Mathematical Excalibur, Volume 9, Number 2

Volume 9, Number 2 May 2004 – July 2004

Inversion

Kin Y. Li

Olympiad Corner

The XVI Asian Pacific Mathematical Olympiad took place on March 2004. Here are the problems. Time allowed: 4 hours.

Problem 1. Determine all finite nonempty sets S of positive integers satisfying

) , (i j

j

i+ _{is an element of S for all i, j in S,}

where (i, j) is the greatest common divisor of i and j.

Problem 2. Let O be the circumcenter and H the orthocenter of an acute triangle

ABC. Prove that the area of one of the

triangles AOH, BOH, COH is equal to the sum of the areas of the other two.

Problem 3. Let a set S of 2004 points in the plane be given, no three of which are collinear. Let ℒ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to color the points of S with at most two colors, such that for any points p, q of S, the number

(continued on page 4)

Editors: ஻ Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK

ଽ υ ࣻ (KO Tsz-Mei)

గ ႀ ᄸ (LEUNG Tat-Wing)

؃ ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST

֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance.

On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is August 9, 2004.

For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI Department of Mathematics

The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected]

In algebra, the method of logarithm

transforms tough problems involving

multiplications and divisions into simpler problems involving additions and subtractions. For every positive number x, there is a unique real number logx in base 10. This is a one-to-one

correspondence between the positive numbers and the real numbers.

In geometry, there are also transformation methods for solving problems. In this article, we will discuss one such method called inversion. To present this, we will introduce the

extended plane, which is the plane

together with a point that we would like to think of as infinity. Also, we would like to think of all lines on the plane will go through this point at infinity! To understand this, we will introduce the

stereographic projection, which can be

described as follow.

Consider a sphere sitting on a point O of a plane. If we remove the north pole N of the sphere, we get a punctured sphere. For every point P on the plane, the line

NP will intersect the punctured plane at a unique point SP. So this gives a one-to-one correspondence between the plane and the punctured sphere. If we consider the points P on a circle in the plane, then the SP points will form a circle on the punctured sphere. However, if we consider the points P on

any line in the plane, then the SP points will form a punctured circle on the sphere with N as the point removed from the circle. If we move a point P on any

line toward infinity, then SP will go

toward the same point N! Thus, in this

model, all lines can be thought of as going to the same infinity.

Now for the method of inversion, let O be a point on the plane and r be a positive number. The inversion with center O and radius r is the function on the extended plane that sends a point X ≠ O to the

image point X′ on the ray OX such that OX·OX′ = r2_.

When X = O, X′ is taken to be the point at infinity. When X is infinity, X′ is taken to be O. The circle with center O and radius r is called the circle of inversion. The method of inversion is based on the following facts.

(1) The function sending X to X′ described above is a one-to-one correspondence between the extended plane with itself. (This follows from checking (X′ )′ = X. )

(2) If X is on the circle of inversion, then

X′ = X. If X is outside the circle of

inversion, then X′ is the midpoint of the chord formed by the tangent points T1, T2

of the tangent lines from X to the circle of inversion. (This follows from

OX·OX′ = (r sec ∠T1OX )(r cos ∠T1OX)

= r2_{. )}

(3) A circle not passing through O is sent to a circle not passing through O. In particular, the images of concyclic points are concyclic. The point O, the centers of the circle and the image circle are collinear. However, the center of the

circle is not sent to the center of the image circle!

(4) A circle passing through O is sent to a line not passing through O. Conversely, a line not passing through O is sent to a circle passing through O.

Mathematical Excalibur, Vol. 9, No. 2, May 04- July 04 Page 2

(5) A line passing through O is sent to itself.

(6) If two curves intersect at a certain angle at a point P ≠ O, then the image curves will also intersect at the same angle at P′. If the angle is a right angle, the curves are said to be orthogonal. So in particular, orthogonal curves are sent to orthogonal curves. Tangent curves are sent to tangent curves.

