Solutions

109.

Suppose that

x^{2} + y^{2} x^{2}  y^{2}

+ 
x^{2}  y^{2} x^{2} + y^{2}

= k . 

Find, in terms of k, the value of the expression

x^{8} + y^{8} x^{8}  y^{8}

+ 
x^{8}  y^{8} x^{8} + y^{8}

. 

Solution 1. Simplifying, we obtain that
k = 
2(x^{4} + y^{4}) x^{4}  y^{4}

, 

and, by extension, that

k^{2} + 4 2k

= 
k 2

+ 
2 k

= 
x^{4} + y^{4} x^{4}  y^{4}

+ 
x^{4}  y^{4} x^{4} + y^{4}

= 
2(x^{8} + y^{8}) x^{8}  y^{8}

. 

Continuing on, we find that

x^{8} + y^{8} x^{8}  y^{8}

+ 
x^{8}  y^{8} x^{8} + y^{8}

= 
2(x^{16} + y^{16}) x^{16}  y^{16}

= 
k^{2} + 4 4k

+ 
4k k^{2} + 4

= 
k^{4} + 24k^{2} + 16 4k(k^{2} + 4)

. 

Comment. R. Barrington Leigh defined a formula
R = 
a+b ab

+ 
ab a+b

= 2· 
a^{2} + b^{2} a^{2}  b^{2}



for a
¹ ±b which he then applied to (a, b) = (x
^{2}, y
^{2}), (x
^{4}, y
^{4}).

110.

Given a triangle ABC with an area of 1.
Let n > 1 be a natural number. Suppose that
M is a point on the side AB with AB = nAM,
N is a point on the side BC with BC = nBN, and
Q is a point on the side CA with CA = nCQ.
Suppose also that {T} = AN ÇCM, {R} = BQ ÇAN and
{S} = CM ÇBQ, where Ç signifies that the singleton
is the intersection of the indicated segments. Find the
area of the triangle TRS in terms of n.
Solution 1. [R. Furmaniak, Y. Ren] The area of a triangle
XYZ will be denoted by [XYZ]. Consider the triangle
ABQ and the line MC that intersects AB at M, BQ at
S and AQ at the external point C. By Menelaus' Theorem
for the triangle ABQ and transversal MC,
1 = 
BM MA

· 
AC QC

· 
QS SB

= (n1)n 
QS SB

Þ 
SB QS

= (n1)n . 

Observe the triangles BSC and QSC. Since the heights from
C to the opposite sides SB and QS coincide, then

[BSC] [QSC]

= 
SB QS

= (n1)nÞ [BSC] = (n1)n[QSC] . 

Examining triangles QSC and QBC, we similarly find that

[QSC] [QBC]

= 
QS QB

= 
QS QS + SB

= 
1 (n1)n + 1

= 
1 n^{2}  n + 1



Þ [QSC] = 
1 n^{2}  n + 1

·[QBC] . 

Since the heights of triangles QBC and ABC from B to QC and
AC coincide, it follows that

[QBC] [ABC]

= 
QC AB

= 
1 n

Þ [QBC] = 
1 n

·[ABC] = 
1 n



Þ [QSC] = 
1 n^{2}  n + 1

· 
1 n

= 
1 n(n^{2}  n + 1)



Þ [BSC] = 
(n1)n n(n^{2}  n + 1)

= 
n1 n^{2}  n + 1

. 

Similarly,
[TAC] = [RBA] = 
n1 n^{2}  n + 1

. 

Now
[TSR] = [ABC]  [BSC]  [TAC]  [RBA] = 1  
3(n1) n^{2}  n + 1

= 
(n2)^{2} n^{2}  n + 1

. 

Solution 2. [M. Butler] Using Menelaus' Theorem with
triangle BQC and transversal ARN, we find that
so that
Thus,

 
= 
é ê
ë

1 + 
(n1)^{2} n


ù ú
û

[ARB] 
 
= 
é ê
ë


n^{2}  n + 1 n


ù ú
û

[ARB] 


whence
[ARB] = 
n1 n^{2}  n + 1

. 

Therefore
[RST] = 1  
3(n1) n^{2}  n + 1

= 
(n2)^{2} n^{2}  n + 1

. 

Solution 3. Let a = [AMT], b = [BNR], c = [CQS],
x = [BRT], y = [CSR], z = [ATS] and d = [RST]. Then
using BM = (n1)AM, [BRM] = (n1)a,
x + d = [BTS] = (n1)[ATS] = (n1)z 

and
nb + y = [BRC] + [CSR] = [BSC] = (n1)[ASC] = (n1)nc . 

