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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) a2 + b2 = c2.
Solution. Conditions (2) and (3) can be written as c - b = 1575 and a2 = c2 - b2. Hence a2 = 1575(c+b) = 32 ·52 ·7 ·(b + c), so that a = 3 ·5 ·7 ·k = 105k. By (1), k 18.

>From 1052 k2 = 1575(c + b), it follows that 7k2 = c + b. Putting this with 1575 = c - b yields that 2c = 7k2 + 1575 and 2b = 7k2 - 1575. Since b is a natural number, 7k2 > 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) = (m2 - n2, 2mn, m2 + n2), 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 non-primitive triples, such as (15, 36, 39) and the triple of this problem. The general form is given by (a, b, c) = (k(m2 - n2), 2kmn, k(m2 + n2)), where k is a positive constant. When you build your solution on such a ``well-known'' 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 n-sided polygon whose diagonals are all of the same length.
Solution 1. [T. Yue] First, we prove that n < 6. Suppose that n 6 and that there exists a convex n-sided polygon with vertices A1, A2, , An whose diagonals are all of the same length. Since the diagonals are all of the same length, we have that A1A5 = A2A5 = A1A4 = A2A4, so that A1A2A5 and A1A2A4 are isosceles triangles and A4 and A5 both lie on the right bisector of A1A2. Since the polynomial is convex, both A4 and A5 must lie on the same side of A1A2. 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 6, and the vertices of the polygon be A1, A2, , An. Suppose that A1An-2 and A2An-1 intersect at O. Then, by the triangle inequality, A1O + OAn-1 > A1An-1 and A2O + OAn-2 > A2An-2, so that


A1An-2 + A2An-1 = A1O + OAn-2 + A2O + OAn-1 > A1An-1 + A2An-2
and so the diagonals AiAj 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) = x3 - (p + 5)x2 - 2(p - 3)(p - 1)x + 4p2 - 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 right-angled triangle with a hypotenuse of 42.
Solution. (a) Observe that


f(x) = [(x - (3 - p)][x2 - 2(p+1)x + 4(p-3)] .

(b) [Y. Sun] From the factorization in (a), we can identify the three roots: x1 = 3 - p,


x2 = (p+1) +   _________
(p-1)2 + 12
 
 ,


x3 = (p+1) -   _________
(p-1)2 + 12
 
 .
Note that x2 - x3 = 2[((p-1)2 + 12)] 2[12] = 43 > 42, so that, by the triangle inequality, x2 and x3 cannot be the legs of a right triangle with hypotenuse 42. On the other hand,


x1 + x3 = 4 -   _________
(p-1)2 + 12
 
4 -   __
12
 
= 2(2 - 3) < 42 ,
so that, by the triangle inequality, x1, x3 and 4 2 cannot be the sides of a triangle. Thus, the only possibility is that x12 + x22 = 32.

Thus, we must have


(3 - p)2 + [(p+1) +   __________
p2 - 2p + 13
 
]2 = 32


2(p+1)   __________
p2 - 2p + 13
 
= -3p2+ 6p + 9 = 3(p+1)(3 - p) .
Therefore, either p = -1 or 2[(p2 - 2p + 13)] = 3(3 - p). The latter possibility leads to


4(p2 - 2p + 13) = 9(p - 3)2 5p2 - 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 - 86)/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 = CAB, b = ABC and g = BCA. Let masses ma, mb, mc 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 = mQ:mP, where mQ and mP 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 hb and hc 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,


mb
mc
= YC
YB
= hc
hb
(*)

(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 (*),


mb
mc
= AY/tang
AY/tanb
= tanb
tang
 .
By symmetry, we see that


ma : mb : mc = tana: tanb: tang .
Since the triangle is acute, the elements of the ratio are positive and it is well-defined.

(a) (ii) Let X = O be the circumcentre of DABC. From (*),


mb
mc
= hc
hb
= OC ·sinCOA
OB ·sinBOA
= sin2b
sin2g
 .
By symmetry, we have that


ma : mb : mc = sin2a: sin2b: sin2g .

