2nd Asian Pacific Mathematics Olympiad 1990 Problems

2nd Asian Pacific Mathematics Olympiad 1990 Problems

A1.  Given θ in the range (0, π) how many (incongruent) triangles ABC have angle A = θ, BC = 1, and the following four points concyclic: A, the centroid, the midpoint of AB and the midpoint of AC?

A2.  x₁, ... , x_n are positive reals. s_k is the sum of all products of k of the x_i (for example, if n = 3, s₁ = x₁ + x₂ + x₃, s₂ = x₁x₂ + x₂x₃ + x₃x₁, s₃ = x₁x₂x₃). Show that s_ks_n-k ≥ (nCk)² s_n for 0 < k < n.

A3.  A triangle ABC has base AB = 1 and the altitude from C length h. What is the maximum possible product of the three altitudes? For which triangles is it achieved?

A4.  A graph with n > 1 points satisfies the following conditions: (1) no point has edges to all the other points, (2) there are no triangles, (3) given any two points A, B such that there is no edge AB, there is exactly one point C such that there are edges AC and BC. Prove that each point has the same number of edges. Find the smallest possible n.

A5.  Show that for any n ≥ 6 we can find a convex hexagon which can be divided into n congruent triangles.

Solution

Given θ in the range (0, π) how many (incongruent) triangles ABC have angle A = θ, BC = 1, and the following four points concyclic: A, the centroid, the midpoint of AB and the midpoint of AC?

Solution

Answer: 1 for θ ≤ 60 deg. Otherwise none.

Let O be the circumcenter of ABC and R the circumradius, let M be the midpoint of BC, and let G be the centroid. We may regard A as free to move on the circumcircle, whilst O, B and C remain fixed. Let X be the point on MO such that MX/MO = 1/3. An expansion by a factor 3, center M, takes G to A and X to O, so G must lie on the circle center X radius R/3.

The circle on diameter OA contains the midpoints of AB and AC (since if Z is one of the midpoints OZ is perpendicular to the corresponding side). So if G also lies on this circle then angle OGA = 90 deg and hence angle MGO = 90 deg, so G must also lie on the circle diameter OM. Clearly the two circles for G either do not intersect in which case no triangle is possible which satisfies the condition or they intersect in one or two points. But if they intersect in two points, then corresponding triangles are obviously congruent (they just interchange B and C). So we have to find when the two circle intersect.

Let the circle center X meet the line OXM at P and Q with P on the same side of X as M. Now OM = R cos θ, so XM = 1/3 R cos θ < 1/3 R = XP, so M always lies inside PQ. Now XO = 2/3 OM = 1/3 R (2 cos θ), so XQ = 1/3 R > XO iff 2 cos θ < 1 or θ > π/3. Thus if θ > π/3, then XQ > XO and so the circle diameter OM lies entirely inside the circle center X radius R/3 and so they cannot intersect. If θ = π/3, then the circles touch at O, giving the equilateral triangle as a solution. If θ < π/3, then the circles intersect giving one incongruent triangle satisfying the condition.

x₁, ... , x_n are positive reals. s_k is the sum of all products of k of the x_i (for example, if n = 3, s₁ = x₁ + x₂ + x₃, s₂ = x₁x₂ + x₂x₃ + x₃x₁, s₃ = x₁x₂x₃). Show that s_ks_n-k ≥ (nCk)² s_n for 0 < k < n.

Solution

Each of s_k and s_n-khave nCk terms. So we may multiply out the product s_ks_n-k to get a sum of (nCk)² terms. We now apply the arithmetic/geometric mean result. The product of all the terms must be a power of s_n by symmetry and hence must be s_n to the power of (nCk)². So the geometric mean of the terms is just s_n. Hence result. 

A triangle ABC has base AB = 1 and the altitude from C length h. What is the maximum possible product of the three altitudes? For which triangles is it achieved?

Solution

Answer: for h ≤ 1/2, maximum product is h², achieved by a triangle with right-angle at C; for h > 1/2, the maximum product is h³/(h² + 1/4), achieved by the isosceles triangle (AC = BC).

Solution by David Krumm

Let AC = b, BC = a, let the altitude from A have length x and the altitude from B have length y. Then ax = by = h, so hxy = h³/ab. But h = a sin B and b/sin B = 1/sin C, so h = ab sin C and the product hxy = h² sin C.

