1. Solve the system of equations:
√3x
Now let z = u + vi; the system then reduces to the single equation
z +1
Let t denote the quantity inside the parentheses; then z = t±p circum-center O. Let G be the centroid of triangle ACD, let E be the midpoint of BG, and let F be the midpoint of AE. Prove that OF is perpendicular to BG if and only if OD is perpendicular to AC.
Solution: We identify points with their vectors originating from the circumcenter, so that A· B = A · C = A · D and A2 = B2 =
C2= D2. Now
(O− F ) · (B − G) = 1
2(A + E)· (B − G)
= 1
4[(2A + B + G)· (B − G)]
= 1
36[18A· B − 6A · (A + C + D) + 9B2− (A + C + D)2]
= 1
36[2A· D − 2C · D].
Therefore OF ⊥ BF if and only if OD ⊥ AC.
3. Determine, as a function of n, the number of permutations of the set {1, 2, . . . , n} such that no three of 1, 2, 3, 4 appear consecutively.
Solution: There are n! permutations in all. Of those, we exclude (n−2)! permutations for each arrangement of 1, 2, 3, 4 into an ordered triple and one remaining element, or 24(n− 2)! in all. However, we have twice excluded each of the 24(n− 3)! permutations in which all four of 1, 2, 3, 4 occur in a block. Thus the number of permutations of the desired form is n!− 24(n − 2)! + 24(n − 3)!.
4. Determine all functions f : N→ N satisfying (for all n ∈ N) f (n) + f (n + 1) = f (n + 2)f (n + 3)− 1996.
Solution: From the given equation, we deduce
f (n)− f(n + 2) = f(n + 3)[f(n + 2) − f(n + 4)].
If f (1) > f (3), then by induction, f (2m−1) > f(2m+1) for all m >
0, giving an infinite decreasing sequence f (1), f (3), . . . of positive integers, a contradiction. Hence f (1)≤ f(3), and similarly f(n) ≤ f (n + 2) for all n.
Now note that
0 = 1996 + f (n) + f (n + 1)− f(n + 2)f(n + 3)
≤ 1996 + f(n + 2) + f(n + 3) − f(n + 2)f(n + 3)
= 1997− [f(n + 2) − 1][f(n + 3) − 1].
In particular, either f (n + 2) = 1 or f (n + 3)≤ 1997, and vice versa.
The numbers f (2m + 1)− f(2m − 1) are either all zero or all posi-tive, and similarly for the numbers f (2m + 2)− f(2m). If they are both positive, eventually f (n + 2) and f (n + 3) both exceed 1997, a
5. Consider triangles ABC where BC = 1 and ∠BAC has a fixed measure α > π/3. Determine which such triangle minimizes the distance between the incenter and centroid of ABC, and compute this distance in terms of α.
Solution: If we fix B and C and force A to lie above the line BC, then A is constrained to an arc. The centroid of ABC is constrained to the image of that arc under a 1/3 homothety at the midpoint of BC. On the other hand, the incenter subtends an angle of (π + α)/2 at BC, so it is also constrained to lie on an arc, but its arc passes through B and C. Since the top of the incenter arc lies above the top of the centroid arc, the arcs cannot intersect (or else their circles would intersect four times). Moreover, if we dilate the centroid arc about the midpoint of BC so that its image is tangent to the incenter arc at its highest point, the image lies between the incenter arc and BC.
In other words, the distance from the incenter to the centroid is al-ways at least the corresponding distance for ABC isosceles. Hence we simply compute the distance in that case. The incenter makes an isosceles triangle of vertex angle (π + α)/2, so its altitude is 1/2 cot(π + α)/4. Meanwhile, the distance of the centroid to BC is 1/3 that of A to BC, or 1/6 cot(α/2). The desired distance is thus
1
2cotπ + α 4 −1
6cotα 2.
6. Let a, b, c, d be four nonnegative real numbers satisfying the condi-tion
2(ab + ac + ad + bc + bd + cd) + abc + abd + acd + bcd = 16.
Prove that
a + b + c + d≥2
3(ab + ac + ad + bc + bd + cd) and determine when equality occurs.
Solution: For i = 1, 2, 3, define si as the average of the products of the i-element subsets of{a, b, c, d}. Then we must show
3s2+ s3= 4⇒ s1≥ s2.
It suffices to prove the (unconstrained) homogeneous inequality 3s22s21+ s3s31≥ 4s32,
as then 3s2+ s3= 4 will imply (s1− s2)3+ 3(s31− s32)≥ 0.
We now recall two basic inequalities about symmetric means of non-negative real numbers. The first is Schur’s inequality:
3s31+ s3≥ 4s1s2, while the second,
s21≥ s2
is a case of Maclaurin’s inequality si+1i ≥ sii+1. These combine to prove the claim:
3s22s21+ s3s31≥ 3s22s21+s22s3
s1 ≥ 4s32.
