Solutions for the December Problems

192.

Let ABC be a triangle, D be the midpoint of AB and E a point on the side AC for which AE = 2EC. Prove that BE bisects the segment CD.

Solution 1. Let G be the midpoint of AE. Then AG = GE = EC and DG || BE. In triangle ADC, DG || FE and GE = EC, from which it follows that DF = FC, as required.

Solution 2. Let u = [ADF] = [BDF] (where [ ¼] denotes area), v = [AFE], w = [CFE] and z = [BFC]. Then 2u + v = 2(w + z) and v = 2w, whence 2u = 2z and u = z. It follows from this (two triangles with the same height and equal collinear bases) that F is the midpoint of CD.

Solution 3. By Menelaus' Theorem, applied to triangle ACD and transversal BFE,

Solution 4. [T. Yue] Let K be the midpoint of AC; then BC = 2DK and BC || DK. Suppose that BE produced and DK produced meet at H. Since triangles EBC and EHK are similar and EC = 2EK, it follows that BC = 2KH and so DH = BC. Thus, DHCB is a parallelogram whose diagonals BH and CD must bisect each other. The result follows.

Solution 5. Place the triangle in the cartesian plane so that B ~ (0, 0), C ~ (3, 0) and A ~ (6a, 6b). Then D ~ (3a, 3b), E ~ (2(a+1), 2b) and the lines BE and and CD have the respective equations y = bx/(a+1) and y = b(x-3)/(a-1). These lines intersect at the point ((3/2)(a+1), (3/2)b), and the result follows.

Solution 6. [L. Chen] [BDE] = [ADE] = ¹/₂[ABE] = [BEC]. Let M and N be the respective feet of the perpendiculars from D and C to BE. Then [BDE] = [BEC]Þ DM = CN. Since DMF and CNF are similar right triangles with DM = CN, they are congruent and so DF = CF.

Solution 7. [F. Chung; Y. Jean] As in the previous solution, [BDE] = [BEC]. Therefore,

Solution 8. [Y. Wei] Let U be a point on BC such that DU || AC. Suppose that DU and BE intersect in V. Then 2EC = AE = 2DV, so that DV = EC. Also ÐVDF = ÐECF abd ÐDFV = ÐECF, so that triangles DVF and CEF are congruent. Hence DF = FC.

Solution 9. Let AF produced meet BC at L. By Ceva's Theorem,

Solution 10. [H. Lee] Let u = [( ®) || DB], v = [( ®) || EC], a = [( ®) || BF], la = [( ®) || FE], b = [( ®) || CF] and mb = [( ®) || FD]. Then

Solution 11. [M. Zaharia] Place masses 1, 1, 2, respectively, at the vertices A, B, C. We locate the centre of gravity of these masses in two ways. Since the masses at A and B have their centre of gravity at D, we can get an equivalent system by replacing the masses at A and B by a mass 2 at the point D. The centre of gravity of the original set-up is equal to the centre of gravity of masses of 2 placed at each of D and C, namely at the midpoint of CD.

On the other hand, the centre of gravity of the masses at A and C is at E. So the centre of gravity of the original set-up is equal to the centre of gravity of a mass 3 located at E and a mass 1 located at B, namely on the segment BE (at the point F for which BF = 3FE). Since both BE and CD contain the centre of gravity of the original set-up, the result follows.

Solution 12. Place the triangle in the complex plane with C at 0, B at 12z and A at 12. Then D is located at 6(z+1) and E at 4. Let P be the midpoint 3(z+1) of CD. Then, BP and PE are collinear since

193.

Determine the volume of an isosceles tetrahedron for which the pairs of opposite edges have lengths a, b, c. Check your answer independently for a regular tetrahedron.

>From the foregoing equations, 2u² = b² + c² - a², 2v² = c² + a² - b² and 2w² = a² + b² - c². (By laying out the tetrahedron flat, we see that the triangle of sides a, b, c is acute and the right sides of these equations are indeed positive.) It follows that the volume of the tetrahedron is

In the case of a regular tetrahedron of side 1, the height is equal to Ö[(2/3)] and the area of a side is equal to Ö3/4, and the formula checks out.

Solution 2. [D. Yu] Let the base of the tetrahedron be triangle ABC, eith a = |BC | = |AD |, b = |AC | = |BD |, c = |AB | = |CD |; let P be the foot of the perpendicular from D to the plane of ABC and let h = |DP |. Then |AP | = Ö[(a² - h²)], |BP | = Ö[(b² - h²)], |CP | = Ö[(c² - h²)].

