Note. The incentre of a triangle is the centre
of the inscribed circle that touches all three sides. A set
is connected if, given two points in the set, it is
possible to trace a continuous path from one to the other
without leaving the set. [X¼Z] refers to the area
of the plane figure X ¼Z.
-
121.
-
Let n be an integer exceeding 1.
Let a1, a2, ¼, an be posive real numbers
and b1, b2, ¼, bn be arbitrary real numbers for which
Prove that åi ¹ j bi bj < 0 .
Solution 1. For the result to hold, we need to
assume that at least one of the bi is nonzero. The condition is that
|
(a1 + a2 + ¼+ an)(b1 + b2 + ¼+ bn) = a1b1 + a2b2 + ¼+ anbn . |
|
Now
|
|
|
= (b1 + b2 + ¼+ bn)2 - (b12 + b22 + ¼+ bn2) |
| |
|
= |
(a1b1 + ¼+ anbn)2
(a1 + ¼+ an)2
|
- (b12 + ¼+ bn2) |
| |
|
£ |
(a12 + ¼+ an2)(b12 + ¼+ bn2)
(a1 + ¼+ an)2
|
- (b12 + ¼+ bn2) |
| |
| = |
(b12 + ¼+ bn2)
(a1 + ¼+ an)2
|
[ (a12 + ¼+ an2) - (a1 + ¼+ an)2] < 0 , |
|
|
from which the desired result follows. The inequality is due to the
Cauchy-Schwarz Inequality.
Solution 2. [R. Barrington Leigh] Suppose that not all the
bi vanish and that b1 ³ b2 ³ ¼ ³ bn (wolog).
Since åi ¹ j aibj = 0, not all the bi
have the same sign, and so b1 > 0 > bn. Wolog, we may assume that
B º b1 + b2 + ¼+ bn ³ 0. (If B < 0,
we can change the signs of all the bi which alters neither
the hypothesis nor the conclusion.) We have that
|
0 = a1(b1 - B) + a2(b2 - B) + ¼+ an(bn - B) < (a1 + a2 + ¼+ an)(b1 - B) , |
|
so that
b1 > B. Hence
|
2 |
å
i ¹ j
|
bi bj = B2 - |
å
| bi2 < B2 - b12 < 0 , |
|
as desired.
-
122.
-
Determine all functions f from the real numbers
to the real numbers that satisfy
|
f(f(x) + y) = f(x2 - y) + 4f(x)y |
|
for any real numbers x, y.
Solution 1. Let y = 1/2(x2 - f(x)). Then
|
f |
æ
ç
è
|
|
f(x) + x2
2
|
|
ö
÷
ø
|
= f |
æ
ç
è
|
|
f(x) + x2
2
|
|
ö
÷
ø
|
+ 2f(x)[x2 - f(x)] , |
|
from which it follows that, for each x, either f(x) = 0
or f(x) = x2. [Note: this does not imply yet that
the same option holds for all x.] In particular, f(0) = 0,
so that f(y) = f(-y) for all y.
Suppose that f(c) = 0. Then, for each real y,
f(y) = f(c2 - y), whence f(c2) = f(0) = 0. Thus, for each
real y, f(y) = f(c4 - y). Suppose that f(y) ¹ 0. Then
|
f(y) = y2 Þ y2 = (c2 - y)2 = (c4 - y)2 Þ 0 = c2(c2 - 2y) = c4(c4 - 2y) . |
|
If c were nonzero, then we would have c2/2 = y = c4/2, so
c = 1 and y = 1/2. But then f(-y) = f(y) = f(1 - y);
substituting y = -1/2 yields 1/4 = f(1/2) = f(3/2), which is false.
Hence c = 0. It follows that, either f(x) º 0
(for all x) or else that f(x) º x2 (for all x).
These solutions can (and should be) checked.
Solution 2. Let y = x2 - f(x). Then
|
f(x2) = f(f(x) + x2 - f(x)) = f(f(x)) + 4f(x)[x2 - f(x)] . |
|
Taking y = 0, we see that f(f(x)) = f(x2), so that
4f(x)[x2 - f(x)] = 0. Hence, for each x, either
f(x) = 0 or f(x) = x2.
Suppose, if possible, that there are two nonzero reals u and
v for which f(u) = 0 and f(v) = v2. Setting (x, y) = (u, v) yields that v2 = f(u2 - v). Since v ¹ 0, we
must have that
|
v2 = f(u2 - v) = u4 - 2u2v + v2Þ 0 = u2(u2 - 2v) Þ v = |
1
2
|
u2 . |
|
This would mean that we could find only one such pair (u, v),
which is false. Hence this case is not possible, so that, either
f(x) = 0 for all x or else that f(x) = x2 for all x.
