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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.
In the following solutions, F is the intersection point of BE and 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,

 CEEA · ABBD · DFFC = -1 ,
so that 1/2 ·(-2) ·(DF/FC) = -1 and DF = FC, as desired.

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,

 DF:FC = [DEF]:[CEF] = [DBF]:[CBF] = ([DEF]+[DBF]):([CEF]+[CBF])

 = [BDE]:[BEC] = 1:1 .

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,

 ADDB · BLLC · CEAE = 1 ,
whence BL = 2LC and, so, LE || AB. Since the triangles ABC and ELC are similar with factor 3, AB = 3EL. Let EL intersect CD at M. Then the triangles AFB and LFE are similar, so that FD = 3FM. But,

 FD + FM + MC = DC = 3MC Þ 2FM = MC ÞFC = FM + MC = 3FM = FD ,
as desired.

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

 a + mb + u = 0
and

 b + la + v = 0 .
Hence

 u = -a - mb    and    v = -b - la .
Therefore, from triangle ABE,

 0
 = (l+ 1)a - 2v + 2u
 = (l+ 1)a + 2b + 2la- 2a - 2mb
 = (3l- 1)a + 2(1 - m)b .
Since { a, b } is a linearly independent set, l = 1/3 and m = 1, yielding the desired result.

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

 12z - 3(z+1) = 3(3z - 1) = 3[3(z+1) - 4] ,
i.e., the vector [( ®) || BP] is a real multiple of [( ®) || PE]. The result follows.

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.
Solution 1. The edges of the tetrahedron can be realized as the diagonals of the six faces of a rectangular parallelepiped with edges of length u, v, w in such a way that a2 = v2 + w2, b2 = u2 + w2 and c2 = u2 + v2. The tetrahedran can be obtained from the parallelepiped by trimming away four triangular pyramids each with three mutually perpendicular faces (surrounding a corner of the parallelepiped) and three pairwise orthogonal edges of lengths u, v, w. Hence the volume of the tetrahedron is equal to

 uvw - 4((1/6)uvw) = (1/3)uvw .

>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

 Ö212 __________________________________Ö(b2 + c2 - a2)(c2 + a2 - b2)(a2 + b2 - c2) .

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

cosa = a2 + c2 - b2
 2a ______Öc2 - h2

cosb = b2 + c2 - a2
 2b ______Öc2 - h2
and

 cos(a+ b) = a2 + b2 - c22ab ,
whence

 a2 + b2 - c22ab =

 (a2 + c2 - b2)(b2 + c2 - a2) - ________________________Ö4a2(c2 - h2) - (a2 + c2 - b2)2 ________________________Ö4b2(c2 - h2) - (b2 + c2 - a2)2

4ab(c2 - h2)
.
Shifting terms and squaring leads to

 [2(a2 + b2 - c2)(c2 - h2) - (a2 + c2 - b2)(b2 + c2 - a2)]2 = [4a2(c2 - h2) - (a2 + c2 - b2)2][4b2(c2 - h2) - (b2 + c2 - a2)2] .
With u = b2 + c2 - a2, v = c2 + a2 - b2, w = a2 + b2 - c2, z = c2 - h2, this can be rendered

 0
 = [2wz - uv]2 - [4a2z - v2][4b2z - u2]
 = z[4(w2 - 4a2b2)z - 4(uvw - a2u2 - b2v2)]
so that

 c2 - h2 = z = a2 u2 + b2 v2 - uvw4a2b2 - w2
and

 h2 = 4a2b2c2 + uvw - a2u2 - b2v2 - c2w24a2b2 - w2 .
Now

 4a2b2 - w2
 = - a4 - b4 - c4 + 2a2b2 + 2a2c2 + 2b2c2
 = (a + b + c)(a + b - c)(b + c - a)(c + a - b)
 = 16S2 ,
where S is the area of triangle ABC.

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

 h2 = 2uvw16S2 = uvw8S2 .
The volume V of the tetrahedron satisfies

 V2 = æç è Sh3 ö÷ ø 2 = S2h29 = uvw72 ,
whence

V =
 ___Öuvw

6 Ö2
= 2uvw
12
.

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 1. Suppose that the lengths of the sides of the triangle are a, b and c, using the conventional notation. Then the distance from B of the following points on the side BC are given by (B, 0), (C, a), (M, a/2), (U, ca/(b+c)), (D, (a+c-b)/2) and (P, ccosB) = (P, (a2 + c2 - b2)/(2a)). One can then verify the desired relation by calculation.

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

 |PU ||DU | = |AP ||ID | = 2sa = 1 + b+ca ,
whence

 |PD ||DU | = b+ca .
Now |PM | = (a/2) - u = (v-u)/2 and |DM | = (b-c)/2. Hence

 |PM ||DM | = v-ub-c .
By Pythagoras' Theorem, c2 - u2 = b2 - v2, whence

 v - ub - c = b + cv + u = b + ca
and the result follows.

