A1. ABC is a triangle. O is the circumcenter, D is the midpoint of AB, and E is the centroid of ACD. Prove that OE is perpendicular to CD iff AB = AC.
A2. The reals w, x, y, z all lie between -π/2 and π/2 and satisfy sin w + sin x + sin y + sin z = 1, cos 2w + cos 2x + cos 2y + cos 2z ≥ 10/3. Prove that they are all non-negative and at most π/6.
A3. Can we find an integer N such that if a and b are integers which are equally spaced either side of N/2 (so that N/2 - a = b - N/2), then exactly one of a, b can be written as 19m + 85n for some positive integers m, n?
A4. There are 1985 people in a room. Each speaks at most 5 languages. Given any three people, at least two of them have a language in common. Prove that there is a language spoken by at least 200 people in the room.
Solutions
2nd Balkan 1985 Problem 1
ABC is a triangle. O is the circumcenter, D is the midpoint of AB, and E is the centroid of ACD. Prove that OE is perpendicular to CD iff AB = AC.
Solution
Use vectors. Take O as the origin. Let OA = A, OB = B, OC = C. Then OD = A/2 + B/2 and the midpoint of AD is 3/4 A + 1/4 B. Hence OE = 2/3 (3/4 A + 1/4 B) + 1/3 C = A/2 + B/2 + C/3, whilst CD = OD - OC = A/2 + B/2 - C. So OE is perpendicular to CD iff (A + B - 2C).(3A + B + 2C) = 0. Expanding and using A2 = B2 = C2, this becomes A.(B - C) = 0, which holds iff OA is perpendicular to BC. Hence result.
Let K be the midpoint of CD, L the midpoint of AC and M the midpoint of BC. Let AM meet CD at G. G is the centroid, so GD = CD/3. K is the midpoint of CD, so KG = CD/6 and KG/DG = 1/2. E is the centroid of ACD, so KE/AE = 1/2. Hence GE is parallel to AD. But OD is perpendicular to AD, so it must be perpendicular to GE. In other words, G lies on the altitude from E in the triangle ODE.
If AB = AC, then O must lie on AM. So GO is the same as AO and is perpendicular to BC and hence to DL. In other words G lies on the altidue from O in the triangle ODE. Thus G is the orthocenter of ODE, so it must lie on the third altitude and hence OE is perpendicular to CD.
Conversely, if OE is perpendicular to CD (or equivalently to DG), then G lies on the altitude from D in the triangle ODE. Hence again it is the orthocenter and lies on the altitude from O. In other words, OG is perpendicular to DE (or DL) and hence to BC. OM is also perpendicular to BC (since O is the circumcenter), so G lies on OM. In other words, O, G, M are collinear. But A, G, M are collinear, so A lies on OM. In other words, A lies on the perpendicular bisector of BC, so AB = AC.
2nd Balkan 1985 Problem 2
The reals w, x, y, z all lie between -π/2 and π/2 and satisfy sin w + sin x + sin y + sin z = 1, cos 2w + cos 2x + cos 2y + cos 2z ≥ 10/3. Prove that they are all non-negative and at most π/6.
Solution
cos 2k = 1 - 2 sin2k. So put a = sin w, b = sin x, c = sin y, d = sin z and we have a + b + c + d = 1, a2 + b2 + c2 + d2 ≤ 1/3. Hence (a - 1/6)2 + (b - 1/6)2 + (c - 1/6)2 + (d - 1/6)2 = a2 + b2 + c2 + d2 - (a + b + c + d)/3 + 1/9 <= 1/9. Hence |a|, |b|, |c|, |d| ≤ 1/6 + 1/3 = 1/2.
Suppose one of a, b, c, d is negative. wlog we may take a < 0. Then b + c + d > 1, b2 + c2 + d2 < 1/3. Hence (b - 1/3)2 + (c - 1/3)2 + (d - 1/3)2 = (b2 + c2 + d2) - 2(b + c + d)/3 + 1/3 < 0. Contradiction. So 0 ≤ a, b, c, d ≤ 1/2. Hence 0 ≤ w, x, y, z ≤ π/6.
2nd Balkan 1985 Problem 3
Can we find an integer N such that if a and b are integers which are equally spaced either side of N/2 (so that N/2 - a = b - N/2), then exactly one of a, b can be written as 19m + 85n for some positive integers m, n?
Solution: Answer: yes, N = 1719,
If 1615 = 19·85 = 19m + 85n with m, n positive, then 19 divides n, so n ≥ 19, so 19m + 85n > 85n ≥ 85·19 = 1615. Contradiction. Hence 1615 cannot be expressed in this form. But all numbers > 1615 can. Adding 1 = 9·19 - 2·85 to 9·85 repeatedly gives: 1615 + 1 = 9·19+17·85, ... , 1615 + 9 = 81·19 + 1·85. Then 1615 + 10 = 90·19 - 85 = 5·19 + 18·85. Continuing to add 1 gives: 1616 + 11 = 14·19 + 16·85, ... , 1615 + 18 = 77·19 + 2·85. Finally, 1615 + 19 = 1·19 + 19·85. Then adding multiples of 19 gives all higher numbers. Obviously 104 = 1·19 + 1·85 and no integer < 104 has the desired form.
So N ≥ 1720 is ruled out, because both 104 and N - 104 have the desired form and are equally spaced about N/2. Similarly, N ≤ 1718 is ruled out, because neither 1615 nor N - 1615 have the desired form. So the only possible candidate for N is 1719.
We have already shown that for N = 1719, just one of a, 1719 - a has the desired form for a ≤ 104 and a ≥ 1615. So it remains to consider 104 < a < 1615. Using 1 = 9,19 - 2·85, we may write a = 19m + 85n with m and n integers (but possibly negative). Then adding and subtracting multiples of 19·85 we may take 1 ≤ m ≤ 85. Then 104 < a < 1615 implies that -17 ≤ n ≤ 18. If 1 ≤ n ≤ 18, then a has the desired form. If not then -17 ≤ n ≤ 0, so 1719 - a = (86·19 + 85) - (19m + 85n) = 19(86 - m) + 85(1 - n), which has the desired form. Thus in any case at least one of a, 1719 - a has the desired form.
Suppose both have the desired form, so that a = 19m + 85n, 1719 - a = 19m' + 85n', with m, n, m', n' all positive integers. Then, adding, 1719 = 19(m + m') + 85(n + n'), so 1615 = 19(m + m' - 1) + 85(n + n' - 1), where m + m' - 1 and n + n' - 1 are positive integers. But we have already shown that is impossible. So a and 1719 - a cannot both have the desired form. So 1719 meets the criterion.
2nd Balkan 1985 Problem 4
There are 1985 people in a room. Each speaks at most 5 languages. Given any three people, at least two of them have a language in common. Prove that there is a language spoken by at least 200 people in the room.
Solution
Take any person. He speaks at most 5 languages. If everyone in the room speaks at least one of those languages but not more than 199 people speak each language, then there are at most 1 + 5·198 = 991 people in the room. But there are 1985 people, so we must be able to find someone who speaks none of the 5 languages. He too speaks at most 5 languages, giving a total of 10 between the two people selected. Now we are told that every other person speaks at least one of these 10 languages. So if no more than 199 people speak each language, then there are at most 2 + 10·198 = 1982 people in the room. But there are 1985 people, so one of the 10 languages must be spoken by 200 or more people.
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