Solutions for November Problems
Notes. A realvalued function f(x) of a real variable is
increasing if and only if u < v implies that f(u)
£ f(v).
The
circumcircle of a triangle is that circle that passes
through its three vertices; its centre is the
circumcentre
of the triangle. The
incircle of a triangle is that circle
that is tangent internally to its three sides; its centre is the
incentre of the triangle.

346.

Let n be a positive integer. Determine the
set of all integers that can be written in the form
where a_{1}, a_{2}, ¼, a_{n} are all positive integers.
Solution 1. The sum cannot exceed
å_{k=1}^{n} k =
^{1}/
_{2}n(n+1). We prove by induction that the set of integers that
can be written in the required form consists of the integers
from 1 to
^{1}/
_{2}n(n+1) inclusive. Observe that 1 is
representable for any n (for example, by making each a
_{k} equal
to kn). Also,
^{1}/
_{2}n(n+1) is representable, by taking
a
_{1} = a
_{2} =
¼ = a
_{n}. Thus, the result holds for n = 1
and n = 2.
Suppose that it holds for n = m ³ 2. Then, by taking a_{m+1} = m + 1 and appending (m+1)/(m+1) to each integer representable by
an mterm sum, we find that each integer between 2 and
^{1}/_{2}m(m+1) + 1 = (m^{2} + m + 2)/2
inclusive can be represented with a (m+1) term
sum. By taking a_{m+1} = 1 and appending m + 1 to
each integer representable by an mterm sum, we find that each
integer between m+2 and ^{1}/_{2}m(m+1) + (m+1) = ^{1}/_{2}(m+1)(m+2) can be represented with an (m+1)term sum.
Since [(m^{2} + m + 2)/2]  (m+2) = ^{1}/_{2}(m^{2}  m  2) = ^{1}/_{2}(m  2)(m+1) ³ 0 for m ³ 2, ^{1}/_{2}(m^{2} + m + 2) ³ m+2. Thus, we can represent all numbers between
1 and ^{1}/_{2}(n+1)(n+2) inclusive. The result follows.
Solution 2. [Y. Zhao] Lemma. For any integer
k with 1 £ k £ ^{1}/_{2}n(n+1), there is a subset
T_{k} of { 1, 2, ¼, n } for which the sum of the numbers
in T_{k} is k.
Proof. T_{k} is the entire set when k = ^{1}/_{2}n(n+1).
For the other values of k, we give a proof by induction on k.
A singleton suffices for 1 £ k £ n. Suppose that the
result holds for k = m  1 < ^{1}/_{2}n(n+1). Then T_{m1} must
lack at least one number. If 1 Ï T_{m1}, let
T_{m} = T_{m1} È{ 1 }. If 1 Î T_{m1}, let i > 1 be the
least number not in T_{m1}. Then let T_{m} = T_{m1} È{ i + 1 } \{ i }. ª
Now to the result. Let 2 £ k £ ^{1}/_{2}n(n+1) be given and
determine T_{k1}. Define a_{i} = 1 when i Î T_{k1} and
a_{i} = ^{1}/_{2}n(n+1)  (k1) when i Ï T_{k1}.
Then


å
 
ì í
î


i
a_{i}

: i Ï T_{k1} 
ü ý
þ



= 
é ë


æ è

n+1
2

ö ø

 (k1) 
ù û

1


å
 { i : i Ï T_{k1} } 
 
 
= 
é ë


æ è

n+1
2

ö ø

 (k1) 
ù û

1


é ë


æ è

n+1
2

ö ø

 
å
 { i : i Î T_{k1} 
ù û

= 1 . 

Since one can give a representation for 1 and since no number
exceeding ((n+1)  2) can be represented, it follows that
the set of representable integers consists of those between 1 and
((n+1)  2) inclusive.

347.

Let n be a positive integer and { a_{1}, a_{2}, ¼, a_{n} } a finite sequence of real numbers which contains
at least one positive term. Let S be the set of indices k
for which at least one of the numbers
a_{k}, a_{k} + a_{k+1}, a_{k} + a_{k+1} + a_{k+2}, ¼,a_{k} + a_{k+1} + ¼+ a_{n} 

is positive. Prove that

å
 { a_{k} : k Î S } > 0 . 

Solution. We prove the result by induction on n.
When n = 1, the result is obvious. Let m
³ 2 and suppose
that the result holds for all n
£ m
1. Suppose a suitable
sequence { a
_{1}, a
_{2},
¼, a
_{m} } is given. If a
_{1} Ï S, then
å{ a
_{k} : k
Î S } > 0, by the induction
hypothesis applied to the (m
1)
element set { a
_{2},
¼, a
_{m}}. Suppose that a
_{1} Î S and that r is the smallest index
for which a
_{1} + a
_{2} +
¼+ a
_{r} > 0. Then, for
1
£ i
£ r
 1, a
_{1} +
¼+ a
_{i} £ 0 and so
(a_{i+1} + ¼+ a_{r}) = (a_{1} + ¼+ a_{r}) (a_{1} + ¼+ a_{i}) > 0 , 

i.e., a
_{2}, a
_{3},
¼, a
_{r} Î S. Hence
å{ a
_{k} : 2
£ k
£ r } > 0. If there are no elements
of S that exceed r. then the desired conclusion follows.
Otherwise, by the induction hypothesis applied to the (m
r)

element set { a
_{r+1},
¼, a
_{m} }, we have that
å{ a
_{k} : k
Î S, r+1
£ k
£ m } > 0. The desired
conclusion follows.
Comment. Most solvers had much more elaborate solutions which
essentially used this idea. No one recognized that the proliferation
of cases could be sidestepped by the technical use of an induction
argument.