Finally, for those who have only seen Schur’s inequality in three variables, note that in general any inequality involving s1, . . . , sk which holds for n ≥ k variables also holds for n + 1 variables, by replacing the variables x1, . . . , xn+1by the roots of the derivative of the polynomial (x− x1)· · · (x − xn+1).
2 1996 Regional Contests:
Problems and Solutions
2.1 Asian Pacific Mathematics Olympiad
1. Let ABCD be a quadrilateral with AB = BC = CD = DA. Let M N and P Q be two segments perpendicular to the diagonal BD and such that the distance between them is d > BD/2, with M ∈ AD, N ∈ DC, P ∈ AB, and Q ∈ BC. Show that the perimeter of the hexagon AM N CQP does not depend on the position of M N and P Q so long as the distance between them remains constant.
Solution: The lengths of AM, M N, N C are all linear in the dis-tance between the segments M N and AC; if this disdis-tance is h, ex-trapolating from the extremes M N = AC and M = N = D gives that
AM + M N + N C = AC +2AB− AC BD/2 h.
In particular, if the segments M N and P Q maintain constant total distance from AC, as they do if their distance remains constant, the total perimeter of the hexagon is constant.
2. Let m and n be positive integers such that n≤ m. Prove that 2nn!≤ (m + n)!
(m− n)! ≤ (m2+ m)n.
Solution: The quantity in the middle is (m+n)(m+n−1) · · · (m−
n + 1). If we pair off terms of the form (m + x) and (m + 1− x), we get products which do not exceed m(m + 1), since the function f (x) = (m + x)(m + 1− x) is a concave parabola with maximum at x = 1/2. From this the right inequality follows. For the left, we need only show (m + x)(m + 1− x) ≥ 2x for x ≤ n; this rearranges to (m− x)(m + 1 + x) ≥ 0, which holds because m ≥ n ≥ x.
3. Let P1, P2, P3, P4be four points on a circle, and let I1be the incenter of the triangle P2P3P4, I2be the incenter of the triangle P1P3P4, I3
be the incentre of the triangle P1P2P4, and I4be the incenter of the
triangle P1P2P3. Prove that I1, I2, I3 and I4 are the vertices of a rectangle.
Solution: Without loss of generality, assume P1, P2, P3, P4 occur on the circle in that order. Let M12, M23, M34, M41be the midpoints of arcs P1P2, P2P3, P3P4, P4P1, respectively. Then the line P3M1 is the angle bisector of ∠P2P3P1 and so passes through I4. Moreover, the triangle M12P2I4 is isosceles because
∠I4M12P2 = ∠P3P1P2
= π− 2∠P1P2I4− 2∠M12P2P1
= π− 2∠M12P2I4.
Hence the circle centered at M passing through P1and P2also passes through I4, and likewise through I3.
From this we determine that the angle bisector of ∠P3M12P4 is the perpendicular bisector of I3I4. On the other hand, this angle bisector passes through M34, so it is simply the line M12M34; by symmetry, it is also the perpendicular bisector of I1I2. We conclude that I1I2I3I4
is a parallelogram.
To show that I1I2I3I4is actually a rectangle, it now suffices to show that M12M34⊥ M23M41. To see this, simply note that the angle be-tween these lines is half the sum of the measures of the arcs M12M23
and M34M41, but these arcs clearly comprise half of the circle.
4. The National Marriage Council wishes to invite n couples to form 17 discussion groups under the following conditions:
(a) All members of the group must be of the same sex, i.e. they are either all male or all female.
(b) The difference in the size of any two groups is either 0 or 1.
(c) All groups have at least one member.
(d) Each person must belong to one and only one group.
Find all values of n, n ≤ 1996, for which this is possible. Justify your answer.
Solution: Clearly n ≥ 9 since each of 17 groups must contain at least one member. Suppose there are k groups of men and 17− k
groups of women; without loss of generality, we assume k≤ 8. If m is the minimum number of members in a group, then the number of men in the groups is at most k(m + 1), while the number of women is at least (k + 1)m. As there are the same number as men as women, we have k(m + 1) ≥ (k + 1)m, so m ≤ k ≤ 8, and the maximum number of couples is k(k + 1)≤ 72. In fact, any number of couples between 9 and 72 can be distributed: divide the men as evenly as possible into 8 groups, and divide the women as evenly as possible into 9 groups. Thus 9≤ n ≤ 72 is the set of acceptable numbers of couples.
5. Let a, b and c be the lengths of the sides of a triangle. Prove that
√a + b− c +√
b + c− a +√
c + a− b ≤√ a +√
b +√ c and determine when equality occurs.
Solution: By the triangle inequality, b + c− a and c + a − b are positive. For any positive x, y, we have
2(x + y)≥ x + y + 2√
xy = (√ x +√
y)2
by the AM-GM inequality, with equality for x = y. Substituting x = a + b− c, y = b + c − a, we get
√a + b− c +√
b + c− a ≤ 2√ a,
which added to the two analogous inequalities yields the desired result. Inequality holds for a + b− c = b + c − a = c + a − b, i.e.
a = b = c.