(7) If points A, B are different from O and points O, A, B are not collinear, then the equation OA·OA′ = r2 = OB·OB′ implies

OA/OB=OB′/OA′. Along with ∠AOB =

∠B′OA′, they imply ∆OAB, ∆OB′A′ are

similar. Then OB OA r OB A O AB B A ⋅ = ′ = ′ ′ 2 so that . 2 AB OB OA r B A ⋅ = ′ ′

The following are some examples that illustrate the powerful method of inversion. In each example, when we do inversion, it is often that we take the point that plays the most significant role and where many circles and lines intersect.

Example 1. (Ptolemy’s Theorem) For

coplanar points A, B, C, D, if they are concyclic, then

AB·CD + AD·BC = AC·BD. Solution. Consider the inversion with

center D and any radius r. By fact (4), the circumcircle of ∆ABC is sent to the line through A′, B′, C′. Since A′B′ + B′C′ =

A′C′, we have by fact (7) that

. 2 2 2 AC CD AD r BC CD BD r AB BD AD r ⋅ = ⋅ + ⋅

Multiplying by (AD·BD·CD)/r2_{, we get}

the desired equation.

Remarks. The steps can be reversed to

get the converse statement that if

AB·CD + AD·BC = AC·BD,

then A,B,C,D are concyclic.

Example 2. (1993 USAMO) Let ABCD

be a convex quadrilateral such that diagonals AC and BD intersect at right angles, and let O be their intersection point. Prove that the reflections of O across AB,

BC, CD, DA are concyclic.

Solution. Let P, Q, R, S be the feet of

perpendiculars from O to AB, BC, CD, DA, respectively. The problem is equivalent to showing P, Q, R, S are concyclic (since they are the midpoints of O to its reflections). Note OSAP, OPBQ, OQCR,

ORDS are cyclic quadrilaterals. Let their

circumcircles be called CA, CB, CC, CD, respectively.

Consider the inversion with center O and any radius r. By fact (5), lines AC and

BD are sent to themselves. By fact (4),

circle CA is sent to a line LA, circle CB is sent to a line LB, circle CC is sent to a line

LC, circle CD is sent to a line LD. Since CA intersect line BD at O and O is sent to infinity, so LA must be parallel to line BD. Similarly, LB is parallel to line AC, LC is parallel to line BD and LD is parallel to line

AC.

Next CA intersects CB at O and P. This implies LA intersects LB at P′. Similarly, LB intersects LC at Q′, LC intersects LD at R′ and LD intersects LA at S′.

Since AC _{⊥ BD, P′Q′R′S′ is a rectangle,} hence cyclic. Therefore, by fact (3), P, Q,

R, S are concyclic.

Example 3. (1996 IMO) Let P be a point

inside triangle ABC such that

∠APB – ∠ACB =∠APC – ∠ABC.

Let D, E be the incenters of triangles APB,

APC, respectively. Show that AP, BD, CE

meet at a point.

Solution. Let lines AP, BD intersect at X,

lines AP, CE intersect at Y. We have to show X = Y. By the angle bisector theorem,

BA/BP = XA/XP. Similarly, CA/CP = YA/YP. As X, Y are on AP, we get X = Y if

and only if BA/BP = CA/CP.

Consider the inversion with center A and any radius r. By fact (7), ∆ABC, ∆AC′B′ are similar, ∆APB, ∆AB′P′ are

similar and ∆APC, ∆AC′P′ are similar. Now ∠B′C′P′ =∠AC′P′ – ∠AC′B′ =∠APC – ∠ABC = ∠APB – ∠ACB = ∠AB′P – ∠AB′C′ =∠C′B′P′. So ∆B′C′P′ is isosceles and P′B′ = P′C′. From ∆APB, ∆AB′P′ similar and ∆APC, ∆AC′P′ similar, we get

. CP CA C P A P B P A P BP BA ₌ ′ ′ ′ = ′ ′ ′ = Therefore, X = Y.

