Solutions to the November problems

185.

Find all triples of natural numbers a, b, c, such
that all of the following conditions hold: (1) a < 1974;
(2) b is less than c by 1575; (3) a^{2} + b^{2} = c^{2}.
Solution. Conditions (2) and (3) can be written as
c
 b = 1575 and a
^{2} = c
^{2}  b
^{2}. Hence
a
^{2} = 1575(c+b) = 3
^{2} ·5
^{2} ·7 ·(b + c), so that
a = 3 ·5 ·7 ·k = 105k. By (1), k
£ 18.
>From 105^{2} k^{2} = 1575(c + b), it follows that 7k^{2} = c + b.
Putting this with 1575 = c  b yields that
2c = 7k^{2} + 1575 and 2b = 7k^{2}  1575. Since b is a natural
number, 7k^{2} > 1575, whence k > 15. Since k is also odd (why?),
the only possibility is for k to be equal to 17. This works and
we obtain the triple (a, b, c) = (1785, 224, 1799).
Comment. This was solved by most participants. Some solutions
used the fact that all primitive pythagorean triples can
be parametrized by (a, b, c) = (m^{2}  n^{2}, 2mn, m^{2} + n^{2}),
where m and n are natural numbers, m > n and the greatest
common divisor of m and n is 1. Note that this representation
is not adequate to get all nonprimitive triples, such as (15, 36, 39)
and the triple of this problem. The general form is given by
(a, b, c) = (k(m^{2}  n^{2}), 2kmn, k(m^{2} + n^{2})), where k is
a positive constant. When you build your solution on such a ``wellknown''
fact, make sure that you recall it correctly and respect all restrictions
and specifics. Otherwise, you risk applying it in a situation that
does not satisfy all the requirements or you find only a subset of
the possible solutions.

186.

Find all natural numbers n such that there exists
a convex nsided polygon whose diagonals are all of the same
length.
Solution 1. [T. Yue] First, we prove that n < 6. Suppose
that n
³ 6 and that there exists a convex n
sided polygon
with vertices A
_{1}, A
_{2},
¼, A
_{n} whose diagonals are
all of the same length. Since the diagonals are all of the same
length, we have that A
_{1}A
_{5} = A
_{2}A
_{5} = A
_{1}A
_{4} = A
_{2}A
_{4}, so that
A
_{1}A
_{2}A
_{5} and A
_{1}A
_{2}A
_{4} are isosceles triangles and A
_{4} and
A
_{5} both lie on the right bisector of A
_{1}A
_{2}. Since the
polynomial is convex, both A
_{4} and A
_{5} must lie on the same
side of A
_{1}A
_{2}. But then they could not be distinct, yielding
a contradiction.
In the case n = 3, there are no diagonals, so the result is
vacuously true. For n = 4, all squares, rectangles and isosceles
trapezoids satisfy the condition. When n = 5, the regular
pentagon is an example. (Is it the only example?)
Solution 2. Suppose that n ³ 6, and the vertices of the
polygon be A_{1}, A_{2}, ¼, A_{n}. Suppose that A_{1}A_{n2}
and A_{2}A_{n1} intersect at O. Then, by the triangle inequality,
A_{1}O + OA_{n1} > A_{1}A_{n1} and A_{2}O + OA_{n2} > A_{2}A_{n2},
so that
A_{1}A_{n2} + A_{2}A_{n1} = A_{1}O + OA_{n2} + A_{2}O + OA_{n1} > A_{1}A_{n1} + A_{2}A_{n2} 

and so the diagonals A
_{i}A
_{j} are not all of the same length.
Hence, n
£ 5 and we can conclude as before.

187.

Suppose that p is a real parameter and that
f(x) = x^{3}  (p + 5)x^{2}  2(p  3)(p  1)x + 4p^{2}  24p + 36 . 



(a) Check that f(3  p) = 0.


(b) Find all values of p for which two of the roots
of the equation f(x) = 0 (expressed in terms of p) can be
the lengths of the two legs in a rightangled triangle with a
hypotenuse of 4Ö2.
Solution. (a) Observe that
f(x) = [(x  (3  p)][x^{2}  2(p+1)x + 4(p3)] . 

