6th Iberoamerican Mathematical Olympiad Problems 1991

A1.  The number 1 or the number -1 is assigned to each vertex of a cube. Then each face is given the product of its four vertices. What are the possible totals for the resulting 14 numbers?

A2.  Two perpendicular lines divide a square into four parts, three of which have area 1. Show that the fourth part also has area 1.

A3.  f is a function defined on all reals in the interval [0, 1] and satisfies f(0) = 0, f(x/3) = f(x)/2, f(1 - x) = 1 - f(x). Find f(18/1991).

B1.  Find a number N with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of N.

B2.  Let p(m, n) be the polynomial 2m² - 6mn + 5n². The range of p is the set of all integers k such that k = p(m, n) for some integers m, n. Find which members of {1, 2, ... , 100} are in the range of p. Show that if h and k are in the range of p, then so is hk.

B3.  Given three non-collinear points M, N, H show how to construct a triangle which has H as orthocenter and M and N as the midpoints of two sides. 

Solutions

Problem 1

The number 1 or the number -1 is assigned to each vertex of a cube. Then each face is given the product of its four vertices. What are the possible totals for the resulting 14 numbers?

Solution

Answer: 14, 6, 2, -2, -6, -10.

If every vertex is 1, we get 14 and that is clearly the highest possible total. The lowest possible total cannot be lower than -14, but we cannot even achieve that because if all the vertices are -1, then all the faces are 1.

If we change a vertex, then we also change three faces. If the vertex and the three faces are all initially the same, then we make a change of ±8. If three are of one kind and one the opposite, then we make a change of ±4. If two are of one kind and two the opposite, then we make no change. Thus any sequence of changes must take us to 14 + 4n for some integer n. But we have already shown that the total is greater than -14 and at most 14, so the only possibilities are -10, -6, -2, 2, 6, 10 and 14.

We show that 10 is not possible. If more than 2 vertices are -1, then the vertex total is at most 2, there are only 6 faces, so the total is less than 10. If all vertices are 1, then the total is 14. If all but one vertex is 1, then the total is 6. So the only possibility for 10 is just two vertices -1. But however we choose any two vertices, there is always a face containing only one of them, so at least one face is -1, so the face total is at most 4 and the vertex total is 4, so the total is less than 10. The other totals are possible, for example:

14: all vertices 1

6: one vertex -1, rest 1

2: three vertices of one face -1, rest 1

-2: all vertices -1

-6: all vertices but one -1

-10: two opposite corners 1, rest -1 

Problem 
4

Find a number N with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of N.

Solution

Answer: 35964

There are 4.3 = 12 numbers with a given digit of n in the units place. Similarly, there are 12 with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit numbers is 12.111 (a + b + c + d + e), where n = abcde. So 8668a = 332b + 1232c + 1322d + 1331e. We can easily see that a = 1 is too small and a = 4 is too big, so a = 2 or 3. Obviously e must be even. 0 is too small, so e = 2, 4, 6 or 8. Working mod 11, we see that 0 = 2b + 2d, so b + d = 11. Working mod 7, we see that 2a = 3b + 6d + e. Using the mod 11 result, b = 2, d = 9 or b = 3, d = 8 or b = 4, d = 7 or b = 5, d = 6 or b = 6, d = 5 or b = 7, d = 4 or b = 8, d = 3 or b = 9, d = 2. Putting each of these into the mod 7 result gives 2a - e = 4, 1, 5, 2, 6, 3, 0, 4 mod 7. So putting a = 2 and remembering that e must be 2, 4, 6, 8 and that all digits must be different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for c. Similarly putting a = 4, gives a, b, d, e = 3, 4, 7, 8 or 3, 5, 6, 4 as the only possibilities. Checking, we find the solution above and no others. 

Problem 5

Let p(m, n) be the polynomial 2m² - 6mn + 5n². The range of p is the set of all integers k such that k = p(m, n) for some integers m, n. Find which members of {1, 2, ... , 100} are in the range of p. Show that if h and k are in the range of p, then so is hk.

Answer

1,2,4,5,8,9,10,13,16,17,18, 20,25,26,29,32,34,36,37, 40,41,45,49,50,52,53,58, 61,64,65,68,72,73,74, 80,81,82,85,89,90,97,98,100

Solution

We have p(m,n) = (m-2n)² + (m-n)², so p(2a-b,a-b) = a² + b². Hence the range of p is just the sums of two squares.

(a² + b²)(c² + d²) = (ac - bd)² + (ad + bc)², which establishes that if h and k are in the range, then so is hk.

Thanks to Johann Peter Gustav

Problem B3

Given three non-collinear points M, N, H show how to construct a triangle which has H as orthocenter and M and N as the midpoints of two sides.

Solution

Take H' so that M is the midpoint of HH'. The circle diameter NH' meets the line through H perpendicular to MN in two points (in general), either of which we may take as A. Then B is the reflection of A in M, and C is the reflection of A in N.

To see that this works, note that M is the midpoint of HH' and AB, so AHBH' is a parallelogram. Hence AH' is parallel to BH and hence perpendicular to AC. In other words ∠NAH' = 90^o, so A lies on the circle diameter NH'. MN is parallel to BC, so A lies on the perpendicular to MN through H.

Thanks to Johann Peter Gustav Lejeune Dirichlet