A1. Solve the following equations for x, y and z:
x + y + z = a; x2 + y2 + z2 = b2; xy = z2
.
What conditions must a and b satisfy for x, y and z to be distinct positive numbers?
A2. Let a, b, c be the sides of a triangle and A its area. Prove that:
a2 + b2 + c2 ≥ 4√3 A
When do we have equality?
A3. Solve the equation cosnx - sinnx = 1, where n is a natural number.
B1. P is inside the triangle ABC. PA intersects BC in D, PB intersects AC in E, and PC intersects AB in F. Prove that at least one of AP/PD, BP/PE, CP/PF does not exceed 2, and at least one is not less than 2.
B2. Construct the triangle ABC, given the lengths AC=b, AB=c and the acute ∠AMB = α, where M is the midpoint of BC. Prove that the construction is possible if and only if
b tan(α/2) ≤ c < b.
When does equality hold?
B3. Given 3 non-collinear points A, B, C and a plane p not parallel to ABC and such that A, B, C are all on the same side of p. Take three arbitrary points A', B', C' in p. Let A'', B'', C'' be the midpoints of AA', BB', CC' respectively, and let O be the centroid of A'', B'', C''. What is the locus of O as A', B', C' vary?
Solutions
Problem A1
Solve the following equations for x, y and z:
x + y + z = a; x2 + y2 + z2 = b2; xy = z2
What conditions must a and b satisfy for x, y and z to be distinct positive numbers?
Solution
A routine slog gives z = (a2 - b2)/2a, x and y = (a2 + b2)/4a ± √(10a2b2 - 3a4 - 3b4)/4a.
A little care is needed with the conditions. Clearly x, y, z positive implies a > 0, and then z positive implies |b| < a. The expression under the root must be positive. It helps if you notice that it factorizes as (3a2 - b2)(3b2 - a2). The second factor is positive because |b| < a, so the first factor must also be positive and hence a < √3 |b|. These conditions are also sufficient to ensure that x and y are distinct, but then z must also be distinct because z2 = xy.
Problem A2
Let a, b, c be the sides of a triangle and A its area. Prove that:
a2 + b2 + c2 ≥ 4√3 A
When do we have equality?
Solution
One approach is a routine slog from Heron's formula. The inequality is quickly shown to be equivalent to a2b2 + b2c2 + c2a2 ≤ a4 + b4 + c4, which is true since a2b2 ≤ (a4 + b4)/2. We get equality iff the triangle is equilateral.
Another approach is to take an altitude lying inside the triangle. If it has length h and divides the base into lengths r and s, then we quickly find that the inequality is equivalent to (h - (r + s)√3/2)2 + (r - s)2 ≥ 0, which is true. We have equality iff r = s and h = (r + s)√3/2, which means the triangle is equilateral.
A third solution is due to Jonathan Mizrahi (somewhat adapted):
We have b2 + c2 >= 2bc with equality iff b = c. Also for any angle x in the range 0o to 180o we have 2bc ≥ 2bc sin(X + 30o) with equality iff X = 60o. So taking X to be the angle between the sides b and c (we cannot call it A because A is already used to mean the area in this question!) we have that b2 + c2 ≥ bc sin(X + 30o) with equality iff the triangle is equilateral. Now 2 sin(X + 30o) = √3 sin X + cos X, so using the cosine rule a2 = b2 + c2 - 2bc cos X, we get the required inequality.
Problem A3
Solve the equation cosnx - sinnx = 1, where n is a natural number.
Solution
Since cos2x + sin2x = 1, we cannot have solutions with n not 2 and 0 < |cos x|, |sin x| < 1. Nor can we have solutions with n=2, because the sign is wrong. So the only solutions have sin x = 0 or cos x = 0, and these are: x = multiple of π, and n even; x even multiple of π and n odd; x = even multiple of π + 3π/2 and n odd.
Problem B1
P is inside the triangle ABC. PA intersects BC in D, PB intersects AC in E, and PC intersects AB in F. Prove that at least one of AP/PD, BP/PE, CP/PF does not exceed 2, and at least one is not less than 2.
Solution
Take lines through the centroid parallel to the sides of the triangle. The result is then obvious.
Problem B2
Construct the triangle ABC, given the lengths AC = b, AB = c and the acute ∠AMB = α, where M is the midpoint of BC. Prove that the construction is possible if and only if
b tan(α/2) ≤ c < b.
When does equality hold?
Answer
Equality holds if ∠BAC = 90o and ∠ACB = α/2
Solution
The key is to take N so that A is the midpoint of NB, then ∠NCB = α.
The construction is as follows: take BN length 2AB. Take circle through B and N such that the ∠BPN = α for points P on the arc BN. Take A as the midpoint of BN and let the circle center A, radius AC cut the arc BN at C. In general there are two possibilities for C.
Let X be the intersection of the arc BN and the perpendicular to the segment BN through A. For the construction to be possible we require AX ≥ AC > AB. But AB/AX = tan α/2, so we get the condition in the question.
Equality corresponds to C = X and hence to ∠BAC = 90o and ∠ACB = α/2.
Problem B3
Given 3 non-collinear points A, B, C and a plane p not parallel to ABC and such that A, B, C are all on the same side of p. Take three arbitrary points A', B', C' in p. Let A'', B'', C'' be the midpoints of AA', BB', CC' respectively, and let O be the centroid of A'', B'', C''. What is the locus of O as A', B', C' vary?
Solution
The key is to notice that O is the midpoint of the segment joining the centroids of ABC and A'B'C'. The centroid of ABC is fixed, so the locus is just the plane parallel to p and midway between p and the centroid of ABC.
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.
, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 
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