(b) [Y. Sun] From the factorization in (a), we can identify the
three roots: x_{1} = 3  p,
x_{2} = (p+1) + 
 _________ Ö(p1)^{2} + 12

, 

x_{3} = (p+1)  
 _________ Ö(p1)^{2} + 12

. 

Note that x
_{2}  x
_{3} = 2
Ö[((p
1)
^{2} + 12)]
³ 2
Ö[12] = 4
Ö3 > 4
Ö2, so that, by the triangle inequality,
x
_{2} and x
_{3} cannot be the legs of a right triangle with
hypotenuse 4
Ö2. On the other hand,
x_{1} + x_{3} = 4  
 _________ Ö(p1)^{2} + 12

£ 4  
 __ Ö12

= 2(2  Ö3) < 4Ö2 , 

so that, by the triangle inequality, x
_{1}, x
_{3} and
4
Ö2 cannot be the sides of a triangle. Thus,
the only possibility is that x
_{1}^{2} + x
_{2}^{2} = 32.
Thus, we must have
(3  p)^{2} + [(p+1) + 
 __________ Öp^{2}  2p + 13

]^{2} = 32 

Þ 2(p+1) 
 __________ Öp^{2}  2p + 13

= 3p^{2}+ 6p + 9 = 3(p+1)(3  p) . 

Therefore, either p =
1 or
2
Ö[(p
^{2}  2p + 13)] = 3(3
 p). The latter possibility leads
to
4(p^{2}  2p + 13) = 9(p  3)^{2} Þ 5p^{2}  46p + 29 = 0 . 

Since 3(3
p) must be positive, we reject one root of this
quadratic for p, and so have only the additional
possibility (23
 8
Ö6)/5.

188.

(a) The measure of the angles of an acute triangle
are a, b and g degrees. Determine (as an
expression of a, b, g) what masses must be
placed at each of the triangle's vertices for the centroid
(centre of gravity) to coincide with (i) the orthocentre of
the triangle; (ii) the circumcentre
of the triangle.


(b) The sides of an arbitrary triangle are a, b, c
units in length. Determine (as an expression of a, b, c) what
masses must be placed at each of the triangle's vertices for the
centroid (centre of gravity) to coincide with (i) the centre of
the inscribed circle of the triangle; (ii) the intersection
point of the three segments joining the vertices of the triangle
to the points on the opposite sides where the inscribed circle
is tangent (be sure to prove that, indeed, the three segments
intersect in a common point).
Solution. [R. Barrington Leigh] Suppose that a, b, c are
the respective lengths of BC, CA and AB of triangle ABC and
that
a =
ÐCAB,
b =
ÐABC and
g =
ÐBCA. Let masses m
_{a}, m
_{b}, m
_{c} be masses placed
at the respective vertices A, B, C. We use the following
lemma, which will be established in the Comments:


Lemma 1. The centroid G of two combined
collections of mass particles is on the line joining the centroids
P and Q of the two collections, and the following ratio holds:
PG:GQ = m_{Q}:m_{P}, where m_{Q} and m_{P} are the totals of the
masses in the two mass collections at Q and P respectively.
It follows that, if X is the centroid of
the masses at the vertices of
DABC and
Y is the intersection point of AX and BC, then Y
is the centroid of the masses at the vertices B and C.
Furthermore, let h
_{b} and h
_{c} be the perpendicular distances
from B and C respectively to the line AY. By the lemma and
the similar right triangles formed by the drawn perpendiculars,

m_{b} m_{c}

= 
YC YB

= 
h_{c} h_{b}


 (*) 
(a) (i) Let Y be the foot of the altitude from A to BC, so that
the orthocentre of the triangle is on AY. By (*),

m_{b} m_{c}

= 
AY/tang AY/tanb

= 
tanb tang

. 

By symmetry, we see that
m_{a} : m_{b} : m_{c} = tana: tanb: tang . 

