Volume 11, Number 2 April 2006 – May 2006
Angle Bisectors Bisect Arcs
Kin Y. Li
Olympiad Corner
Below was the Find Round of the 36th Austrian Math Olympiad 2005. Part 1 (May 30, 2005)
Problem 1. Show that an infinite number of multiples of 2005 exist, in which each of the 10 digits 0,1,2,…,9 occurs the same number of times, not counting leading zeros.
Problem 2. For how many integer
values of a with |a| ≤ 2005 does the system of equations x2 = y + a, y2 = x + a have integer solutions?
Problem 3. We are given real numbers a, b and c and define sn as the sum sn = an + bn + cn of their n-th powers for non-negative integers n. It is known that s1 = 2, s2 = 6 and s3 = 14 hold. Show that
8 | | 2− 1⋅ 1 = + − n n n s s s
holds for all integers n > 1.
Problem 4. We are given two equilateral triangles ABC and PQR with parallel sides, “one pointing up” and “one pointing down.” The common area of the triangles’ interior is a hexagon. Show that the lines joining opposite corners of this hexagon are concurrent.
(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 16, 2006.
For individual subscription for the next five issues for the 05-06 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 general, angle bisectors of a triangle do not bisect the sides opposite the angles. However, angle bisectors always bisect the arcs opposite the angles on the circumcircle of the triangle! In math competitions, this fact is very useful for problems concerning angle bisectors or incenters of a triangle involving the circumcircle. Recall that the incenter of a triangle is the point where the three angle bisectors concur. Theorem. Suppose the angle bisector of ∠BAC intersect the circumcircle of ∆ABC at X ≠ A. Let I be a point on the line segment AX. Then I is the incenter of ∆ABC if and only if XI = XB = XC.
A
B C
X I
Proof. Note ∠BAX =∠CAX =∠CBX. So XB = XC. Then
I is the incenter of ∆ABC
⇔ ∠CBI =∠ABI
⇔ ∠IBX −∠CBX =∠BIX −∠BAX
⇔ ∠IBX = ∠BIX ⇔ XI = XB = XC.
Example 1. (1982 Australian Math
Olympiad) Let ABC be a triangle, and let the internal bisector of the angle A meet the circumcircle again at P. Define Q and R similarly. Prove that AP + BQ + CR > AB + BC + CA.
A
B
C
P
R
I
Q
Solution. Let I be the incenter of ∆ABC. By the theorem, we have 2IR = AR + BR > AB and similarly 2IP > BC, 2IQ > CA. Also AI + BI > AB, BI + CI > BC and CI + AI > CA. Adding all these
nequalities together, we get i
2(AP + BQ + CR) > 2(AB + BC + CA). Example 2. (1978 IMO) In ABC, AB =
AC. A circle is tangent internally to the circumcircle of ABC and also to the sides AB, AC at P, Q, respectively. Prove that the midpoint of segment PQ is the center of the incircle of ∆ABC.
A
B C
X
P Q
I
Solution. Let I be the midpoint of line segment PQ and X be the intersection of the angle bisector of ∠BAC with the arc BC not containing A.
By symmetry, AX is a diameter of the circumcircle of ∆ABC and X is the midpoint of the arc PXQ on the inside circle, which implies PX bisects
QPB
∠ . Now ∠ABX = 90˚ = ∠PIX
so that X, I, P, B are concyclic. Then
∠IBX =∠IPX =∠BPX =∠BIX.
So XI = XB. By the theorem, I is the incenter of ∆ABC.
Example 3. (2002 IMO) Let BC be a diameter of the circle Γ with center O. Let A be a point on Γ such that 0˚ <
AOB
∠ < 120˚. Let D be the midpoint of the arc AB not containing C. The line through O parallel to DA meets the line AC at J. The perpendicular bisector of OA meets Γ at E and at F. Prove that J is the incenter of the triangle CEF.
