157.

Prove that if the quadratic equation x² + ax + b + 1 = 0 has nonzero integer solutions, then a² + b² is a composite integer.

Comment. You should make sure that you are familiar with the relationship between the coeffients and roots of polynomials; a poor way to solve this problem is to use the quadratic formula. Note that, if, say, u = 0, then a² + b² can be prime; for example, you get the value 5 when (u, v) = (0, 2).

158.

Let f(x) be a polynomial with real coefficients for which the equation f(x) = x has no real solution. Prove that the equation f(f(x)) = x has no real solution either.

Thus, either g(x) > 0 for all x or else g(x) < 0 for all x. Then

Solution 2. Suppose, if possible, that f(f(a)) = a. Let b = f(a). Then f(b) = a. By hypothesis, b ¹ a. Wolog, suppose that a < b. Then f(a) - a > 0 and f(b) - b < 0. Since f(x) - x is a polynomial, it is continuous and so the Intermediate Value Theorem applies on the closed interval [a, b]. As it has opposite signs at the endpoints of [a, b], it must vanish somewhere in the interior of the interval. But then this contradicts the hypothesis. Hence f(f(x)) = x can have no real solutions.

Comment. It is important to highlight that f(x) is a continuous function, as this is key to the result. (Can you construct a counterexample where it is not assumed that f is a polynomial or continuous?) Several of you tried to give a geometric argument for this, and apart from mangling the terminology (e.g., not distinguishing between a function and its graph, points and values), operated at too intuitive a level. Notice that the statement ``f(x) > 0 or f(x) < 0 for all x'' lacks a certain precision, as it could be interpreted to mean that at any individual point x, one of the two alternatives occurs (the function does not vanish), but that different alternative might occur at different points.

159.

Let 0 £ a £ 4. Prove that the area of the bounded region enclosed by the curves with equations

y = 1 - |x - 1 |

and

y = |2x - a |

cannot exceed ¹/₃.

Let 1 £ a £ 3. In this case, the bounded region is a quadrilateral with the four vertices (a/3, a/3), (a/2, 0), ((a+2)/3, (4-a)/3) and (1, 1). Noting that this quadrilateral is the result of removing two smaller triangles from a larger one (draw a diagram!), we find that its area is

The case 3 £ a £ 4 is the symmetric image of the case 0 £ a £ 1 and we find that the area of the bounded region cannot exceed 1/4.

Comment. This is not a difficult problem, but it does require a lot of care in keeping the situation straight and the computations sound. Always keep in mind completion of the square when it is a matter of optimizing a quadratic. Use of calculus is more cumbersome in such situations, and begs the question as to whether the optimum is a maximum or a minimum. (You can lose points for not explicitly resolving this issue.) R. Barrington Leigh solved the problem by looking at the area between the graph of y = 1 - |x - 1 |- |2x - a | and the x-axis. The equation of this curve with the absolute value signs is equal to y = 3x - a for x £ a/2, to y = a - x when a/2 £ x £ 1 and a + 2 - 3x when x £ 1, with suitable modification when a ³ 2. Because at each value of x, the difference in the ordinates remains the same as in the original problem, the area here is the same as the area between the two graphs of the problem. You might want to try the problem in this way.

160.

Let I be the incentre of the triangle ABC and D be the point of contact of the inscribed circle with the side AB. Suppose that ID is produced outside of the triangle ABC to H so that the length DH is equal to the semi-perimeter of DABC. Prove that the quadrilateral AHBI is concyclic if and only if angle C is equal to 90^°.

Solution 1. The quadrilateral AHBI is concyclic if and only if DI:DH = DB:DA, or equivalently,

Solution 2. The concyclic condition is equivalent to

Comment. We can end the solution, keeping the equivalences to the end, by noting that

Solution 3. [A. Feizmohammadi]

On the other hand, suppose that AHBI is concyclic. Then

161.

Let a, b, c be positive real numbers for which a + b + c = 1. Prove that

a3 a2 + b2 + b3 b2 + c2 + c3 c2 + a2 ³ 1 2  .

Comment. R. Furmaniak noted that

162.

Let A and B be fixed points in the plane. Find all positive integers k for which the following assertion holds:

among all triangles ABC with |AC | = k |BC |, the one with the largest area is isosceles.