Since the triangle is acute, the elements of the ratio are
positive and it is welldefined.
(a) (ii) Let X = O be the circumcentre of DABC. From (*),

m_{b} m_{c}

= 
h_{c} h_{b}

= 
OC ·sinÐCOA OB ·sinÐBOA

= 
sin2b sin2g

. 

By symmetry, we have that
m_{a} : m_{b} : m_{c} = sin2a: sin2b: sin2g . 

(b) (i) Let X = I, the incentre of the triangle ABC; this is
the intersection of the angle bisectors. Then ÐBAI = ÐCAI, and, by (*),

m_{b} m_{c}

= 
h_{c} h_{b}

= 
b ·sinÐCAI c ·sinÐBAI

= 
b c

. 

By symmetry, we find that m
_{a} : m
_{b} : m
_{c} = a : b : c.
(b) (ii) Let the incircle of ABC meet BC at D,
CA at E and AB at F. Let x, y, z be the respective
lengths of AE = AF, BD = BF, CD = CE, so that y + z = a,
z + x = b, x + y = c. Then 2x = b + c  a, 2y = c + a  b
and 2z = a + b  c. The semiperimeter s of the triangle is
equal to (a + b + c)/2, so that x = s  a, y = s  b and
z = s  c. Using Ceva's Theorem, we see from

AF FB

· 
BD DC

· 
CE EA

= 
x y

· 
y z

· 
z x

= 1 

that AD, BE and CF concur at a point X. From (*),
we have that

m_{b} m_{c}

= 
CD BD

= 
z y

= 
1/(sb) 1/(sc)



so that, from symmetry,
m_{a} : m_{b} : m_{c} = 
1 s  a

: 
1 s  b

: 
1 s  c

. 

Comment. A couple of correct solutions were received, most
based on the lemma and the derived statement (*). This solution
was chosen because of its excellently organized and clear presentation.
Some students came up with other expressions in the ratios,
equivalent after some simple trigonometric transformations to the
expressions here. Two of the students, A. Feiz Mohammadi and
J.Y. Zhoa used an interesting an helpful fact:
Lemma 2. Let X be the centroid of triangle ABC and
denote by S_{a}, S_{b} and S_{c} the respective areas of the
triangles XBC, XCA and XAB. Then m_{a} : m_{b} : m_{c} = S_{a} : S_{b} : S_{c}.
Lemma 2 can be used to obtain a beautiful and elegant solution to
each of the four parts of the problem. Let us turn to the proofs
of these lemmata.
Proof of Lemma 1. We statement can be replaced by the
equivalent form: given two point masses m_{P} and m_{Q}
at the respective points P and Q, their centroid is on
the line segment PQ at the point G for which PG:GQ = m_{Q}:m_{P}. From the definition of a centroid, if O is
any point selected as the origin of vectors,

®
OG

= 
m_{P} + m_{Q}

= 
m_{P} m_{P} + m_{Q}


®
OP

+ 
m_{Q} m_{P} + m_{Q}


®
OQ

. 
 (1) 
In general, consider a point K on the segment PQ for which
PK : KQ = k : (1
 k), for 0 < k < 1. Then

®
OK

= 
®
OP

+ 
®
PK

= 
®
OP

+ k 
®
PQ

= 
®
OP

+ k( 
®
OQ

 
®
OP

) = (1  k) 
®
OP

+ k 
®
OQ

. 
 (2) 
Now choose k = m
_{Q}/(m
_{P} + m
_{Q}), so that
1
 k = m
_{P}/(m
_{P} + m
_{Q}). From (1) and (2), we have that
G = K and we can complete the proof of the lemma.
Proof of Lemma 2. We prove that m_{a} : m_{b} = S_{a} : S_{b},
whence the rest follows from symmetry. Let K be the intersection of
CX and AB. Then m_{a} : m_{b} = BK : AK. Since triangles
AKC and BKC have the same altitudes from C,
[AKC]:[BKC] = AK : BK. Similarly, [AXK]:[BXK] = AK:BK.
Now S_{a} = [BKC]  [BXK] and S_{b} = [AKC]  [AXK], so that
S_{a} : S_{b} = BK : AK and the result is proven.

