Solutions

318.

Solve for integers x, y, z the system
1 = x + y + z = x^{3} + y^{3} + z^{2} . 

[Note that the exponent of z on the right is 2, not 3.]
Solution 1. Substituting the first equation into the
second yields that
x^{3} + y^{3} + [1  (x + y)]^{2} = 1 

which holds if and only if


= (x + y)(x^{2}  xy + y^{2}) + (x + y)^{2}  2(x + y) 
 
 
= (x + y)(x^{2}  xy + y^{2} + x + y  2) 
 
 
= (1/2)(x + y)[(x  y)^{2} + (x + 1)^{2} + (y + 1)^{2}  6] . 

It is straightforward to check that the only possibilities are that
either y =
x or (x, y) = (0,
2), (
2, 0) or
(x, y) = (
3,
2), (
2,
3) or (x, y) = (1, 0), (0, 1). Hence
(x, y, z) = (t, t, 1), (1, 0, 0), (0, 1, 0), (2, 3, 6),(3, 2, 6), (2, 0, 3), (0, 2, 3) 

where t is an arbitrary integer. These all check out.
Solution 2. As in Solution 1, we find that either
x + y = 0, z = 1 or x^{2} + (1  y)x + (y^{2} + y 2) = 0.
The discriminant of the quadratic in x is
3y^{2}  6y + 9 = 3(y + 1)^{2} + 12 , 

which is nonnegative when
y + 1
 £ 4.
Checking out the possibilities leads to the solution.
Solution 3.


= 1  z^{2} = x^{3} + y^{3} 
 
 
= (x + y)[(x + y)^{2}  3xy] = (1  z)[(1  z)^{2}  3xy] , 

whence either z = 1 or 3xy = (1
 2z + z
^{2})
 (1 + z) = z(z
 3).
The former case yields (x, y, z) = (x,
x, 1) while the latter
yields
x + y = 1  z xy = 
1
3

z(z  3) . 

Thus, we must have that z
º 0 (mod 3) and that x, y
are roots of the quadratic equation
t^{2}  (1  z)t + 
z(z  3)
3

= 0 . 

The discriminant of this equation is [12
 (z
 3)
^{2}]/3.
Thus, the only possibilities are that z = 0, 3, 6; checking these
gives the solutions.

319.

Suppose that a, b, c, x are real numbers for which
abc ¹ 0 and

xb + (1  x)c
a

= 
xc + (1  x)a
b

= 
xa + (1  x)b
c

. 

Is it true that, necessarily, a = b = c?
Comment. There was an error in the original formulation of this
problem, and it turns out that the three numbers a, b, c are
not necessarily equal. Note that in the problem, a, b, c, x all
have the same status. Some solvers, incorrectly, took the
given conditions as an identity in x, so that they assumed that
the equations held for some a, b, c and all x.
Solution 1. Suppose first that a + b + c ¹ 0.
Then the three equal fractions are equal to the sum of
their numerators divided by the sum of the denominators [why?]:

x(a + b + c) + (1  x)(a + b + c)
a + b + c

= 1 . 

Hence a = xb + (1
 x)c, b = xc + (1
 x)a, c = xa + (1
 x)b,
from which x(b
 c) = (a
 c), x(c
 a) = (b
 a), x(a
 b) = (c
 b). Multiplying these three equations together yields that
x
^{3}(b
 c)(c
 a)(a
 b) = (a
 c)(b
 a)(c
 b). Therefore,
either x =
1 or at least two of a, b, c are equal.
If x = 1, then a + b = 2c, b + c = 2a and c + a = 2b.
This implies for example that a  c = 2(c  a), whence a = c.
Similarly, a = b and b = c. Suppose on the other hand that, say,
a = b; then b = c and c = a.
The remaining case is that a + b + c = 0. Then each entry and sum
of pairs of entries is nonzero, and

xa + (1  x)b
(a + b)

= 
x(ab) + (1  x)a
b



Þ xab + (1  x)b^{2} = x(a + b)^{2}  (1  x)(a^{2} + ab) 

Þ (1  x)(a^{2} + ab + b^{2}) = x(a^{2} + ab + b^{2}) . 

