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 y^{2} + 2y = x^{3}  x^{2}  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
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(x^{2}  y^{2}) 

for all real
x and
y.
View solution
 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 . 

View solution
 48.

For vectors in threedimensional 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 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
p^{2} = s^{2} + (a + s)^{2} + 2s(a+s)cosB 

and
r^{2} = s^{2} + (c + s)^{2} + 2s(c+s)cosA . 

Since
a >
c and cos
B ³ cos
A, 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 = t^{2} so that
y + 1 = (t^{2}  2)t. Thus, we obtain the solution
(x, y) = (t^{2}  1, t^{3}  2t  1) . 

With these polynomials,
both sides of the equation are equal to
t^{6}  4
t^{4} + 4
t^{2}  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
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



and
The slope of
MD is (sin
q)/(cos
q 1). The slope
of
ND is (
u cos
q 1)/((
u+2)sin
q). Equating these
two leads to the equation
u(cos^{2} q sin^{2} q cosq) = 2sin^{2} q+ cosq 1 

which reduces to
(u + 1)(2cos^{2} q cosq 1) = 0 . 

Since
u + 1 > 0, we have that
0 = 2cos
^{2} q cos
q 1 = (2cos
q+ 1)(cos
q1). Hence cos
q = 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) = 
æ ç
è


1 2

(u + v), 
1 2

(u  v) 
ö ÷
ø

. 

From the functional equation, we find that
vf(
u) 
uf(
v) = (
u^{2} 
v^{2})
uv, whence

f(u) u

 u^{2} = 
f(v) v

 v^{2} . 

Thus (
f(
x)/
x) 
x^{2} must be some constant
a, so that
f(
x) =
x^{3} +
ax. This checks out for any constant
a.
47.
 47.

First solution.


 
Þ (x + y)(x + z) = x^{2} + x(y + z) + yz ³ x(y + 2 
 __ Öyz

+ z) = x(Öy + Öz)^{2} . 

 

Hence

x

£ 
x x + Öx(Öy + Öz)

= 
Öx Öx + Öy + Öz

. 

Similarly
and
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
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 ^{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.