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. [XZ] refers to the area of the plane figure XZ.

Let n be an integer exceeding 1. Let a1 , a2 ,, an be posive real numbers and b1 , b2 ,, bn be arbitrary real numbers for which

ij ai bj =0.

Prove that ij 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 ++ an )( b1 + b2 ++ bn )= a1 b1 + a2 b2 ++ an bn .


2 ij bi bj =( b1 + b2 ++ bn )2 -( b1 2 + b2 2 ++ bn 2 ) = ( a1 b1 ++ an bn )2 ( a1 ++ an )2 -( b1 2 ++ bn 2 ) ( a1 2 ++ an 2 )( b1 2 ++ bn 2 ) ( a1 ++ an )2 -( b1 2 ++ bn 2 ) = ( b1 2 ++ bn 2 ) ( a1 ++ an )2 [( a1 2 ++ an 2 )-( a1 ++ 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 b2 bn (wolog). Since ij ai bj =0, not all the bi have the same sign, and so b1 >0> bn . Wolog, we may assume that B b1 + b2 ++ bn 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)++ an ( bn -B)<( a1 + a2 ++ an )( b1 -B),

so that b1 >B. Hence

2 ij bi bj = B2 - bi 2 < B2 - b1 2 <0,

as desired.

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 ( f(x)+ x2 2 )=f ( f(x)+ x2 2 )+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)0. Then

f(y)= y2 y2 =( c2 -y)2 =( c4 -y)2 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)0 (for all x) or else that f(x) 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 v0, we must have that

v2 =f( u2 -v)= u4 -2 u2 v+ v2 0= u2 ( u2 -2v)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(x) =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+4 x2 -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.

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 2 abd.
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 2 b and two having as base a triangle with base b and height 1 2 a. 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 ( abd 12 )= 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 2 ad. The tetrahedron ABCD is the union of the nonoverlapping tetrahedra ABFC and ABFD, each with ΔABF as ``base'' and perpendicular height along CD. Hence the volume of ABCD is equal to

1 3 [ABF](FC+FD)= 1 3 ( 1 2 ad )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 AEBC. Then AD must lie in the plane containing AE and perpendicular to BC. Let F lie on AD produced with EFAD. Note that EF=d. Let G be the foot of the perpendicular from D to AE produced. Then

[ADE]= 1 2 ADEF= 1 2 bd= 1 2 AEGD.

It follows that the volume of ABCD is equal to

1 3 [ABC]GD= 1 6 AEBCGD= 1 6 abd.

Prove that

( 14 + 1 4 )( 34 + 1 4 )( 54 + 1 4 )( 114 + 1 4 ) ( 24 + 1 4 )( 44 + 1 4 )( 64 + 1 4 )( 124 + 1 4 ) = 1 313 .

Solution. The left side can be written as

Π{4 x4 +1:xodd,1x11}÷Π{4 x4 +1:xeven,2x12}.


4 x4 +1 =4 x4 +4 x2 +1-4 x2 =(2 x2 +1)2 -(2x)2 =(2 x2 -2x+1)(2 x2 +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 )][( 102 + 112 )( 112 + 122 )] [( 12 + 22 )( 22 + 32 )][( 32 + 42 )( 42 + 52 )][( 112 + 122 )( 122 + 132 )] = 12 122 + 132 = 1 313 .

Comment. In searching for factors, note that any common divisor of 4 n4 +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)(2 n2 +2n+1),

so that we can try either 2n+1 (which does not work) or 2 n2 +2n+1 to find that 4 n4 +1=(2 n2 +2n+1)(2 n2 -2n+1) and

4(n+1)4 +1=(2 n2 +2n+1)(2 n2 +6n+5)=(2 n2 +2n+1)[2(n+2)2 -2(n+2)+1].

Determine the set of complex numbers z which satisfy

Im( z4 )=(Re( z2 ))2 ,

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 =5 y2 , so that the locus consists of the points z on the lines determined by the equations y=x, y=-x, y=(5-2)x, y=(-5-2)x.
Solution 2. Let z=r(cosθ+isinθ); then z2 = r2 (cos2θ+isin2θ) and z4 = r4 (cos4θ+isin4θ). The condition is equivalent to

r4 sin4θ=( r2 cos2θ)2 &lrArr;2sin2θcos2θ=cos2 2θ.

Hence cos2θ=0 or tan2θ= 1 2 . The latter possibility leads to tan2 θ+4tanθ-1=0 or tanθ=-2±5. This yields the same result as in Solution 1.
Solution 3. Let z=x+yi. Then z2 = x2 - y2 +2xyi and z4 =( x4 -6 x2 y2 + 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±5)x.

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 Γ be the set of circles and S be the set of points belonging to at least two of them. For CΓ and sSC, define f(s,C)=1/k, where k is the number of circles passing through s, including C. For CΓ and s\notC, define f(s,C)=0. Observe that, for each sS,

CΓ f(s,C)=1.

Let CΓ; select sSC for which f(s,C)=1/k is minimum. Let C= C1 , C2 ,, Ck be the circles that contain s. These circles (apart from C) meet C in distinct points, so that

sS f(s,C) 1 k + k-1 k =1.

Hence the number of points in S is equal to

sS CΓ f(s,C)= CΓ sS f(s,C)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.