Mathematical Excalibur, Vol. 11, No. 2, Apr. 06 - May 06 Page 2 O C B A E F D J
Solution. The condition ∠ AOB < 120˚ ensures I is inside ∆CEF (when ∠AOB increases to 120˚, I will coincide with C). Now radius OA and chord EF are perpendicular and bisect each other. So EOFA is a rhombus. Hence A is the midpoint of arc EAF. Then CA bisects ∠ECF. Since OA = OC, ∠AOD = 1/2∠AOB = ∠OAC. Then DO is parallel to AJ. Hence ODAJ is a parallelogram. Then AJ = DO = EO = AE. By the theorem, J is the incenter of ∆CEF.
Example 4. (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 and CE meet at a point.
A B C F G P H I J D E K
Solution. Let lines AP, BP, CP intersect the circumcircle of ∆ABC again at F, G, H respectively. Now
∠APB −∠ACB =∠FPG −∠AGB
=∠FAG.
Similarly, ∠APC − ∠ABC = ∠FAH. So AF bisects ∠HAG. Let K be the incenter of ∆HAG. Then K is on AF and lines HK, GK pass through the midpoints I, J of minor arcs AG, AH respectively. Note lines BD, CE also pass through I, J as they bisect ∠ABP, ∠ACP respectively.
Applying Pascal’s theorem (see vol.10, no. 3 of Math Excalibur) to B, G, J, C,
H, I on the circumcircle, we see that P=BG∩CH, K=GJ∩HI and BI∩CJ= BD∩CE are collinear. Hence, BD∩CE is on line PK, which is the same as line AP. Example 5. (2006 APMO) Let A, B be two distinct points on a given circle O and let P be the midpoint of line segment AB. Let O1 be the circle tangent to the line AB at P and tangent to the circle O. Let ℓ be the tangent line, different from the line AB, to O1 passing through A. Let C be the intersection point, different from A, of ℓ and O. Let Q be the midpoint of the line segment BC and O2 be the circle tangent to the line BC at Q and tangent to the line segment AC. Prove that the circle O2 is tangent to the circle O.
A P B N L C Z Q M J K
Solution. Let the perpendicular to AB through P intersect circle O at N and M with N and C on the same side of line AB. By symmetry, segment NP is a diameter of the circle of O1 and its midpoint L is the center of O1. Let line AL intersect circle O again at Z. Let line ZQ intersect line CM at J and circle O again at K.
Since AB and AC are tangent to circle O1, AL bisects ∠ CAB so that Z is the midpoint of arc BC. Since Q is the midpoint of segment BC, ∠ZQB = 90˚ = ∠LPA and ∠JQC = 90˚ =∠MPB. Next
∠ZBQ =∠ZBC =∠ZAC =∠LAP.
So ∆ZQB, ∆LPA are similar. Since M is the midpoint of arc AMB,
∠JCQ =∠MCB =∠MCA =∠MBP.
So ∆JQC, ∆MPB are similar.
By the intersecting chord theorem, AP·BP = NP·MP = 2LP·MP. Using the similar triangles above, we have
. 2 1 CQ BQ JQ ZQ BP AP MP LP ⋅ ⋅ = ⋅ ⋅ =
By the intersecting chord theorem, KQ·ZQ = BQ·CQ so that
KQ = (BQ·CQ)/ZQ = 2JQ.
This implies J is the midpoint of KQ. Hence the circle with center J and diameter KQ is tangent to circle O at K and tangent to BC at Q. Since J is on the bisector of ∠BCA, this circle is also tangent to AC. So this circle is O2. Example 6. (1989 IMO) In an acute-angled triangle ABC the internal bisector of angle A meets the circumcircle of the triangle again at A1. Points B1 and C1 are defined similarly. Let A0 be the point of intersection of the line AA1 with the external bisectors of angles B and C. Points B0 and C0 are defined similarly. Prove that:
(i) the area of the triangle A0B0C0 is twice the area of the hexagon AC1BA1CB1,
(ii) the area of the triangle A0B0C0 is at least four times the area of the triangle ABC. C0 B0 C A0 B A I A1 B1 C1
Solution. (i) Let I be the incenter of ∆ABC. Since internal angle bisector and external angle bisector are perpendicular, we have ∠B0BA0 = 90˚. By the theorem, A1I = A1B. So A1 must be the midpoint of the hypotenuse A0I of right triangle IBA0. So the area of ∆BIA0 is twice the area of ∆BIA1. Cutting the hexagon AC1BA1CB1 into six triangles with common vertex I and applying a similar area fact like the last statement to each of the six triangles, we get the conclusion of (i).
