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.

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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  .

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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  .

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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.

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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.

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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 ~ æ ç è u, 1 - ucosq sinq ö ÷ ø

N ~ æ ç è 1 2 u, 1 - ucosq 2sinq 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^°.

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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) = æ ç è 1 2 (u + v), 1 2 (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.

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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.

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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.