4th All Soviet Union Mathematical Olympiad 1970 Problems & Solutions

1.  Given a circle, diameter AB and a point C on AB, show how to construct two points X and Y on the circle such that (1) Y is the reflection of X in the line AB, (2) YC is perpendicular to XA.

2.  The product of three positive numbers is 1, their sum is greater than the sum of their inverses. Prove that just one of the numbers is greater than 1.

3.  What is the greatest number of sides of a convex polygon that can equal its longest diagonal?

4.  n is a 17 digit number. m is derived from n by taking its decimal digits in the reverse order. Show that at least one digit of n + m is even.

5.  A room is an equilateral triangle side 100 meters. It is subdivided into 100 rooms, all equilateral triangles with side 10 meters. Each interior wall between two rooms has a door. If you start inside one of the rooms and can only pass through each door once, show that you cannot visit more than 91 rooms. Suppose now the large triangle has side k and is divided into k² small triangles by lines parallel to its sides. A chain is a sequence of triangles, such that a triangle can only be included once and consecutive triangles have a common side. What is the largest possible number of triangles in a chain?

6.  Given 5 segments such that any 3 can be used to form a triangle. Show that at least one of the triangles is acute-angled.

7.  ABC is an acute-angled triangle. The angle bisector AD, the median BM and the altitude CH are concurrent. Prove that angle A is more than 45 degrees.

8.  Five n-digit binary numbers have the property that every two numbers have the same digits in just m places, but no place has the same digit in all five numbers. Show that 2/5 ≤ m/n ≤ 3/5.

9.  Show that given 200 integers you can always choose 100 with sum a multiple of 100.

10.  ABC is a triangle with incenter I. M is the midpoint of BC. IM meets the altitude AH at E. Show that AE = r, the radius of the inscribed circle.

11.  Given any positive integer n, show that we can find infinitely many integers m such that m has no zeros (when written as a decimal number) and the sum of the digits of m and mn is the same.

12.  Two congruent rectangles of area A intersect in eight points. Show that the area of the intersection is more than A/2.

13.  If the numbers from 11111 to 99999 are arranged in an arbitrary order show that the resulting 444445 digit number is not a power of 2.

14.  S is the set of all positive integers with n decimal digits or less and with an even digit sum. T is the set of all positive integers with n decimal digits or less and an odd digit sum. Show that the sum of the kth powers of the members of S equals the sum for T if 1 ≤ k < n.

15.  The vertices of a regular n-gon are colored (each vertex has only one color). Each color is applied to at least three vertices. The vertices of any given color form a regular polygon. Show that there are two colors which are applied to the same number of vertices. 

Solutions

Problem 1

Given a circle, diameter AB and a point C on AB, show how to construct two points X and Y on the circle such that (1) Y is the reflection of X in the line AB, (2) YC is perpendicular to XA.

Solution

AB is a diameter, so BX is perpendicular to AX and hence parallel to YC. YX is perpendicular to BC, so YC = YB. Hence X and Y lie on the perpendicular bisector of BC. 

Problem 2

The product of three positive numbers is 1, their sum is greater than the sum of their inverses. Prove that just one of the numbers is greater than 1.

Solution

The product of the numbers is 1, so they cannot all be greater than 1 or all less than 1. If all equalled 1, then the sum would not be greater than the sum of the inverses. So we must have either one or two greater than 1. Thus it is sufficient to show that we cannot have two of the numbers greater than 1.

Suppose that a, b > 1. Then since a + b + c > 1/a + 1/b + 1/c, we have a + b + 1/ab > 1/a + 1/b + ab, and hence (1 - 1/a)(1 - 1/b) > (a - 1)(b - 1). Dividing by (a - 1)(b -1) gives ab < 1. Contradiction. 

Problem 2

What is the greatest number of sides of a convex polygon that can equal its longest diagonal?

Solution

Answer: 2, except for the equilateral triangle.

It is easy to find two. Take the two sides to be AB and AC with angle BAC =60 deg, and take the other vertices on the minor arc of the circle center A radius AB between B and C.

Let the longest diagonal have length k. Suppose there are three sides with length k. Extend them (if necessary) so they meet at A, B, C. Suppose ∠A > 60^o. Take the vertices on side AB to be P, Q (where we may have P = A, or Q = B, or both). Take the vertices on side AC to be R, S (where we may have R = A, or S = C, or both). Then AQ ≥ k, AS ≥ k, so QS > k. Contradiction. Hence angle A ≤ 60^o. The same is true for ∠B and ∠C. Hence ∠A = ∠B = ∠C = 60^o. But now QS > k unless A = P = R. Similarly, B and C must be vertices of the convex polygon, so that it is just an equilateral triangle. 

Problem 4

n is a 17 digit number. m is derived from n by taking its decimal digits in the reverse order. Show that at least one digit of n + m is even.

Solution

Let the number be n with digits d₁d₂ ... d₁₇, so that the reversed number n' has digits d₁₇d₁₆ ... d₁. Let the digits of n + n' be a₀a₁ ... a₁₇, where a₀ may be zero. Let the carry forward when adding digits to get a_i be c_i-1, so that, in general, c_i + d_i + d_18-i = a_i + 10 c_i-1. Obviously c_i is 0 or 1.

Suppose all the digits a_i are odd (except that a₀ may be zero). Now c₉ + 2d₉ = a₉ + 10 c₈. Since a₉ is odd, c₉ must be 1. But if we consider c₁₀ + d₁₀ + d₈ = a₁₀ + 10 c₉, we see that since a₁₀ is odd it is at least 1 and hence d₈ + d₁₀ is at least 10. Hence there must be a non-zero carry c₉ in c₁₀ + d₁₀ + d₈ = a₁₀ + 10 c₉ irrespective of the value of c₁₀.

We can now iterate and conclude successively that c₁₂, c₁₄, c₁₆ must be non-zero.

Problem 6

Given 5 segments such that any 3 can be used to form a triangle. Show that at least one of the triangles is acute-angled.

Solution

Let the segments have length a ≤ b ≤ c ≤ d ≤ e. Then if all triangles are obtuse we have e² > c² + d², d² > b² + c², c² > a² + b². Adding e² > a² + 2b² + c² > a² + 3b². But e ≤ a + b, so e² ≤ a² + 2ab + b² ≤ a² + 3b². Contradiction.

Problem 7

ABC is an acute-angled triangle. The angle bisector AD, the median BM and the altitude CH are concurrent. Prove that angle A is more than 45 degrees.

Solution

We use Ceva's theorem. Since AD, BM, CH are concurrent, we have (BD/DC).(CM/MA).(AH/BH) = 1. But CM = MA and since AD is the angle bisector BD/DC = AB/AC, so (AB/AC).(AH/BH) = 1. Hence AH/AC = BH/AB < 1. So angle HAC > angle HCA. But angle AHC = 90 deg, so angle A > 45 deg. 

Problem 13

If the numbers from 11111 to 99999 are arranged in an arbitrary order show that the resulting 444445 digit number is not a power of 2.

Solution

Let the set of numbers be S. Define the function f on S as follows. Replace each digit i in n by 9-i for 0 < i < 9. This gives f(n). Then f( f(n) ) = n, so f is a bijection. The fixed points have only the digits 0 and 9 and so are all divisible by 9. The other points divide into pairs (n, f(n)) and the sum of each pair is divisible by 9. Hence the sum of all the numbers in S is divisible by 9.