Solutions

262.

Let ABC be an acute triangle. Suppose that P and U are points on the side BC so that P lies between B and U, that Q and V are points on the side CA so that Q lies between C and V, and that R and W are points on the side AB so that R lies between A and W. Suppose also that

ÐAPU = ÐAUP = ÐBQV = ÐBVQ = ÐCRW = ÐCWR .

The lines AP, BQ and CR bound a triangle T₁ and the lines AU, BV and CW bound a triangle T₂. Prove that all six vertices of the triangles T₁ and T₂ lie on a common circle.

Let K be the intersection of AP and BQ (a vertex of T₁). Since ÐKPC + ÐKQC = 180^°, CQKP is concyclic. Hence ÐAKB + ÐAZB = ÐPKQ + ÐPCQ = 180^° and AKBZ is concyclic. Since AH ^BC and BH ^AC, the angle between AH and BH is equal to ÐACB = ÐXZY, so that AHBZ is also concyclic. Thus, A, H, K, B, Z lie on a common circle, so that ÐHKZ = ÐHBC = 90^°.

Now q = ÐAPC = ÐZAK = ÐZHK, so that, in the right triangle HKZ, |HK | = rcosq. Similarly, it can be shown that the distance from each vertex of triangle T₁ and T₂ from H is rcosq and the result follows.

Solution 2. [R. Dan] Let H be the orthocentre of triangle ABC, let AP and BQ intersect at D, and let AU and BQ intersect at E. Triangle APU is isosceles with AP = AU, and AH a bisector of ÐPAU and a right bisector of PU. Suppose X = AH ÇBC, Y = BH ÇAC and Z = CH ÇAB.

Since triangles APU and BQV are similar, ÐPAH = ÐQBH, so that ABDH is concyclic and ÐADH = ÐABH. Similarly, ACEH is concyclic and ÐAEH = ÐACH. Since the quadrilateral BZYC has right angles at Z and Y, BZYC is concyclic and

Since AH is common, ÐADH = ÐAEH and ÐDAH = ÐEAH, triangles ADH and AEH are congruent (ASA) and HD = HE. Thus, H is equidistant from the intersections of BQ with both AP and AU. Similarly, H is equidistant from the intersection of AP and both BQ and BV. Following around, we can show that H is equidistant from all the vertices of triangle T₁ and T₂, and the result follows.

---

263.

The ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are each used exactly once altogether to form three positive integers for which the largest is the sum of the other two. What are the largest and the smallest possible values of the sum?

A four-digit sum can arise either as the sum of two three-digit numbers or as the sum of a four-digit and a two-digit number. In the former case, the sum must exceed 1000 and be less than 2000 and, in the latter case, it must be at least 2000.

Thus, the smallest possible sum must be obtained by adding two three-digit numbers to get a four-digit sum. Since the digits of the sum are all distinct, the smallest possible sum is at least 1023. Since 589 + 437 = 1026, the smallest sum is at most 1026. We may assume that each digit in the first summand exceeds the corresponding digit in the second summand. The only possibilities for a lower sum are

For the largest sum, let the first summand have four digits and the second two. The hundreds digit of the first summand is 9 and the thousands digit of the sum exceeds the thousands digit of the first summand by 1. Since 5987 + 34 = 6021, the largest sum is at least 6021. The only possibilities to consider for a larger sum are

Thus, the smallest sum is 1026 and the largest is 6021.

Solution 2. [C. Shen] As in Solution 1, we eliminate the possibility of a five-digit sum. Suppose that we have

The smallest sum is at least 1023 and at most 1026 = 589 + 437. Suppose that

---

264.

For the real parameter a, solve for real x the equation

x =   æ Ö a + Ö a + x    .

A complete answer will discuss the circumstances under which a solution is feasible.

We check that these solutions work. When a ³ -1/4, x = ¹/₂(1 + Ö{1 + 4a}) and

Solution 2. x = Ö{a + Ö{a + x}} Þ x² - a = Ö{a + x}

Solution 3. For a solution, we require that x ³ 0. By squaring twice, we are led to the equation

In the second case,

Solution 4. [J. Zhao] For any real x, one of the following must hold:

Comment. Many solvers did not pay attention to the feasibility of the solutions. Solution 3 was particularly insidious, because it was easy to skip the analysis that only one of the values of x gave a solution when a > 0. Surd equations is a dandy topic for students to lose points they should gain because of carelessness or a superficial treatment.

---

265.

Note that 959² = 919681, 919 + 681 = 40²; 960² = 921600, 921 + 600 = 39²; and 961² = 923521, 923 + 521 = 38². Establish a general result of which these are special instances.

Consider first the situation b = 10. When k = 1, we have that u = 4 and we get the case 6² = 36, 3 + 6 = 3². When k = 2, we have that 10 £ u £ 14 and find that 86² = 7396, 73 + 96 = 13² and so on up to 90² = 8100,81 + 0 = 9². When k = 3, we have that 32 £ u £ 44, and find that 956² = 913936, 913 + 936 = 43² and so on up to 968² = 937024, 937 + 24 = 31².