The locus of possible positions for C is the line parallel to AB and a distance h from it. [Or strictly the pair of such lines.] If h ≤ 1/2, then there is a point on that line with angle ACB = 90 deg, so in this case we can obtain hxy = h² by taking angle ACB = 90 deg and that is clearly the best possible.

If h > 1/2, then there is no point on the line with angle ACB = 90 deg. Let L be the perpendicular bisector of AB and let L meet the locus at C. Then C is the point on the locus with the angle C a maximum. For if D is any other point of the line then the circumcircle of ABD also passes through the corresponding point D' on the other side of C and hence C lies inside the circumcircle. If L meets the circumcircle at C', then angle ADB = angle AC'B > angle ACB. Evidently sin C = 2 sin C/2 cos C/2 = h/(h² + 1/4), so the maximum value of hxy is h³/(h² + 1/4).

My original, less elegant, solution is as follows.

Take AP perpendicular to AB and length h. Take Q to be on the line parallel to AB through P so that BQ is perpendicular to AB. Then C must lie on the line PQ (or on the corresponding line on the other side of AB). Let a(A) be the length of the altitude from A to BC and a(B) the length of the altitude from B to AC. If C maximises the product h a(A) a(B), then it must lie on the segment PQ, for if angle ABC is obtuse, then both a(A) and a(B) are shorter than for ABQ. Similarly if BAC is obtuse. So suppose PC = x with 0 ≤ x ≤ 1. Then AC = √(x² + h²), so a(B) = h/√(x² + h²). Similarly, a(A) = h/√( (1-x)² + h²). So we wish to minimise f(x) = (x² + h²)( (1-x)² + h²) = x⁴ - 2x³ + (2h² + 1)x² - 2h²x + h⁴ + h². We have f '(x) = 2(2x-1)(x² - x + h²), which has roots x = 1/2, 1/2 ± √(1/4 - h²).

Thus for h >= 1/2, the minimum is at x = 1/2, in which case CA = CB. For h < 1/2, the minimum is at x = 1/2 ± √(1/4 - h²). But if M is the midpoint of AB and D is the point on AB with AD = 1/2 ± √(1/4 - h²), then DM = √(1/4 - h²). But DC = h, and angle CDM = 90, so MC = 1/2 and hence angle ACB = 90.

A graph with n points satisfies the following conditions: (1) no point has edges to all the other points, (2) there are no triangles, (3) given any two points A, B such that there is no edge AB, there is exactly one point C such that there are edges AC and BC. Prove that each point has the same number of edges. Find the smallest possible n.

Solution

Answer: 5.

We say A and B are joined if there is an edge AB. For any point X we write deg X for the number of points joined to X. Take any point A. Suppose deg A = m. So there are m points B₁, B₂, ... , B_m joined to A. No B_i, B_j can be joined for i ≠ j, by (2), and a point C ≠ A cannot be joined to B_i and B_j for i ≠ j, by (3). Hence there are deg B_i - 1 points C_ij joined to B_i and all the C_ij are distinct.

Now the only points that can be joined to C_ij, apart from B_i, are other C_hk, for by (3) any point of the graph is connected to A by a path of length 1 or 2. But C_ij cannot be joined to C_ik, by (2), and it cannot be joined to two distinct points C_kh and C_kh' by (3), so it is joined to at most one point C_kh for each k ≠ i. But by (3) there must be a point X joined to both B_k and C_ij (for k ≠ i), and the only points joined to B_k are A and C_kh. Hence C_ij must be joined to at least one point C_kh for each k ≠ i. Hence deg C_ij = m.

But now if we started with B_i instead of A and repeated the whole argument we would establish that deg B_i is the same as the deg C_hk, where C_hk is one of the points joined to C_i1. Thus all the points have the same degree.

Suppose the degree of each point is m. Then with the notation above there is 1 point A, m points B_i and m(m-1) points C_jk or m² + 1 in all. So n = m² + 1. The smallest possible m is 1, but that does not yield a valid graph because if does not satisfy (1). The next smallest possibility is m = 2, giving 5 points. It is easy to check that the pentagon satisfies all the conditions. 

Show that for any n ≥ 6 we can find a convex hexagon which can be divided into n congruent triangles.

Solution

We use an isosceles trianglea as the unit. The diagram shows n = 4 and n = 5. We can get any n ≥ 4 by adding additional rhombi in the middle.