 13.
 ABC is a triangle and D is a point on AB produced beyond B such that BD = AC, and E is a point on AC produced beyond C such that CE = AB. The right bisector of BC meets DE at P. Prove that ÐBPC = ÐBAC.
 (Note: This is Problem 2244 in CM+MM, May, 1997.)
 14.
 (a) Let a > b > c > d > 0 and a + d = b + c. Show that ad < bc.
 (b) Let a, b, p, q, r, s be positive integers for which

 15.
 The Fibonacci sequence { F_{n} } is defined by F_{1} = F_{2} = 1 and F_{n+2} = F_{n+1} + F_{n} for n = 0, ±1, ±2, ±3, ¼. The real number t is the positive solution of the quadratic equation x^{2} = x + 1.
 (a) Prove that, for each positive integer n, F_{n} = (1)^{n+1} F_{n}.
 (b) Prove that, for each integer n, t^{n} = F_{n} t+ F_{n1}.
 (c) Let G_{n} be any one of the functions F_{n+1}F_{n}, F_{n+1}F_{n1} and F_{n}^{2}. In each case, prove that G_{n+3} + G_{n} = 2(G_{n+2} + G_{n+1}).
 16.
 (a) Show that, if


 (b) Give an example of numbers a and b that satisfy the condition in (a) and check that both equations hold.
 17.
 The sequence of functions { P_{n} } satisfies the following relations:


View solution
 18.
 Each point in the plane is coloured with one of three distinct colours. Prove that there are two points that are unit distant apart with the same colour.
 13.
 First solution. Let the lengths a, b, c, u and the angles a, b, g, l, m, n be as indicated in the diagram.
 In the solution, we make use of the fact that if p/q = r/s, then both fractions are equal to (p+r)/(q + s). Since ÐDBP = 90^{°} + l 2b, it follows that



 Since a = 2usinl and, by the Law of Sines,


 13.
 Second solution. Let M be the midpoint of BC. A rotation of 180^{°} about M interchanges B and C and takes E to G, D to F and P to Q. Then AB = CE = BG and AC = BD = CF. Join GA and FA. Let 2a = ÐBAC. Since AE  BG and AB is a transversal, ÐGBA = ÐBAC = 2a. Since AB = BG, ÐBGA = 90^{°} a. But ÐBGF = ÐCED = 90^{°}  a. Thus, G, A, F are collinear.
 Since GF and DE are equidistant from M, we can use Cartesian coordinates with the origin at M, the line y = 1 as GF and the line y = 1 as DE. Let A ~ (a, 1), B ~ (u, mu), C ~ (u, mu). Then P ~ (m, 1), Q ~ (m, 1),

 The slope of AE is 1/u and of AD is 1/u, so that


 13.
 Third solution. [M. Boase] Let XAY be drawn parallel to DE.
 Since M is the midpoint of BC, the distance from M to DE is the average of the distances from B and C to DE. Similarly, the distance from M to XY is the average of the distances from B and C to XY. The distance of B (resp. C) to DE equals the distance of C (resp. B) to XY. Hence, M is equidistant from DE and XY. If PM produced meets XY in Q, then PM = MQ and so ÐBQC = ÐBPC.
 Select R on MQ (possibly produced) so that ÐBAC = ÐBRC. Since DADE  DRBC, ÐRBC = ÐRCB = ÐADE. Since BARC is a concyclic quadrilateral, ÐBAR = 180^{°}  ÐRCB = 180^{°}  ÐADE = 180^{°} ÐXAD = ÐBAQ from which it follows that R = Q and so ÐBPC = ÐBQC = ÐBRC = ÐBAC.
 13.
 Fourth solution. [Jimmy Chui] Set coordinates: A ~ (0, (m+n)b), B ~ (ma, nb), C ~ (na, mb) D ~ ((m+n)a, 0) and E ~ ((m+n)a, 0) where m = AB , n = AC  and a^{2} + b^{2} = 1. Then the line BC has the equation





 13.
 Fifth solution. Determine points L and N on DE such that BL  AE and LN = NE. Now


 13.
 Sixth solution. [C. So] Let F, N, G be the feet of the perpendiculars dropped from B, M, C respectively to DE. Note that FN = NG, so that MF = MG. Let ÐADE = ÐAED = q, AB  = c, AC  = b and h be the altitude of DADE. Then


 13.
 Seventh solution. [A. Chan] Let ÐADE = ÐAED = q, so ÐBAC = 180^{°}  2q. Suppose that ÐACB = f, ÐCPE = s and ÐBCP = r. By the Law of Sines for triangles ABC and PCE, we find that




 Hence s q = f r, so, with q+ s = f+ r, we have that

 14.
 (a) First solution. Since c = a + d  b, we have that

 (a) Second solution. Let a + d = b + c = u. Then

 (a) Third solution. Let x = a  b > 0. Since a  b = c  d, we have that a = b + x and d = c  x. Hence

 (a) Fourth solution. Since Öa > Öb > Öc > Öd, Öa  Öd > Öb  Öc. Squaring and using a + d = b + c yields 2Ö[bc] > 2Ö[ad], whence the result.
 14.
 (b) First solution. Since all variables represent integers,


 15.
 (a) First solution. Since F_{0} = F_{2}  F_{1} = 0, the result holds for n = 0. Since F_{1} = F_{1}  F_{0} = 1, the result holds for n = 1. Suppose that we have established the result for n = 0, 1, 2, ¼r. Then