Solution 2. As in Solution 1, we need consider only the case that k ³ 1, and we can dispose of the case k = 1. Let D be the foot of the perpendicular from C to AB (possibly produced; note that D and B lie on the same side of A) and let the respective lengths of AB, AD and CD be c, x and v. We wish to maximize v. The given condition implies that

Solution 3. As before, we can deal with the k = 1 case. Let k ¹ 1. Let the sides of the triangle be (a, ka, c) where c is the length of AB; let C be the angle opposite c. Then

Alternatively, the case c = a corresponds to cosC = k/2, and equating k/2 and 2k/(k² + 1) leads to k² = 3. The case c = ka corresponds to cosC = 1/2k, and this leads to k² = 1/3. We conclude as before.

Solution 4. Dispose of the k = 1 case as before. Let k ¹ 1. Let c = |AB | be fixed, and let the other two sides have length a and la. Sixteen times the square of the area of the triangle is, by Heron's rule,

Comments. Several students committed the fallacy of setting, for example, AC = AB, noting that the locus of C in this case was a circle with diameter containing AB, claiming that the area was maximized when ÐC = 90^° and so BC was equal to Ö2AC. They deduced that Ö2 and 1/Ö2 were the only values of k that allowed the area to be maximized by an isosceles triangle. However, this puts things the wrong way around. The value k is fixed, and then we look at all appropriate triangles. All the solvers did was to maximize the area over all isosceles triangles for which AC = AB; as C traces out the set of points satisfying these conditions, of course, the ratio of the lengths of AC and BC vary. Those solvers who made this mistake are invited to investigate the case k = Ö2 in more detail, and this will help them understand where they went wrong.

If we coordinatize the situation with A ~ (0, 0) and B ~ (1, 0), we see that the locus of C is given by

163.

Let R_i and r_i re the respective circumradius and inradius of triangle A_i B_i C_i (i = 1, 2). Prove that, if ÐC₁ = ÐC₂ and R₁r₂ = r₁R₂, then the two triangles are similar.

Comment. There were many variations using various trigonometric identities.

164.

Let n be a positive integer and X a set with n distinct elements. Suppose that there are k distinct subsets of X for which the union of any four contains no more that n - 2 elements. Prove that k £ 2^n-2.

Note that 2^n-2 < k £ |A_i ||B_i |, since each set of S is the union of a set in A_i and a set in B_i. (|·| refers to the number of elements.) Therefore, either |A_i | > 2^i-1 or |B_i | > 2^n-i-1. In the former case, A_i must contain a complementary pair of subsets (use the pigeonhole principle on complementary pairs of sets) of { x₁, ¼, x_i }; the the latter, B_i must contain a complementary pair of sets of { x_i+1, ¼, x_n }.

By our choice of x₁, we see that A₁ = { Æ, x₁ }, so that |A₁ | = 2. Therefore there are values of the index i for which |A_i | > 2^i-1. If it turns out that |A_n-1 | > 2^n-2, then A_n-1 contains two sets whose union is { x₁, ¼, x_n-1 } and so S contains two sets whose union has at least n-1 elements. In this case, S fails to satisfy the hypotheses of the problem.

Suppose that |A_n-1 | £ 2^n-2. Then select the smallest index k for which |A_k | £ 2^k-1. Then k ³ 2 and |A_k-1 | > 2^k-2 while |B_k | > 2^n-k-1. Therefore, A_k-1 contains two sets whose union is { x₁, x₂, ¼, x_k-1 } and B_k contains two sets whose union is { x_k+1, ¼,x_n }. We conclude that S must contain four sets whose union contains the n-1 elements of X apart from x_k, and so S again fails to satisfy the hypothesis of the problem.

We conclude that any family of more than 2^n-2 sets fails so satisfy the hypothesis of the problem, and so that every family that does satisfy the hypothesis must have at more 2^n-2 elements.

Solution 2. Let P be a class of k subsets for which the union of no four contains more than n-2 elements. Suppose that A₁ and A₂ are two members for which the cardinality m = |A₁ ÈA₂ | of the union of a pair is maximum. Since m < n, we can construct a set Y = A₁ ÈA₂ È{ y }, where y Ï A₁ ÈA₂; thus |Y | = m + 1.