Suppose that a = ÐBCP and b = ÐACP. Then using the Law of Cosines on triangles BCP, ACP and ABC, we obtain that

Now consider the numerator of h². Its value when w = a² + b² - c² is set equal to 0 is 4a²b²c² - a²u² - b²v² = 4a²b²c² - a²(2b²) - b²(2a²) = 0, so that w divides the numerator. So also do u and v. Hence the numerator of degree 6 in a, b, c must be a multiple of uvw, also of degree 6 in a, b, c. Hence the numerator is a multiple of uvw. Comparing the coefficients of a⁶ (say) gives that the numerator must be 2uvw. Hence

The checking for the tetrahedron proceeds as before.

194.

Let ABC be a triangle with incentre I. Let M be the midpoint of BC, U be the intersection of AI produced with BC, D be the foot of the perpendicular from I to BC and P be the foot of the perpendicular from A to BC. Prove that

|PD ||DM | = |DU ||PM | .

Solution 2. [L. Chen] Let the side lengths of the triangle be a, b, c, as conventional, and, wolog, suppose that c < b. Let u = |BP | and v = |PC |. Then, equating two expressions for the area of the triangle, with r = |ID | as the inradius, we find that |AP | = 2rs/a. From similar triangle, we have that

195.

Let ABCD be a convex quadrilateral and let the midpoints of AC and BD be P and Q respectively, Prove that

|AB |2 + |BC |2 + |CD |2 +|DA |2 = |AC |2 + |BD |2 +4 |PQ |2  .

Solution 2. [T. Yin] We use the result that for any parallelogram KLMN, 2|KL |²+ 2|LM |² = |KM |² + |LN |². This is straightforward to verify using the Law of Cosines, for example. Let W, X, Y, Z be the respective midpoints of the sides AB, BC, CD, DA. Using the fact that all of WXYZ, PXQZ and PWQY are parallelograms, we have that

196.

Determine five values of p for which the polynomial x² + 2002x - 1002p has integer roots.

(4; 2, -2004), (784; 336, -2338), (1780; 668, -2670), (3004; 2002, -3004), (3012; 1004, -3006),

(4460; 1338, -3340), (8012; 2004, -4006), (8024; 2006, -4008), (-556; -334, -1668),

(-1000; -1000, -1002).

Solution 1. If x satisfies the equation x² + 2002x - 1002p = 0, then we must have p = x(x + 2002)/(1002). If we choose integers x for which x(x + 2002) is a multiple of 1002, then this value of p will be an integer that yields a quadratic with two integer roots, namely x and -2002-x. One way to do this is to select either x º 0 or x º 2 (mod 1002). Observing that 1002 = 2 ×3 ×167, we can also try to make x º 0 (mod 167) and x º 2 (mod 6). For example, x = 668 works. We can also try x º 2 (mod 167) and x º 0 (mod 6); in this case, x = 336 works.

Solution 2. The discriminant of the quadratic is 4 times 1001² + 1002p. Suppose that p is selected to make this equal to a square q². Then we have that

197.

Determine all integers x and y that satisfy the equation x³ + 9xy + 127 = y³.

Solution 2. The equation can be rewritten

198.

Let p be a prime number and let f(x) be a polynomial of degree d with integer coefficients such that f(0) = 0 and f(1) = 1 and that, for every positive integer n, f(n) º 0 or f(n) º 1, modulo p. Prove that d ³ p - 1. Give an example of such a polynomial.

Now

By Fermat's Little Theorem, the polynomial x^p-1 satisfies the condition.

Solution 2. [M. Guay-Paquet] Let

Let x \not º 0 (mod p). Then (x, 2x, 3x, ¼, (p-1)x) is a permutation of (1, 2, 3, ¼, p-1), so that h(x) º h(1) (mod p).

Suppose that g(x) = h(x) - h(1). The degree of g is equal to d, g(0) º -h(1) \not º 0 (mod p) and g(x) º 0 whenever x \not º 0 (mod p). Therefore, g(x) differs from a polynomial of the form k(x-1)(x-2)¼(x-[`(p-1)]) by a polynomial whose coefficients are multiples of p. Since k \not º 0 (mod p) (check out the value at 0), the coefficient of x^k-1 must be nonzero, and so d ³ p-1, as desired.