Solutions

171.

Let n be a positive integer. In a round-robin match, n teams compete and each pair of teams plays exactly one game. At the end of the match, the ith team has x_i wins and y_i losses. There are no ties. Prove that

x12 + x22 + ¼+ xn2 = y12 + y22 + ¼+ yn2 .

Solution 2. Since x_i + y_i = n-1 for each i and x₁ + x₂ + ¼+ x_n = y₁ + y₂ + ¼+ y_n = (n || 2) (the number of games played), we find that

172.

Let a, b, c, d. e, f be different integers. Prove that

(a - b)2 + (b - c)2 + (c - d)2 + (d - e)2 + (e - f)2+ (f - a)2 ³ 18 .

Solution 2. [A. Critch] We prove a more general result. Let n be a positive integer exceeding 1. Let (t₁, t₂, ¼, t_n-1, t_n) be an n-tple of distinct integers, and suppose that the smallest of these is t_n. Define t₀ = t_n, and wolog suppose that t₀ = t_n = 0. Suppose that, for 1 £ i £ n, s_i = |t_i - t_i-1 |. Let the largest integer be t_r; since the integers are distinct, we must have

How can this lower bound be achieved? Since it is equal to 2² (n - 1) + 1² + 1², we can have n-2 differences equal to 2 and 2 differences equal to 1. Thus, we can start by going up the odd integers, and then come down via the even integers to 0. In the case of n = 6, this yields the 6-tple (1, 3, 5, 4, 2, 0).

173.

Suppose that a and b are positive real numbers for which a + b = 1. Prove that

æ ç è a + 1 a ö ÷ ø 2   + æ ç è b + 1 b ö ÷ ø 2   ³ 25 2  .

Determine when equality holds.

Solution 1. By the root-mean-square, arithmetic mean inequality, we have that

Solution 2. By the RMS-AM inequality and the harmonic-arithmetic means inequality, we have that

Solution 3.

Solution 4. [F. Feng] From the Cauchy-Schwarz and arithmetic-geometric means inequalities, we find that

174.

For which real value of x is the function

(1 - x)5 (1 + x)(1 + 2x)2

maximum? Determine its maximum value.

We adopt the same strategy to consider the situation when -1 £ x £ -1/2. For convenience, let u = -x, so that the want to maximize

It remains to show whether the value of the function when x = -7/8 exceeds its value when x = 0. Recall the Bernoulli inequality: (1 + x)ⁿ > 1 + nx for -1 < x, x ¹ 0 and positive integer n. This can be established by induction (do it!). Using this, we find that

Solution 2. Let f(x) = (1 - x)⁵(1 + x)(1 + 2x)². Then f¢(x) = -2(1 - x)⁴(1 + 2x)x(7 + 8x). We see that f¢(x) > 0 if and only if x < -7/8 or -1/2 < x < 0. Hence f(x) has relative maxima only when x = -7/8 and x = 0. Checking these two candidates tells us that the absolute maximum of 3⁷ ×5⁵ ×2^-22 occurs when x = -7/8.

175.

ABC is a triangle such that AB < AC. The point D is the midpoint of the arc with endpoints B and C of that arc of the circumcircle of DABC that contains A. The foot of the perpendicular from D to AC is E. Prove that AB + AE = EC.

Solution 2. Note that the length of the shorter arc AD is less than the length of the shorter arc DC. Locate a point H on the chord AC so that AD = HD. Consider triangles ABD and HCD. We have that AD = DH, BD = CD and ÐABD = ÐHCD. This is a case of SSA congruence, the ambiguous case. Since angles BAD and DHC are both obtuse (why?), they must be equal rather the supplementary, and the triangles ABD and HCD are congruent. (Congruence can also be established using the Law of Sines.) In particular, AB = HC. Since triangle ADH is isosceles, E is the midpoint of AH, so that AB + AE = HC + EH = EC.