Example 4. (1995 Israeli Math Olympiad)

Let PQ be the diameter of semicircle H. Circle O is internally tangent to H and tangent to PQ at C. Let A be a point on H and B a point on PQ such that AB _{⊥ PQ} and is tangent to O. Prove that AC bisects ∠PAB.

Solution. Consider the inversion with

center C and any radius r. By fact (7), ∆CAP, ∆CP′A′ similar and ∆CAB, ∆CB′A′ similar. So AC bisects PAB if and only if ∠CAP =∠CAB if and only if ∠CP′A′ =

∠CB′A′.

By fact (5), line PQ is sent to itself. Since circle O passes through C, circle O is sent to a line O′ parallel to PQ. By fact (6), since H is tangent to circle O and is orthogonal to line PQ, H is sent to the semicircle H′ tangent to line O′ and has diameter P′Q′. Since segment AB is tangent to circle O and is orthogonal to

PQ, segment AB is sent to arc A′B′ on the

semicircle tangent to line O′ and has diameter CB’. Now observe that arc A′Q′ and arc A′C are symmetrical with respect to the perpendicular bisector of CQ′ so we get ∠CP′A′ = ∠CB′A′.

In the solutions of the next two examples, we will consider the nine-point circle and the Euler line of a triangle. Please consult Vol. 3, No. 1 of Mathematical Excalibur for discussion if necessary.

n

Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li,

Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for

submitting solutions is August 9, 2004. Problem 201. (Due to Abderrahim

OUARDINI, Talence, France) Find

hich nonright triangles ABC satisfy w

tan A tan B tan C

> [tan A] + [tan B] + [tan C], where [t] denotes the greatest integer less than or equal to t. Give a proof. Problem 202. (Due to LUK Mee Lin,

La Salle College) For triangle ABC, let D, E, F be the midpoints of sides AB, BC, CA, respectively. Determine

which triangles ABC have the property that triangles ADF, BED, CFE can be folded above the plane of triangle DEF to form a tetrahedron with AD coincides with BD; BE coincides with

CE; CF coincides with AF.

Problem 203. (Due to José Luis

DÍAZ-BARRERO, Universitat Politec- nica de Catalunya, Barcelona, Spain)

Let a, b and c be real numbers such that

a + b + c ≠ 0. Prove that the equation

(a+b+c)x2 _{+ 2(ab+bc+ca)x + 3abc = 0}

has only real roots.

Problem 204. Let n be an integer with

n > 4. Prove that for every n distinct

integers taken from 1, 2, …, 2n, there always exist two numbers whose least common multiple is at most 3n + 6. Problem 205. (Due to HA Duy Hung,

Hanoi University of Education, Vietnam) Let a, n be integers, both

greater than 1, such that an _{– 1 is} divisible by n. Prove that the greatest common divisor (or highest common factor) of a – 1 and n is greater than 1.

*****************

Solutions

****************

Problem 196. (Due to John

PANAGEAS, High School “Kaisari”,

Athens, Greece) Let be

positive real numbers with sum equal to 1. Prove that for every positive integer m,

x x x1, 2,..., ). ... ( ₁m ₂m _nm m _{x x x} n n≤ + + +

Solution. CHENG Tsz Chung (La Salle

College, Form 5), Johann Peter Gustav Lejeune DIRICHLET (Universidade de Sao Paulo – Campus Sao Carlos), KWOK Tik Chun (STFA Leung Kau Kui College, Form 6), POON Ming Fung (STFA Leung Kau Kui College, Form 6), Achilleas P. PORFYRIADIS (American College of Thessaloniki “Anatonia”, Thessaloniki, Greece), SIU Ho Chung (Queen’s College, Form 5) and YU Hok Kan (STFA Leung Kau

ui College, Form 6). K

Applying Jensen’s inequality to f (x) =

xm _{on [0, 1] or the power mean inequality,} we have . ) ( 1 1 n x x n x x m n m m n _≤ + + + +_{L L}

Using

x

+ L

+

x

Other commended solvers: TONG Yiu

Wai (Queen Elizabeth School, Form 6), YEUNG Wai Kit (STFA Leung Kau Kui College, Form 3) and YEUNG Yuen Chuen (La Salle College, Form 4).

