1st Brazilian Mathematical Olympiad Problems 1979

1st Brazilian Mathematical Olympiad Problems 1979

1.  Show that if a < b are in the interval [0, π/2] then a - sin a < b - sin b. Is this true for a < b in the interval [π, 3π/2]?

2.  The remainder on dividing the polynomial p(x) by x² - (a+b)x + ab (where a and b are unequal) is mx + n. Find the coefficients m, n in terms of a, b. Find m, n for the case p(x) = x²⁰⁰ divided by x² - x - 2 and show that they are integral.

3.  The vertex C of the triangle ABC is allowed to vary along a line parallel to AB. Find the locus of the orthocenter.

4.  Show that the number of positive integer solutions to x₁ + 2³x₂ + 3³x₃ + ... + 10³x₁₀ = 3025 (*) equals the number of non-negative integer solutions to the equation y₁ + 2³y₂ + 3³y₃ + ... + 10³y₁₀ = 0. Hence show that (*) has a unique solution in positive integers and find it.

5.(i)   ABCD is a square with side 1. M is the midpoint of AB, and N is the midpoint of BC. The lines CM and DN meet at I. Find the area of the triangle CIN. (ii)   The midpoints of the sides AB, BC, CD, DA of the parallelogram ABCD are M, N, P, Q respectively. Each midpoint is joined to the two vertices not on its side. Show that the area outside the resulting 8-pointed star is 2/5 the area of the parallelogram.

(iii)   ABC is a triangle with CA = CB and centroid G. Show that the area of AGB is 1/3 of the area of ABC.

(iv)   Is (ii) true for all convex quadrilaterals ABCD? 

Solutions

Problem 1

Show that if a < b are in the interval [0, π/2] then a - sin a < b - sin b. Is this true for a < b in the interval [π, 3π/2]?

Answer

Yes.

Solution

We have b = (b+a)/2 + (b-a)/2, a = (b+a)/2 - (b-a)/2, so sin b - sin a = (sin((b+a)/2) cos((b-a)/2) + sin((b-a)/2) cos((b+a)/2) ) - (sin((b+a)/2) cos((b-a)/2) - sin((b-a)/2) cos((b+a)/2) ) = 2 sin((b-a)/2) cos((b+a)/2) ≤ 2 sin((b-a)/2) < 2 (b-a)/2 = b-a.

The second case is trivial because both x and -sin x are increasing in the interval [π, 3π/2].

Problem 2

The remainder on dividing the polynomial p(x) by x² - (a+b)x + ab (where a and b are unequal) is mx + n. Find the coefficients m, n in terms of a, b. Find m, n for the case p(x) = x²⁰⁰ divided by x² - x - 2 and show that they are integral.

Answer

m = (2²⁰⁰-1)/3, n = (2²⁰⁰+2)/3.

Solution

p(x) = q(x)(x-a)(x-b) + mx + n. So putting x = a, b we get p(a) = ma + n, p(b) = mb + n. Solving, m = (p(a)-p(b))/(a-b), n = (p(b)a-p(a)b)/(a-b).

In the case given a = 2, b = -1, so m = (2²⁰⁰-1)/3, n = (2²⁰⁰+2)/3. Note that 2 = -1 mod 3, so 2²⁰⁰ = 1 mod 3 and hence 2²⁰⁰ - 1 is a multiple of 3, so m is integral. n = m+1, so n is also integral.

Problem 3

The vertex C of the triangle ABC is allowed to vary along a line parallel to AB. Find the locus of the orthocenter.

Answer

A parabola.

Solution

Take axes so that A is (-a,0), B is (a,0) and C is (k,b). Then the orthocenter lies on the line x = k. The line AC has gradient b/(k+a), so the perpendicular has gradient -(k+a)/b. Hence the altitude from B has equation y + (x-a)(k+a)/b = 0. So the intersection is x = k, y = -(k-a)(k+a)/b. So the locus is all or part of the parabola by = a²-x². But we can get an orthocenter with any x-coordinate (by taking C to have the same x-coordinate), so we can get all points on the parabola.

Problem 4

Show that the number of positive integer solutions to x₁ + 2³x₂ + 3³x₃ + ... + 10³x₁₀ = 3025 (*) equals the number of non-negative integer solutions to the equation y₁ + 2³y₂ + 3³y₃ + ... + 10³y₁₀ = 0. Hence show that (*) has a unique solution in positive integers and find it.

Answer

x₁ = x₂ = ... = x₁₀ = 1.

Solution

We have 1³ + 2³ + ... + 10³ = 3025. Now x_i is a positive integer solution to (*) iff y_i = x_i - 1 are all non-negative and satisfy (y₁+1) + 2³(y₂+1) + 3³(y₃+1) + ... + 10³(y₁₀+1) = 3025 and hence y₁ + 2³y₂ + 3³y₃ + ... + 10³y₁₀ = 0. But that clearly has the unique solution y_i = 0, so the unique solution to (*) is x_i = 1.

Problem 5

(i)   ABCD is a square with side 1. M is the midpoint of AB, and N is the midpoint of BC. The lines CM and DN meet at I. Find the area of the triangle CIN.

(ii)   The midpoints of the sides AB, BC, CD, DA of the parallelogram ABCD are M, N, P, Q respectively. Each midpoint is joined to the two vertices not on its side. Show that the area outside the resulting 8-pointed star is 2/5 the area of the parallelogram.

(iii)   ABC is a triangle with CA = CB and centroid G. Show that the area of AGB is 1/3 of the area of ABC.

(iv)   Is (ii) true for all convex quadrilaterals ABCD?

Answer

(i) 1/20, (iv) no.

Solution

(i) CIN is similar to DCN, so area CIN = (CN/DN)² area DCN = ( (1/2)/(√5/2) )² 1/4 = 1/20.

(ii) If we stretch the plane parallel to one of the squares sides then all areas are increased by the same factor and hence the ratio (area CIN/area ABCD) is unchanged. If we now shear parallel to one of the sides, areas are unchanged, so the ratio (area CIN/area ABCD) remains 1/20. Thus the result holds for parallelograms. The area outside the star is made up of 8 small triangles, each area 1/20, so it is 2/5.

(iii) Let the median be AM. Then AGB and ABC have the same base AB, so area AGB/area ABC = GM/AM = 1/3. [This is trivial, but they wanted to give a hint for part (iv)]

(iv) If we take A and B close together, then we get the same figure as in (iii) and so the ratio tends to 1/3. Hence the 2/5 result is not true for all convex quadrilaterals.