Solutions

227.

Let n be an integer exceeding 2 and let a₀, a₁, a₂, ¼, a_n, a_n+1 be positive real numbers for which a₀ = a_n, a₁ = a_n+1 and

ai-1 + ai+1 = ki ai

for some positive integers k_i, where 1 £ i £ n.

Prove that

2n £ k1 + k2 + ¼+ kn £ 3n .

As for the other inequality, since the expression has cyclic symmetry, there is no loss in generality in supposing that a_n ³ a₁ and a_n ³ a_n-1 with inequality in at least one case, so that 2a_n > a_n-1 + a₁. Therefore, k_n = 1 and a_n = a_n-1 + a₁.

We establish the right inequality by induction. For the case n = 3, we may suppose that

Now suppose the result holds when the index is n - 1 ³ 3. Then, supposing that k_n = 1 and substituting for a_n, we obtain the n-1 equations

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241.

Determine sec40^° + sec80^°+ sec160^°.

Solution 1. The values 40^°, 80^° and 160^° all satisfy cos3q = 3D -1/2, or 8 cos³ q- 6cosq+ 1 = 3D 0. Thus, cos40^°. cos80^° and cos160^° are the roots of the cubic equation 8x³ - 6x + 1 = 3D 0, so that their reciprocals sec40^°, sec80^° and sec160^° are the roots of the cubic equation x³ - 6x² + 8 = 3D 0. The sum of the roots of this cubic is

sec40° + sec80° + sec160° = 3D 6 .

Solution 2. Let z = 3D cos40^° + i sin 40^°. Then z⁹ = 3D 1. In fact, since z⁹ - 1 = 3D (z - 1)(z² + z + 1) (z⁶ + z³ + 1) and the first two factors fail to vanish, z⁶ + z³ + 1 = 3D 0. Also 1 + z + z² + ¼+ z⁸ = 3D (1 + z + z²)(1 + z³ + z⁶) = 3D 0. Observe that cos40^° = 3D ¹/₂(z + ¹/_z), cos80^° = 3D ¹/₂(z² + [ 1/(z²)]) and cos160^° = 3D ¹/₂(z⁴ + [ 1/(z⁴)]), so that the given sum is equal to

	2 é ë  z 1 + z2 +  z2 1 + z4 +  z4 1 + z8 ù û	=3D 2 é ë  z 1 + z2 +  z2 1 + z4 +  z5 1 + z ù û
		=3D 2 é ë  z(1 + z + z4 + z5) + z2(1 + z + z2 + z3) + z5(1 + z2 + z4 + z6) (1 + z)(1 + z2)(1 + z4) ù û
		=3D 2 é ë  z7 + z6 + 3z5 + z4 + z3 + 3z2 + z + 1 (1 + z)(1 + z2)(1 + z4) ù û
		=3D 2 é ë  (z+1)(z6 + z3 + 1) + 3z2 (z3 + 1) (1 + z)(1 + z2)(1 + z4) ù û
		=3D 2 é ë   0 - 3z8 1 + z + z2 + z3 + z4 + z5 + z6 + z7 ù û =3D 2 é ë  -3z8 -z8 ù û =3D 6 .

Solution 3. [T. Liu]

	sec40° + sec80° + sec 160°	=3D  cos40° + cos80° cos40° cos80° +  1 cos 160°
		=3D  2cos60° cos20° cos40°cos80° +  1 cos160°
		=3D  cos20° cos160° + cos40° cos80° cos40° cos80° cos 160°
		=3D  cos180° + cos140° + cos120° + cos40° cos40°(cos240° + cos80°)
		=3D  -1 - 1/2 (1/2)(-cos40° + cos120°+ cos40°) =3D  -3/2 -1/4 =3D 6 .

Solution 4. Let x = 3D cos40^°, y = 3D cos 80^° and z = 3D cos160^°. Then

x + y + z = 3D 2cos60°cos20° - cos20° = 3D 0

and

	xy + yz + zx	=3D  1 2 [cos120° + cos140° + cos240° + cos 80° + cos200° + cos120°]
		=3D  1 2 é ë -  3 2 + x + y + z ù û =3D -  3 4  .

Now

	8 sin40° cos40° cos80° cos160°	=3D 4 sin80° cos80° cos 160°
		=3D 2 sin160° cos160° = 3D sin 320° = 3D -sin40°

so that xyz = 3D -1/8. Then the sum of the problem is equal to (xy + yz + zx)/(xyz) = 3D 6.

