Write your solutions to be read. You may have to rework and re-edit your solution to make the argument clear, close gaps in the reasoning, define symbols used and present ideas in logical order. Always check your work. It is too easy to miscopy or misrepresent a symbol and to make a computational mistake. In a competition, your goal is to maximize the number of points you receive. In the following solutions, [A ¼Z] represents the area of the figure A ¼Z, dist (A, BC) is the perpendicular distance from point A to line BC, and dist (a, b) is the perpendicular distance between parallel lines a and b.

13.

Suppose that x₁, x₂, ¼,x_n are nonnegative real numbers for which x₁ + x₂ + ¼+ x_n < ¹/₂. Prove that

Comments. Some of you tried a ``moving variables'' argument, or an ``extreme case'' argument, i.e., if we make x_i as large as possible, we reduce the product, so that if it works for (x₁, x₂, ¼, x_n) = (¹/₂,0, 0, ¼, 0), then we are in business. Some felt that if it works for the extreme case with each x_i = 1/2n, then we are ok. Unless you back this up with solid argument and detailed analysis that relates the general case to the extreme case, then it is worthless. ``Moving variable'' arguments are always risky and best avoided. Both approaches often muddy the situation rather than clarify it.

Solution 1. If 0 < u, v < 1, then

Comments. You should carry out the necessary induction argument. Note that the problem asks for > rather than ³ .

Solution 2. [R. Barrington Leigh] Let

14.

Given a convex quadrilateral, is it always possible to determine a point in its interior such that the four line segments joining the point to the midpoints of the sides divide the quadrilateral into four regions of equal area? If such a point exists, is it unique?

Solution 1. Observe that [APS] = ¹/₄[ABD] and [CRQ] = ¹/₄[CDB], so that [APS] + [CRQ] = ¹/₄[ABCD] (why?). Since

In a similar way, we can draw a line n parallel to and between PQ and SR for which dist (n, PQ) = dist (D, SR) and dist (n, SR) = dist (B, PQ), so that, for any point N on n, [NPQ] = [DSR], [NRS] = [BRP] and [NPBQ] = [DSNR] = ¹/₄[ABCD].

The lines m and n are not parallel, and, so, intersect in a unique point O for which

We show that O is the only point with this property. Let L be another point inside ABCD. Then L lies in one of the four partitioning quadrilaterals, say [OQCR], so that [LQCR] < [OQCR] = ¹/₄[ABCD]. Hence, L will satisfy the conditions of the problem.

Comments. A careful solution will contain the observation that m and n are not parallel, and will intersect within the quadrilateral. The fact that m and n have a unique point of intersection does not, in and of itself, establish that the requisite point cannot be found in some other way. Therefore, uniqueness needs to be explicitly handled, either by showing that the required point must lie on m and n or by some other argument.

Solution 2. Analysis. Suppose such a point O exists. Then [OPBQ] = [OCQR] = ¹/₄[ABCD], so that

Synthesis. Construct lines h and k to satisfy the foregoing conditions. They must intersect in a unique point O within ABCD. Now

15.

Determine all triples (x, y, z) of real numbers for which

Comments. In solving a system of equations, one begins by assuming a solution and determining what properties it must have. Since this may involve one-way reasoning, such properties may not necessarily yield a solution and extraneous solutions may arise. Thus, when you have solved the equations, for a complete solution, you should check that the solution is valid.

If, in your manipulations, you divide by a certain quantity or find a quantity in the denominator of a fraction, you should explore the possibility that the quantity could vanish, Remember that you cannot divide by zero. When you write up your final solution, it is often a good idea to deal with this possibility ahead of time and get it out of the way.

Solution 1. Suppose one of the variable, say x, is 0. Then z(x + 1) = 0, so z = 0, and y(z + 1) = 0 so y = 0. Suppose one of the variables, say x is -1. The x(y + 1) = 0, so y = -1 and y(z + 1) = 0, so z = -1. Hence, if any variable assumes either of the values -1 and 0, then all are equal. Henceforth, we assume that none of them have either value.

>From the equations, we find that

If y = z, then z + 1 = x + 1. so that z = x and x = y = z. Conversely, the system is satisfied when x = y = z.