Analogously, from CN = (n
1)BN and AQ = (n
1)CQ, we get
x + d = (n1)z, y + d = (n1)x, z + d = (n1)y 

and
nb + y = (n^{2}  n)c, nc + z = (n^{2}  n)a, na + x = (n^{2}  n)b , 

whence
x + y + x + 3d = (n1)(x + y + z) or 3d = (n2)(x + y + z) 

and
n(a + b + c) + (x + y + z) = (n^{2}  n)(a + b + c) or x + y + z = (n^{2}  2n)(a + b + c) . 

>From 1 = n[na + x + b] = n[nb + y + c] = n[nc + z + a],
we find that

= n^{2}(a + b + c) + n(x +y + z) + n(a + b + c) 
 
= n(n+1)(a + b + c) + n(x + y + z) 
 
= 
é ê
ë


n+1 n2

+ n 
ù ú
û

(x + y + z) 
 
= 
é ê
ë


n^{2}  n + 1 n  2


ù ú
û

(x + y + z) , 


whence
d = 
(n2)^{2} n^{2}  n + 1

. 

Solution 4. Since the ratio of areas remains invariant
under shear transformations and dilations, we may assume that
the triangle is right isosceles. Assign coordinates:
A ~ (0, 0), B ~ (0, 1), C = (1, 0),
M ~ (0, 1/n),
Then
R ~ 
æ ç
è


n1 n^{2}  n + 1

, 
(n1)^{2} n^{2}  n + 1


ö ÷
ø



S ~ 
æ ç
è


(n1)^{2} n^{2}  n + 1

, 
1 n^{2}  n + 1


ö ÷
ø



and
T ~ 
æ ç
è


1 n^{2}  n + 1

, 
n1 n^{2}  n + 1


ö ÷
ø

. 

A computation of the area of
DRST now yields the result.
Comment. Considering how reasonable the result it, H. Lee
noted that when n > 1, then (n2)^{2} < n^{2}  n + 1, so that
[TSR] < 1 as expected, and also noted that when n = 2, we
get the special case of the medians that intersect in a common
point and yield [TSR] = 0.

111.

(a) Are there four different numbers, not exceeding
10, for which the sum of any three is a prime number?


(b) Are there five different natural numbers such
that the sum of every three of them is a prime number?
Solution. [R. Barrington Leigh] (a) Yes, there are four
such numbers. The threemember sums of the set { 1, 3, 7, 9 }
are the primes 11, 13, 17, 19.
(b) No. We prove the statement by contradiction. Suppose that there
are five different natural numbers for which every sum of three is
prime. As the numbers are distinct and positive, each such sum must
be at least 1 + 2 + 3 = 6, and so cannot be a multiple of 3.
Consider the five numbers, modulo 3. If there are three in the same
congruent class, their sum is a multiple of 3. If there is one
each congruent to 0, 1, 2 modulo 3, then the sum of these three is
a multiple of 3. Otherwise, there are only two congruence classes
represented with at most two numbers in each, an impossibility.
Hence in all cases, there must be three who sum to a multiple of 3.
Comment. Part (b) need not be framed as a contradiction.
One could formulate it as follows: Let five positive integers be
given. Argue that either each congruent class modulo 3 is
represented or that some class is represented by at least three
of the numbers. Then note that therefore some three must sum to
a multiple of 3. Observe that 3 itself is not a possible sum.
Hence, among every five positive integers, there are three who
add to a nonprime multiple of 3, and simply say that the answer to
the question is ``no''.

112.

Suppose that the measure of angle BAC in the
triangle ABC is equal to a. A line passing through
the vertex A is perpendicular to the angle bisector of
ÐBAC and intersects the line BC at the point M.
Find the other two angles of the triangle ABC in terms of
a, if it is known that BM = BA + AC.
Solution. Let q be the line through A perpendicular to
the bisector of angle BAC; this line bisects the external angle
at A. The possibility that q is parallel to BC is precluded
by the condition that it intersects BC at M. Let
ÐABC =
b and
ÐACB =
g, Since p is not parallel
to BC,
b is not equal to
g.
Case i. Suppose that b > g. Then M intersects
BC so that B lies between M and C. Let BA be produced to
D so that AC = AD. Since MB = BA + AC = BA + AD = BD,
ÐDMB = ÐMDB and so ÐMDB = ^{1}/_{2}ÐDBC = ^{1}/_{2}b (exterior angle). Since AD = AC,
ÐADC = ÐACD = ^{1}/_{2}ÐBAC = ^{1}/_{2}a.
Since MA produced bisects ÐDAC, MA produced right
bisects DC and so MD = MC. Therefore
g+ 
a 2

= ÐMCD = ÐMDC = 
b 2

+ 
a 2

, 

whence
b = 2
g. Therefore, 180
^{°} =
a+
b+
g =
a+ 3
g and
g = 
180^{°}  a 3

and b = 
360^{°}  2a 3

. 