Solution 3. From (x, y) = (0, 0), we have that
f(f(0)) = f(0). From (x, y) = (0, -f(0)), we have that
f(0) = f(f(0)) - 4f(0)2, whence f(0) = 0. From x = 0, we
have that f(y) = f(-y) for all y. Finally, taking y = x2
and y = -f(x), we get
|
f(x2 + f(x)) = f(0) + 4f(x)x2 = f(x2 + f(x)) - 4f(x)2+ 4f(x)x2 . |
|
so that 0 = 4f(x)[x2 - f(x)]. We can finish as in the other
solutions.
Solution 4. [R. Barrington Leigh] Taking y = x - f(x) and
then y = x2 - x yields that
|
|
|
= f(f(x) + x - f(x)) = f(x2 - x + f(x)) + 4f(x)(x - f(x)) |
| |
|
= f(x2 - (x2 - x)) + 4f(x)(x2 - x) + 4f(x)(x - f(x)) |
| |
| = f(x)[1 + 4x2 - 4x + 4x - 4f(x)] = f(x) + 4f(x)[x2 - f(x)] , |
|
|
so that for each x, either f(x) = 0 or f(x) = x2. The solution
can be completed as before.
-
123.
-
Let a and b be the lengths of two opposite
edges of a tetrahedron which are mutually perpendicular and
distant d apart. Determine the volume of the tetrahedron.
Solution 1. Construct parallel planes distant d apart that
contain the edges of lengths a and b. In the planes, congruent
parallelograms can be constructed whose diagonals are of lengths
a and b and right bisect each other, and each of which has
an edge of the tetrahedron as a diagonal. Each parallelogram can
be obtained from the other by a translation relating their centres,
so the two parallelograms bound a prism with opposite faces distant
d apart. The volume of this prism is 1/2abd.
The prism is the disjoint union of the given tetrahedron and four
tetrahedra, all of height d, two having as base a triangle with
base a and height 1/2b and two having as base a triangle
with base b and height 1/2a. Each of these latter four
tetrahedra have volume 1/3(1/2·[ab/2])d = [abd/12]. Hence, the volume of the given tetrahedron is
|
|
abd
2
|
- 4 |
æ
ç
è
|
|
abd
12
|
|
ö
÷
ø
|
= |
1
6
|
(abd) . |
|
Solution 2. Suppose that ABCD is the tetrahedron with
opposite edges AB of length a and CD of length b
orthogonal and at distance d from each other.
Case (i). Suppose that AB and CD are oriented so that
there are points E and F on AB and CD respectively for which
EF is perpendicular to both AB and CD. Then |EF | = d and [ABF] = 1/2ad. The tetrahedron ABCD is the union
of the nonoverlapping tetrahedra ABFC and ABFD, each with
DABF as ``base'' and perpendicular height along CD.
Hence the volume of ABCD is equal to
|
|
1
3
|
[ABF](|FC |+ |FD |) = |
1
3
|
|
æ
ç
è
|
|
1
2
|
ad |
ö
÷
ø
|
|CD | = |
1
6
|
abd . |
|
Case (ii). Suppose that E and F are on AB possibly
produced and on CD produced, say, with EF perpendicular to
AB and CD. Then we can argue in a way similar to that in
Case (i) that the volume of ABCD is equal to the volume
of ABFC less the volume of ABFD to obtain the answer (1/6)abd.
Solution 3. [C. Lau; H. Lee]
Let ABCD be the given tetrahedron with
|BC | = a and |AD | = b. Suppose E lies on
BC, possibly produced, with AE ^BC. Then AD must lie
in the plane containing AE and perpendicular to BC. Let
F lie on AD produced with EF ^AD. Note that |EF | = d. Let G be the foot of the perpendicular from D to AE
produced. Then
|
[ADE] = |
1
2
|
|AD ||EF | = |
1
2
|
bd = |
1
2
|
|AE ||GD | . |
|
It follows that the volume of ABCD is equal to
|
|
1
3
|
[ABC]|GD | = |
1
6
|
|AE ||BC ||GD | = |
1
6
|
abd . |
|
-
124.