189.

There are n lines in the plane, where n is an
integer exceeding 2. No three of them are concurrent and no two
of them are parallel. The lines divide the plane into regions;
some of them are closed (they have the form of a convex polygon);
others are unbounded (their borders are broken lines consisting of
segments and rays).


(a) Determine as a function of n the number of unbounded
regions.


(b) Suppose that some of the regions are coloured, so that
no two coloured regions have a common side (a segment or ray). Prove
that the number of regions coloured in this way does not exceed
^{1}/_{3}(n^{2} + n).
Solution 1. (a) Draw a circle big enough to contain in its
interior all the intersection points of the n lines. Since there
are n lines and each of them intersects the circle in two points,
they divide the circle into 2n arcs. Each of the arcs corresponds to
one unbounded region, so that there are 2n unbounded regions.
(b) [M. Butler] Let k regions be coloured. Construct a circle
around all the bounded regions, big enough to contain part of
each of the unbounded regions as well as the intersection points
of the lines. Regions that were unbounded before are now bounded
(in part by an arc of the circle). Denote the k coloured
regions by R_{1}, R_{2}, ¼, R_{k}, and let R_{i} have
v_{i} vertices (including the intersections of the lines and
the circles). Every region has at least three vertices, so that
v_{i} ³ 3, and v_{1} + v_{2} + ¼+ v_{k} ³ 3k (1).
On the other hand, we can tabulate v_{1} + v_{2} + ¼+ v_{k} by
adding up the numer of coloured regions meeting at an intersection,
for all intersection points. Since no three lines have a common
point, there are (n  2) intersection points among the
lines and each of them is a vertex of at most two coloured regions.
So there are at most 2 (n  2) coloured regions counted
so far. In addition, there are 2n intersections between the
circles and the lines with at most one coloured region at
each of them. Taking (1) into consideration, we deduce that
2 
æ ç
è

n
2

ö ÷
ø

+ 2n ³ v_{1} + v_{2} + ¼+ v_{k} ³ 3k 

so that n
^{2} + n
³ 3k and the result follows.
Solution 2. (b) Let m_{2}, m_{3}, ¼
m_{k} denote the number
of coloured regions with respectively 2, 3, ¼, k sides (rays
or segments). Regions with two sides are angles, and hence unbounded.
Since no two adjecent regions are coloured, m_{2} £ (1/2)2n = n.
On the other hand, each line is divided by the other n1 lines into
n parts, so that the number of parts of all the lines is n^{2}.
Therefore
2m_{2} + 3m_{3} + 4m_{4} + ¼+ km_{k} £ n^{2} . 

The total number k of all the coloured regions satisfies

= m_{1} + m_{2} + ¼+ m_{k} = (1/3)m_{2} + (2/3)m_{2} + m_{3} + ¼+ m_{k} 
 
£ (1/3)m_{2} + (1/3)(2m_{2} + 3m_{3} + 4m_{4} + ¼+ km_{k} 
 


as desired.

190.

