1 + t) = (s + 1)/s and z = t = 1/(1 + s).
Check that x and z can similarly be expressed in terms of
y = r.
Solution 2. From the equations, we find that
x(y  z) = z  x , y(z  x) = x  y , and z(x  y) = y  z . 

Multiplying these equations yields
xyz(y  z)(z  x)(x  y) = (z  x)(x  y)(y  z) . 

Hence, either two variables (and so all) are equal or xyz = 1.
The system is satisfied when all variables are equal.
Suppose that xyz = 1. Then
xy + x + 1 = xy + x + xyz = x(yz + y + 1) 

and
yz + y + 1 = yz + y + xyz = y(zx + z + 1) . 

Since xy + x = yz + y = zx + z, either x = y = z = 1 or
xy + x + 1 = yz + y + 1 = zx + z + 1 = 0. We need to check
that there are solutions of this type. Select arbitrary
x
¹ 0, 1, then y to satisfy xy + x + 1 = 0 and
z to satisfy xyz = 1. Then
zx + z + 1 = zx + z + xyz = z(xy + x + 1) = 0 

and
yz + y + 1 = yz + y + xyz = y(zx + z + 1) = 0 

so that x(y + 1) = y(x + 1) = z(x + 1) = 1 as desired.
Solution 3. As before, we can check that
(x, y, z) = (0, 0, 0) is the only solution in which any
variable vanishes. Henceforth, suppose that xyz ¹ 0.
Let z = vx. Then y + 1 = v(x + 1), whence
y = vx + v  1. Therefore
x(vx + v) = x(y + 1) = y(z + 1) = (vx + v  1)(vx + 1) 

so that
v(v  1)x^{2} + v(v  1)x + (v  1) = 0 . 

Either v = 1 and we are led to x = y = z, which works, or
Hence
(x, y, z) = 
æ ç
è


2v

, 
(v  2) ± 
 ______ Öv^{2}  4v

2

, 
2


ö ÷
ø

. 

where v < 0 or v
³ 4. It can be checked that these
values satisfy the equations. (
Exercise: Check that
this colution is consistent with the other solutions.)
Solution 4. As in the foregoing solutions, we can check
that (x, y, z) = (0, 0, 0), (1, 1, 1) are the only solutions
in which any variable assumes either of the values 0 or 1.
Henceforth, suppose x, y, z all differ from 0 and 1.
Let x (y + 1) = y(z + 1) = z(x + 1) = k. Then
z = k/(x + 1) and y = (k/x)  1 = (k  x)/x. Therefore,


= y(z + 1) = 
æ ç
è


k  x x


ö ÷
ø


æ ç
è


k + x + 1 x + 1


ö ÷
ø


 
Þ k(x^{2} + x) = k^{2}  x^{2} + k  x = k(k + 1)  (x^{2} + x) 
 
Þ (k + 1)(x^{2} + x  k) = 0 
 
Þ k = 1 or x^{2} + x = k . 

 

Suppose that x
^{2} + x = k. Then x(x + 1) = z(x + 1). Since
x
¹ 1, x = z, and so y + 1 = x + 1. Thus, x = y = z.
Suppose that k = 1. Then z = 1/(x + 1) and y = (x + 1)/x.
It can be checked that this works.
Comment. It is not hard to check independently that the
only nonzero solution in which x, y, z have the same sign
is in fact given by x = y = z. For in this case, the
ratio of any pair is positive, and the system can be
rewritten
whence
3 = (x/y) + (y/z) + (z/x). By the ArithmeticGeometric
Means Inequality (applicable since the three summands are
positive), (x/y) + (y/z) + (z/x)
³ 3 with
equality if and only if x/y = y/z = z/x or x = y = z.
 16.

Suppose that ABCDEZ is a
regular octahedron whose pairs of opposite
vertices are (A, Z), (B, D) and (C, E).
The points F, G, H are chosen on the segments
AB, AC, AD respectively such that
AF = AG = AH.


(a) Show that EF and DG must intersect
in a point K, and that BG and EH must intersect
in a point L.


