Solutions and comments.

79.

Let x₀, x₁, x₂ be three positive real numbers. A sequence { x_n } is defined, for n ³ 0 by

xn+3 = xn+2 + xn+1 + 1 xn  .

Determine all such sequences whose entries consist solely of positive integers.

Without loss of generality, we can assume that the smallest entry in the sequence is y. Then,

Suppose that y = 1. Then x £ y + z + 1 = z + 2 £ y + x + 3 = x + 4. Since x divides x and y + z + 1, it must divide their difference, which cannot exceed 4. Hence x = y + z + 1 or x must be one of 1, 2, 3, 4. Similarly, z = x + y + 1 or z must be one of 1, 2, 3, 4.

If x = y + z + 1 = z + 2, then x + y + 1 = z + 4 and so z divides 4. We get the periods

Suppose that y = 2. Then x £ y + z + 1 £ z + 3 £ x + 6, so that x must divide some number not exceeding 6. Similarly, z cannot exceed 6. If x = 2, then x + y + 1 = 5 and so z = 5; this yields the period (2, 2, 5, 4, 5, 2, 2, 1) already noted. If x = 3, then z must be 2, 3 or 6, and we obtain (3, 2, 3, 2, 3, 2, 3, 2); the possibilities z = 2, 6 do not work. If x = 4, z = 7, which does not work. If x = 5, then z = 2, 4 and we get a period already noted. If x = 6, then z = 3, which does not work.

Hence there are five possible cycles and the sequence can begin at any term in the cycle:

Solution 2. [O. Ivrii] We show that the sequence contains a term that does not exceed 2. Suppose that none of x₀, x₁ and x₂ is less than 3. Let k = max(x₁, x₂). Then, noting that k is an integer and that k ³ 3, we deduce that

Suppose, from some point on in the sequence, there is no term equal to 1. Then there are three consecutive terms a, 2, b. The previous term is (a+3)/b and the following term is (b+3)/a, so that a divides b + 3 and b divides a + 3. Since a - 3 £ b £ a + 3, b divides two numbers that differ by at most 6; similarly with a. Hence, neither a nor b exceeds 6. Testing out possibilities leads to (a, b) = (6, 3), (5, 2), (3, 3).

Finally, we suppose that the sequence has three consecutive terms a, 1, b and by similar arguments are led to the sequences obtained in the first solution.

Comment. C. Lau established that x_n x_n+4 = x_n+2 x_n+6 and thereby obtained the periodicity. R. Barrington Leigh showed that, if all terms of the sequence were at least equal to 2, then the sequence { y_n } defined by y_n = max(x_n, x_n+1) satisfies y_n+1 £ y_n. Since the same recursion defines the sequence ``going backwards'', we also have y_n-1 £ y_n, for all n. Hence { y_n } is a constant sequence, and so { x_n } is either constant or has period 2. It is straightforward to rule out the constant possibility. If the periodic segment of the sequence is (a, b), then b = (a + b + 1)/a or (a-1)(b-1) = 2 and we are led to the segment (2, 3). Otherwise, there is a 1 in the sequence and we can conclude as before.

80.

Prove that, for each positive integer n, the series

¥ å k=1  kn 2k

converges to twice an odd integer not less than (n+1)!.

For nonnegative integers n, let

Observe that S₀ = 1, S₁ = 2, S₂ = 6 and S₃ = 26. We prove by induction that, for each n ³ 0,

Solution 2. Define S_n as in the foregoing solution. Then, for n ³ 1,

For n ³ 1,

Solution 3. [of the first part using an idea of P. Gyrya] Let f(x) be a differentiable function and let D be the differentiation operator. Define the operator L by

It is straightforward to verify that

81.

Suppose that x ³ 1 and that x = ëx û+ { x }, where ëx û is the greatest integer not exceeding x and the fractional part { x } satisfies 0 £ x < 1. Define

f(x) = Ö ëx û + Ö {x } Öx   .

(a) Determine the smallest number z such that f(x) £ z for each x ³ 1.

(b) Let x₀ ³ 1 be given, and for n ³ 1, define x_n = f(x_n-1). Prove that lim_{n ® ¥} x_n exists.

(b) In determining the fate of { x_n }, note that after the first entry, the sequence lies in the interval [1, 2). So, without loss of generality, we may assume that 1 £ x₀ < 2. If x_n = 1, then each x_n = 1 and the limit is 1. For the rest, note that f(x) simplifies to (1 + Ö[(x-1)])/Öx on (1, 2). The key point now is to observe that there is exactly one value v between 1 and 2 for which f(v) = v, f(x) > x when 1 < x < v and f(x) < x when v < x < 2. Assume these facts for a moment. A derivative check reveals that f(x) is strictly increasing on (1, 2), so that for 1 < x < v, x < f(x) < f(v) = v, so that the iterates { x_n } constitute a bounded, increasing sequence when 1 < x₀ < v which must have a limit. (In fact, this limit must be a fixed point of f and so must be v.) A similar argument shows that, if v < x₀ < 2, then the sequence of iterates constitute a decreasing convergent sequence (with limit v).

