Solutions
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318.
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Solve for integers x, y, z the system
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1 = x + y + z = x3 + y3 + z2 . |
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[Note that the exponent of z on the right is 2, not 3.]
Solution 1. Substituting the first equation into the
second yields that
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x3 + y3 + [1 - (x + y)]2 = 1 |
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which holds if and only if
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= (x + y)(x2 - xy + y2) + (x + y)2 - 2(x + y) |
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= (x + y)(x2 - xy + y2 + x + y - 2) |
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| = (1/2)(x + y)[(x - y)2 + (x + 1)2 + (y + 1)2 - 6] . |
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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
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(x, y, z) = (t, -t, 1), (1, 0, 0), (0, 1, 0), (-2, -3, 6),(-3, -2, 6), (-2, 0, 3), (0, -2, 3) |
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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 x2 + (1 - y)x + (y2 + y- 2) = 0.
The discriminant of the quadratic in x is
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-3y2 - 6y + 9 = -3(y + 1)2 + 12 , |
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which is nonnegative when |y + 1 | £ 4.
Checking out the possibilities leads to the solution.
Solution 3.
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| = (x + y)[(x + y)2 - 3xy] = (1 - z)[(1 - z)2 - 3xy] , |
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whence either z = 1 or 3xy = (1 - 2z + z2) - (1 + z) = z(z - 3).
The former case yields (x, y, z) = (x, -x, 1) while the latter
yields
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x + y = 1 - z xy = |
1
3
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z(z - 3) . |
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Thus, we must have that z º 0 (mod 3) and that x, y
are roots of the quadratic equation
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t2 - (1 - z)t + |
z(z - 3)
3
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= 0 . |
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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.
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319.
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Suppose that a, b, c, x are real numbers for which
abc ¹ 0 and
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xb + (1 - x)c
a
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= |
xc + (1 - x)a
b
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= |
xa + (1 - x)b
c
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. |
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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?]:
|
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x(a + b + c) + (1 - x)(a + b + c)
a + b + c
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= 1 . |
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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
x3(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
|
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xa + (1 - x)b
-(a + b)
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= |
x(-a-b) + (1 - x)a
b
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|
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Þ xab + (1 - x)b2 = x(a + b)2 - (1 - x)(a2 + ab) |
|
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Þ (1 - x)(a2 + ab + b2) = x(a2 + ab + b2) . |
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Since 2(a2 + ab + b2) = (a + b)2 + a2 + b2 > 0, 1 - x = x
and x = 1/2. But in this case, the equations become
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b+c
2a
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= |
c+ a
2b
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= |
a+b
2c
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|
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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
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xb + (1 - x)c = da ; xc + (1 - x)a = db ; xa + (1 - x)b = dc . |
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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
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xb2 + (1 - x)bc = xac + (1 - x)a2 |
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or
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x(a2 + b2) - x(a + b)c = a2 - bc . |
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Similarly
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x(c2 + a2) - x(c + a)b = b2 - ca ; |
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x(b2 + c2) - x(b + c)a = c2 - ab . |
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Adding these three equations together leads to
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2x[(a - b)2 + (b - c)2 + (c - a)2] = (a - b)2 + (b - c)2 + (c - a)2 . |
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Hence, either a = b = c or x = 1/2.
If x = 1/2, then for some constant k,
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b+c
a
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= |
c + a
b
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= |
a + b
c
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= k , |
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whence
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-ka + b + c = a - kb + c = a + b - kc = 0 . |
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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.
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320.
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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
Solution 1. Since ÐLCM = 90° and
CL = CM, we have that ÐCLM = ÐCML = 45°.
Let ÐACB = 2q. Then ÐCAB = 45° -q and ÐCBA = 45° + q.
It follows that
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= (2R sinÐCAB)2 + (2R sinÐCBA)2 |
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= 4R2 (sin2 (45° - q) + sin2 (45° + q)) |
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| = 4R2 (sin2 (45° - q) + cos2 (45° - q)) = 4R2 . |
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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
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4R2 = |AC |2 + |CD |2 = |AC |2+ |BC |2 . |
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321.
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Determine all positive integers k for which
k1/(k-7) is an integer.
Solution. When k = 1, the number is an integer.
Suppose that 2 £ k £ 6. Then k - 7 < 0 and so
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0 < k1/(k-7) = 1/(k1/7-k) < 1 |
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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
101/3, which is not an integer [why?].
Suppose that k ³ 11. We establish by induction that
k < 2k-7. This is clearly true when k = 11. Suppose it
holds for k = m ³ 11. Then
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m + 1 < 2m-7 + 2m - 7 = 2(m+1) - 7 ; |
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the desired result follows by induction. Thus, when k ³ 11,
1 < k1/(k-7) < 2 and the number is not an integer.
Thus, the number is an integer if and only if k = 1, 8, 9.
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322.
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The real numbers u and v satisfy
and
Determine u + v.
