12th International Mathematical Olympiad 1970 Problems & Solutions

A1.  M is any point on the side AB of the triangle ABC. r, r₁, r₂ are the radii of the circles inscribed in ABC, AMC, BMC. q is the radius of the circle on the opposite side of AB to C, touching the three sides of AB and the extensions of CA and CB. Similarly, q₁ and q₂. Prove that r₁r₂q = rq₁q₂.

A2.  We have 0 ≤ x_i < b for i = 0, 1, ... , n and x_n > 0, x_n-1 > 0. If a > b, and x_nx_n-1...x₀ represents the number A base a and B base b, whilst x_n-1x_n-2...x₀ represents the number A' base a and B' base b, prove that A'B < AB'.

A3.  The real numbers a₀, a₁, a₂, ... satisfy 1 = a₀ ≤ a₁ ≤ a₂ ≤ ... . b₁, b₂, b₃, ... are defined by b_n = ∑_1≤k≤n (1 - a_k-1/a_k)/√a_k. (a)  Prove that 0 ≤ b_n < 2.

(b)  Given c satisfying 0 ≤ c < 2, prove that we can find a_n so that b_n > c for all sufficiently large n.

B1.  Find all positive integers n such that the set {n, n+1, n+2, n+3, n+4, n+5} can be partitioned into two subsets so that the product of the numbers in each subset is equal.

B2.  In the tetrahedron ABCD, angle BDC = 90^o and the foot of the perpendicular from D to ABC is the intersection of the altitudes of ABC. Prove that:       (AB + BC + CA)² ≤ 6(AD² + BD² + CD²).

When do we have equality?

B3.  Given 100 coplanar points, no 3 collinear, prove that at most 70% of the triangles formed by the points have all angles acute. 

Solutions

Problem A1

M is any point on the side AB of the triangle ABC. r, r₁, r₂ are the radii of the circles inscribed in ABC, AMC, BMC. q is the radius of the circle on the opposite side of AB to C, touching the three sides of AB and the extensions of CA and CB. Similarly, q₁ and q₂. Prove that r₁r₂q = rq₁q₂.

 

Solution

We need an expression for r/q. There are two expressions, one in terms of angles and the other in terms of sides. The latter is a poor choice, because it is both harder to derive and less useful. So we derive the angle expression.

Let I be the center of the in-circle for ABC and X the center of the external circle for ABC. I is the intersection of the two angle bisectors from A and B, so c = r (cot A/2 + cot B/2). The X lies on the bisector of the external angle, so angle XAB is 90^o - A/2. Similarly, angle XBA is 90^o - B/2, so c = q (tan A/2 + tan B/2). Hence r/q = (tan A/2 + tan B/2)/(cot A/2 + cot B/2) = tan A/2 tan B/2.

Applying this to the other two triangles, we get r₁/q₁ = tan A/2 tan CMA/2, r₂/q₂ = tan B/2 tan CMB/2. But CMB/2 = 90^o - CMA/2, so tan CMB/2 = 1/tan CMA/2. Hence result. 

Problem A2

We have 0 ≤ x_i < b for i = 0, 1, ... , n and x_n > 0, x_n-1 > 0. If a > b, and x_nx_n-1...x₀ represents the number A base a and B base b, whilst x_n-1x_n-2...x₀ represents the number A' base a and B' base b, prove that A'B < AB'.

 

Solution

We have aⁿb^m > bⁿa^m for n > m. Hence aⁿB' > bⁿA'. Adding aⁿbⁿ to both sides gives aⁿB > bⁿA. Hence x_naⁿB > x_nbⁿA. But x_naⁿ = A - A' and x_nbⁿ = B - B', so (A - A')B > (B - B')A. Hence result.

Note that the only purpose of requiring x_n-1 > 0 is to prevent A' and B' being zero. 