Problem 197. In a rectangular box, the lengths of the three edges starting at the same vertex are prime numbers. It is also given that the surface area of the box is a power of a prime. Prove that exactly one of the edge lengths is a prime number of the form 2k _{−1. (Source: KöMaL Gy.3281)}

Solution. CHAN Ka Lok (STFA Leung

Kau Kui College, Form 4), KWOK Tik Chun (STFA Leung Kau Kui College, Form 6), John PANAGEAS (Kaisari High School, Athens, Greece), POON Ming Fung (STFA Leung Kau Kui College, Form 6), Achilleas P. PORFYRIADIS (American College of Thessaloniki “Anatonia”, Thessaloniki, Greece), SIU Ho Chung (Queen’s College, Form 5), TO Ping Leung (St. Peter’s Secondary School), YEUNG Wai Kit (STFA Leung Kau Kui College, Form 3), YEUNG Yuen Chuen (La Salle College, Form 4) and YU Hok Kan

STFA Leung Kau Kui College, Form 6). (

Let the prime numbers x, y, z be the lengths of the three edges starting at the same vertex. Then 2(xy + yz + zx) = pn _for some prime p and positive integer n. Since the left side is even, we get p = 2. So xy +

yz + zx = 2n─1_{. Since x, y, z are at least 2,} the left side is at least 12, so n is at least 5. If none or exactly one of x, y, z is even, then xy + yz + zx would be odd, a contradiction. So at least two of x, y, z are even and prime, say x = y = 2. Then z =

2n─3_{−1. The result follows.}

Other commended solvers: NGOO Hung

Wing (Valtorta College).

Problem 198. In a triangle ABC, AC =

BC. Given is a point P on side AB such

that ∠ACP = 30○_{. In addition, point Q}

outside the triangle satisfies ∠CPQ = ∠CPA + ∠APQ = 78○_{. Given that all}

angles of triangles ABC and QPB, measured in degrees, are integers, determine the angles of these two triangles. (Source: KöMaL C. 524)

Solution. CHAN On Ting Ellen (True

Light Girls’ College, Form 4), CHENG Tsz Chung (La Salle College, Form 5), POON Ming Fung (STFA Leung Kau Kui College, Form 6), TONG Yiu Wai (Queen Elizabeth School, Form 6), YEUNG Yuen Chuen (La Salle College, Form 4) and YU Hok Kan (STFA Leung

au Kui College, Form 6). K

As ∠ACB >∠ACP = 30○_{, we get}

∠CAB = ∠CBA < (180○_{− 30}○_{) / 2 = 75}○_.

Hence ∠CAB ≤ 74○_{. Then}

∠CPB = ∠CAB + ∠ACP ≤ 74○_{+ 30}○ _{= 104}○_. Now ∠QPB = 360○ _{– ∠QPC − ∠CPB} ≥ 360○ _{– 78}○ _{– 104}○ _{= 178}○_.

Since the angles of triangle QPB are positive integers, we must have

∠QPB = 178○_{, ∠PBQ = 1}○ _=∠PQB

and all less-than-or-equal signs must be equalities so that

∠CAB = ∠CBA = 74○ _{and ∠ACB = 32}○_.

Other commended solvers: CHAN Ka Lok

(STFA Leung Kau Kui College, Form 4), KWOK Tik Chun (STFA Leung Kau Kui College, Form 6), Achilleas P. PORFYRIADIS (American College of Thessaloniki “Anatonia”, Thessaloniki, Greece), SIU Ho Chung (Queen’s College, Form 5), YEUNG Wai Kit (STFA Leung Kau Kui College, Form 3), Richard YEUNG Wing Fung (STFA Leung Kau Kui College, Form 6) and YIP Kai Shing (STFA Leung Kau Kui College, Form 4).