(ii) Using (i), we only need to show the area of hexagon AC1BA1CB1 is at least twice the area of ∆ABC.
B
C
A
A
1D
A
2H
(continued on page 4)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 16, 2006.
Problem 251. Determine with proof the largest number x such that a cubical gift of side x can be wrapped completely by folding a unit square of wrapping paper (without cutting). Problem 252. Find all polynomials f(x) with integer coefficients such that for every positive integer n, 2n − 1 is divisible by f(n).
Problem 253. Suppose the bisector of ∠BAC intersect the arc opposite the angle on the circumcircle of ∆ABC at A1. Let B1 and C1 be defined similarly. Prove that the area of ∆A1B1C1 is at least the area of ∆ABC.
Problem 254. Prove that if a, b, c > 0, hen t 2 ) ( ) ( a b c a b c abc + + + + + ≥4 3abc(a+b+c). Problem 255. Twelve drama groups are to do a series of performances (with some groups possibly making repeated performances) in seven days. Each group is to see every other group’s performance at least once in one of its
ay-offs. d
Find with proof the minimum total number of performances by these groups.
*****************
Solutions
****************
Problem 246. A spy plane is flying at the speed of 1000 kilometers per hour along a circle with center A and radius 10 kilometers. A rocket is fired from A at the same speed as the spy plane such that it is always on the radius from A to the spy plane. Prove such a path for the rocket exists and find how long it takes for the rocket to hit the spy plane.
(Source: 1965 Soviet Union Math Olympiad)
Solution. Jeff CHEN (Virginia, USA), Koyrtis G. CHRYSSOSTOMOS (Larissa, Greece, teacher), G.R.A. 20 Math Problem Group (Roma, Italy) and Alex O Kin-Chit (STFA Cheng Yu Tung
econdary School). S
O
P
R
Q
A
L
Let the spy plane be at Q when the rocket was fired. Let L be the point on the circle obtained by rotating Q by 90˚ in the forward direction of motion with respect to the center A. Consider the semicircle with diameter AL on the same side of line AL as Q. We will show the path from A to L along the semicircle satisfies the conditions.
For any point P on the arc QL, let the radius AP intersect the semicircle at R. Let O be the midpoint of AL. Since
∠QAP =∠RLA = 1/2∠ROA
and AL = 2AO, the length of arc AR is the same as the length of arc QP. So the conditions are satisfied.
Finally, the rocket will hit the spy plane at L after 5π/1000 hour it was fired.
Comments: One solver guessed the path should be a curve and decided to try a circular arc to start the problem. The other solvers derived the equation of the path by a differential equation as follows: using polar coordinates, since the spy plane has a constant angular velocity of 1000/10 = 100 rad/sec, so at time t, the spy plane is at (10, 100t) and the rocket is at (r(t), θ(t)). Since the rocket and the spy plane are on the same radius, so θ(t) = 100t. Now they have the same speed, so
. 10 )) ( ' ) ( ( )) ( ' (r t 2+ rtθ t 2= 6 Then . 100 ) ( 100 ) ( ' 2 = − tr t r
Integrating both sides from 0 to t, we get the equation r = 10 sin(100t) = 10 sin θ, which describes the path above.
Problem 247. (a) Find all possible positive integers k ≥ 3 such that there are k positive integers, every two of them are
not relatively prime, but every three of them are relatively prime.
(b) Determine with proof if there exists an infinite sequence of positive integers satisfying the conditions in (a)
bove. a
(Source: 2003 Belarussian Math Olympiad)
Solution. G.R.A. 20 Math Problem Group (Roma, Italy) and YUNG Fai. (a) We shall prove by induction that the conditions are true for every positive integer k ≥ 3.
For k = 3, the numbers 6, 10, 15 satisfy the conditions. Assume it is true for some k ≥ 3 with the numbers being a1, a2, …, ak. Let p1, p2, …, pk be distinct prime numbers such that each pi is greater than a1a2…ak. For I = 1 to k, let bi
= aipi and let bk+1= p1p2…pk. Then gcd(bi, bj)=gcd(ai, aj) >1 for 1≤ i < j ≤k, gcd(bi, bk+1) = pi > 1 for 1 ≤ i ≤ k, gcd(bh, bi, bj) = gcd(ah, ai, aj) = 1
for 1≤ h ≤ i < j ≤ k and gcd(bi, bj, bk+1) = 1 for 1 ≤ i < j ≤ k, completing the induction.