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 1. Let X denote the vector from an origin to a point X. Then, vectorially, it can be verified that

 (A - B)·(A - B)
 + (B - C)·(B - C)+ (C - D)·(C - D) + (D - A)·(D - A)
 -(A - C)·(A - C) - (B - D)·(B - D)
 = -2A ·B - 2B ·C - 2C ·D - 2D ·A +2A ·C + 2B ·D + A2 + B2 + C2 + D2
 = 4 æç è A + C2 - B + D2 ö÷ ø · æç è A + C2 - B + D2 ö÷ ø ,
which yields the desired result.

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

 |AB |2 + |BC |2+ |CD |2 + |DA |2
 = 4[ |PX |2 + |PW |2 +|PZ |2 + |PY |2 ]
 = 2[ |PQ |2 + |XZ |2+ |PQ |2 + |WY |2 ]
 = 4|PQ |2 + 2 [ |XZ |2 + |WY |2 ]
 = 4|PQ |2 + 4 [ |WZ |2 + |WX |2 ]
 = 4 |PQ |2 + |BD |2 + |AC |2  .

196.
Determine five values of p for which the polynomial x2 + 2002x - 1002p has integer roots.
Answer. Here are some values of (p; u, v) with u and v the corresponding roots: (0; 0, -2002),

(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

 1002p = q2 - 10012 = (q - 1001)(q + 1001) .
We select q so that either q - 1001 or q + 1001 is divisible by 1002. For example q = 2003, 1, 3005, 4007 all work. We can also make one factor divisible by 667 and the other by 6.

197.
Determine all integers x and y that satisfy the equation x3 + 9xy + 127 = y3.
Solution 1. Let x = y + z. Then the equation becomes (3z + 9)y2 + (3z2 + 9z)y + (z3 + 127) = 0, a quadratic in y whose discriminant is equal to

 (3z + 9)2 z2
 - 4(3z + 9)(z3 + 127)
 = (3z + 9) [ z2 (3z + 9) - 4(z3 + 127) ]
 = - (3z + 9)(z3 - 9z2 + 508) .
Note that z3 - 9z2 + 508 = z2(z - 9) + 508 is nonnegative if and only if z ³ -5 (z being an integer) and that 3z + 9 is nonnegative if and only if z ³ -3. Hence the discriminant is nonnegative if and only if z = -3, -4, -5. >From the quadratic equation, we have that z3 + 127 º 0 (mod 3). The only possibility is z = -4 and this leads to the equation 0 = -3y2 + 12y + 63 = -3(y - 7)(y + 3) and the solutions (x, y) = (3, 7),(-7, -3).

Solution 2. The equation can be rewritten

 (x - y)[(x - y)2 + 3xy] + 9xy = -127
or

 u3 + 3v(u + 3) = -127
where u = x - y and v = xy. Hence

 3v = - u3 + 127u + 3 = - éê ë (u2 - 3u + 9) + 100u + 3 ùú û .
Therefore, u3 + 127 º 0 (mod 3), so that u º 2 (mod 3), and u + 3 divides 100. The candidates are

 u = -103, -28, -13, -7, -4, -1, 2, 17, 47 .
Checking these out leads to the posible solutions.

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.
Solution. Since the polynomial is nonconstant, d ³ 1, so that the result holds for p = 2. Henceforth, assume that p is an odd prime. Let 0 £ k £ p - 2. Consider the polynomial

 pk (x) = x(x-1)(x-2)¼(x - k + 1)(x - k - 1)¼(x - p + 2)k!(p-k-2)!(-1)p-k .
We have that pk(k) = 1 and pk(x) = 0 when x = 0, 1, 2, ¼,k - 1, k + 1, ¼, p - 2. Let

 g(x) = p-2å k=0 f(k)pk (x) .
Then the degree of g(x) does not exceed p - 2 and g(x) = f(x) for x = 0, 1, 2, ¼, p-2; in fact, g(x) is the unique polynomial of degree less than p-1 that agrees with f at these p-1 points (why?).

Now

 g(p - 1) = p-2å k=0 (-1)p-k (p-1)!k!(p-k-1)! f(k) = p-2å k=0 (-1)p-k æç è p-1 k ö÷ ø f(k) .
Since ((p-1) || k) = (p || k) - ((p-1) || (k-1)) and (p || k) º 0 (mod p) for 1 £ k £ p-1, and induction argument yields that ((p-1) || k) º (-1)k for 1 £ k £ p-1, so that

 g(p-1) º (-1)p p-2å k=0 f(k)
(mod p). Since f(0) = 0 and f(1) = 1, it follows that åk=0p-2 f(k) is congruent to some number between 1 and p-2 inclusive, so that g(p-1) \not º 0 and g(p-1) \not º 1 (mod p). Hence f(p-1) ¹ g(p-1), so that f and g are distinct polynomials. Thus, the degree of g exceeds p-2 as desired.

By Fermat's Little Theorem, the polynomial xp-1 satisfies the condition.

Solution 2. [M. Guay-Paquet] Let

 h(x) = f(x) + f(2x) + ¼+ f((p-1)x) .
Then h(1) \not º 0 (mod p) and h(0) = 0. The degree of h is equal to d, the degree of f.

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 xk-1 must be nonzero, and so d ³ p-1, as desired.

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