Since 2(a
^{2} + ab + b
^{2}) = (a + b)
^{2} + a
^{2} + b
^{2} > 0, 1
 x = x
and x = 1/2. But in this case, the equations become

b+c
2a

= 
c+ a
2b

= 
a+b
2c



each member of which takes the value
1/2 for all a, b, c for
which a + b + c = 0.
Hence, the equations hold if and only if either a = b = c and
x is arbitrary, or x = 1/2 and a + b + c = 0.
Comment. On can get the first part another way. If d
is the common value of the three fractions, then
xb + (1  x)c = da ; xc + (1  x)a = db ; xa + (1  x)b = dc . 

Adding these yeilds that a + b + c = d(a + b + c), whence
d = 1 or a + b + c = 0.
Solution 2. The first inequality leads to
xb^{2} + (1  x)bc = xac + (1  x)a^{2} 

or
x(a^{2} + b^{2})  x(a + b)c = a^{2}  bc . 

Similarly
x(c^{2} + a^{2})  x(c + a)b = b^{2}  ca ; 

x(b^{2} + c^{2})  x(b + c)a = c^{2}  ab . 

Adding these three equations together leads to
2x[(a  b)^{2} + (b  c)^{2} + (c  a)^{2}] = (a  b)^{2} + (b  c)^{2} + (c  a)^{2} . 

Hence, either a = b = c or x = 1/2.
If x = 1/2, then for some constant k,

b+c
a

= 
c + a
b

= 
a + b
c

= k , 

whence
ka + b + c = a  kb + c = a + b  kc = 0 . 

Add the three left members to get
Therefore, k = 2 or a + b + c = 0. If k = 2, then
a = b = c, as in Solution 1. If a + b + c = 0, then k =
1
for any relevant values of a, b, c. Hence, either a = b = c or
x = 1/2 and a + b + c = 0.

320.

Let L and M be the respective intersections of
the internal and external angle bisectors of the triangle
ABC at C and the side AB produced. Suppose that CL = CM
and that R is the circumradius of triangle ABC. Prove that
AC ^{2} + BC ^{2} = 4R^{2} . 

Solution 1. Since
ÐLCM = 90
^{°} and
CL = CM, we have that
ÐCLM =
ÐCML = 45
^{°}.
Let
ÐACB = 2
q. Then
ÐCAB = 45
^{°} q and
ÐCBA = 45
^{°} +
q.
It follows that


= (2R sinÐCAB)^{2} + (2R sinÐCBA)^{2} 
 
 
= 4R^{2} (sin^{2} (45^{°}  q) + sin^{2} (45^{°} + q)) 
 
 
= 4R^{2} (sin^{2} (45^{°}  q) + cos^{2} (45^{°}  q)) = 4R^{2} . 

Solution 2. [B. Braverman] ÐABC is obtuse [why?].
Let AD be a diameter of the circumcircle of triangle ABC.
Then ÐADC = ÐCBM = 45^{°} + ÐLCB
(since ABCD is concyclic). Since ÐACD = 90^{°},
ÐDAC = 45^{°}  ÐLCB = ÐCAB. Hence,
chords DC and CB, subtending equal angles at the circumference
of the circumcircle, are equal. Hence
4R^{2} = AC ^{2} + CD ^{2} = AC ^{2}+ BC ^{2} . 


321.

Determine all positive integers k for which
k^{1/(k7)} is an integer.
Solution. When k = 1, the number is an integer.
Suppose that 2
£ k
£ 6. Then k
 7 < 0 and so
0 < k^{1/(k7)} = 1/(k^{1/7k}) < 1 

and the number is not an integer. When k = 7, the expression
is undefined.
When k = 8, the number is equal to 8, while if k = 9, the
number is equal to 3. When k = 10, the number is equal to
10^{1/3}, which is not an integer [why?].
Suppose that k ³ 11. We establish by induction that
k < 2^{k7}. This is clearly true when k = 11. Suppose it
holds for k = m ³ 11. Then
m + 1 < 2^{m7} + 2^{m  7} = 2^{(m+1)  7} ; 

the desired result follows by induction. Thus, when k
³ 11,
1 < k
^{1/(k7)} < 2 and the number is not an integer.
Thus, the number is an integer if and only if k = 1, 8, 9.

322.

The real numbers u and v satisfy
u^{3}  3u^{2} + 5u  17 = 0 

and
v^{3}  3v^{2} + 5v + 11 = 0 . 

Determine u + v.
Solution 1. The equations can be rewritten
u^{3}  3u^{2} + 5u  3 = 14 , 

v^{3}  3v^{2} + 5v  3 = 14 . 