Solution 3. Let DM be a diameter of the circumcircle, so that M is the midpoint of one of the arcs BC. Let H be that point on the chord AC for which DA = DH and let DH be produced to meet the circle again in K. Since ÐMAD = ÐDEA = 90^°, it follows that ÐMAC = 90^° - ÐEAD = ÐADE. Since AM and DE are both angle bisectors, ÐBAC = ÐADH.

Because ADCK is concyclic, triangles ADH and KCH are similar, so that HC = CK. From the equality of angles BAC and ACK, we deduce the equality of the arcs BAC and ACK, and so the equality of the arcs BA and CK. Hence AB = CK = HC. Therefore AB + AE = HC + EH = EC.

Solution 4. [R. Shapiro] Since A lies on the short arc BD, ÐBAD is obtuse. Hence the foot of the perpendicular from D to BA produced is outside of the circumcircle of triangle ABC. In triangles KBD and ECD, ÐBKD = ÐDEC = 90^°, ÐKBD = ÐABD = ÐACD and BD = CD. Hence the triangles KBD and ECD are congruent, so that DK = DE and BK = EC. Since the triangle ADK and ADE are right with a common hypotenuse AD and equal legs DK and DE, they are congruent and AK = AE. Hence EC = BK = BA + AK = BA + AE, as desired.

176.

Three noncollinear points A, M and N are given in the plane. Construct the square such that one of its vertices is the point A, and the two sides which do not contain this vertex are on the lines through M and N respectively. [Note: In such a problem, your solution should consist of a description of the construction (with straightedge and compasses) and a proof in correct logical order proceeding from what is given to what is desired that the construction is valid. You should deal with the feasibility of the construction.]

However, the construction does not work if A actually lies on the circle with diameter MN. In this case, A and C would coincide and the situation degenerates. If A lies on the right bisector of MN, then C can be the other point where the right bisector intersects the circle, and M and N can be the other two vertices of the square. If A is not on the right bisector, then there is no square; all of the points A, M, C, N would have to be on the circle, and AM and AN would have to subtend angles of 45^° at C, which is not possible.

Proof. If C and O are on the same side of KN, then ÐKCN = ¹/₂ÐKON = 45^°, so that CN makes an angle of 45^° with AC produced, and so CN produced contains a side of the square. Similarly, CM produced contains a side of the square. If C and O are on opposite sides of KN, then ÐKCN = 135^°, and CN still makes an angle of 45^° with AC produced; the argument can be completed as before.

Solution 2. Construction. Construct circle of diameter MN. Draw AM = MR (with the segment MR intersecting the interior of the circle) and AM ^MR. Construct the circle AMR. Let this circle intersect the given circle at C. Then construct the square with diagonal AC. If A lies on the circle, then the candidates for C are A and M. We cannot take C = A, as the situation degererates; if we take C = M, then the angle ACM and segment CM degenerate. We can complete the analysis as in the first solution.

Proof. Since DARM is right isosceles, ÐARM = 45^°. Hence the circle is the locus of points at which AM subtends an angle equal to 45^° or 135^°. Hence the lines AC and CM intersect at an angle of 45^°. Since ÐMCN = 90^°, the lines AC and CN also intersect at 45^°. It follows that the remaining points on the square with diagonal AC must lie on the lines CM and CN.

177.

Let a₁, a₂, ¼, a_n be nonnegative integers such that, whenever 1 £ i, 1 £ j, i + j £ n, then

ai + aj £ ai+j £ ai + aj + 1 .

(a) Give an example of such a sequence which is not an arithmetic progression.

(b) Prove that there exists a real number x such that a_k = ëkx û for 1 £ k £ n.

Solution. [A. Critch] The assertion to be proved is that all the semi-closed intervals [a_k/k, a_k+1/k) have a point in common. Suppose, if possible, that this fails. Then there must be a pair (p, q) of necessarily distinct integers for which a_q/q ³ (a_p + 1)/p. This is equivalent to pa_q ³ qa_p + q. Suppose that the sum of these two indices is as small as possible.

Suppose that p > q, so that p = q + r for some positive r. Then

Suppose that p < q, so that q = p + s for some positive s. Then