 15.
 (b) First solution. The result holds for n = 0, n = 1 and n = 2. Suppose that it holds for n = 0, 1, 2, ¼,r. Then


 15.
 (b) Second solution. The result holds for n = 1. Suppose that it holds for n = r ³ 0. Then




 15.
 (c) First solution. Let G_{n} = F_{n} F_{n+1}. Then

 Let G_{n} = F_{n+1}F_{n1}. Then

 Let G_{n} = F_{n}^{2}. Then

 Comments. Since F_{n}^{2} = F_{n} F_{n1} + F_{n} F_{n2}, the third result of (c) can be obtained from the first two. J. Chui observed that, more generally, we can take G_{n} = F_{n+u}F_{n+v} where u and v are integers. Then

 16.
 (a) First solution. Let



 16.
 (a) Second solution. [M. Boase]



 16.
 (a) Third solution. [A. Birka] Let cosa = x and cosb = y. Then





 16.
 (a) Fourth solution. [J. Chui] Note that the given equation implies that sin2b = 2sin(a+ b) and that the numerator of



 16.
 (a) Fifth solution. [A. Tang] From the given equation, we have that




 16.
 (a) Sixth solution. [D. Arthur] The given equations yield 2 sin(a+ b) = sin2b, cosasinb = cosb(sina+ sinb) and sinacosb = sinb(cosa+ cosb). Hence

 16.
 (b) First solution. The given equation is equivalent to 2 sin(a+ b) + sin2b = 0. Try b = 45^{°} so that sin(a 45^{°}) = ^{1}/_{2}. We take a = 75^{°}. Now


 17.
 First solution. Taking x = 1, 2, 3, ¼ yields the respective sequences


 From the definition of P_{n}, we find that



 17.
 Second solution. [By inspection, we make the conjecture that P_{n} (x) = x^{2} P_{n1}(x)  P_{n2}. Rather than prove this directly from the rather awkward condition on P_{n}, we go through the back door.] Define the sequence { Q_{n} } for n = 0, 1, 2, ¼ by

 Lemma: Q_{n}^{2} (x)  Q_{n+1}Q_{n1} = x^{2} for n ³ 1.
 Proof: This result holds for n = 1. Assume that it holds for n = k1 ³ 1. Then

 From the lemma, we find that



 Comment: It can also be established that P_{n+1}^{2} + P_{n}^{2} = (1 + P_{n} P_{n+1})x^{2} for each n ³ 0.
 17.
 Third solution. [I. Panayotov] First note that the sequence { P_{n} (x) } is defined for all values of x, i.e., the denominator 1 + P_{n1}(x)P_{n}(x) never vanishes for n and x. Suppose otherwise, and let n be the least number for which there exists u for which 1 + P_{n1}(u)P_{n} (u) = 0. Then n ³ 3 and

 We now prove by induction that P_{n+1} = x^{2} P_{n}  P_{n1}. Suppose that P_{k} = x^{2} P_{k1}  P_{k2} for 3 £ k £ n, so that in particular we know that P_{k} is a polynomial for 1 £ k £ n. Substituting for P_{k} yields



 18.
 First solution. Suppose that the points in the plane are coloured with three colours. Select any point P.
 Consider the diagrammed configuration, in which all segments have length 1. If P, Q, R are all coloured differently, then either the result holds or S must have the same colour as P. If P, U, V are all coloured differently, then either the result holds or W must have the same colour as P. Hence, either one of the triangles PQR and PUV has two vertices the same colour, or else S and W must be coloured the same.
 18.
 Second solution. Suppose, if possible, the planar points can be coloured without two points unit distance apart being coloured the same. Then if A and B are distant Ö3 apart, then there are distinct points C and D such that ACD and BCD are equilateral triangles (ABCD is a rhombus). Since A and B must be coloured differently from the two colours of C and D, A and B must have the same colour. Hence, if O is any point in the plane, every point on the circle of radius Ö3 consists of points coloured the same as O. But there are two points on this circle unit distant apart, and we get a contradiction of our initial assumption.
 18.
 Third solution. Suppose we can colour the points of the plane with three colours, red, blue and yellow so that the result fails. We show that three collinear points at unit distance are coloured with three different colours. Let P, Q, R be three such points, and embed P, Q, R in the hexagon ABPCDR as indicated with all indicated segments of unit length.
 If, say, Q is red, B and A must be coloured differently, as are A and R, R and D, D and C, C and P, P and B. Thus, B, R, C, are one colour, say, blue, and A, D, P the other, say yellow. The preliminary result follows.
 Now consider any isosceles triangle UVW with UV  = UW  = 3 and VW  = 2. It follows from the preliminary result that U and V must have the same colour, as do U and W. But V and W cannot have the same colour and we reach a contradiction.
 18.
 Fourth solution. [D. Arthur] Suppose that the result is false. Let A, B be two points with AB  = 3. Within the segment AB select PQ with AP  = PQ  = QB  = 1, and suppose that R and S are points on the same side of AB with DRAP and DSPQ equilateral. Then RS  = 1. Suppose if possible that A and Q have the same colour. Then P must have a second colour and R and S the third, leading to a contradiction. Hence A must be coloured differently from both P and Q. Similarly B must be coloured differently from both P and Q. Since P and Q are coloured differently, A and B must have the same colour.
 Now consider a trapezoid ABCD with CB  = AB = AD  = 3 and CD  = 1. By the foregoing observation, C, A, B, D must have the same colour. But this yields a contradiction. The result follows.