Problem 199. Let R+ denote the

positive real numbers. Suppose is a strictly decreasing function such that for all ,

+ +_{→ R} R f : + ∈ R y x, f (x + y) + f (f (x) + f (y)) = f (f (x + f (y)) + f (y + f (x))). Prove that f (f (x)) = x for every x > 0.

Source: 1997 Iranian Math Olympiad)

(

Solution. Johann Peter Gustav Lejeune

DIRICHLET (Universidade de Sao Paulo – Campus Sao Carlos) and Achilleas P. PORFYRIADIS (American College of

Mathematical Excalibur, Vol. 9, No. 2, May 04- July 04 Page 4

Thessaloniki “Anatonia”, Thessaloniki, Greece).

Setting y = x gives

f (2x) + f (2f (x)) = f (2f ( x + f (x))).

Setting both x and y to f(x) in the given equation gives

f (2f (x)) + f (2f (f (x)))

= f (2f (f (x) + f (f (x)))).

Subtracting this equation from the one above gives

f (2f (f (x))) – f (2x)=f (2f ( f (x) + f (f (x))))

– f ( 2f (x + f (x))). Assume f (f (x)) > x. Then 2f (f (x)) > 2x. Since f is strictly decreasing , we have

f(2f (f (x))) < f (2x). This implies the left

side of the last displayed equation is negative. Hence,

f (2f ( f (x) + f ( f (x)))) < f ( 2f ( x + f (x))).

Again using f strictly decreasing, this inequality implies

2f ( f (x) + f ( f (x))) > 2f ( x + f (x)), which further implies

f (x) + f (f (x)) < x + f (x).

Canceling f (x) from both sides leads to the contradiction that f (f (x)) < x.

Similarly, f (f (x)) < x would also lead to a contradiction as can be seen by reversing all inequality signs above. Therefore, we must have f (f (x)) = x.

Problem 200. Aladdin walked all over the equator in such a way that each moment he either was moving to the west or was moving to the east or applied some magic trick to get to the opposite point of the Earth. We know that he travelled a total distance less than half of the length of the equator altogether during his westward moves. Prove that there was a moment when the difference between the distances he had covered moving to the east and moving to the west was at least half of the length of the equator. (Source:

KöMaL F. 3214) Solution.

Let us abbreviate Aladdin by A. At every moment let us consider a twin, say ∀, of

A located at the opposite point of the

position of A. Now draw the equator

circle. Observe that at every moment

either both are moving east or both are

moving west. Combining the movement swept out by A and ∀, we get two continuous paths on the equator. At the same moment, each point in one path will have its opposite point in the other path. Let N be the initial point of A in his travel and let P(N)denote the path beginning with N. Let W be the westernmost point on P(N). Let N’ and W’ be the opposite points of N and W respectively. By the westward travel condition on A, W cannot be as far as N’.

Assume the conclusion of the problem is false. Then the easternmost point reached by P(N) cannot be as far as N’. So P(N) will not cover the inside of minor arc WN’ and the other path will not cover the inside of minor arc W’N. Since A _{have walked} over all points of the equator (and hence A and∀ together walked every point at least twice), P(N) must have covered every point of the minor arc W’N at least twice. Since P(N) cannot cover the entire equator, every point of minor arc W’N must be traveled westward at least once by A or ∀. Then A travelled westward at least a distance equal to the sum of lengths of minor arcs W’N and NW, i.e. half of the equator. We got a contradiction.

Other commended solvers: POON Ming

Fung (STFA Leung Kau Kui College, Form 6).

Olympiad Corner

(continued from page 1)

of lines in _{ℒ which separate p from q is} odd if and only if p and q have the same color.

Note: A line _{ℓ separates two points p}

and q if p and q lie on opposite sides of _ℓ with neither point on _ℓ.

Problem 4. For a real number x, let

⎣ ⎦

⎥ ⎦ ⎥ ⎢ ⎣ ⎢ + − ) 1 ( )! 1 ( n n n

is even for every positive integer n.

Problem 5. Prove that

(a2 _{+2) (b}2 _{+2) (c}2 _{+2) ≥ 9 (ab+bc+ca)}

for all real numbers a, b, c > 0.

