13th Balkan Mathematical Olympiad Problems 1996

13th Balkan Mathematical Olympiad Problems 1996

A1.  Let d be the distance between the circumcenter and the centroid of a triangle. Let R be its circumradius and r the radius of its inscribed circle. Show that d² ≤ R(R - 2r).

A2.  p > 5 is prime. A = {p - n² where n² < p}. Show that we can find integers a and b in A such that a > 1 and a divides b.

A3.  In a convex pentagon consider the five lines joining a vertex to the midpoint of the opposite side. Show that if four of these lines pass through a point, then so does the fifth.

A4.  Can we find a subset X of {1, 2, 3, ... , 2¹⁹⁹⁶-1} with at most 2012 elements such that 1 and 2¹⁹⁹⁶-1 belong to X and every element of X except 1 is the sum of two distinct elements of X or twice an element of X? 

Solutions

13th Balkan 1996 Problem 1

Let d be the distance between the circumcenter and the centroid of a triangle. Let R be its circumradius and r the radius of its inscribed circle. Show that d² ≤ R(R - 2r).

Solution

Use vectors. Take the origin as the circumcenter O. Let A be the vector OA, B be the vector OB and C be the vector OC. Then the vector OG = (A + B + C)/3. Hence 9 OG² = A² + B² + C² + 2(A.B + B.C + C.A). Now A² = B² = C² = R². We have A.B = 2R² cos 2C = 2R² - AB² = 2R² - c² (with the usual notation AB = c, BC = a, CA = b). Hence 9 OG² = 9 R² - (a² + b² + c²). By the arithmetic geometric mean inequality we have (a² + b² + c²) ≥ 3 (abc)^2/3. Hence R² - OG² ≥ (abc)^2/3/3.

Now the area of ABC = area IAB + area IBC + area ICA = rc/2 + ra/2 + rb/2, where I is the incenter. It is also 1/2 ab sin C. But considering the triangle ABO, we have c = 2R sin C and hence 2 area ABC = r(a + b + c) = abc/(2R). So 2rR = abc/(a + b + c) ≤ 1/3 (abc)^2/3. Hence R² - OG² ≥ 2rR.

13th Balkan 1996 Problem 2

5 is prime. A = {p - n² where n² < p}. Show that we can find integers a and b in A such that a > 1 and a divides b.

Solution

Take m² to be the largest square less than p. Let p - m² = k. Then p - (m - k)² = (m² + k) - (m² - 2km + k²) = k(2m - k + 1). So p - m² divides p - |m - k|² (*).

Note that we cannot have k = m because then m would divide p. We cannot have (m - k) = m because then p = m². We cannot have (k - m) = m, because then p = m² + 2m = m(m+2). So (*) provides suitable a and b unless k = 1.

If k = 1, them m must be even (otherwise p = m² + 1 would be even), hence p - (m-1)² = 2m divides p - 1 = m². 

13th Balkan 1996 Problem 3

In a convex pentagon consider the five lines joining a vertex to the midpoint of the opposite side. Show that if four of these lines pass through a point, then so does the fifth.

Solution

Let the pentagon be ABCDE. Use vectors. Take the origin O to be the point at which the lines from each of A, B, C D to the midpoint of the opposite side meet. Take the vector OA to be a, OB to be b etc. The side opposite A is CD. Its midpoint is (c + d)2. So we have a x (c + d) = 0 and hence a x c = d x a. Similarly, b x d = e x b, c x e = a x c, d x a = b x d. Hence e x b = b x d = d x a = a x c = c x e, which shows that the line from E also passes through the origin. 

13th Balkan 1996 Problem 4

Can we find a subset X of {1, 2, 3, ... , 2¹⁹⁹⁶-1} with at most 2012 elements such that 1 and 2¹⁹⁹⁶-1 belong to X and every element of X except 1 is the sum of two distinct elements of X or twice an element of X?

Solution Answer: yes.

In the sequence 2ⁿ - 1, 2ⁿ⁺¹ - 2, 2ⁿ⁺² - 2², ... , 2²ⁿ - 2ⁿ, 2²ⁿ - 1 each number is double its predecessor, except for the last number which is the sum of its predecessor and the first number. Thus by adding n elements we can extend a sequence from 2ⁿ - 1 to 2²ⁿ - 1 whilst preserving the property that each element (except the first) is the sum of two previous elements or is double a previous elements.