(b) Let EG meet the plane of AKL in M.
Show that AKML is a square.
Comment. Many students had complicated arguments
involving similar triangles. You should try to envisage
the situation in terms of transformations, as this gives
you a better sense of what is going on. Of course, if
a synthetic argument cannot be found, you always have
recourse to the ``refuge of the destitute'', coordinate
geometry and the hope that the computations will not
be too horrendous. In this case, they are not bad at all.
Solution 1. (a) Since AF:AB = AG:GC, it follows that
FG  BC  ED, while FG < BC = ED. Hence, FGDE is a
coplanar isosceles trapezoid and so EF and DG must
intersect in a point K. A 90^{°} rotation
about the axis AZ takes B ® C,
F ® G, C ® D, G ® H,
D ® E, E ® B. Hence EF ® BG
and DG ® EH, so that BG and EH must
intersect in a point L, which is the image of K under
the rotation.
(b) KE and AB intersect in F, so that the two lines are
coplanar. Also KF:KE = FG:ED = FG:BC = AF:FB so that
DKAF ~ DEFB and AK  EB.
Hence K lies in a plane
through A parallel to BCDE. Because the 90^{°}
rotation about the axis AZ (which is perpendicular to the
planes BCDE and AKL takes K ® L, AK = AL and
ÐKAL = 90^{°}.
Consider a dilation with centre E and factor AB /FB . Let I be on AE with AI = AF.
The the dilation takes F ® K, H ® L,
I ® A and the plane of FGHI to the parallel plane
AKL. The image of G under this dilation is the intersection of
EG and the plane of AKL, namely M. Thus the square FGHI goes
to KMLA which must also be a square.
Solution 2. The dilationreflection perpendicular to a plane
through FG perpendicular to BE and CD with factor
AF /FB  takes B ® E,
C ® D, and fixes F and G. The lines BF and
CG with intersection A gets carried to lines EF and DG
which intersect in a point K for which AK is perpendicular
to the plane and so parallel to BE and CD, and the distance from
K to the plane is AF /FB  times
the distance from A to the plane.
Similarly, considering a dilationreflection to a plane through GH
perpendicular to BC and DE with the same factor produces the point L with
AL perpendicular to this plane and so parallel to BC and ED.
Thus ÐKAL = 90^{°}.
The reflection in the plane AECZ fixes A, C, E, G and interchanges the
points in each of the pairs (B, D) and (F, H). Hence the line pairs
(EF, DG) and (EH, BG) are interchanged as is the pair K and L. Thus
AK = AL and KL is perpendicular to AECZ. The triangle AKL is in
a plane through A parallel to BCDE. The proof that AKML is a
square can be completed as in Solution 1.
Solution 3. Assign solid coordinates: A ~ (0, 0, 1),
B ~ (0, 1, 0), C ~ (1, 0, 0), D ~ (0, 1, 0),
E ~ (1, 0, 0). For some t with 0 < t < 1,
F ~ (0, (1t), t), G ~ (1t, 0, t), H ~ (0, 1t, t).
The line EF consists of points
(u, (1  u)(1  t), (1  u)t), with real u and the line DG
consists of points ((1  v)(1  t), v, (1  v)t) with real v.
They meet in K ~ ((1t)/t, (t1)/t, 1). Similarly,
L ~ ((1t)/t, (1t)/t, 1), so A, K, L lie on the plane
z = 1. The line EG consists of points
(w + (1  t)(1  w), 0, t(1  w)) with real w, and this
intersects the plane z = 1 in the point (2(1  t)/t, 0, 1).
The desired result can now be verified.
 17.

Suppose that r is a real number.
Define the sequence x_{n} recursively by
x_{0} = 0, x_{1} = 1, x_{n+2} = rx_{n+1}  x_{n}
for n ³ 0. For which values of r is it true
that
x_{1} + x_{3} + x_{5} + ¼+ x_{2m1} = x_{m}^{2} 

for m = 1, 2, 3, 4,
¼.
Answer. The property holds for all values of r.
Solution 1. Define x_{1} = 1; the recurrence still
holds with this extension of the index. Note that, for
n ³ 0,

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

(x_{n} x_{n+2}  x_{n1} x_{n+1}) = x_{n+1}(x_{n+1} + x_{n1})  x_{n} (x_{n} + x_{n+2}) 
 
= x_{n+1} (rx_{n})  x_{n} (rx_{n+1}) = 0 

 

so that x
_{n+1}^{2}  x
_{n}^{2} = x
_{n} x
_{n+2}  x
_{n1} x
_{n+1}.
We prove by induction that for each nonnegative integer m,
x_{2m} = x_{m} (x_{m+1}  x_{m1}) 

x_{2m+1} = x_{m+1}^{2}  x_{m}^{2} = x_{m} x_{m+2}  x_{m1}x_{m+1} . 