These can be rewritten as
(u  1)^{3} + 2(u  1) = 14 , 

(v  1)^{3} + 2(v  1) = 14 . 

Adding these equations yields that


= (u  1)^{3} + (v  1)^{3} + 2(u + v  2) 
 
 
= (u + v  2)[(u  1)^{2}  (u  1)(v  1) + (v  1)^{2} + 2] . 

Since the quadratic t
^{2}  st + s
^{2} is always positive [why?], we
must have that u + v = 2.
Solution 2. Adding the two equations yields


= (u^{3} + v^{3})  3(u^{2} + v^{2}) + 5(u + v)  6 
 
 
= (u + v)[(u + v)^{2}  3uv]  3[(u + v)^{2}  2uv] + 5(u + v)  6 
 
 
= [(u + v)^{3}  3(u + v)^{2} + 5(u + v)  6]  3uv(u + v  2) 
 
 
= 
1
2

(u + v  2)[(u  v)^{2} + (u  1)^{2} + (v  1)^{2} + 4] . 

Since the second factor is positive, we must have that u + v = 2.
Solution 3. [N. Horeczky] Since x^{3}  3x^{2} + 5x = (x  1)^{3} + 2(x  1) + 3 is an increasing function of x
(since x  1 is increasing), the equation
x^{3}  3x^{2} + 5x  17 = 0 has exactly one real solution, namely
x = u. But


= v^{3}  3v^{2} + 5v + 11 
 
 
= (v  2)^{3} + 3(v  2)^{2} + 5(v  2) + 17 
 
 
= [(2  v)^{3}  3(2  v)^{2} + 5(2  v)  17] . 

Thus x = 2
 v satisfies x
^{3}  3x
^{2} + 5x
 17 = 0,
so that 2
 v = u and u + v = 2.
Comment. One can see also that each of the two given equations
has a unique real root by noting that the sum of the squares of
the roots, given by the cofficients, is equal to 3^{2}  2×5 = 1.
Solution 4. [P. Shi] Let m and n be determined by
u + v = 2m and u  v = 2n. Then u = m + n, v = m  n,
u^{2} + v^{2} = 2m^{2} + 2n^{2}, u^{2}  v^{2} = 4mn, u^{2} + uv + v^{2} = 3m^{2} + n^{2}, u^{2}  uv + v^{2} = m^{2} + 3n^{2}, u^{3} + v^{3} = 2m(m^{2} + 3n^{2}) and u^{3}  v^{3} = 2n(3m^{2} + n^{2}).
Adding the equations yields that


= (u^{3} + v^{3})  3(u^{2} + v^{2}) + 5(u + v)  6 
 
 
= 2m^{3} + 6mn^{2}  6m^{2}  6n^{2} + 10m  6 
 
 
= 6(m  1)n^{2} + 2(m^{3}  3m^{2} + 5m  3) 
 
 
= 6(m  1)n^{2} + 2(m  1)(m^{2}  2m + 3) 
 
 
= 2(m  1)[3n^{2} + (m  1)^{2} + 2] . 

Hence m = 1.

323.

Alfred, Bertha and Cedric are going from their
home to the country fair, a distance of 62 km. They have
a motorcycle with sidecar that together accommodates at most 2 people
and that can travel at a maximum speed of 50 km/hr. Each can
walk at a maximum speed of 5 km/hr. Is it possible for all
three to cover the 62 km distance within 3 hours?
Solution 1. We consider the following regime. A begins
by walking while B and C set off on the motorcycle for a time
of t
_{1} hours. Then C dismounts from the motorcycle and
continues walking, while B drives back to pick up A for a
time of t
_{2} hours. Finally, B and A drive ahead until
they catch up with C, taking a time of t
_{3} hours. Suppose
that all of this takes t = t
_{1} + t
_{2} + t
_{3} hours.
The distance from the starting point to the point where B
picks up A is given by
5(t_{1} + t_{2}) = 50(t_{1}  t_{2}) 

km, and the distance from the point where B drops off C
until the point where they all meet again is given by
5(t_{2} + t_{3}) = 50(t_{3}  t_{2}) . 

Hence 45t
_{3} = 45t
_{1} = 55t
_{2}, so that t
_{1} = t
_{3} = (11/9)t
_{2}
and so t = (31/9)t
_{2} and
t_{1} = 
11
31

t , t_{2} = 
9
31

t , t_{3} = 
11
31

t . 