348.

Suppose that f(x) is a realvalued function defined
for real values of x. Suppose that f(x)  x^{3} is an increasing
function. Must f(x)  x  x^{2} also be increasing?
Solution. The answer is
no. Consider f(x) = x
^{3} + x.
Then x = f(x)
 x
^{3} is increasing, but g(x) = f(x)
 x
 x
^{2} = x
^{3}  x
^{2} is not. Indeed, g(0) = 0 while g(
^{1}/
_{2}) =
^{1}/
_{8} < 0.
Comment. See Problem 348.(b) included with the February set.

349.

Let s be the semiperimeter of triangle ABC.
Suppose that L and N are points on AB and CB produced
(i.e., B lies on segments AL and CN) with
AL  = CN  = s. Let K be the point
symmetric to B with respect to the centre of the circumcircle
of triangle ABC. Prove that the perpendicular from K to the
line NL passes through the incentre of triangle ABC.
Let the incentre of the triangle be I.
Solution 1. Let P be the foot of the perpendicular from
I to AK, and Q the foot of the perpendicular from
I to CK. Since BK is a diameter of the circumcircle of
triangle ABC, ÐBAK = ÐBCK = 90^{°} and
IP  BA, IQ  BC. Now IP  = s  a = BN ,
IQ  = s  c = BL  and
ÐPIQ = ÐABC = ÐNBL, so that
DIPQ º DBNL (SAS). Select R on IP and
S on IQ (possibly produced) so that IR = IQ, IS = IP.
Thus, DISR º DBNL and RS  NL (why?).
Since IPKQ is concyclic, ÐKIP + ÐIRS = ÐKIP+ ÐIQP = ÐKIP + ÐIKP = 90^{°}. Therefore
IK is perpendicular to RS, and so to NL.
Solution 2. Lemma. Let W, X, Y, Z be four points in
the plane. Then WX ^YZ if and only ig WY ^{2} WZ ^{2} = XY ^{2}  XZ ^{2}.
Proof. Note that
2 ( 
®
W

 
®
X

)·( 
®
Z

 
®
Y

) = ( 
®
Y

 
®
W

)^{2}  ( 
®
Z

 
®
W

)^{2} ( 
®
Y

 
®
X

)^{2}+ ( 
®
Z

 
®
X

)^{2} . ª . 

Since BK is a diameter of the circumcircle, ÐLAK = ÐNCK = 90^{°}. We have that


= (KA ^{2} + AL ^{2})  (KC ^{2} +CN ^{2}) = KA^{2}  KC ^{2} 
 
 
= (BK ^{2}  AB ^{2})  (BK ^{2} BC ^{2}) = BC ^{2}  AB ^{2} . 

Let U and V be the respective feet of the perpendiculars from
I to BA and BC. Observe that
AU
 =
BN
 = s
 a,
CV
 =
BL
 = s
 c and
BU
 =
BV
 = s
 b,
so that
UL
 =
BC
 = a,
VN
 =
AB
 = c. Then
IL ^{2}  IN ^{2} = (IU ^{2} + UL ^{2})  (IV ^{2}+ VN ^{2}) = UL ^{2}  VN ^{2} = BC ^{2}  AB ^{2} , 

which, along with the lemma, implies the result.

350.

Let ABCDE be a pentagon inscribed in a circle with
centre O. Suppose that its angles are given by
ÐB = ÐC = 120^{°}, ÐD = 130^{°},
ÐE = 100^{°}. Prove that BD, CE and AO are
concurrent.
Solution 1. [P. Shi; Y. Zhao] The vertices ABCDE are the
vertices A
_{1}, A
_{5}, A
_{7}, A
_{11}, A
_{12} of a regular
18
gon. (Since A,B,C,D,E lie on a circle, the position of
the remaining vertices are determined by that of A are the
angle sizes. Alternatively, look at the angles subtended at the
centre by the sides of the pentagon.) Since the sum of the angles
of a pentagon is 540
^{°},
ÐA = 70
^{°}. Since
ÐAED > 90
^{°}, D and E lie on the same side of the
diameter through A, and B and C lie on the other side. Thus,
the line AO produced intersects the segment CD. Consider the
triangle ACD for which AO produced, BD and CE are cevians.
We apply the trigonometric version of Ceva's theorem.
We have that ÐCAO = 30^{°}, ÐADB = 40^{°},
ÐDCE = 10^{°}, ÐOAD = 10^{°},
ÐBDC = 20^{°} and ÐECA = 70^{°}. Hence

sinÐCAO ·sinÐADB ·sinÐDCE
sinÐOAD ·sinÐBDC ·sinÐECA

= 
( 
1
2

)sin40^{°} sin10^{°} 
sin10^{°} sin20^{°} sin70^{°}

= 
cos20^{°}
sin70^{°}

= 1 . 