Let Q = { Y ÇA : A Î P }. No two subsets in Q are complementary (i.e. have union equal to Y), so that |Q | £ 2^m. Let Z = X \Y and R = { Z ÇA : A Î P }. Then |Z | = n - (m + 1) = n - m - 1. We prove that R has no two sets that are complementary, i.e., whose union is Z. For, otherwise, suppose that (A₃ ÇZ) È(A₄ ÇZ) = Z, for two sets A₃ and A₄ in P. Then Z = (A₃ ÈA₄) ÇZ, so that

Since each set in P is uniquely determined by its intersections with Y and Z, we have that

Comment. Several solvers pointed out that if you took all the subsets of n-2 of the elements of X, then we got an ``extreme'' case of a family of sets that satisfied the hypothesis and the conclusion, and essentially said that if we tried to stuff in any more sets, we would contradict the hypothesis. However, this begs the question as to whether you could drop some of these sets and replace than by a larger number of sets while still keeping the hypothesis.

165.

Let n be a positive integer. Determine all n-tples { a₁, a₂, ¼, a_n } of positive integers for which a₁ + a₂ + ¼+ a_n = 2n and there is no subset of them whose sum is equal to n.

Solution 2. [J.Y. Jin] Wolog, suppose that a₁ £ a₂ £ ¼ £ a_n. Note that a₁ cannot exceed 2. If a₁ = 2, then each a_i is equal to 2 and n is odd. Suppose that a₁ = 1. Then a_n ³ 3. Let s_k = a₁ + ¼+ a_k for 1 £ k £ n. Then s_n-1 = 2n - a_n £ 2n - 3. By the hypothesis, we see that the set of the s_k with 1 £ k £ n-1 contains at most one member from each of the n-1 sets:

Comment. Some of the solvers tried an induction argument. However, the process was faulty, since they generally started with an n-tple and then tried to build from that an (n+1)-tple. However, for an induction argument to work, you need to start with a general (n+1)-tple that satisfies the condition, and then indicate how to reduce it to the result for a smaller set.

166.

Suppose that f is a real-valued function defined on the reals for which

f(xy) + f(y - x) ³ f(y + x)

for all real x and y. Prove that f(x) ³ 0 for all real x.

Comment. O. Bormashenko noted that when 2x = Ö[(z² + 4)] + 2 - z and 2y = Ö[(z² + 4)] + 2 + z, then xy = x+y and y - x = z.

167.

Let u = (Ö5 - 2)^1/3 - (Ö5 + 2)^1/3 and v = (Ö[189] - 8)^1/3 - (Ö[189] + 8)^1/3. Prove that, for each positive integer n, uⁿ + vⁿ⁺¹ = 0.

Comment. Some students observed that

168.

Determine the value of

cos5° + cos77° + cos149° +cos221° + cos293° .

A very useful relation is 4cos36^° cos72^° = 1. This can be checked geometrically. Let PQS be a triangle for which ÐP = ÐS = 36^° and ÐPQS = 108^°. Let R be a point on the side PS for which ÐPQR = 72^° and ÐSQR = 36^°. Then PQ = PR, PQ = QS and QR = RS; let r be the common length of PQ, PR, QS and let s be the common length of QR and RS. Then cos72^° = s/2r and cos36^° = r/2s and the desired result follows. An algebraic derivation of this result can also be given.

We can make use of complex numbers. Let z = cos72^°+ i sin72^° so that z is a nonreal root of

Solution 1. Let cos36^° = t. Then

Solution 2. [C. Huang]

Comment. There were other solutions with similar manipulations. A quick solution can also be realized using complex numbers. The given expression is equal to the real part of

169.

Prove that, for each positive integer n exceeding 1,

1 2n + 1 21/n < 1 .

Solution 2. [R. Furmaniak] By the arithmetic-geometric means inequality, we have that

Solution 3. [S. Wong] By the arithmetic-geometric means inequality

170.

Solve, for real x,

x ·21/x + 1 x ·2x = 4 .

Solution 2. As before, we see that x must be positive. The equation is equivalent to