CMS/SMC
Société mathématique du Canada
www.smc.math.ca
Société mathématique du Canada
  location: 
       
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.
View solution
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)).
View solution
45.
ABC is a triangle with circumcentre O such that A 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 A.
View solution
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.

View solution

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

View solution

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  .

View solution




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 b c, with a > c. Then, for the internal angles of DABC, A B 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 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 ~

u, 1 - ucosq
sinq


N ~

1
2
u, 1 - ucosq
2sinq


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 A = 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) =

1
2
(u + v), 1
2
(u - v)

  .
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 -   __
yz
 
)2 0
x2 + yz 2x   __
yz
 
(x + y)(x + z) = x2 + x(y + z) + yz x(y + 2   __
yz
 
+ z) = x(y + z)2  .
Hence
x
x +   ________
(x+y)(x+z)
 
x
x + x(y + z)
= x
x + y + z
  .
Similarly
y
y +   ________
(y+z)(y+x)
 
y
x + y + z
and
z
z +   ________
(z+x)(z+y)
 
z
x + y + z
  .
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 - ca]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)bc
= |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.


© Société mathématique du Canada, 2014 : http://smc.math.ca/