Suppose that yz + y + 1 = 0. Then x(y + 1) = z(x + 1) = y(z + 1) = -1, and so xy + x + 1 = xz + z + 1 = 0. Suppose z = t, Then x = -(t + 1)/t and y = -1/(t + 1). Conversely, it is straightforward to check that the system is satisfied by

Comment. The solution with x, y, z unequal may look non-symmetrical. Verify that, if x = s = -(t + 1)/t, then y = -1/(1 + t) = -(s + 1)/s and z = t = -1/(1 + s). Check that x and z can similarly be expressed in terms of y = r.

Solution 2. From the equations, we find that

Suppose that xyz = 1. Then

Solution 3. As before, we can check that (x, y, z) = (0, 0, 0) is the only solution in which any variable vanishes. Henceforth, suppose that xyz ¹ 0. Let z = vx. Then y + 1 = v(x + 1), whence y = vx + v - 1. Therefore

Solution 4. As in the foregoing solutions, we can check that (x, y, z) = (0, 0, 0), (-1, -1, -1) are the only solutions in which any variable assumes either of the values 0 or -1. Henceforth, suppose x, y, z all differ from 0 and -1.

Let x (y + 1) = y(z + 1) = z(x + 1) = k. Then z = k/(x + 1) and y = (k/x) - 1 = (k - x)/x. Therefore,

Comment. It is not hard to check independently that the only nonzero solution in which x, y, z have the same sign is in fact given by x = y = z. For in this case, the ratio of any pair is positive, and the system can be rewritten

16.

Suppose that ABCDEZ is a regular octahedron whose pairs of opposite vertices are (A, Z), (B, D) and (C, E). The points F, G, H are chosen on the segments AB, AC, AD respectively such that AF = AG = AH.

(a) Show that EF and DG must intersect in a point K, and that BG and EH must intersect in a point L.

(b) Let EG meet the plane of AKL in M. Show that AKML is a square.

Solution 1. (a) Since AF:AB = AG:GC, it follows that FG || BC || ED, while FG < BC = ED. Hence, FGDE is a coplanar isosceles trapezoid and so EF and DG must intersect in a point K. A 90^° rotation about the axis AZ takes B ® C, F ® G, C ® D, G ® H, D ® E, E ® B. Hence EF ® BG and DG ® EH, so that BG and EH must intersect in a point L, which is the image of K under the rotation.

(b) KE and AB intersect in F, so that the two lines are coplanar. Also KF:KE = FG:ED = FG:BC = AF:FB so that DKAF ~ DEFB and AK || EB. Hence K lies in a plane through A parallel to BCDE. Because the 90^° rotation about the axis AZ (which is perpendicular to the planes BCDE and AKL takes K ® L, AK = AL and ÐKAL = 90^°.

Consider a dilation with centre E and factor |AB |/|FB |. Let I be on AE with AI = AF. The the dilation takes F ® K, H ® L, I ® A and the plane of FGHI to the parallel plane AKL. The image of G under this dilation is the intersection of EG and the plane of AKL, namely M. Thus the square FGHI goes to KMLA which must also be a square.

Solution 2. The dilation-reflection perpendicular to a plane through FG perpendicular to BE and CD with factor |AF |/|FB | takes B ® E, C ® D, and fixes F and G. The lines BF and CG with intersection A gets carried to lines EF and DG which intersect in a point K for which AK is perpendicular to the plane and so parallel to BE and CD, and the distance from K to the plane is |AF |/|FB | times the distance from A to the plane.

Similarly, considering a dilation-reflection to a plane through GH perpendicular to BC and DE with the same factor produces the point L with AL perpendicular to this plane and so parallel to BC and ED. Thus ÐKAL = 90^°.

The reflection in the plane AECZ fixes A, C, E, G and interchanges the points in each of the pairs (B, D) and (F, H). Hence the line pairs (EF, DG) and (EH, BG) are interchanged as is the pair K and L. Thus AK = AL and KL is perpendicular to AECZ. The triangle AKL is in a plane through A parallel to BCDE. The proof that AKML is a square can be completed as in Solution 1.