Inversion

(continued from page 2)

Example 5. (1995 Russian Math Olympiad) Given a semicircle with

diameter AB and center O and a line, which intersects the semicircle at C and D and line AB at M (MB < MA, MD < MC). Let K be the second point of intersection of the circumcircles of triangles AOC and

DOB. Prove that ∠MKO = 90○_.

Solution. Consider the inversion with

center O and radius r = OA. By fact (2), A,

B, C, D are sent to themselves. By fact

(4), the circle through A, O, C is sent to line AC and the circle through D, O, B is sent to line DB. Hence, the point K is sent to the intersection K′ of lines AC with DB and the point M is sent to the intersection

M′ of line AB with the circumcircle of

∆OCD. Then the line MK is sent to the circumcircle of OM′K′.

To solve the problem, note by fact (7), ∠MKO=90○ _{if and only if ∠K′M′O= 90}○_.

Since BC_{⊥AK′, AD⊥BK′ and O is the} midpoint of AB, so the circumcircle of ∆OCD is the nine-point circle of ∆ABK′, which intersects side AB again at the foot of perpendicular from K′ to AB. This point is M′. So ∠K′M′O = 90○ _{and we are}

done.

Example 6. (1995 Iranian Math Olympiad) Let M, N and P be points of

intersection of the incircle of triangle

ABC with sides AB, BC and CA

respectively. Prove that the orthocenter of ∆MNP, the incenter of ∆ABC and the circumcenter of ∆ABC are collinear.

Solution. Note the incircle of ∆ABC is

the circumcircle of ∆MNP. So the first two points are on the Euler line of ∆MNP.

C′ of PM, MN, NP, respectively. The

circumcenter of ∆A′B′C′ is the center of the nine point circle of ∆MNP, which is on the Euler line of ∆MNP. By fact (3), the circumcircle of ∆ABC is also on the Euler line of ∆MNP.

. Solution. We claim that S = (2) is the only such set.

Let S be a finite nonempty subset of positive integers satisfying (l), and let s E S. Then 2 =, s E S. [.l mark]

Furthermore, since *. _‘r

2+2 2 (2,2)= ’ we see that S = {2} is a solution. [l mark]

Now we suppose that S contains at ieast one other element. Case (i): S contains some odd positive integer a. Then

af2=$S,

where a + 2 is aIso odd, and by inductian S contains infinitely many odd integers, a contradiction. [2 marks] Case (ii): S contains no odd integers but at Ieast one even integer greater than 2. Then Iet 2b, where b > 1, be the, smallest such even integer in S. Then

b,l=g&S. __

But 2 < b + 1 < 26, which contradicts ouz assumption (if 6 + 1 is odd) or the minimality of 2b (if b + 1 is even).

This completes the proof. [3 marks]

lhbl- 2.

Solution If 0 = H (that is, .4BC is equilateral), then all three areas are 0, and the result holds. We now assume that 0 #$ H, and separate the proof into two cases.

__ __ ._._ __-&* ~~~.e&nm+mr&~_._.f.~ r* _v. F, *

CH _L AB. S&e 0 is the circumcentre, O”lies on the perpendicular bisec,tor of A.?

_-_.--

CH J_ _-LB, hence CH is the perpendicular bisector of AB. Hence AC = BC. Thus AAOH E ABOH,

so the sreas [AOH] and [BOH] of these triangles are equal. Also, in this case [COH] = 0, so [AOH] .= [BOH] -t [COH]. [2 marks]

Case (ii). Suppose OH meets two sides of the triangle, say it meets AB and AC. Then the centroid G lies on the Buler line OH. Extend _4G, meering BC at Dj then D is the midpoint of BC. Connect DO and DR. From the property of the centroid, AG = 2DG, heace [AOa = S[DOH]. [2 marks] Also, because D is

the midpoint of BC, we observe rhat the sum of the distxxes from B and C to OH equals twice the distance

from D to OH. Hence [BOH] f [COH] = 2(DOH]. [2 marks] Together with [AOH] = 2[DOH], we get the desired result. [1 mark]

