4th Brazilian Mathematical Olympiad Problems 1982

4th Brazilian Mathematical Olympiad Problems 1982

1.  The angles of the triangle ABC satisfy ∠A/∠C = ∠B/∠A = 2. The incenter is O. K, L are the excenters of the excircles opposite B and A respectively. Show that triangles ABC and OKL are similar.

2.  Any positive integer n can be written in the form n = 2^b(2c+1). We call 2c+1 the odd part of n. Given an odd integer n > 0, define the sequence a₀, a₁, a₂, ... as follows: a₀ = 2ⁿ-1, a_k+1 is the odd part of 3a_k+1. Find a_n.
3.  S is a (k+1) x (k+1) array of lattice points. How many squares have their vertices in S?
4.  Three numbered tiles are arranged in a tray as shown:

Show that we cannot interchange the 1 and the 3 by a sequence of moves where we slide a tile to the adjacent vacant space.
5.  Show how to construct a line segment length (a⁴ + b⁴)^1/4 given segments length a and b.
6.  Five spheres of radius r are inside a right circular cone. Four of the spheres lie on the base of the cone. Each touches two of the others and the sloping sides of the cone. The fifth sphere touches each of the other four and also the sloping sides of the cone. Find the volume of the cone.

Solutions

Problem 1
The angles of the triangle ABC satisfy ∠A/∠C = ∠B/∠A = 2. The incenter is O. K, L are the excenters of the excircles opposite B and A respectively. Show that triangles ABC and OKL are similar.
Solution

∠AOC = 180^o - ∠A/2 - ∠C/2. But ∠KAO = ∠KCO = 90^o, so ∠AKC = ∠A/2 + ∠C/2. So considering triangle AKL, ∠ALK = 90^o - ∠A/2 - ∠C/2 = ∠B/2. Similarly, ∠ BLC = ∠B/2 + ∠C/2, so ∠BKL = ∠A/2.
Now we use the given facts that ∠C = ∠A/2 and ∠A = ∠B/2. So ∠OKL = ∠BKL (same angle) = ∠C, and ∠OLK = ∠ALK (same angle) = ∠A. Hence triangles OKL and BCA are similar.

Problem 2
Any positive integer n can be written in the form n = 2^b(2c+1). We call 2c+1 the odd part of n. Given an odd integer n > 0, define the sequence a₀, a₁, a₂, ... as follows: a₀ = 2ⁿ-1, a_k+1 is the odd part of 3a_k+1. Find a_n.
Answer
(3ⁿ - 1)/2
Solution
A simple induction shows that a_k = 3^k2^n-k - 1 for k ≤ n-1. It is certainly true for k = 0. Suppose it is true for k < n-1. Then 3a_k + 1 = 3^k+12^n-k - 2. Since n-k > 1, the odd part is 3^k+12^n-(k+1) - 1, so the result is true for k+1. That gets us as far as a_n-1 = 3^n-12 - 1. Now we want the odd part of 2(3ⁿ - 1). Certainly 3ⁿ - 1 is even. We have 3ⁿ - 1 = (-1)ⁿ - 1 = 2 mod 4 for n odd, so for n odd it is not divisible by 4. Hence for n odd we have a_n = (3ⁿ - 1)/2.

Problem 3
S is a (k+1) x (k+1) array of lattice points. How many squares have their vertices in S?
Answer
k(k+1)²(k+2)/12
Solution

The key is to consider how many squares have their vertices on the perimeter of given n+1 x n+1 array whose side are parallel to the sides of the array.
The diagram shows that there are n such squares. There are (k+1-n)² such arrays. So the total number of squares is k·1² + (k-1)2² + ... + 1·k² = ∑₁^k (k+1-i)i² = (k+1)k(k+1)(2k+1)/6 - k²(k+1)²/4 = k(k+1)²(k+2)/12.

Problem 4
Three numbered tiles are arranged in a tray as shown:
Show that we cannot interchange the 1 and the 3 by a sequence of moves where we slide a tile to the adjacent vacant space.
Solution
Write down the order of the tiles reading clockwise around the perimeter, starting at 1. We get 123 and no move changes that, so we will always get 123 after any sequence of moves. But the desired arrangement would give 132, so it is not possible.

Problem 5
Show how to construct a line segment length (a⁴ + b⁴)^1/4 given segments length a and b.
Solution

We show first how to get the square and the square root. Take ABC with ∠A = 90^o, altitude AD length a and BD = 1. Then by similar triangles BC/AB = AB/BD, so BC = AB²/BD = AB² = a² + 1. Hence CD = a².

Conversely, we can take BD = 1, CD = a and then construct A (as the intersection of the perpendicular at D and the circle diameter BC) length √a.
So now given a, b construct a², b². Then take a right-angled triangle with those lengths as its shorter sides and the hypoteneuse is √(a⁴ + b⁴). Finally, take the square root to get the required length.

Problem 6
Five spheres of radius r are inside a right circular cone. Four of the spheres lie on the base of the cone. Each touches two of the others and the sloping sides of the cone. The fifth sphere touches each of the other four and also the sloping sides of the cone. Find the volume of the cone.
Answer (1/3)πr³(2√2+1)³.
Solution

 

The left-hand diagram shows the four spheres on the base. Evidently AC = 2r√2. The right-hand diagram shows a vertical section through A, C and the center O of the top sphere. P is the apex of the cone and QR is a diameter of its base. Evidently PQR is similar to OAC and its sides are parallel and a distance r outside the corresponding sides of OAC.
Also AOC is congruent to ABC, so ∠AOC = 90^o. Hence ∠QPR = 90^o also and so OP = r√2. The altitude from O in AOC has length AC/2 = r√2. Hence the altitude from P in PQR has length OP + r√2 + r = (2√2+1)r. Thus radius of the cone's base is also (2√2+1)r and its volume is (1/3)πr³(2√2+1)³.