Problem A3

The real numbers a₀, a₁, a₂, ... satisfy 1 = a₀ <= a₁ ≤ a₂ <= ... . b₁, b₂, b₃, ... are defined by b_n = sum for k = 1 to n of (1 - a_k-1/a_k)/√a_k.

(a)  Prove that 0 ≤ b_n < 2.

(b)  Given c satisfying 0 ≤ c < 2, prove that we can find a_n so that b_n > c for all sufficiently large n.

 

Solution

(a)  Each term of the sum is non-negative, so b_n is non-negative. Let c_k = √a_k. Then the kth term = (1 - a_k-1/a_k)/√a_k = c_k-1²/c_k (1/a_k-1 - 1/a_k) = c_k-1²/c_k (1/c_k-1 + 1/c_k)(1/c_k-1 - 1/c_k). But c_k-1²/c_k (1/c_k-1 + 1/c_k) ≤ 2, so the kth term ≤ 2(1/c_k-1 - 1/c_k). Hence b_n <= 2 - 2/c_n < 2.

(b)  Let c_k = d^k, where d is a constant > 1, which we will choose later. Then the kth term is (1 - 1/d²)1/d^k, so b_n = (1 - 1/d²)(1 - 1/dⁿ⁺¹)/(1 - 1/d) = (1 + 1/d)(1 - 1/dⁿ⁺¹). Now take d sufficiently close to 1 that 1 + 1/d > c, and then take n sufficiently large so that (1 + 1/d)(1 - 1/dⁿ⁺¹) > c.

Problem B1

Find all positive integers n such that the set {n, n+1, n+2, n+3, n+4, n+5} can be partitioned into two subsets so that the product of the numbers in each subset is equal.

 

Solution

The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be {1, 2, 3, 4, 5, 6}, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets. 

Problem B2

In the tetrahedron ABCD, angle BDC = 90^o and the foot of the perpendicular from D to ABC is the intersection of the altitudes of ABC. Prove that:

      (AB + BC + CA)² ≤ 6(AD² + BD² + CD²).

When do we have equality?

 

Solution

The first step is to show that angles ADB and ADC are also 90^o. Let H be the intersection of the altitudes of ABC and let CH meet AB at X. Planes CED and ABC are perpendicular and AB is perpendicular to the line of intersection CE. Hence AB is perpendicular to the plane CDE and hence to ED. So BD² = DE² + BE². Also CB² = CE² + BE². Subtracting: CB² - BD² = CE² - DE². But CB² - BD² = CD², so CE² = CD² + DE², so angle CDE = 90^o. But angle CDB = 90^o, so CD is perpendicular to the plane DAB, and hence angle CDA = 90^o. Similarly, angle ADB = 90^o.

Hence AB² + BC² + CA² = 2(DA² + DB² + DC²). But now we are done, because Cauchy's inequality gives (AB + BC + CA)² ≤ 3(AB² + BC² + CA²).

We have equality iff we have equality in Cauchy's inequality, which means AB = BC = CA. 

Problem B3

Given 100 coplanar points, no 3 collinear, prove that at most 70% of the triangles formed by the points have all angles acute.

 

Solution

Improved and corrected by Gerhard Wöginger, Technical University Graz

At most 3 of the triangles formed by 4 points can be acute. It follows that at most 7 out of the 10 triangles formed by any 5 points can be acute. For given 10 points, the maximum no. of acute triangles is: the no. of subsets of 4 points x 3/the no. of subsets of 4 points containing 3 given points. The total no. of triangles is the same expression with the first 3 replaced by 4. Hence at most 3/4 of the 10, or 7.5, can be acute, and hence at most 7 can be acute.

The same argument now extends the result to 100 points. The maximum number of acute triangles formed by 100 points is: the no. of subsets of 5 points x 7/the no. of subsets of 5 points containing 3 given points. The total no. of triangles is the same expression with 7 replaced by 10. Hence at most 7/10 of the triangles are acute.

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.