Solution 3. Assign solid coordinates: A ~ (0, 0, 1), B ~ (0, -1, 0), C ~ (1, 0, 0), D ~ (0, 1, 0), E ~ (-1, 0, 0). For some t with 0 < t < 1, F ~ (0, -(1-t), t), G ~ (1-t, 0, t), H ~ (0, 1-t, t). The line EF consists of points (-u, -(1 - u)(1 - t), (1 - u)t), with real u and the line DG consists of points ((1 - v)(1 - t), v, (1 - v)t) with real v. They meet in K ~ ((1-t)/t, (t-1)/t, 1). Similarly, L ~ ((1-t)/t, (1-t)/t, 1), so A, K, L lie on the plane z = 1. The line EG consists of points (-w + (1 - t)(1 - w), 0, t(1 - w)) with real w, and this intersects the plane z = 1 in the point (2(1 - t)/t, 0, 1). The desired result can now be verified.

17.

Suppose that r is a real number. Define the sequence x_n recursively by x₀ = 0, x₁ = 1, x_n+2 = rx_n+1 - x_n for n ³ 0. For which values of r is it true that

Answer. The property holds for all values of r.

Solution 1. Define x_-1 = -1; the recurrence still holds with this extension of the index. Note that, for n ³ 0,

We prove by induction that for each nonnegative integer m,

Hence

Solution 2. [R. Mong] We prove by induction that, for m ³ 1,

Solution 3. The recursion x_n+2 = rx_n+1 - x_n has characteristic polynomial t² - rt + 1 with discriminant r² - 4. Its roots are distinct as long as r ¹ ±2, so we deal with r = ±2 separately.

The solution for r = 2 is x_n = n and for r = -2 is x_n = (-1)^n-1 n, and it is straightforward to establish the desired relation in these cases. When r ¹ ±2, the characteristic polynomial has distinct roots s and 1/s, where s+ 1/s = r, i.e. s² = rs- 1. Note that s ¹ ±1 since r ¹ ±2. The solution of the recursion is

Comment. Solvers who used the approach of Solution 3 failed to consider the case in which the characteristic polynomial had a double root.

18.

Let a and b be integers. How many solutions in real pairs (x, y) does the system

Comments. Several of the solvers got all mixed up with the status of the variables in the problem, and, for example, found infinitely many solutions to the equations. a and b are fixed in advance; they are parameters, and your final answer will be conditioned by the the characteristics of various pairs (a, b). Rather than rush blindly into the problem, it is a good strategy to gain some understanding of the situation by looking at particular cases. For example, if one takes (a, b) = (0, 0), it is not too hard to see that (x, y) = (0, 0) is the only solution. Similarly, if (a, b) = (1, 1), one arrives at the sole solution (x, y) = (¹/₂,¹/₂). However, if (a, b) = (1, 0), we find two distinct solutions, (x, y) = (-¹/₂, 1),(0, ¹/₂). This not only clarifies the situation, but gives you a couple of examples against which you can check your final answer. Get in the habit of using examples to help understanding. It is an excellent exercise to plot, on the same axes, the graphs of the two equations for various values of a and b.

Solution 1. Since 2x and 2y are the difference of two integers, x and y must have the form m + ^u/₂ and n + ^v/₂ respectively, where m and n are integers and u and v each take one of the values 0 and 1. Plugging these in yields

(i) Let a + b º 0 (mod 3). Then, for a solution, we must have

(ii) Let a + b º 1 (mod 3). Then, for a solution, we must have

(iii) Let a + b º 2 (mod 3). Then

Hence, when a + b º 0 or a + b º 2 (mod 3), there is one solution to the system, while when a + b º 1 (mod 3), there are two solutions.

Solution 2. Observe that x and y must be half integers. Since

(i) Let a + b º 0 (mod 3). Then x + y is the integer (a + b)/3. If x and y are not themselves integers, then

(ii) Let a + b º 1 (mod 3). Then 2(a + b) + 1 º 0 (mod 3), and so

(iii) a + b º 2 (mod 3). Then x + y = (a + b + 1)/3, an integer. In this case, we can check that x and y cannot be integers, so that x = ëx û+ 1/2 and y = ëy û+ 1/2, and so