-
Prove that
|
|
|
(14 + |
1
4
|
)(34 + |
1
4
|
)(54 + |
1
4
|
)¼(114 + |
1
4
|
) |
|
(24 + |
1
4
|
)(44 + |
1
4
|
)(64 + |
1
4
|
)¼(124 + |
1
4
|
) |
|
= |
1
313
|
. |
|
Solution. The left side can be written as
|
|
Õ
| { 4x4 + 1 : x odd, 1 £ x £ 11 }¸ |
Õ
| { 4x4 + 1 : x even, 2 £ x £ 12 } . |
|
Now
|
|
|
= 4x4 + 4x2 + 1 - 4x2 = (2x2 + 1)2 - (2x)2 |
| |
| = (2x2 - 2x + 1)(2x2 + 2x + 1) = [(x-1)2 + x2][x2 + (x+1)2] . |
|
|
>From this, we see that the left side is equal to
|
|
[12(12 + 22)][(22 + 32)(32 + 42)] ¼[(102 + 112)(112 + 122)]
[(12 + 22)(22 + 32)][(32 + 42)(42 + 52)] ¼[(112 + 122)(122 + 132)]
|
= |
12
122 + 132
|
= |
1
313
|
. |
|
Comment. In searching for factors, note that any common
divisor of 4n4 + 1 and 4(n+1)4 + 1 must divide the difference
|
[(n4 + 1)4 - n4] = [(n+1)2 - n2][(n+1)2 + n2] = (2n+1)(2n2 + 2n + 1) , |
|
so that we can try either 2n + 1 (which does not work)
or 2n2 + 2n + 1 to find that
4n4 + 1 = (2n2 + 2n + 1)(2n2 - 2n + 1) and
|
4(n+1)4 + 1 = (2n2 + 2n + 1)(2n2 + 6n + 5) = (2n2 + 2n + 1)[2(n+2)2 - 2(n+2) + 1] . |
|
-
125.
-
Determine the set of complex numbers z which
satisfy
and sketch this set in the complex plane.
(Note: Im and Re refer respectively to the imaginary
and real parts.)
Solution 1. Let z = x + yi and z2 = u + vi. Then
u = x2 - y2, v = 2xy and z4 = (u2 - v2) + 2uvi.
Im (z4) = (Re (z2))2 implies that
2uv = u2. Thus, u = 0 or u = 2v. These reduce to
x2 = y2 or (x - 2y)2 = 5y2, so that the locus consists
of the points z on the lines determined by the equations
y = x, y = -x, y = (Ö5 - 2)x, y = (-Ö5 - 2)x.
Solution 2. Let z = r(cosq+ isinq); then
z2 = r2(cos2q+ isin2q) and
z4 = r4(cos4q+ isin4q). The condition is
equivalent to
|
r4 sin4q = (r2 cos2q)2Û 2sin2qcos2q = cos2 2q . |
|
Hence cos2q = 0 or tan2q = 1/2. The
latter possibility leads to tan2 q+ 4tanq- 1 = 0
or tanq = -2 ±Ö5. This yields the same result as
in Solution 1.
Solution 3. Let z = x + yi. Then z2 = x2 - y2 + 2xyi
and z4 = (x4 - 6x2y2 + y4) + 4xy(x2 - y2)i.
Then the condition in the problem is equivalent to
|
4xy(x2 - y2) = (x2 - y2)2 , |
|
which in turn is equivalent to y = ±x or
y2 + 4xy - x2 = 0, i.e., y = (-2 ±Ö5)x.
-
126.
-
Let n be a positive integer exceeding 1, and
let n circles (i.e., circumferences)
of radius 1 be given in the plane such that
no two of them are tangent and the subset of the plane formed
by the union of them is connected. Prove that the number of
points that belong to at least two of these circles is at least
n.
Solution. Let G be the set of circles and S be
the set of points belonging to at least two of them. For
C Î G and s Î S ÇC, define f(s, C) = 1/k, where
k is the number of circles passing through s, including C.
For C Î G and s Ï C, define f(s, C) = 0.
Observe that, for each s Î S,
Let C Î G; select s Î S ÇC for which
f(s, C) = 1/k is minimum. Let C = C1, C2, ¼, Ck
be the circles that contain s. These circles (apart
from C) meet C in distinct points, so that
|
|
å
s Î S
|
f(s, C) ³ |
1
k
|
+ |
k-1
k
|
= 1 . |
|
Hence the number of points in S is equal to
|
|
å
s Î S
|
|
å
C Î G
|
f(s, C) = |
å
C Î G
|
|
å
s Î S
|
f(s, C) ³ n . |
|
Comment. The full force of the connectedness condition is
not needed. It is required only that each circle intersect with
at least one other circle.