(b) Assume there are infinitely many positive integers a1, a2, a3, … satisfying the conditions in (a). Let a1 have exactly m prime divisors. For i = 2 to m + 2, since each of the m + 1 numbers gcd(a1, ai) is divisible by one of these m primes, by the pigeonhole principle, there are i, j with 2 ≤ i < j ≤ m + 2 such that gcd(a1, ai) and gcd(a1, aj) are divisible by the same prime. Then gcd(a1, ai, aj) > 1, a contradiction. Commended solvers: CHAN Nga Yi (Carmel Divine Grace Foundation Secondary School, Form 6) and CHAN Yat Sing (Carmel Divine Grace Foundation Secondary School, Form 6).
Problem 248. Let ABCD be a convex quadrilateral such that line CD is tangent to the circle with side AB as diameter. Prove that line AB is tangent to the circle with side CD as diameter if and only if lines BC and AD are parallel.
Solution. Jeff CHEN (Virginia, USA) and Koyrtis G. CHRYSSOSTOMOS (Larissa, Greece, teacher).
Mathematical Excalibur, Vol. 11, No. 2, Apr. 06 - May 06 Page 4 E A B D C F
Let E be the midpoints of AB. Since CD is tangent to the circle, the distance from E to line CD is h1 = AB/2. Let F be the midpoint of CD and let h2 be the distance from F to line AB. Observe that the areas of ∆CEF and ∆DEF = CD·AB/8. Now
line AB is tangent to the circle with side CD as diameter ⇔ h2=CD/2
⇔ areas of ∆AEF, ∆BEF, ∆CEF and ∆DEF are equal to AB·CD/8 ⇔ AD∥EF, BC∥EF
⇔ AD∥BC.
Problem 249. For a positive integer n, if a1,⋯, an, b1, ⋯, bn are in [1,2] and then prove that , 2 2 1 2 2 1 an b bn a +L+ = +L+ ). ( 10 17 2 2 1 3 1 3 1 n n n a a b a b a + + ≤ + +L L
Solution. Jeff CHEN (Virginia, USA). For x, y in [1,2], we have 1/2 ≤ x/y ≤2 ⇔ y/2 ≤ x ≤ 2y ⇔ (y/2 − x)(2y − x) ≤ 0 ⇔ x 2 + y2 ≤ 5xy/2.
Let x = ai and y = bi, then ai2 + bi2 ≤ 5aibi/2. Summing and manipulating, we get . 5 4 ) ( 5 2 1 2 2 1 2 1
∑
∑
∑
= = = − = + − ≤ − n i i i n i i n i i ib a b a aLet x = (ai3/bi)1/2 and y = (aibi)1/2. Then x/y = ai/bi in [1,2]. So ai3/bi + aibi≤ 5ai2/2. Summing, we get . 2 5 1 2 1 1 3
∑
∑
∑
= = = ≤ + n i i n i i i n i i i ab a b aAdding the two displayed inequalities, we get ). ( 10 17 2 2 1 3 1 3 1 n n n a a b a b a + + ≤ + +L L
Problem 250. Prove that every region with a convex polygon boundary cannot be dissected into finitely many regions with nonconvex quadrilateral boundaries. Solution. YUNG Fai.
Assume the contrary that there is a dissection of the region into nonconvex quadrilateral R1, R2, …, Rn. For a nonconvex quadrilateral Ri, there is a vertex where the angle is θi > 180˚, which we refer to as the large vertex of the quadrilateral. The three other vertices, where the angles are less than 180˚ will be referred to as small vertices.
Since the boundary of the region is a convex polygon, all the large vertices are in the interior of the region. At a large vertex, one angle is θi > 180˚, while the remaining angles are angles of small vertices of some of the quadrilaterals and add up to 360˚ − θi. Now
∑
= − n i i 1 ) 360 ( o θaccounts for all the angles associated with all the small vertices. This is a contradiction since this will leave no more angles from the quadrilaterals to form the angles of the region.