Solution 1. The equations can be rewritten
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v3 - 3v2 + 5v - 3 = -14 . |
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These can be rewritten as
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(u - 1)3 + 2(u - 1) = 14 , |
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(v - 1)3 + 2(v - 1) = -14 . |
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Adding these equations yields that
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|
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= (u - 1)3 + (v - 1)3 + 2(u + v - 2) |
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| = (u + v - 2)[(u - 1)2 - (u - 1)(v - 1) + (v - 1)2 + 2] . |
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Since the quadratic t2 - st + s2 is always positive [why?], we
must have that u + v = 2.
Solution 2. Adding the two equations yields
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= (u3 + v3) - 3(u2 + v2) + 5(u + v) - 6 |
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= (u + v)[(u + v)2 - 3uv] - 3[(u + v)2 - 2uv] + 5(u + v) - 6 |
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= [(u + v)3 - 3(u + v)2 + 5(u + v) - 6] - 3uv(u + v - 2) |
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| = |
1
2
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(u + v - 2)[(u - v)2 + (u - 1)2 + (v - 1)2 + 4] . |
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Since the second factor is positive, we must have that u + v = 2.
Solution 3. [N. Horeczky] Since x3 - 3x2 + 5x = (x - 1)3 + 2(x - 1) + 3 is an increasing function of x
(since x - 1 is increasing), the equation
x3 - 3x2 + 5x - 17 = 0 has exactly one real solution, namely
x = u. But
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|
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| |
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= (v - 2)3 + 3(v - 2)2 + 5(v - 2) + 17 |
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| = -[(2 - v)3 - 3(2 - v)2 + 5(2 - v) - 17] . |
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Thus x = 2 - v satisfies x3 - 3x2 + 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 32 - 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,
u2 + v2 = 2m2 + 2n2, u2 - v2 = 4mn, u2 + uv + v2 = 3m2 + n2, u2 - uv + v2 = m2 + 3n2, u3 + v3 = 2m(m2 + 3n2) and u3 - v3 = 2n(3m2 + n2).
Adding the equations yields that
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= (u3 + v3) - 3(u2 + v2) + 5(u + v) - 6 |
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= 2m3 + 6mn2 - 6m2 - 6n2 + 10m - 6 |
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= 6(m - 1)n2 + 2(m3 - 3m2 + 5m - 3) |
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= 6(m - 1)n2 + 2(m - 1)(m2 - 2m + 3) |
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| = 2(m - 1)[3n2 + (m - 1)2 + 2] . |
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Hence m = 1.
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323.
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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 t1 hours. Then C dismounts from the motorcycle and
continues walking, while B drives back to pick up A for a
time of t2 hours. Finally, B and A drive ahead until
they catch up with C, taking a time of t3 hours. Suppose
that all of this takes t = t1 + t2 + t3 hours.
The distance from the starting point to the point where B
picks up A is given by
km, and the distance from the point where B drops off C
until the point where they all meet again is given by
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5(t2 + t3) = 50(t3 - t2) . |
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Hence 45t3 = 45t1 = 55t2, so that t1 = t3 = (11/9)t2
and so t = (31/9)t2 and
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t1 = |
11
31
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t , t2 = |
9
31
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t , t3 = |
11
31
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t . |
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The total distance travelled in the t hours is equal to
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50t1 + 5(t2 + t3) = |
650
31
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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
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d + |
9d
11
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+ (62 - |
2d
11
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) = |
18d + 682
11
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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
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62 - (47/45)50
5
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+ |
47
45
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= 3 |
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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
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62 - 5 |
æ
è
|
|
47
45
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+ |
47
55
|
|
ö
ø
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= 62 - 47 |
æ
è
|
|
20
99
|
|
ö
ø
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= |
5198
99
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|
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km from the fair, which they will reach in a further
hours. The total travel time for A and B is
|
|
|
+ |
47
55
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+ |
1
50
|
|
é
ë
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62 - 5 |
æ
è
|
|
47
45
|
+ |
47
55
|
|
ö
ø
|
|
ù
û
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|
| |
| |
| = |
9 ×47
10 ×5
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é
ë
|
|
1
9
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+ |
1
11
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ù
û
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+ |
31
25
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= |
517 + 423 + 682
550
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= |
811
275
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|
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hours. This is less than three hours.
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324.
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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
1/2Ö{2b2 + c2} [why?]; similarly,
|DM | = 1/2Ö{2a2 + c2}.
Also, |BD | = Ö{a2 + b2}.
Let q = ÐBMD. Then,
by the law of Cosines,
whence
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sinq = |
|
|
Ö
|
4c2(a2 + b2) - (a2 - b2)2
|
|
. |
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The required area is
|
|
1
2
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|BM ||DM |sinq = |
1
8
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|
Ö
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4c2(a2 + b2) - (a2 - b2)2
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. |
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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
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= |MD |2 - |
æ
è
|
|
d
2
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+ x |
ö
ø
|
2
|
|
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whence
If follows that
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|MN |2 = |
2a2b2 - a4 - b4 + 4a2c2 + 4b2c2
16(a2 + b2)
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. |
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