These equations check out for m = 0. Suppose they hold for
m = k. Then


= rx_{2k+1}  x_{2k} = r(x_{k+1}^{2}  x_{k}^{2})  x_{k}(x_{k+1}  x_{k1}) 
 
= x_{k+1} (rx_{k+1}  x_{k})  x_{k} (rx_{k}  x_{k1}) 
 
= x_{k+1}x_{k+2}  x_{k} x_{k+1} = x_{k+1}(x_{k+2}  x_{k}) 

 



= rx_{2k+2}  x_{2k+1} = r(x_{k+1} x_{k+2}  x_{k+1} x_{k})  (x_{k+1}^{2}  x_{k}^{2}) 
 
= x_{k+1} (rx_{k+2}  x_{k+1})  x_{k} (rx_{k+1}  x_{k}) 
 
= x_{k+1} x_{k+3}  x_{k} x_{k+2} = x_{k+2}^{2}  x_{k+1}^{2} 

 

so the result holds for m = k+1.
Hence
x_{1} + x_{3} + ¼+ x_{2m1} = (x_{1}^{2}  x_{0}^{2}) + (x_{2}^{2}  x_{1}^{2})+ ¼+ (x_{m}^{2}  x_{m1}^{2}) = x_{m}^{2} 

as desired.
Solution 2. [R. Mong] We prove by induction that, for m ³ 1,
x_{1} + x_{3} + ¼+ x_{2m1} = x_{m}^{2} 

x_{2} + x_{4} + ¼+ x_{2m} = x_{m} x_{m+1} . 

These hold for m = 1. Assume they hold for 1
£ k
£ m. Then


+ ¼+ x_{2k+1} = 1 + (rx_{2}  x_{1}) + (rx_{4}  x_{3}) + ¼+ (rx_{2k}  x_{2k1}) 
 
= 1 + r[x_{2} + x_{4} + ¼+ x_{2k}]  [1 +r(x_{2} + x_{4} + ¼+ x_{2k2})  (x_{1} + x_{3} + ¼+ x_{2k3})] 
 
= r(x_{k} x_{k+1}  x_{k1} x_{k}) + x_{k1}^{2} = rx_{k} x_{k+1}  x_{k1}(rx_{k}  x_{k1}) 
 
= rx_{k} x_{k+1}  x_{k1} x_{k+1} = x_{k+1} (rx_{k}  x_{k1}) = x_{k+1}^{2} 

 



+ ¼+ x_{2k+2} = (rx_{1}  x_{0}) + (rx_{3}  x_{2}) + ¼+ (rx_{2k+1}  x_{2k}) 
 
= rx_{k+1}^{2}  x_{k} x_{k+1} = x_{k+1} (rx_{k+1}  x_{k}) = x_{k+1}x_{k+2} . 

 

Hence the result holds for all m and the result follows.
Solution 3. The recursion x_{n+2} = rx_{n+1}  x_{n} has
characteristic polynomial t^{2}  rt + 1 with discriminant
r^{2}  4. Its roots are distinct as long as r ¹ ±2,
so we deal with r = ±2 separately.
The solution for r = 2 is x_{n} = n and for r = 2 is
x_{n} = (1)^{n1} n, and it is straightforward to establish the
desired relation in these cases. When r ¹ ±2, the characteristic
polynomial has distinct roots s and 1/s, where
s+ 1/s = r, i.e. s^{2} = rs 1.
Note that s ¹ ±1 since r ¹ ±2. The solution of
the recursion is
x_{n} = 
s s^{2}  1

( s^{n}  s^{n} ) . 

Then


 
= 
s s^{2}  1

(s+ s^{3} + ¼+ s^{2m1})  
s s^{2}  1

(s^{1} + s^{3} + ¼+ s^{(2m1)}) 
 
= 
s^{2} s^{2}  1


é ê
ë


s^{2m}  1 s^{2}  1


ù ú
û

 
1 s^{2}  1


é ê
ë


s^{2m}  1 s^{2}  1


ù ú
û


 
= 
s^{2} (s^{2}  1)^{2}

[ s^{2m}  1  1 + s^{2m} ] = 
s^{2} (s^{2}  1)^{2}

(s^{m}  s^{m})^{2} = x_{m}^{2} . 

 

Comment. Solvers who used the approach of Solution 3 failed to
consider the case in which the characteristic polynomial had a double root.
 18.

