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Problem Set 8

43.
The sides BC, CA, AB of triangle ABC are produced to the poins R, P, Q respectively, so that CR = AP = BQ. Prove that triangle PQR is equilateral if and only if triangle ABC is equilateral.
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44.
Determine polynomials a(t), b(t), c(t) with integer coefficients such that the equation y2 + 2y = x3 - x2 - x is satisfied by (x, y) = (a(t)/c(t), b(t)/c(t)).
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45.
ABC is a triangle with circumcentre O such that ÐA exceeds 90° and AB < AC. Let M and N be the midpoints of BC and AO, and let D be the intersection of MN and AC. Suppose that AD = 1/2(AB + AC). Determine ÐA.
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46.
Determine all functions f from the set of reals to the set of reals which satisfy the functional equation
 (x - y)f(x + y) - (x + y)f(x - y) = 4xy(x2 - y2)
for all real x and y.

47.
Let x, y and z be positive real numbers. Show that
x
 x + __________Ö(x + y)(x + z)
+ y
 y + __________Ö(y + z)(y + x)
+ z
 z + __________Ö(z + x)(z + y)
£ 1  .

48.
For vectors in three-dimensional real space, establish the identity
 [a ×(b - c)]2 + [b ×(c - a)]2 + [c ×(a - b)]2 = (b ×c)2 + (c ×a)2+ (a ×b)2 + (b ×c + c ×a + a ×b)2  .

Solutions to Problem Set 8

43.

43.
First solution. Suppose that triangle ABC is equilateral. A rotation of 60° about the centroid of DABC will rotate the points R, P and Q. Hence DPQR is equilateral. On the other hand, suppose, wolog, that a ³ b ³ c, with a > c. Then, for the internal angles of DABC, A ³ B ³ C. Suppose that |PQ | = r, |QR | = p and |PR | = q, while s is the common length of the extensions. Then
 p2 = s2 + (a + s)2 + 2s(a+s)cosB
and
 r2 = s2 + (c + s)2 + 2s(c+s)cosA .
Since a > c and cosB ³ cosA, we find that p > r, and so DPQR is not equilateral.

44.

44.
First solution. The equation can be rewritten (y + 1)2 = (x - 1)2(x + 1). Let x + 1 = t2 so that y + 1 = (t2 - 2)t. Thus, we obtain the solution
 (x, y) = (t2 - 1, t3 - 2t - 1) .
With these polynomials, both sides of the equation are equal to t6 - 4t4 + 4t2 - 1.

45.

45.
First solution. Assign coordinates: A ~ (0, 0), B ~ (2 cosq, 2 sinq), C ~ (2u, 0) where 90° < q < 180° and u > 1. First, we determine O as the intersection of the right bisectors of AB and AC. The centre of AB has coordinates (cosq, sinq) and its right bisector has equation
 (cosq)x + (sinq)y = 1 .
The centre of segment AC has coordinates (u, 0) and its right bisector has equation x = u. Hence, we find that
 O ~ æç è u, 1 - ucosqsinq ö÷ ø
 N ~ æç è 12 u, 1 - ucosq2sinq ö÷ ø
 M ~ (u + cosq, sinq)
and
 D ~ (u + 1, 0)  .
The slope of MD is (sinq)/(cosq- 1). The slope of ND is (u cosq- 1)/((u+2)sinq). Equating these two leads to the equation
 u(cos2 q- sin2 q- cosq) = 2sin2 q+ cosq- 1
which reduces to
 (u + 1)(2cos2 q- cosq- 1) = 0 .
Since u + 1 > 0, we have that 0 = 2cos2 q- cosq- 1 = (2cosq+ 1)(cosq-1). Hence cosq = -1/2 and so ÐA = 120°.

46.

46.
First solution. Let u and v be any pair of real numbers. We can solve x + y = u and x - y = v to obtain
 (x, y) = æç è 12 (u + v), 12 (u - v) ö÷ ø .
From the functional equation, we find that vf(u) - uf(v) = (u2 - v2)uv, whence
 f(u)u - u2 = f(v)v - v2  .
Thus (f(x)/x) - x2 must be some constant a, so that f(x) = x3 + ax. This checks out for any constant a.

47.

47.
First solution.
 (x - __Öyz )2 ³ 0
 Þ x2 + yz ³ 2x __Öyz
 Þ (x + y)(x + z) = x2 + x(y + z) + yz ³ x(y + 2 __Öyz + z) = x(Öy + Öz)2  .
Hence
x
 x + ________Ö(x+y)(x+z)
£ x
x + Öx(Öy + Öz)
= Öx
Öx + Öy + Öz
.
Similarly
y
 y + ________Ö(y+z)(y+x)
£ Öy
Öx + Öy + Öz
and
z
 z + ________Ö(z+x)(z+y)
£ Öz
Öx + Öy + Öz
.
Adding these inequalities yields the result.

48.

48.
First solution. Let u = b×c, v = c×a and w = a×b. Then, for example, a×(b - c) = a×b - a×c = a×b + c×a = v + w. The left side is equal to
 (v + w)·(v + w) +(u + w)·(u + w) +(u + v)·(u + v) = 2[(u·u) + (v·v) +(w·w) + (u·v) + (v·w) + (w·u)]
while the right side is equal to
 (u·u) + (v·v) + (w ·w) + (u + v + w)2
which expands to the final expression for the left side.

48.
Second solution. For vectors u, v, w, we have the identities
 (u ×v) ×w = (u ·w) v - (v ·w)u
and
 u ·(v ×w) = (u ×v) ·w .
Using these, we find for example that
 [a ×(b - c)] ·[a ×(b - c)]
 = [a ×(b - c) ×a]·(b - c)
 = { (a·a)(b - c) -[(b - c)·a]a } ·(b - c)
 = |a |2 [|b |2 +|c |2 - 2(b·c)]- [(b·a - c·a]2
 = |a |2 [|b |2 +|c |2 - 2(b·c)]- (b·a)2 - (c ·a)2+ 2(b·a)(c·a)  .
Also
 (b×c)·(b ×c)
 = [(b·b)c - (c·b)b]·c
 = |b |2 |c |2 -(c ·b)2
and
 (b ×c) ·(c ×a) = [(b ×c) ×c] ·a = (b ·c)(c ·a) -(c ·c)(b ·a) .
From these the identity can be checked.

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