?&blew

Solution Suppose first that the ser of 2004 points forms a con\-ex 2004gon P. Colou the vertices

alternately black and white. If we delete two vertices v snd u..,, the rest of the polygon fa&, into two pieces of i

and j vertices where i + j = 2002, .SO that ,i and j are the same parity Cnossibly one of i or j equals zero). The

number of lines in L: which separate v and w is ij, which is even if v and w are Merent colours and odd if u and w are the ssme colour. Thus we have a “good” colouring, that is, one which sat&r&s the amditions of the question. [I mark]

Now given an arbitrary con&uration S of 2004 points with no three collinear, we start with P and s~cce+ sively move each point of P to a point of S, bPing M&UI that the moving point. CTOSSCS lines in I: only one at a time. After 2004 such moves, we will have a good colouring of S, so long sa we maintain “goodness,, during

the moves. [2 marks]

To do this, whenever a point A being moved crosses a Line defined by two points B and C, switch the colours of A, B and C. [2 marks] We claim this maintains the goodness of the colouring. Note that A ends

up on the

or C) if ad only if BC did not separate A from P before the move. Since we have changed the colour of A but not P, A aad P are still coloured correctly as far ss line BC is concerned. Similar remarks hold for point

B with respect to line AC, and point C with respect to line AB. The relative positions of other points or lines. are not affected by the movement of point A. Thus the new colouring is still good. This completes the proof. [2 marks]

_. .

f%-vb\eHi4-.

Solution Let

fi- = (n - I)! n(n f 1) *

If n = 1,2,3,4, then. direct calculation shows that [NJ = 0. Let n > 4. We rewrite N as

*

N _{= -} (n-l)! _j

_{& -}

n [I mark]

(1)

where the third term is aIways even.

The first two terms of (1) are both of the form (m - l)!/m for m > 3. If m is not a prime or a square of a prime, then m = ab for 1 < a < b < m, and thus (m - l)! contains-both factors a, b and and at least one additional even factor. ‘Thus (m -l)!/m is an even integer. If m = p2 for an odd prime’p, then 1 < p < 2p < m, and again (m - l)!/m is aa even integer. [2 marks]

Now let m be a prime. By W&on’s Theorem, (m - l)! s -1 mod m, whence (m-l)!-+1 [l mark], and this integer must be odd as its numerator is odd. [l mark] Therefore m

is an integer

(m-l)! = (m-1)!i-1 _-- . 1

m m m

- ._ . . . -._ - - - -. . . _ - _ . -._ _. _ _ - - _.- - - -. - . _. ______ _ _,

has an even integral part. [I. mark] Since n and n + 1 cannot both be prime, we have proved that at Ieast two of the three terms of _?i in (1) are even integers and the other has an even integral part. Therefore the integral. part of LV is even. [l mark]

Solution Let

x=a+b+c, y=abiac+bc, z = abc.

Then

a2 + b2 + c2 = x2 - 2y, a2b2 + a2c2 + b22 = y2 - 2x2, a2b2r? - .z2 - , [I mark]

so the inequality to be proved becomes

.z2 + 2(y2 - 22.2) + 4(x2 - 2~) + 8 1 %I or

t2 + 2ys - 422 + 4z2 -17yi8>0. [l mark] Nowfroma2+b2+c?>ab+ac+bc=yweget

x2 =a2+b2+c2+2y>3y. [l mark]

Also

a2b2 + a2c2 + b2c2 = (ab)2 + (ac)2 + (bc)’ > ab . oc + ac . bc f bc . ad = (a f b + c)abc = xz, and thus

Y2 = c;r2b2 + a2c2 + b2c2 + 222 > 3x2 - * [l mark]

Hence

z2 + 2y’ -422+4x2 - 17y + 8 = (z - ;)2 i ;(y - 3)2 + +j+y’ - 3~2) + :(x2 - 3y) 2 0

__ _.. _ . . . ._ _._

as required. [3 marks]