Olympiad Corner
(continued from page 1)
Part 2, Day 1 (June 8, 2005)
Problem 1. Determine all triples of positive integers (a,b,c), such that a + b +c is the least common multiple of a, b and c. Problem 2. Let a, b, c, d be positive real numbers. Prove . 1 1 1 1 3 3 3 3 b c d a abcd d c b a+ + + ≤ + + +
Problem 3. In an acute-angled triangle ABC, circle k1 with diameter AC and k2 with diameter BC are drawn. Let E be the foot of B on AC and F be the foot of A on BC. Furthermore, let L and N be the points in which the line BE intersects with k1 (with L lying on the segment BE) and K and M be the points in which the line AF intersects with k2 (with K on the segment AF). Prove that KLMN is a cyclic quadrilateral.
Part 2, Day 2 (June 9, 2005)
Problem 4. The function f is defined for all integers {0, 1, 2, …, 2005}, assuming non-negative integer values in each case. Furthermore, the following conditions are fulfilled for all values of x for which the function is
efined: d
f(2x + 1) = f(2x), f(3x + 1) = f(3x) and f(5x + 1) = f(5x).
How many different values can the unction assume at most?
f
Problem 5. Determine all sextuples (a,b,c,d,e,f) of real numbers, such that the following system of equations is
ulfilled: f
4a=(b+c+d+e)4, 4b=(c+d+e+f)4, 4c=(d+e+f+a)4, 4d=(e+f+a+b)4, 4e=(f+a+b+c)4, 4f=(a+b+c+d)4. Problem 6. Let Q be a point in the interior of a cube. Prove that an infinite number of lines passing through Q exists, such that Q is the mid-point of the line-segment joining the two points P and R in which the line and the cube intersect.
Angle Bisectors Bisect Arcs
(continued from page 2)
Let H be the orthocenter of ∆ABC. Let line AH intersect BC at D and the circumcircle of ∆ABC again at A2. Note ∠ A2BC = ∠A2AC
= ∠DAC = 90˚ −∠ACD = ∠HBC.
Similarly, we have ∠A2CB = ∠HCB. Then ∆BA2C ≅ ∆BHC. Since A1 is the midpoint of arc BA1C, it is at least as far from chord BC as A2. So the area of ∆ BA1C is at least the area of ∆ BA2C. Then the area of quadrilateral BA1CH is at least twice the area of ∆BHC. Cutting hexagon AC1BA1CB1 into three quadrilaterals with common vertex H and comparing with cutting ∆ABC into three triangles with common vertex H in terms of areas, we get the conclusion of (ii).
Remarks. In the solution of (ii), we saw the orthocenter H of ∆ABC has the property that ∆BA2C ≅ ∆BHC (hence, also HD = A2D). These are useful facts for problems related to the orthocenters involving the circumcircles.
____,_ _...- --. -._I --.. ___..._
Solution: We first note that 2005 = 5 - 401. Letting M := 1234678905 aud
Nk := M - (lo’“.(k-‘) + Jo’+2) + . . . + 10’0 + 1) )
we see that each Nk ends in 5 and’is therefore divisible by 5, and also that each Nk is composed of the same number of each of the ten digits 0,. . . ,9.
Since we have an infinite number of such numbers Nk but only a finite number of congruence classes modulo 401, some congruence class modulo 401 must contain an infinite number of numbers Nb. Let N,,, be the smallest among these numbers. Taking all values of Nb from this congruence class, we obtain au iniinite number.of numbers of the form
Nk - IV,,., 5: NL_,,, . !O *oh ,
all of which must be divisible by 401.. Since 10 and 401 are relatively prime, all resulting numbers Nk_,,, must be divisible by 401, and since each of them ends in the digit 5, they must all be divisible by. 2005, yielding an in&rite number pf numbers with, the required properties. Solution: If (z, y) is au integer solution of the given system of equations, subtracting the second equation from the first shows us that .
x1-yz=.y-z w (z- y)(z+y)+(z--y)=O w (z-y)(z+Y+l)=O must hold. We therefore consider two cares.