Find all integer values of the parameter a for which
the equation
has exactly one integer among its solutions.
Solution 1. [H. Li] To deal with the absolute values, we
consider the equation for different ranges of x.
If x <
1/2, the equation becomes
3x + 1 = a (E
_{1}).
If
1/2
£ x < 2, the equation is x + 3 = a (E
_{2}).
If x
³ 2, the equation is 3x
 1 = a (E
_{3}).
Thus, the given equation is the union of three linear equations.
The graph of a as a function of x is a broken line consisting
of two rays u
_{1} and u
_{3}
(corresponding to E
_{1} and E
_{3}) and a segment u
_{2}
(corresponding to E
_{2}); please graph it before continuing to
read. The minimum possible value of a is 2.5. For an integer
a to admit more than one integer solution to the equation, the
horizontal line y = a must intersect the graph in at least two
lattice points.
When x < 1/2, a > 2.5 and when 1/2 £ x < 2, then
2.5 £ a < 5. The only lattice points on segment u_{2} are
(0, 3) and (1, 4) and the second one has a corresponding
lattice point on u_{1}. So, for a = 4, there are two integer
solutions, 1 and 1, to the equation. Let x be an
integer. When x ³ 2, a ³ 5
and we want to consider lines y = a intersecting rays u_{1} and
u_{3}. When x < 1/2, then
3x + 1 º 1 (mod 3), while if x ³ 2, 3x  1 º 2 (mod 3), so that none of the horizontal lines can
intersect both of them at a lattice point.
Hence, in conclusion, the given equation has exactly one integer
solution x for the following values of a: a = 3,
a ³ 5 and a º 1 (mod 3) and a ³ 5 and a º 2
(mod 3).
Solution 2. First, solve the equation by considering the
three cases:
· x ³ 2: Then x = [(a+1)/3] and the equation is
solvable in this range Û a ³ 5;
· 1/2 < x < 2: Then x = a  3 and the equation is solvable
in this range Û 5/2 < a < 5;
· x £ 1/2: Then x = (1  a)/3 and the equation is solvable
in this range Û a ³ 5/2.
Thus, summing up, we see that the equation has
· no solution when a < 5/2,
· the unique solution x = 1/2 when a = 5/2,
· two solutions x = a  3 and x = (1a)/3 when
5/2 < a < 5, of which both are integers when a = 4 and
one is an integer when a = 3,
· two solutions x = 2 and x = 4/3 when a = 5, and
· two solutions x = (1 ±a)/3 when a > 5.
When a > 5 we can check that there is no solution when
a º 0, and exactly one solution when a \not º 0
(mod 3), and we obtain the set in the first solution.

191.

In Olymonland the distances between every two
cities is different. When the transportation program of the country
was being developed, for each city, the closest of the other cities
was chosen and a highway was built to connect them. All highways
are line segments. Prove that


(a) no two highways intersect;


(b) every city is connected by a highway to no more than
5 other cities;


(c) there is no closed broken line composed of highways only.
Solution. (a) Assume that the highways AC and BD
intersect at a point O. From the existence of AC, either
C is the closest city to A or A is the closest city to
C. A similar statement holds for BD. Wolog, suppose that
C is the closest city to A and D the closest city to B.
Hence, AD > AC, BC > BD so that AD + BC > AC + BD.
On the other hand, from the triangle inequality,
AO + OD > AD & BO + OC > BC ÞAC + BD = (AO + OC) + (BO + OD) > AD + BC , 

which contradicts the earlier inequality. Thus, no two
highways intersect.
(b) Consider any three cities A, B, X. If X is connected by
a highway to both A and B, then AB must be the longest side
in triangle ABX. To prove it, let us suppose, if possible, that
one of the other sides, say AX, is longest. Then AX > AB and
AX > XB, so that AX would not exist as B is closer to A than
X is, and B is closer to X than A is.
So AB is the longest side, which implies that ÐAXB is
the greatest angle in the triangle. Thus, ÐAXB > 60^{°}.
Assume that a city X is connect to six other cities A, B,C, D, E, F by highways. Then each of the angles with vertices at
X with these cities must exceed 60^{°}, so the sum of
the angles going round X from one highway back to it must exceed
360^{°}, which yields a contradiction. Therefore, every city
is connected by a highway to no more than five other cities.
(c) Suppose that there are n cities A_{1}, A_{2}, ¼, A_{n} that
are connected in this order by a closed broken line of highways.
Since all distances between pairs of cities are distinct, there
must be a longest distance between a pair of adjacent cities, say
A_{1} and A_{n}. Then A_{1}A_{n} > A_{1}A_{2}, so A_{n} is not the closest
city to A_{1}. Also A_{1}A_{n} > A_{n1}A_{n}, so A_{1} is not the closest
city to A_{n}. Therefore, the highway A_{1}A_{n} must not exist and
we get a contradiction. So there is no broken line composed only
of highways.