Case ii. Supppose that b < g. Then M intersects
BC so that C lies between B and M. Let BA be produced to
D so that AD = AC. Since AM bisects ÐDAC, it right
bisects CD and so triangle MDC is isosceles. Then
ÐADM = ÐADC + ÐMDC = ÐACD + ÐMCD = ÐACM. Since BM = BA + AC = BD,
ÐACM = ÐADM = ÐBDM = ÐBMD. Since
ÐACM = a+ b (exterior angle),
180^{°} = ÐDBM + 2ÐBMD = b+ 2(a+ b),
so that b = ^{1}/_{3}(180^{°}  2a) and
g = 180^{°}  ÐACM = 180^{°}  (a+b) = 120^{°}  ^{1}/_{3}a = (360^{°}  a)/3.
Question. Why cannot you just say the second case can be
handled as the first case, by symmetry?

113.

Find a function that satisfies all of the following
conditions:


(a) f is defined for every positive integer n;


(b) f takes only positive values;


(c) f(4) = 4;


(d)

1 f(1)f(2)

+ 
1 f(2)f(3)

+ ¼+ 
1 f(n)f(n+1)

= 
f(n) f(n+1)

. 

Solution. [R. Barrington Leigh] The function for which
f(n) = n for every positive integer n satisfies the condition.
[
Exercise: establish this by induction.] We now show that
this is the only example. Substituting n = 1 into (d) and
noting that f(2)
¹ 0, we find that f(1)
^{2} = 1, whence
f(1) = 1. Applying (d) to two consecutive values of the argument
yields that

f(n) f(n+1)

 
f(n1) f(n)

= 
1 f(n)f(n+1)

, 

whence
[f(n)]^{2}  1 = f(n1)f(n+1) . 

Substituting n = 2 and n = 3 into this and noting that
f(1) = 1 and f(4) = 4, we find that
and
whence
0 = [f(2)]^{4}  2[f(2)]^{2}  4f(2) = f(2)[f(2)  2]([f(2)]^{2} + 2f(2) + 2) . 

Since the first and third factors are positive for all postive
possibilities for f(2), we must have f(2) = 2. As
f(n+1) = 
[f(n)]^{2}  1 f(n1)

, 

we can prove by induction that f(n) = n for all positive integers n.

114.

A natural number is a multiple of 17. Its binary
representation (i.e., when written to base 2) contains
exactly three digits equal to 1 and some zeros.


(a) Prove that there are at least six digits equal
to 0 in its binary representation.


(b) Prove that, if there are exactly seven digits equal
to 0 and three digits equal to 1, then the number must be even.
Solution 1. (a) If there are fewer than six digits equal to
0 in its binary representation, then the number must have at
most eight digits and be of the form 2
^{a} + 2
^{b} + 2
^{c}
where 0
£ a < b < c
£ 7. The first eight powers of
2 with nonnegative exponent are congruent to 1, 2, 4, 8,
1,
2,
4,
8 modulo 17, and the sum of any three of these
cannot equal to zero and must lie between
14 and 14.
Hence it is not possible for three powers of 2 among the first
eight to sum to a multiple of 17. Hence, the number must have
at least nine digits, including three zeros.
(b) Suppose that the number is equal to 2^{a} + 2^{b} + 2^{c} where
0 £ a < b < c £ 9. If this number has exactly 10 digits
and is odd, then a = 0 and c = 9, so that the number is
equal to 1 + 2^{b} + 2^{9} = 513 + 2^{b} º 3 + 2^{b} (mod 17).
But there is no value of b that will make this vanish, modulo
17. Hence, a 10digit number divisible by 17 must be even.
An example is 2 ×17^{2} = 2 ×(1 + 2^{5} + 2^{8}) = (1001000010)_{2}.
Solution 2. [R. Furmaniak] (a) Since 17_{10} = 10001_{2}, any
binary number abcd_{2} with four or fewer digits multiplied by 17
will yield abcdabcd_{2}. Since the first and last four digits are
the same, there must be an even number of 1s. Thus, any multiple
of 17 with exactly three binary digits must be a product of 17 and
a number that has at least 5 binary digits. Every such product must
have at least 9 digits, and so at least three digits equal to 0.
(b) As in Solution 1.