Examples from base 3 are 5² = (221)₃, (2 + 21)₃ = 3²; 6² = (1100)₃, (11 + 0)₃ = 2²; 20² = (112211)₃, (112 + 211)₃ = 6²; 21² = (121100)₃, (121 + 100)₃ = 5².

Comment. In the above system, we could replace u - 1 by u - d and get other instances. For example, with (b, k) = (10, 2), we can get the instances (27² = 729, 7 + 29 = 6²), (29² = 841, 8 + 41 = 7²), (30², 3²), (39², 6²), (40², 4²), (50², 5²), (57², 9²). (60², 6²), (70², 7²), (75², 9²), (78², 12²), (80², 8²) and (98², 10²).

Another formulation is to note that the numerical equations are special instances of the system n² = b^k x + y; (n - x)² = x + y with 0 £ y £ b^k - 1 and 0 £ n < b^k, where (n, b, x, y) = (959, 10, 919, 681), (960, 10, 921, 600), (961, 10, 923, 521). These equations imply that x(2n - x) = (b^k - 1)x, whence x = 2n - (b^k - 1). Thus,

This question was, on the whole, badly done. In describing the generalization, one needs to provide a road map whereby one can make the appropriate subsitutions to obtain further examples. Many solvers were content to write down some equations of which the numerical examples were an instance without analyzing the conditions under which the equations could be used to obtain further examples. In effect, no further information was provided to show where other examples might be found.

---

266.

Prove that, for any positive integer n, (2n || n) divides the least common multiple of the numbers 1, 2, 3, ¼, 2n-1, 2n.

Let p be a prime divisor of (2n || n), so that p £ 2n. Suppose that p^k is the highest power of p that divides an integer not exceeding 2n. Then p^k £ 2n < p^k+1. The exponent of p in the prime factorization of (2n || n) is equal to

---

267.

A non-orthogonal reflection in an axis a takes each point on a to itself, and each point P not on a to a point P¢ on the other side of a in such a way that a intersects PP¢ at its midpoint and PP¢ always makes a fixed angle q with a. Does this transformation preserves lines? preserve angles? Discuss the image of a circle under such a transformation.

Suppose that the axis of reflection is the y axis. Let A and B be mutual images with A to the left and B to the right of the axis, AB meeting the axis at P and the upper right (and lower left) angles of intersection being q. If A ~ (x, y) (with x £ 0), then P ~ (0, y - x cotq) and B ~ (-x, y - 2x cotq). If B ~ (u, v) (with u ³ 0), then P ~ (0, v - ucotq) and A ~ (-u, v - 2u cotq). Thus the transformation is given by

Comment. A synthetic way of analyzing the image of a circle is to note that two chords of a circle that bisect each other must be diameters, and so have the same length. Using this, one can argue that a circle with one diameter along the axis and other perpendicular to the axis does not go to a circle.

---

268.

Determine all continuous real functions f of a real variable for which

f(x + 2f(y)) = f(x) + y + f(y)

for all real x and y.

Setting y = x yields

Suppose that the former holds. Then, for all x ³ 0, we have that g(x) = 3x and so x + 2f(x) = 3x, whence f(x) = x. Therefore, when x is arbitrary and y ³ 0, f(x + 2y) = f(x) + 2y. In particular, 0 = f(-2y + 2y) = f(-2y) + 2y so that f(-2y) = -2y. Hence, for all x, we must have that f(x) = x. A similar argument can be followed to show that f(x) º x when a < 0. Therefore, the only two solutions are f(x) = x and f(x) = -x/2.

Solution 2. [S. Eastwood] Setting x = y, we find that f(x + 2f(x)) = x + 2f(x) for all real x. Let

Suppose that A = { 0 }. Then, for all real x, x + 2f(x) = 0 and so f(x) = -x/2. This is a valid solution.

Suppose that A has a nonzero element a. Then

Solution 3. [J. Zhao] Suppose that f(u) = f(v). Then, for each x,

Suppose that f is increasing. Then

Suppose that f is decreasing. Suppose, if possible, that p and q are such that p + 2f(p) < q + 2f(q). Then f(p + 2f(p)) > f(q + 2f(q)). But, we get a contradiction since f(p + 2f(p)) = p + 2f(p) and f(q + 2f(q)) = q + 2f(q). Hence, there is a constant c such that, for all real x, x + 2f(x) = c. Hence, f(x) = -¹/₂(-x + c). Plugging this into the functional equation, we find that c = 0, and so we obtain the solution f(x) = -x/2.

Comments. If we assume that f(x) is a polynomial, then it can be shown that its degree must be 1. Assuming a solution f(x) = cx + d for constants c and d leads to the equations d = 0 = 2c² - c - 1 = (2c + 1)(c - 1). Thus, it is not hard to get a partial solution.

There were a number of approaches to ascertaining that f(0) = 0. A. Critch began with the observation that f(2f(0)) = f(0 + 2f(0)) = f(0) + 0 + f(0) = 2f(0). Let a = 2f(0), so that f(a) = a. Furthermore,

R. Dan noted that f(2f(y)) = y + f(y), and then went on to derive