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


mb
mc
= hc
hb
= b ·sinCAI
c ·sinBAI
= b
c
 .
By symmetry, we find that ma : mb : mc = 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 semi-perimeter 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


mb
mc
= CD
BD
= z
y
= 1/(s-b)
1/(s-c)
so that, from symmetry,


ma : mb : mc = 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 Sa, Sb and Sc the respective areas of the triangles XBC, XCA and XAB. Then ma : mb : mc = Sa : Sb : Sc.

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 mP and mQ at the respective points P and Q, their centroid is on the line segment PQ at the point G for which PG:GQ = mQ:mP. From the definition of a centroid, if O is any point selected as the origin of vectors,



OG
 
=
mP
OP
 
+ mQ
OQ
 

mP + mQ
= mP
mP + mQ

OP
 
+ mQ
mP + mQ

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 = mQ/(mP + mQ), so that 1 - k = mP/(mP + mQ). 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 ma : mb = Sa : Sb, whence the rest follows from symmetry. Let K be the intersection of CX and AB. Then ma : mb = BK : AK. Since triangles AKC and BKC have the same altitudes from C, [AKC]:[BKC] = AK : BK. Similarly, [AXK]:[BXK] = AK:BK. Now Sa = [BKC] - [BXK] and Sb = [AKC] - [AXK], so that Sa : Sb = 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(n2 + 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 R1, R2, , Rk, and let Ri have vi vertices (including the intersections of the lines and the circles). Every region has at least three vertices, so that vi 3, and v1 + v2 + + vk 3k (1).

On the other hand, we can tabulate v1 + v2 + + vk 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

n
2


+ 2n v1 + v2 + + vk 3k
so that n2 + n 3k and the result follows.

Solution 2. (b) Let m2, m3, mk denote the number of coloured regions with respectively 2, 3, , k sides (rays or segments). Regions with two sides are angles, and hence unbounded. Since no two adjecent regions are coloured, m2 (1/2)2n = n. On the other hand, each line is divided by the other n-1 lines into n parts, so that the number of parts of all the lines is n2. Therefore


2m2 + 3m3 + 4m4 + + kmk n2 .
The total number k of all the coloured regions satisfies


k
= m1 + m2 + + mk = (1/3)m2 + (2/3)m2 + m3 + + mk
(1/3)m2 + (1/3)(2m2 + 3m3 + 4m4 + + kmk
(1/3)(n + n2) ,
as desired.



190.
Find all integer values of the parameter a for which the equation


|2x + 1 |+ |x - 2 | = a
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 (E1). If -1/2 x < 2, the equation is x + 3 = a (E2). If x 2, the equation is 3x - 1 = a (E3). 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 u1 and u3 (corresponding to E1 and E3) and a segment u2 (corresponding to E2); 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 u2 are (0, 3) and (1, 4) and the second one has a corresponding lattice point on u1. So, for a = 4, there are two integer solutions, -1 and 1, to the equation. Let x be an integer. When x 2, a 5 and we want to consider lines y = a intersecting rays u1 and u3. When x < -1/2, then -3x + 1 1 (mod 3), while if x 2, 3x - 1 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 5 and a 1 (mod 3) and a 5 and a 2 (mod 3).

Solution 2. First, solve the equation by considering the three cases:

x 2: Then x = [(a+1)/3] and the equation is solvable in this range a 5;

-1/2 < x < 2: Then x = a - 3 and the equation is solvable in this range 5/2 < a < 5;

x 1/2: Then x = (1 - a)/3 and the equation is solvable in this range a 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 = (1-a)/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 0, and exactly one solution when a \not 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   AC + 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 AXB is the greatest angle in the triangle. Thus, AXB > 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 A1, A2, , An 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 A1 and An. Then A1An > A1A2, so An is not the closest city to A1. Also A1An > An-1An, so A1 is not the closest city to An. Therefore, the highway A1An must not exist and we get a contradiction. So there is no broken line composed only of highways.


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