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248.

Find all real solutions to the equation

æ Ö x + 3 - 4 Ö x - 1   +   æ Ö x + 8 - 6 Ö x - 1   = 1 .

Solution 2. [Z. Wu] For a solution to exist, we require that x ³ 1 and that both 0 £ x + 3 - 4Ö{x - 1} £ 1 and 0 £ x + 8 - 6Ö{x - 1} £ 1. These two conditions lead to (x + 2)² £ 16(x - 1) and (x + 7)² £ 36(x - 1), which in turn leads to

On the other hand, 5 £ x £ 10 implies that 2 £ Ö{x-1} £ 3, so that (as in Solution 1) we find that the equation is equivalent to (Ö{x - 1} - 2) + (3 - Ö{x-1}) = 1, which is an identity. Thus, the equation holds exactly when 5 £ x £ 10.

Comment. Your first observation should be that, in order for the equation to make sense, we require that x ³ 1. It is important not just to write down a lot of algebraic equations, but to indicate the logical relationships between them; which equations imply which other equations? which pairs of equations are equivalent? This is especially desirable when surd equations are involved, where the operations that lead from one equation to another are not logically reversible and extraneous solutions might be introduced.

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249.

The non-isosceles right triangle ABC has ÐCAB = 90^°. Its inscribed circle with centre T touches the sides AB and AC at U and V respectively. The tangent through A of the circumscribed circle of triangle ABC meets UV in S. Prove that:

(b) |d₁ - d₂ | = r , where r is the radius of the inscribed circle, and d₁ and d₂ are the respective distances from S to AC and AB.

Ad (b), let P and Q be the respective feet of the perpendiculars from S to AB and AC. Note that SQAP is a rectangle so that ÐPUS = ÐPSU = 45^° and so PU = PS. Then |QS |- |PS | = |AP |- |PU | = r.

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250.

In a convex polygon P, some diagonals have been drawn so that no two have an intersection in the interior of P. Show that there exists at least two vertices of P, neither of which is an enpoint of any of these diagonals.

On the other side of d, select a diagonal g which has the smallest number of vertices between its endpoints on the side opposite to the side of d. By an argument similar to the above, there is at least one vertex on the side of g opposite to d from which no diagonal has been drawn.

Solution 2. [S. King] The result is vacuously true for triangles. Suppose that the polygon has at least four sides. Suppose that a (possibly void) collection of diagonals as specified in the problem is given. We continue adding diagonals one at a time such that each new diagonal does not cross any previous one in the interior of the polygon. At each stage, the polygon P is partitioned into polygons with fewer sides all of whose vertices are vertices of the polygon P. As long as any of the subpolygons has more than three sides, we can add a new diagonal. However, the process will eventually terminate with a triangulation of P, i.e., a partitioning of P into n - 2 triangles all of whose vertices are vertices of P. (Exercise. Explain why the number of triangles is n - 2. One way to do this is to note that the sum of all the angles of the triangles is equal to the sum of the angles in the polygon.)

Each triangle must have at most two sides in common with the given polygon. Since there are n sides and n - 2 triangles, at least two triangles have two sides in common with P. In each case, the vertex common to the two sides has no diagonal emanating from it (neither an original diagonal nor an added diagonal), and the result follows.

Comment. Many solvers failed to appreciate that the collection of diagonals is given, and that the problem is to establish the desired property no matter what the collection is. A lot of arguments had the students constructing diagonals without indicating how the ones constructed might have anything to do with a given set; in effect, they were giving a particular situation in which the result obtained. Several solvers tried induction, using one diagonal to split P into two, but did not handle well the possibility that the loose vertices in the subpolygons might be at the ends of the subdividing diagonal. One way around this is to make the result stronger, and show that one can find two non-adjacent vertices that are not the endpoints of diagonals. This is certainly true for quadrilaterals, and using this an induction hupothesis yielded a straightforward argument for polygons of higher order.

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251.

Prove that there are infinitely many positive integers n for which the numbers { 1, 2, 3, ¼, 3n } can be arranged in a rectangular array with three rows and n columns for which (a) each row has the same sum, a multiple of 6, and (b) each column has the same sum, a multiple of 6.

We now show that, for each n of this form, we can actually construct an array with the desired property. Starting with the magic square, we derive the following array for n = 9:

We generalize this for n = 12k + 9, for k a nonnegative integer. Suppose that an array is possible. Then the sum of all the elements in the array is (36k + 27)(18k + 14) = 18(4k + 3)(9k + 7). The sum of the elements in each column is 6(9k + 7) = 54k + 42 and the sum in each row is 6(4k + 3)(9k + 7) = (4k + 3)(54k + 42). If we can achieve this with distinct entries, then we have constructed the array.