The total distance travelled in the t hours is equal to
50t_{1} + 5(t_{2} + t_{3}) = 
650
31



kilometers. In three hours, they can travel
1950/31 = 60 + (90/31) > 62 kilometers in this way, so that
all will reach the fair before the three hours are up.
Solution 2. Follow the same regime as in Solution 1.
Let d be the distance from the start to the point where
B drops C in kilometers. The total time for
for C to go from start to finish,
namely
hours, and we wish this to be no greater than 3. The condition is that
d
³ 470/9.
The time for B to return to pick up A after dropping C is
9d/550 hours in which he covers a distance of 9d/11 km.
The total distance travelled by the motorcycle is
d + 
9d
11

+ (62  
2d
11

) = 
18d + 682
11



km, and this is covered in
hours. To get A and B to their destinations on time, we wish this
to not exceed 3; the condition for this is that d
£ 484/9.
Thus, we can get everyone to the fair on time if
Thus, if d = 53, for example, we can achieve the desired journey.
Solution 3. [D. Dziabenko] Suppose that B and C take
the motorcycle for exactly 47/45 hours while A walks after them.
After 47/45 hours, B leaves C to walk the rest of the way,
while B drives back to pick up A. C reaches the destination
in exactly

62  (47/45)50
5

+ 
47
45

= 3 

hours. Since B and A start and finish at the same time,
it suffices to check that that B reaches the fair on time.
When B drops C off, B and A are 47 km apart. It takes
B 47/55 hours to return to pick up A. At this point, they
are now
62  5 
æ è


47
45

+ 
47
55


ö ø

= 62  47 
æ è


20
99


ö ø

= 
5198
99



km from the fair, which they will reach in a further
hours. The total travel time for A and B is


+ 
47
55

+ 
1
50


é ë

62  5 
æ è


47
45

+ 
47
55


ö ø


ù û


 
 
= 
9 ×47
10 ×5


é ë


1
9

+ 
1
11


ù û

+ 
31
25

= 
517 + 423 + 682
550

= 
811
275



hours. This is less than three hours.

324.

The base of a pyramid ABCDV is a rectangle
ABCD with AB  = a, BC  = b and
VA  = VB  = VC  = VD  = c. Determine the area of the intersection
of the pyramid and the plane parallel to the edge VA that
contains the diagonal BD.
Solution 1. A dilation with centre C and factor 1/2 takes
A to S, the centre of the square and V to M, the midpoint
of VC. The plane of intersection is the plane that contains
triangle BMD. Since BM is a median of triangle
BVC with sides c, c, b, its length is equal to
^{1}/
_{2}Ö{2b
^{2} + c
^{2}} [why?]; similarly,
DM
 =
^{1}/
_{2}Ö{2a
^{2} + c
^{2}}.
Also,
BD
 =
Ö{a
^{2} + b
^{2}}.
Let
q =
ÐBMD. Then,
by the law of Cosines,
cosq = 
c^{2}  a^{2}  b^{2}
 Ö 
2b^{2} + c^{2}

 Ö 
2a^{2} + c^{2}


, 

whence
sinq = 
 Ö 
4c^{2}(a^{2} + b^{2})  (a^{2}  b^{2})^{2}

 Ö 
2b^{2} + c^{2}

 Ö 
2a^{2} + c^{2}


. 

The required area is

1
2

BM DM sinq = 
1
8

 Ö

4c^{2}(a^{2} + b^{2})  (a^{2}  b^{2})^{2}

. 

Comment. One can also use Heron's formula to get the
area of the triangle, but this is more labourious. Another method
is to calculate (1/2)BD MN , where
N is the foot of the perpendicular from M to BD, Note
that, when a ¹ b, N is not the same as S [do you see why?].
If d = BD  and x = SN  and, say
MB  £ MD , then
MN ^{2} = MB ^{2}  
æ è


d
2

 x 
ö ø

2

= MD ^{2}  
æ è


d
2

+ x 
ö ø

2



whence
x = 
MD ^{2}  MB ^{2}
2d

. 

If follows that
MN ^{2} = 
2a^{2}b^{2}  a^{4}  b^{4} + 4a^{2}c^{2} + 4b^{2}c^{2}
16(a^{2} + b^{2})

. 