Hence the three lines AO, BD and CE are concurrent as desired.
Solution 2. [C. Sun] Let BD and CE intersect at P.
We can compute the following angles:
ÐEAB = ÐEBA = 70^{°}, ÐAEB = 40^{°},
ÐBEP = ÐPDC = 20^{°}, ÐEBP = ÐPCD = 10^{°}, ÐPBC = ÐPED = 40^{°},
ÐBCP = ÐEDP = 110^{°} and ÐBPC = ÐEPD = 30^{°}. Since triangle ABE is acute, the circumcentre
O of it (and the pentagon) lie in its interior, and
ÐAOB = ^{1}/_{2}ÐAEB = 80^{°}. Since triangle
ABE is isosceles, ÐBAO = ÐABO = 50^{°}.
Let Q be the foot of the perpendicular from E to AB and R the
foot of the perpendicular from B to EP produced. Since
DEQB º DERB (ASA), QB = RB. Since
ÐBPR = 30^{°}, ÐPBR = 60^{°} and
PB = 2RB = 2QB = AB. Hence ABP is isosceles with apex
ÐABP = 80^{°}. Thus, ÐBAP = ÐBPA = 50^{°}.
Hence ÐBAO = ÐBAP = 50^{°}, so that AO must pass
through P and the result follows.

351.

Let { a_{n} } be a sequence of real numbers for which
a_{1} = 1/2 and, for n ³ 1,
a_{n+1} = 
a_{n}^{2}
a_{n}^{2}  a_{n} + 1

. 

Prove that, for all n, a_{1} + a_{2} + ¼+ a_{n} < 1.
Solution. Let b
_{n} = 1/a
_{n} for n
³ 1, whence
b
_{1} = 2 and b
_{n+1} = b
_{n}^{2}  b
^{n} + 1 = b
_{n}(b
_{n}  1) + 1
for n
³ 1. Then { b
_{n} } is an increasing sequence of
integers and

1
b_{n}

= 
1
b_{n}  1

 
1
b_{n+1}  1



for n
³ 1. Hence


= 
1
b_{1}

+ 
1
b_{2}

+ ¼+ 
1
b_{n}


 
 
= 
é ë


1
b_{1}  1

 
1
b_{2}  1


ù û

+ 
é ë


1
b_{2}  1

 
1
b_{3}  1


ù û

+ ¼+ 
é ë


1
b_{n}  1

 
1
b_{n+1}  1


ù û


 
 
= 
1
b_{1}  1

 
1
b_{n+1}  1

= 1  
1
b_{n+1}  1

< 1 . 


352.

Let ABCD be a unit square with points M and N
in its interior. Suppose, further, that MN produced does not
pass through any vertex of the square. Find the smallest value of
k for which, given any position of M and N, at least one
of the twenty triangles with vertices chosen from the set
{ A, B, C, D, M, N } has area not exceeding k.
Solution 1. Wolog, suppose that M lies in the interior
of triangle ABN. Then
[ABM] + [AMN] + [BMN] + [CND] = [ABN] + [CND] = 
1
2

, 

so that at least one of the four triangles on the left has area
not exceeding 1/8. Hence k
£ 1/8. We give a configuration
for which each of the twenty triangles has area not less than
1/8, so that k = 1/8.
Suppose that M and N are both located on the line joining
the midpoints of AD and BC with M distant 1/4 from the
side AD and N distant 1/4 from the side BC. Then

1
4

= [ABM] = [CDM] = [ABN] = [CDN] 


1
2

= [ABC] = [BCD] = [CDA] = [DAB] 


1
8

= [DAM] = [BCN] = [AMN] = [BMN] = [CMN] = [DMN] = [AMC] = [ANC] = [BMD] = [BND] . 

Solution 2. [L. Fei] Suppose that all triangle have
area exceed k. Then M and N must be in the interior of the
square distant more than 2k from each edge to ensure that areas
of triangles like ABM exceed k. Similarly, M and
N must be distant more than kÖ2 from the diagonals
of the square. For points M and N to be available that satisfy
both conditions, we need to find a point that is distant at least
2k from the edges and kÖ2 from the diagonal; such a
point would lie on a midline of the square. The condition is that
2k + Ö2(kÖ2) < ^{1}/_{2} or k < ^{1}/_{8}.
On the other hand, we can give a configuration in which each area is at
least equal to k and some areas are exactly ^{1}/_{8}. This
would have M and N on the same midline, each equidistant from
an edge and the centre of the square.