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# Solutions

318.
Solve for integers x, y, z the system
 1 = x + y + z = x3 + y3 + z2 .
[Note that the exponent of z on the right is 2, not 3.]
Solution 1. Substituting the first equation into the second yields that
 x3 + y3 + [1 - (x + y)]2 = 1
which holds if and only if
 0
 = (x + y)(x2 - xy + y2) + (x + y)2 - 2(x + y)
 = (x + y)(x2 - xy + y2 + 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 x2 + (1 - y)x + (y2 + y- 2) = 0. The discriminant of the quadratic in x is

 -3y2 - 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)(1 + z)
 = 1 - z2 = x3 + y3
 = (x + y)[(x + y)2 - 3xy] = (1 - z)[(1 - z)2 - 3xy] ,
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
 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
 t2 - (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 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

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

 Þ xab + (1 - x)b2 = x(a + b)2 - (1 - x)(a2 + ab)

 Þ (1 - x)(a2 + ab + b2) = x(a2 + ab + b2) .
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
 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

 xb2 + (1 - x)bc = xac + (1 - x)a2
or
 x(a2 + b2) - x(a + b)c = a2 - bc .
Similarly
 x(c2 + a2) - x(c + a)b = b2 - ca ;

 x(b2 + c2) - x(b + c)a = c2 - ab .
 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
 (2 - k)(a + b + c) = 0 .
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 = 4R2 .
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
 |BC |2 + |AC |2
 = (2R sinÐCAB)2 + (2R sinÐCBA)2
 = 4R2 (sin2 (45° - q) + sin2 (45° + q))
 = 4R2 (sin2 (45° - q) + cos2 (45° - q)) = 4R2 .

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

 4R2 = |AC |2 + |CD |2 = |AC |2+ |BC |2 .

321.
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
 0 < k1/(k-7) = 1/(k1/7-k) < 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 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

 m + 1 < 2m-7 + 2m - 7 = 2(m+1) - 7 ;
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.

322.
The real numbers u and v satisfy
 u3 - 3u2 + 5u - 17 = 0
and
 v3 - 3v2 + 5v + 11 = 0 .
Determine u + v.
Solution 1. The equations can be rewritten
 u3 - 3u2 + 5u - 3 = 14 ,

 v3 - 3v2 + 5v - 3 = -14 .
These can be rewritten as
 (u - 1)3 + 2(u - 1) = 14 ,

 (v - 1)3 + 2(v - 1) = -14 .
 0
 = (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 t2 - st + s2 is always positive [why?], we must have that u + v = 2.

Solution 2. Adding the two equations yields

 0
 = (u3 + v3) - 3(u2 + v2) + 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 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

 0
 = v3 - 3v2 + 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 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

 0
 = (u3 + v3) - 3(u2 + v2) + 5(u + v) - 6
 = 2m3 + 6mn2 - 6m2 - 6n2 + 10m - 6
 = 6(m - 1)n2 + 2(m3 - 3m2 + 5m - 3)
 = 6(m - 1)n2 + 2(m - 1)(m2 - 2m + 3)
 = 2(m - 1)[3n2 + (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 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

 5(t1 + t2) = 50(t1 - t2)
km, and the distance from the point where B drops off C until the point where they all meet again is given by
 5(t2 + t3) = 50(t3 - t2) .
Hence 45t3 = 45t1 = 55t2, so that t1 = t3 = (11/9)t2 and so t = (31/9)t2 and
 t1 = 11 31 t ,   t2 = 9 31 t ,   t3 = 11 31 t .
The total distance travelled in the t hours is equal to
 50t1 + 5(t2 + t3) = 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

 d 50 + 62 - d 5
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
 18d + 682 550
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
 470 9 £ d £ 484 9 .
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 æè 47 45 + 47 55 öø = 62 - 47 æè 20 99 öø = 5198 99
km from the fair, which they will reach in a further
 5198 99 ×50 = 2599 2475
hours. The total travel time for A and B is
 47 45
 + 47 55 + 1 50 éë 62 - 5 æè 47 45 + 47 55 öø ùû
 = 9 ×47 10 ×5 éë 1 9 + 1 11 ùû + 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 1/2Ö{2b2 + c2} [why?]; similarly, |DM | = 1/2Ö{2a2 + c2}. Also, |BD | = Ö{a2 + b2}. Let q = ÐBMD. Then, by the law of Cosines,
cosq =  c2 - a2 - b2

 Ö 2b2 + c2 Ö 2a2 + c2
,
whence
sinq =
 Ö 4c2(a2 + b2) - (a2 - b2)2

 Ö 2b2 + c2 Ö 2a2 + c2
.
The required area is
 1 2 |BM ||DM |sinq = 1 8 Ö 4c2(a2 + b2) - (a2 - b2)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 - æè d 2 - x öø 2 = |MD |2 - æè d 2 + x öø 2
whence
 x = |MD |2 - |MB |2 2d .
If follows that
 |MN |2 = 2a2b2 - a4 - b4 + 4a2c2 + 4b2c2 16(a2 + b2) .
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