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Solutions

109.
Suppose that


x2 + y2
x2 - y2
+ x2 - y2
x2 + y2
= k .
Find, in terms of k, the value of the expression


x8 + y8
x8 - y8
+ x8 - y8
x8 + y8
 .
Solution 1. Simplifying, we obtain that


k = 2(x4 + y4)
x4 - y4
 ,
and, by extension, that


k2 + 4
2k
= k
2
+ 2
k
= x4 + y4
x4 - y4
+ x4 - y4
x4 + y4
= 2(x8 + y8)
x8 - y8
 .
Continuing on, we find that


x8 + y8
x8 - y8
+ x8 - y8
x8 + y8
= 2(x16 + y16)
x16 - y16
= k2 + 4
4k
+ 4k
k2 + 4
= k4 + 24k2 + 16
4k(k2 + 4)
 .

Comment. R. Barrington Leigh defined a formula


R = a+b
a-b
+ a-b
a+b
= 2· a2 + b2
a2 - b2
for a b which he then applied to (a, b) = (x2, y2), (x4, y4).



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 CM, {R} = BQ AN and {S} = CM BQ, where 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
= (n-1)n QS
SB
SB
QS
= (n-1)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
= (n-1)n [BSC] = (n-1)n[QSC] .
Examining triangles QSC and QBC, we similarly find that


[QSC]
[QBC]
= QS
QB
= QS
QS + SB
= 1
(n-1)n + 1
= 1
n2 - n + 1


[QSC] = 1
n2 - 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
[QBC] = 1
n
·[ABC] = 1
n


[QSC] = 1
n2 - n + 1
· 1
n
= 1
n(n2 - n + 1)


[BSC] = (n-1)n
n(n2 - n + 1)
= n-1
n2 - n + 1
 .
Similarly,


[TAC] = [RBA] = n-1
n2 - n + 1
 .
Now


[TSR] = [ABC] - [BSC] - [TAC] - [RBA] = 1 - 3(n-1)
n2 - n + 1
= (n-2)2
n2 - n + 1
 .

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


BR
RQ
· QA
AC
· CN
NB
= -1
so that


BR
RQ
· 1-n
n
· n-1
1
= -1


BR = n
(n-1)2
·RQ .
Thus,


n-1
n
= [ABQ] = [ARB] + [ARQ]
=

1 + (n-1)2
n


[ARB]
=

n2 - n + 1
n


[ARB]
whence


[ARB] = n-1
n2 - n + 1
 .
Therefore


[RST] = 1 - 3(n-1)
n2 - n + 1
= (n-2)2
n2 - n + 1
 .

Solution 3. Let a = [AMT], b = [BNR], c = [CQS], x = [BRT], y = [CSR], z = [ATS] and d = [RST]. Then using BM = (n-1)AM, [BRM] = (n-1)a,


x + d = [BTS] = (n-1)[ATS] = (n-1)z
and


nb + y = [BRC] + [CSR] = [BSC] = (n-1)[ASC] = (n-1)nc .
Analogously, from CN = (n-1)BN and AQ = (n-1)CQ, we get


x + d = (n-1)z,    y + d = (n-1)x,    z + d = (n-1)y
and


nb + y = (n2 - n)c,   nc + z = (n2 - n)a,   na + x = (n2 - n)b ,
whence


x + y + x + 3d = (n-1)(x + y + z)   or   3d = (n-2)(x + y + z)
and


n(a + b + c) + (x + y + z) = (n2 - n)(a + b + c)  or    x + y + z = (n2 - 2n)(a + b + c) .
>From 1 = n[na + x + b] = n[nb + y + c] = n[nc + z + a], we find that


3
= n2(a + b + c) + n(x +y + z) + n(a + b + c)
= n(n+1)(a + b + c) + n(x + y + z)
=

n+1
n-2
+ n

(x + y + z)
=

n2 - n + 1
n - 2


(x + y + z) ,
whence


d = (n-2)2
n2 - 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),


N ~

1
n
, n-1
n



Q ~

n-1
n
, 0

 .
Then


R ~

n-1
n2 - n + 1
, (n-1)2
n2 - n + 1



S ~

(n-1)2
n2 - n + 1
, 1
n2 - n + 1


and


T ~

1
n2 - n + 1
, n-1
n2 - n + 1


  .
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 (n-2)2 < n2 - 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 three-member 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 BAC 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 ABC = b and ACB = 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, DMB = MDB and so MDB = 1/2DBC = 1/2b (exterior angle). Since AD = AC, ADC = ACD = 1/2BAC = 1/2a. Since MA produced bisects DAC, MA produced right bisects DC and so MD = MC. Therefore


g+ a
2
= MCD = MDC = b
2
+ a
2
 ,
whence b = 2g. Therefore, 180 = a+ b+ g = a+ 3g 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 DAC, it right bisects CD and so triangle MDC is isosceles. Then ADM = ADC + MDC = ACD + MCD = ACM. Since BM = BA + AC = BD, ACM = ADM = BDM = BMD. Since ACM = a+ b (exterior angle), 180 = DBM + 2BMD = b+ 2(a+ b), so that b = 1/3(180 - 2a) and g = 180 - ACM = 180 - (a+b) = 120 - 1/3a = (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)
+ + 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) 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(n-1)
f(n)
= 1
f(n)f(n+1)
 ,
whence


[f(n)]2 - 1 = f(n-1)f(n+1) .
Substituting n = 2 and n = 3 into this and noting that f(1) = 1 and f(4) = 4, we find that


[f(2)]2 - 1 = f(3)
and


[f(3)]2 - 1 = 4f(2) ,
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(n-1)
 ,
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 2a + 2b + 2c 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 2a + 2b + 2c 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 + 2b + 29 = 513 + 2b 3 + 2b (mod 17). But there is no value of b that will make this vanish, modulo 17. Hence, a 10-digit number divisible by 17 must be even. An example is 2 ×172 = 2 ×(1 + 25 + 28) = (1001000010)2.

Solution 2. [R. Furmaniak] (a) Since 1710 = 100012, any binary number abcd2 with four or fewer digits multiplied by 17 will yield abcdabcd2. 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.


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