Thus starting with 1 = 2¹ - 1, we can construct a sequence of 1 + 2 + 3 + 5 + 9 + 17 + 33 + 65 + 129 = 264 terms which has the property and last term 2²⁵⁶ - 1. With 243 doublings, we can extend the sequence to last term 2⁴⁹⁹ - 2²⁴³. The sequence already includes the 6 terms 2²⁴³ - 2¹¹⁵ (115 doublings after 2¹²⁸ - 1), 2¹¹⁵ - 2⁵¹, 2⁵¹ - 2¹⁹, 2¹⁹ - 2³, 2³ - 2, and 1, so by successively adding these we reach the 264 + 243 + 6 = 513rd term 2⁴⁹⁹ - 1.

We can now extend this to 2⁹⁹⁸ - 1 with a further 500 terms, then to 2¹⁹⁹⁶ - 1 with a further 999 terms, giving 2012 in all.

[To be completely explicit the set is:
1, 2=1+1, 3=1+2, 6=3+3, 12= 6+6, 15=12+3, 30=15+15, 60=30+30, 120=60+60, 240=120+120
255=240+15, 2⁸⁺ⁿ-2ⁿ=(2^8+n-1-2^n-1)+(2^8+n-1-2^n-1), n = 1, 2, ... , 8
2¹⁶-1=(2¹⁶-2⁸)+(2⁸-1), 2¹⁶⁺ⁿ-2ⁿ=(2^16+n-1-2^n-1)+(2^16+n-1-2^n-1), n = 1, 2, ... , 16
2³²-1=(2³²-2¹⁶)+(2¹⁶-1), 2³²⁺ⁿ-2ⁿ=(2^32+n-1-2^n-1)+(2^32+n-1-2^n-1), n = 1, 2, ... , 32
2⁶⁴-1=(2⁶⁴-2³²)+(2³²-1), 2⁶⁴⁺ⁿ-2ⁿ=(2^64+n-1-2^n-1)+(2^64+n-1-2^n-1), n = 1, 2, ... , 64
2¹²⁸-1=(2¹²⁸-2⁶⁴)+(2⁶⁴-1), 2¹²⁸⁺ⁿ-2ⁿ=(2^128+n-1-2^n-1)+(2^128+n-1-2^n-1), n = 1, 2, ... , 128
2²⁵⁶-1=(2²⁵⁶-2¹²⁸)+(2¹²⁸-1), 2²⁵⁶⁺ⁿ-2ⁿ=(2^256+n-1-2^n-1)+(2^256+n-1-2^n-1), n = 1, 2, ... , 243
2⁴⁹⁹-2¹¹⁵=(2⁴⁹⁹-2²⁴³)+(2²⁴³-2¹¹⁵), 2⁴⁹⁹-2⁵¹)=(2⁴⁹⁹-2¹¹⁵)+(2¹¹⁵-2⁵¹), 2⁴⁹⁹-2¹⁹=(2⁴⁹⁹-2⁵¹)+(2⁵¹-2¹⁹), 2⁴⁹⁹-2³=(2⁴⁹⁹-2¹⁹)+(2¹⁹-2³), 2⁴⁹⁹-2=(2⁴⁹⁹-2³)+6, 2⁴⁹⁹-1=(2⁴⁹⁹-2)+1
2⁴⁹⁹⁺ⁿ-2ⁿ=(2^499+n-1-2^n-1)+(2^499+n-1-2^n-1), n = 1, ... , 499
2⁹⁹⁸-1=(2⁹⁹⁸-2⁴⁹⁹)+(2⁴⁹⁹-1), 2⁹⁹⁸⁺ⁿ-2ⁿ=(2^998+n-1-2^n-1)+(2^998+n-1-2^n-1), n = 1, 2, ... , 998
2¹⁹⁹⁶-1=(2¹⁹⁹⁶-2⁹⁹⁸)+(2⁹⁹⁸-1) ].