It remains to show that a unique fixed point v exists. Let x = 1 + u with u > 0. Then it can be checked that f(x) = x if and only if 1 + 2Öu + u = 1 + 3u + 3u² + u³ or u⁵ + 6u⁴ + 13u³ + 12u² + 4u - 4 = 0. Since the left side is strictly increasing in u, takes the value -4 when u = 0 and the value 32 when u = 1, the equation is satified for exactly one value of u in (0, 1); now let v = 1 + u. The value of V turns out to be about 1.375, (Note that f(x) > x if and only if x < u.)

82.

(a) A regular pentagon has side length a and diagonal length b. Prove that

b2 a2 + a2 b2 = 3 .

(b) A regular heptagon (polygon with seven equal sides and seven equal angles) has diagonals of two different lengths. Let a be the length of a side, b be the length of a shorter diagonal and c be the length of a longer diagonal of a regular heptagon (so that a < b < c). Prove that:

a2 b2 + b2 c2 + c2 a2 = 6

and

b2 a2 + c2 b2 + a2 c2 = 5 .

(b) Let A, B, C, D, E be consecutive vertices of the regular heptagon. Let AB, AC and AD have respective lengths a, b, c, and let ÐBAC = q. Then q = p/7, the length of BC, of CD and of DE is a, the length of AE is c, ÐCAD = ÐDAE = q, since the angles are subtended by equal chords of the circumcircle of the heptagon, ÐADC = 2q, ÐADE = ÐAED = 3q and ÐACD = 4q. Triangles ABC and ACD can be glued together along BC and DC (with C on C) to form a triangle similar to DABC, whence

Solution 2. (b) [R. Barrington Leigh] Let the heptagon be ABCDEFG; let AD and BG intersect at P, and BF and CG intersect at Q. Observe that |PD | = |GE | = b, |AP | = c - b, |GP | = |DE | = a, |BP | = b - a, |GQ | = |AB | = a, |CQ | = c - a. From similarity of triangles, we obtain the following:

Solution 3. (b) [of the second result by J. Chui] Let the heptagon be ABCDEFG and q = p/7. Using the Law of Cosines in the indicated triangles ACD and ABC, we obtain the following:

Examining triangles ABC and ADE, we find that cosq = b/2a and cosq = (2c² - a²)/(2c²) = 1 - (a²/2c²), so that

Adding equations (1), (2), (3) yields

Solution 4. [of the second result by X. Jin] By considering isosceles triangles with side-base pairs (a, b), (c, a) and (b, c), we find that b² = 2a²(1 - cos5q), a² = 2c²(1 - cosq), c² = 2b²(1 - cos3q), where q = p/7. Then

Solution 5. (b) There is no loss of generality in assuming that the vertices of the heptagon are placed at the seventh roots of unity on the unit circle in the complex plane. Then z = cos(2p/7) + isin(2p/7) be the fundamental seventh root of unity. Then z⁷ = 1, 1 + z+ z² + ¼+ z⁶ = 0 and (z, z⁶), (z², z⁵), (z³,z⁴) are pairs of complex conjugates. We have that

Solution 6. (b) Suppose that the circumradius of the heptagon is 1. By considering isosceles triangles with base equal to the sides or diagonals of the heptagon and apex at the centre of the circumcircle, we see that

83.

Let \frak C be a circle with centre O and radius 1, and let \frak F be a closed convex region inside \frak C. Suppose from each point on \frak C, we can draw two rays tangent to \frak F meeting at an angle of 60^°. Describe \frak F.

Hence C = D = P and ABC is an equilateral triangle containing \frak F with all its sides tangent to \frak F. Since A is arbitrary, \frak F is contained within the intersection of all such triangles, namely the circle \frak D with centre O and radius 1/2, and every chord of the given circle tangent to \frak F has length Ö3. If \frak F were a proper subset of \frak D, there would be a point Q on the circumference of \frak D outside \frak F. and a tangent of \frak F separating \frak F from Q. This tangent chord would intersect the interior of \frak D and so be longer than Ö3, yielding a contradiction. Hence \frak F must be the circle \frak D.

84.

Let ABC be an acute-angled triangle, with a point H inside. Let U, V, W be respectively the reflected image of H with respect to axes BC, AC, AB. Prove that H is the orthocentre of DABC if and only if U, V, W lie on the circumcircle of DABC,

Now suppose that U, V, W lie on the circumcircle. Let \frak C₁, \frak C₂, \frak C₃ be the respective reflections of the circumcircle about the axes BC, CA, AB. These three circles intersect in the point H. If H¢ is the orthocentre of the triangle, then by the first part of the solution, the reflective image of H¢ about the three axes lies on the circumcircle, so that H¢ belongs to \frak C₁, \frak C₂, \frak C₃ and H = H¢ or else HH¢ is a common chord of the three circles. But the latter does not hold, as the common chords AH, BH and CH of pairs of the circles intersect only in H.

Solution 2. Let H be the orthocentre, and P, Q, R the pedal points as defined in the first solution. Since ARHQ is concyclic

[A. Lin] Suppose that U, V, W are on the circumcircle. >From the reflection about BC, ÐBCU = ÐBCH. >From the reflections about BA and BC, we see that BW = BH = BU, and so, since the equal chords BW and BU subtend equal angles at C, ÐBCW = ÐBCU. Hence ÐBCW = ÐBCH, with the result that C, H, W are collinear and CW is an altitude. Similarly, AU and BV are altitudes that contain H, and so their point H of intersection must be the orthocentre.