We build the array by juxtaposing horizontally 4k + 3 square 3 ×3 blocks of the form:

For example, when n = 45, k = 3, there are 15 blocks and the choice of a, b, c for these blocks can be read along the rows of

Solution 2. [Y. Zhao] We can form the 3 ×9 array:

Solution 3. [J. Zhao] For the time being, neglect the conditions involving divisibility by 6, and focus only on the condition that the numbers 1, 2, ¼, 3n be used and that the row sums and the column sums be each the same. Then, when n = 3, a magic square will serve.

Suppose that, for some k ³ 1, we have found a suitable 3 ×3^k matrix M. Let A be the 3 ×3^k+1 matrix obtained by placing three copies of M side by side and B the 3 ×3^k+1 matrix determined by placing side by side the 3 ×3^k matrices B₁, B₂, B₃ where each column of B₁ is (the transpose of) (0, 1, 2), of B₂ is (1, 2, 0), and of B₃ is (2, 0, 1). Each of A and B has constant row sums and constant column sums.

Let N = A + 3^k+1B. Then N not only has constant row and column sums, but consists of the numbers 1, 2, ¼, 3^k+2 (why?). The row sums of M are each (1/6)(3^k+1)(3^k+1 + 1), so that the row sums of N are each

We now require that each of (1/6)(3^k+1)(3^k+1 + 1) and (3/2)(3^k+1 + 1) be divisible by 6. This will occur exactly when 3^k+1 + 1 º 0 (mod 4), so that k must be even. Thus, we can obtain an array as desired when n = 9^m for some positive integer m. (Note that 9^m º 9 (mod 12).)

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252.

Suppose that a and b are the roots of the quadratic x² + px + 1 and that c and d are the roots of the quadratic x² + qx + 1. Determine (a - c)(b - c)(a + d)(b + d) as a function of p and q.

Solution 2. Using a + b = -p, c + d = -q and ab = cd = 1, we obtain that

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253.

Let n be a positive integer and let q = p/(2n+1). Prove that cot² q, cot² 2q, ¼, cot² nq are the solutions of the equation

æ è 2n + 1 1 ö ø xn - æ è 2n + 1 3 ö ø xn-1 + æ è 2n+1 5 ö ø xn-2 -¼ = 0 .

Solution 2. [Y. Zhao] Observe that, for each complex a,

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254.

Determine the set of all triples (x, y, z) of integers with 1 £ x, y, z £ 1000 for which x² + y² + z² is a multiple of xyz.

Since

The case kx = 2 leads to x² + (y - z)² = 0 which has no solutions as specified. Hence kx = 3 and k = 1 or k = 3. For these two cases, we find that the base solutions are respectively (x, y, z) = (3, 3, 3) and (x, y, z) = (1, 1, 1).

Suppose that k = 1. Modulo 3, any square is congruent to 0 or 1. If, say, x º 0 (mod 3), then y² + z² º 0 (mod 3); this can occur only if y and z are multiples of 3. Hence (x, y, z) = (3u, 3v, 3v) for some integers u, v, w. But then 9u² + 9v² + 9w² = 27uvw, or u² + v² + w² = 3uvw. Contrarily, any solution (u, v, w) of this equation gives rise to a solution (x, y, z) of x² + y² + z² = xyz. Therefore there is a one-one correspondence between solutions of x² + y² + z² = xyz where all numbers are multiples of 3 and solutions of x² + y² + z² = 3xyz. We will obtain these solutions below.

The only other possibility is that none of x, y, z is divisible by 3. But then, x² + y² + z² would be a multiple of 3 and xyz not a multiple of 3; thus there are no solutions of this type.

Suppose that k = 3. From the above, we know that every solution arises from a solution for which 1 £ x £ y £ z £ 3xy/2 and for such a solution x = 1. Let (x, y, z) = (1, y, ty) where 1 £ y £ 3/2. Then 1 + (1 + t²)y² = 3ty², so that

To get a handle on the situation, fix x = u and consider a sequence of solutions (x, y, z) = (u, v_n-1, v_n). The solution w = v_n-1 satisfies the quadratic equation

Using the recursion v_n+1 = 3uv_n - v_n-1, we get the following sequences for various values of u: