39th Canadian Mathematical Olympiad

Wednesday, March 28, 2007

Solutions to the 2007 CMO paper

Solution to 1. Identify ve subsets A; B; C; D; E of the board, where C consists of the squares occupied by the six dominos
already placed, B is the upper right corner, D is the lower left corner, A consists of the squares above and to the left of
those in B [ C [ D and E consists of the squares below and to the right of those in B [ C [ D. The board can be coloured
checkerboard fashion so that A has 13 black and 16 white squares, B a single white square, E 16 black and 13 white squares
and D a single black square. Each domino beyond the original six must lie either entirely in A [ B [ D or C [ B [ D, either of
which contains at most 14 dominos. Thus, altogether, we cannot have more that 2  14 + 6 = 34 dominos. This is achievable,
by placing 14 dominos in A [ D and 14 in E [ B .
Solution to 2. If the triangles are isosceles, then they must be congruent and the desired ratio is 1. For, if they share
equal side lengths, at least one of these side lengths on one triangle corresponds to the same length on the other. And if they
share unequal side lengths, then either equal sides correspond or unequal sides correspond in both directions and the ratio is
1. This falls within the bounds.
Let the triangles be scalene. It is not possible for the same length to be an extreme length (largest or smallest) of both
triangles. Therefore, we must have a situation in which the corresponding side lengths of the two triangles are (x; y; z ) and
(y; z; u) with x < y < z and y < z < u. We are given that y=x = z=y = u=z = r > 1. Thus, y = rx andpz = ry = r2 x. From
the triangle inequality z < x + y, we have that r2 < 1 + r. Since r2 r 1 < p 0 and r > 1, 1 < r < 12 ( 5 + 1). The ratio of
the dimensions from the smaller to the larger triangle is 1=r which satis es 2 ( 5 1) < 1=r < 1. The result follows.
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Solution to 3. (a) Let f (x) = x2 + 4. Then

f (xy ) + f (y x) f (y + x) = (x2 y2 + 4) + (y x)2 + 4 (y + x)2 4
= (xy)2 4xy + 4 = (xy 2)2  0 : (1)
Thus, f (x) = x2 + 4 satis es the condition.
(b) Consider (x; y) for which xy = x + y. Rewriting this as (x 1)(y 1) = 1, we nd that this has the general solution
(x; y) = (1 + t 1 ; 1 + t), for t 6= 0. Plugging this into the inequality, we get that f (t t 1 )  0 for all t 6= 0. For arbitrary real

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u,the equation t t 1 = u leads to the quadratic t2 ut 1 = 0 which has a positive discriminant and so a real solution.
Hence f (u)  0 for each real u.
Comment. The substitution v = y x, u = y + x whose inverse is x = 12 (u v), y = 12 (u + v) renders p the condition as
f ( (u
1
4
2
v )) + f (v )  f (u). The same strategy as in the foregoing solution leads to the choice u = 2 + v 2 + 4 and f (v )  0
2

for all v.
Solution to 4 (b). It is straightforward to verify that a  1 = 1 for a 6= 1, so that once 1 is included in the list, it can never
by removed and so the list terminates with the single value 1.
Solution to 4 (a). There are several ways of approaching (a). It is important to verify that the set fx : 0 < x < 1g is
closed under the operation  so that it is always de ned.
If 0 < a; b < 1, then
a + b 2ab
0< <1:
1 ab
The left inequality follows from
a + b 2ab = a(1 b) + b(1 a) > 0
and the right from
a + b 2ab (1 a)(1 b)
1 = >0:
1 ab 1 ab
Hence, it will never happen that a set of numbers will contain a pair of reciprocals, and the operation can always be performed.
Solution 1. It can be shown by induction that any two numbers in any of the sets arise from disjoint subsets of S .

Use an induction argument on the number of entries that one starts with. At each stage the number of entries is reduced
by one. If we start with n numbers, the nal result is
1 22 + 33


+ ( 1)n 1 nn
;
1 2 + 23 34 +


+ ( 1)n 1 (n 1)n
n

where i is the symmetric sum of all i i fold products of the n elements xi in the list.
Solution 2. De ne
a+b 2ab
ab= :
1 ab
This operation is commutative and also associative:
a+b+c 2(ab + bc + ca) + 3abc
a  (b  c) = (a  b)  c = :
1 (ab + bc + ca) + 2abc
Since the nal result amounts to a  product of elements of S with some arrangement of brackets, the result follows.
Solution 3. Let (x) = x=(1 x) for 0 < x < 1. This is a one-one function from the open interval (0; 1) to the half line
(0; 1). For any numbers a; b 2 S , we have that
 
a+b 2ab a+b 2ab a+b 2ab
 = ab) (a+b 2ab) = 1 a b+ab
1 ab (1

= a + b = (a) + (b) : (2)

1 a 1 b

Let T = f(s) : s 2 S g. Then replacing a; b in S as indicated corresponds to P

replacing (a) and (b) in T by (a) + (b) to
get a new pair of sets related by . The nal result is the inverse under  of f(s) : s 2 S g.
Solution 4. Let f (x) = (1 x) be de ned for positive x unequal to 1. Then f (x) > 1 if and only if 0 < x < 1. Observe
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that
1 xy 1 1
f (x  y ) = = + 1:
1 x y + xy 1 x 1 y
If f (x) > 1 and f (y) > 1, then also f (x  y) > 1. It follows that if x and y lie in the open interval (0; 1), so does x  y. We
also note that f (x) is a one-one function.

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 To each list L, we associate the function g(L) de ned by
X
g ( L) =ff (x) : x 2 Lg :
Let Ln be the given list, and let the subsequent lists be Ln 1 ; Ln 2 ;


; L1 , where Li has i elements. Since f (x  y) =
f (x) + f (y ) 1, g(Li ) = g(Ln ) (n i) regardless of the choice that creates each list from its predecessors. Hence
g (L1 ) = g (Ln ) (n 1) is xed. However, g(L1 ) = f (a) for some number a with 0 < a < 1. Hence a = f 1 (g(Ln ) (n 1))
is xed.
Solution to 5 (a). Let I be the incentre of triangle ABC . Since the quadrilateral AEIF has right angles at E and F , it
is concyclic, so that passes through I . Similarly, 2 and 3 pass through I , and (a) follows.
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Solution to 5 (b). Let ! and I denote the incircle and incentre of triangle ABC , respectively. Observe that, since AI
bisects the angle F AE and AF = AE , then AI right bisects the segment F E . Similarly, BI right bisects DF and CI right
bisects DE .
We invert the diagram through !. Under this inversion, let the image of A be A0 , etc. Note that the centre I of inversion
is collinear with any point and its image under the inversion. Under this inversion, the image of 1 is EF , which makes A0
the midpoint of EF . Similarly, B 0 is the midpoint of DF and C 0 is the midpoint of DE . Hence, 0 , the image of under
this inversion, is the circumcircle of triangle A0 B 0 C 0 , which implies that 0 is the nine-point circle of triangle DEF .
Since P is the intersection of and 1 other than A, P 0 is the intersection of 0 and EF other than A0 , which means
that P 0 is the foot of the altitude from D to EF . Similarly, Q0 is the foot of the altitude from E to DF and R0 is the foot
of the altitude from F to DE .
Now, let X , Y and Z be the midpoints of arcs BC , AC and AB on respectively. We claim that X lies on P D.
Let X 0 be the image of X under the inversion, so I , X and X 0 are collinear. But X is the midpoint of arc BC , so A,
A0 , I , X 0 and X are collinear. The image of line P D is the circumcircle of triangle P 0 ID, so to prove that X lies on P D, it
suces to prove that points P 0 , I , X 0 and D are concyclic.
We know that B 0 is the midpoint of DF , C 0 is the midpoint of DE and P 0 is the foot of the altitude from D to EF .
Hence, D is the re ection of P 0 in B 0 C 0 .
Since IA0 ? EF , IB 0 ? DF and IC 0 ? DE , I is the orthocentre of triangle A0 B 0 C 0 . So, X 0 is the intersection of the
altitude from A0 to B 0 C 0 with the circumcircle of triangle A0 B 0 C 0 . From a wellknown fact, X 0 is the re ection of I in B 0 C 0 .
This means that B 0 C 0 is the perpendicular bisector of both P 0 D and IX 0 , so that the points P 0 , I , X 0 and D are concyclic.
Hence, X lies on P D. Similarly, Y lies on QE and Z lies on RF . Thus, to prove that P D, QE and RF are concurrent,
it suces to prove that DX , EY and F Z are concurrent.
To show this, consider tangents to at X , Y and Z . These are parallel to BC , AC and AB , respectively. Hence, the
triangle
that these tangents de ne is homothetic to the triangle ABC . Let S be the centre of homothety. Then the
homothety taking triangle ABC to
takes ! to , and so takes D to X , E to Y and F to Z . Hence DX , EY and F Z
concur at S .
Comment. The solution uses the following result: Suppose ABC is a triangle with orthocentre H and that AH intersects
BC at P and the circumcircle of ABC at D. Then HP = P D. The proof is straightforward: Let BH meet AC at Q. Note
that AD ? BC and BQ ? AC . Since \ACB = \ADB ,
\HBC = \QBC = 90 \QCB = 90 \ACB = 90 \ADB = \DBP ;
from which follows the congruence of triangle HBP and DBP and equality of HP and P D.
Solution 2. (a) Let 2 and 3 intersect at J . Then BDJF and CDJE are concyclic. We have that

\F JE = 360 (\DJF + \DJE )

= 360 (180 \ABC + 180 \ACB )

= \ABC + \ACB = 180 \F AE : (3)
Hence AF JE is concyclic and so the circumcircles of AEF , BDF and CED pass through J .
(b) [Y. Li] Join RE, RD, RA and RB. In 3 , \ERD = \ECD = \ACB and \REC = \RDC . In , \ARB = \ACB .
Hence, \ERD = \ARB =) \ARE = \BRD. Also,
\AER = 180 \REC = 180 \RDC = \BDR :
Therefore, triangle ARE and BRD are similar, and AR : BR = AE : BD = AF : BF . If follows that RF bisects angle
ARB , so that RF passes through the midpoint of minor arc AB on . Similarly, P D and QE are respective bisectors of
angles BP C and CQA and pass through the midpoints of the minor arc BC and CA on ..

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 Let O be the centre of circle , and U , V , W be the respective midpoints of the minor arc BC , CA, AB on this circle,
so that P U contains D, QV contains E and RW contains F . It is required to prove that DU , EV and F W are concurrent.
Since ID and OU are perpendicular to BC , IDkOU . Similarly, IE kOV and IF kOW . Since jIDj = jIE j = jIF j = r (the
inradius) and jOU j = jOV j = jOW j = R (the circumradius), a translation IO ! followed by a dilatation of factor R=r takes
triangle DEF to triangle U V W , so that these triangles are similar with corresponding sides parallel.
Suppose that EV and F W intersect at K and that DU and F W intersect at L. Taking account of the similarity of the
triangles KEF and KV W , LDF and LU W , DEF and U V W , we have that
KF : F W = EF : V W = DF : U W = LF : LW ;
so that K = L and the lines DU , EV and F W intersect in a common point K , as desired.