14th Vietnamese Mathematical Olympiad 1976 Problems

14th Vietnamese Mathematical Olympiad 1976 Problems

A1.  Find all integer solutions to m^m+n = n¹², n^m+n = m³.

A2.  Find all triangles ABC such that (a cos A + b cos B + c cos C)/(a sin A + b sin B + c sin C) = (a + b + c)/9R, where, as usual, a, b, c are the lengths of sides BC, CA, AB and R is the circumradius.

A3.  P is a point inside the triangle ABC. The perpendicular distances from P to the three sides have product p. Show that p ≤ 8 S³/(27abc), where S = area ABC and a, b, c are the sides. Prove a similar result for a tetrahedron. 

B1.  Find all three digit integers abc = n, such that 2n/3 = a! b! c!  

B2.  L, L' are two skew lines in space and p is a plane not containing either line. M is a variable line parallel to p which meets L at X and L' at Y. Find the position of M which minimises the distance XY. L" is another fixed line. Find the line M which is also perpendicular to L". 

B3.  Show that 1/x₁ⁿ + 1/x₂ⁿ + ... + 1/x_kⁿ ≥ kⁿ⁺¹ for real numbers x_i with sum 1.  

Solution

14th VMO 1976

Problem A1

Find all integer solutions to m^m+n = n¹², n^m+n = m³.

Answer

(m,n) = (4,2)

Solution

Suppose a prime p divides n. Then from the first equation it must also divide m. So suppose pa is the highest power of p dividing m, and pb is the highest power of p dividing n. Then the first equation gives 12b = a(m+n), and the second gives 3a = b(m+n). Hence 36b = 3a(m+n) = b(m+n)², so m+n = 6 and a = 2b. Hence m = 4, n = 2.

Thanks to Suat Namli 

14th VMO 1976

Problem B1

Find all three digit integers abc = n, such that 2n/3 = a! b! c!

Answer

432

Solution

2n/3 < 1000 < 7!, so a, b, c ≤ 6. Hence n ≤ 666, so 2n/3 ≤ 444, so a ≤ 4. Also 6! > 444, so b, c ≤ 5.

Consider first the case a = 4. Then (2n/3)/a! ≤ 2·455/(3·24) < 13. So b!c! ≤ 12, so b, c ≤ 3. Also 2n is divisible by a!3, so n is divisible by 9. So {b,c} = {2,3}, so n = (3/2)4!3!2! = 432.

Now suppose a = 3. Then n = (3/2) 6 b! c! = 9 b! c!, so we need 300 < 9 b! c! < 355, so 33 < b!c! < 40. Hence b, c ≤ 4. But 4! < 33 and 2·4! > 40, so b, c ≤ 3. Hence b = c = 3. But (3/2)3!3!3! = 324 ≠ 333, so there are no solutions with a = 3.

Suppose a = 2. Then n = 3b!c!, so 200 < 3b!c! < 255. But 3·5! > 255 and 3·3!3! < 200, so at least one of b, c must be 4. If the other is x, then 200/72 ≤ x! ≤ 255/72, so x! = 3. Contradiction, so there are no solutions with a = 2.

Finally, suppose a = 1. Then n = (3/2)b!c!, so (2/3)100 ≤ b!c! ≤ (2/3)155, and so 67 ≤ b!c! ≤ 103. But 5! > 103 and 3!3! < 67, so one of b, c must be 4. If the other is x, then 2.7 < x! < 4.3, so x! = 3 or 4. Contradiction.

Thanks to Suat Namli