Let a and b be integers. How many solutions
in real pairs (x, y) does the system
have?
Comments. Several of the solvers got all mixed up with the status
of the variables in the problem, and, for example, found infinitely
many solutions to the equations. a and b are fixed in advance;
they are parameters, and your final answer will be conditioned by
the the characteristics of various pairs (a, b). Rather than
rush blindly into the problem, it is a good strategy to gain some
understanding of the situation by looking at particular cases. For example,
if one takes (a, b) = (0, 0), it is not too hard to see that
(x, y) = (0, 0) is the only solution. Similarly, if
(a, b) = (1, 1), one arrives at the sole solution
(x, y) = (^{1}/_{2},^{1}/_{2}). However, if (a, b) = (1, 0),
we find two distinct solutions, (x, y) = (^{1}/_{2}, 1),(0, ^{1}/_{2}). This not only clarifies the situation, but gives
you a couple of examples against which you can check your final
answer. Get in the habit of using examples to help understanding.
It is an excellent exercise to plot, on the same axes, the graphs
of the two equations for various values of a and b.
Solution 1. Since 2x and 2y are the difference of two integers,
x and y must have the form m + ^{u}/_{2} and n + ^{v}/_{2}
respectively, where m and n are integers and u and v each take
one of the values 0 and 1. Plugging these in yields
Solving for m and n gives
For a viable solution, u and v must be such that the right side of
each equation is a multiple of 3 (and this is where the characteristics
of the parameters a and b enter in). We consider three cases:
(i) Let a + b º 0 (mod 3). Then, for a solution, we must have
0 º 2b  a + v  2u = 3b  (a + b) + v  2u º v  2u 

and
0 º 2a  b + u  2v º u  2v 

modulo 3. Since { u, v }
Í { 0, 1 }, we must have
u = v = 0, and we obtain the solution
(x, y) = 
æ ç
è


2b  a 3

, 
2a  b 3


ö ÷
ø

. 

This checks out.
(ii) Let a + b º 1 (mod 3). Then, for a solution, we must have
0 º 2b a + v  2u º 1 + v  2u 

and
0 º 2a  b + u  2v º 1 + u  2v 

modulo 3. The only solutions of u  2v
º v  2u
º 0
(mod 3) with u and v equal to 0 or 1 are given by
(u, v) = (1, 0), (0, 1), so, either
(x, y) = (m + 
1 2

, n) with 3m = 2b  a  2, 3n = 2a  b + 1 

or
(x, y) = (m, n + 
1 2

) with 3m = 2b  a + 1, 3n = 2a  b  2 . 

Thus, we get two solutions
(x, y) = 
æ ç
è


4b  2a  1 6

, 
2a  b + 1 3


ö ÷
ø

, 
æ ç
è


2b  a + 1 3

, 
4a  2b  1 6


ö ÷
ø

. 

These check out.
(iii) Let a + b º 2 (mod 3). Then
0 º 2b  a + v  2u º 2 + v  2u 

and
0 º 2a  b + u  2v º 2 + u  2v 

modulo 3. For a solution, we must have
v  2u
º u  2v
º 2 (mod 3), so that
(u, v) = (1, 1) and (x, y) = (m +
^{1}/
_{2},n +
^{1}/
_{2}) with 3m = 2b  a  1 and
3n = 2a  b  1. Hence there is a unique solution
(x, y) = 
æ ç
è


4a  2a + 1 6

, 
4a  2b + 1 6


ö ÷
ø

. 

This checks out.
Hence, when a + b º 0 or a + b º 2 (mod 3),
there is one solution to the system, while when
a + b º 1 (mod 3), there are two solutions.
Solution 2. Observe that x and y must be half integers.
Since
ëx û+ ëy û+ 2x + 2y = a + b 

and x 
^{1}/
_{2} £ ëx
û £ x,
y 
^{1}/
_{2} £ ëy
û £ y, we must have
that
a + b £ 3(x + y) £ a + b + 1 . 

Hence, x + y, being a half integer, must have exactly one
of the values
depending on the divisibility of the numerator of the
fractions by 3. Consider cases:
(i) Let a + b º 0 (mod 3). Then x + y is the
integer (a + b)/3. If x and y are not themselves
integers, then
ëx û+ ëy û = (x  
1 2

) + (y  
1 2

) = x + y  1 

so that
modulo 3, a contradiction. Hence, x and y are both
integers and
(x, y) = 
æ ç
è


2b  a 3

, 
2a  b 3


ö ÷
ø

. 

This checks out.
(ii) Let a + b º 1 (mod 3). Then 2(a + b) + 1 º 0
(mod 3), and so
x + y = 
3

= 
2(a + b) + 1 6

, 

a halfinteger. Hence there are integers m and n for which
(x, y) = (m + 
1 2

, n) or (m + 1, n  
1 2

) 

where m + n = (a + b  1)/3. We find that
(x, y) = 
æ ç
è


4b  2a  1 6

, 
2a  b + 1 3


ö ÷
ø

, 
æ ç
è


2b  a + 1 3

, 
4a  2b  1 6


ö ÷
ø

. 

These check out.
(iii) a + b º 2 (mod 3). Then x + y = (a + b + 1)/3,
an integer. In this case, we can check that x and y cannot be integers,
so that x = ëx û+ 1/2 and y = ëy û+ 1/2,
and so
(x, y) = 
æ ç
è


4a  2a + 1 6

, 
4a  2b + 1 6


ö ÷
ø

. 

This checks out.