CaseI:z-y=O
In this c&r we have z = y = m. Substituting in one of the given equations yields o=me- m=m(m-l),
aud we see that a is the product of two consecutive integers. As such, a must be non-negative, and all such numbers not greater than 2005 yield solutions of this type. Since 45.44’= 1980 < 2005 and 46.45 = 2070 > 2005 there are 45 numbers we can substitute for 4, since m can
sssume any value 1 5 m 5 45, yielding a different value of a in each case. (Note that negative values of m yield the same values for a.)
CaseII:z+y+l=O _
In this caee we have z = m and y = -(m + l), and substituting yields ~=mGr+1=m(mf1)+1.
In this case, a is one greater than the product of two consecutive integers, and we see that, every value of a from Case I yields a second possible value of 4 by adding 1. This means that there are also 45 values of this type that yield integer solutions of,the system of equations, making the total number of such values a to be 90.
~--
Solution: Considering symmetric functions yields the values for a, 6 and c. Since a+b+c = s,=2,
ab+&+ca = +(s;- sg)=-l aud ubc = -f * (8; -53 -3s,(a6+k+co))=0, a, 6 and c are the roots of the cubic equation
2s - 222 - I = 0.
f;;fi’Y
I&= -1 and therefore
.' ".. .,
%I-lhti ",+ (f-l-+b~-~)(~"+'+~~+l),=B~+b~+4"-1P-~(.'+~~j
= 3, - '&?b" i-(cd)."-'as2 =si- 2.(ab)“- 6 .(ab)” = 8: -8.(-I)“, which yields the required result
@ ., Solution:
_ _ . _ -.
v
R
:
Numbering the vertices of the hexagon as shown in tlmfigure, we see that sides 12 and 45 of the hexagon lie on paraKe liner AC and RQ. Since the altitudes PI and BII of the trikrglm P12 aud 845 each result by subtr&ting the’distance between the parallel lines AC and RiJ from the equal altitudes PI11 and BIV of tria&+a PQR and ABC respectively, they are of equal length. It therefore follows that the sidea of the equilateral triangles P12 and B45 are also of equal length, and 12 and 45 sre therefore ‘not only parallel, but also of equal length. The quadrilateral 1245 is therefore a parallelogram, and its diagonals intersect in their common mid-point. It therefore follows that the diagonal 14 of the hexagon pssses through themid-point
of the diagonal 25. ‘.
Analogously, the quadrilateral 2356 must also be a parallelogram, and the diagonal 36 must therefore also pa& through the mid-point of 25, which is therefore the common point of all.
three diagonals of the hexagon.
Solution: We first note that all three numbers cannot be equal, since this would imply lcm(a, b, c) = a, which contradicts Icm(a, b, c) = a + b + c. We assume o 5 6 -< c without loss of generality. It follows that a + b < 2c must hold, aud we therefore have
c<a+b+c<3c,
and therefore a + b + c = 2c (since krn(a, 6, c) must be a multiple of c) and a + b = c. Since b divides Icm(a, b, c) = 24 + 26, we see that b divides 24. From a 5 b we conclude that either b = a or 6 =.2a must hold.
If b = a, we have c = a’+ 6 = 24, aud therefore fcm(a, 6, c) = Icm(a, a, 2a) = 24, which is certainlynotequaltoa+b+c=a+a+2a=4a.
IQn(a,b,c)=Zcm(a;2a,30)~~~=+22a+3a=a’+b+c
.for aii positive integers’ a.
We see that aii .possible triples with. the required property are those of the form
.(a,+,34 with a>$.
0
2 Solution: ,We Iirst note that the left baud side can be written asaf&+c+d 0 6.c d 1’1 1.1
=-&+~+-&‘d’+--&=~+~+-J-p& 1 -1’ -abed’
If we now consider the sequence ( ately yields
f, f, ;, t) tbree times, the rearrangement &@miity immedi- _. _ . . : 88 required.
0 3
Solution: k0
. .---Since LBEC = 90” and BC is the diameter of ke., E is a coinmon point of AC and L. Similarly, F is a common point of BC and ki. Aiso, since ZAEB F LAFB = 900, both E and F lib on the
circle with diameter AB, as shown in the figure. It therefore follows that CE - CA = CF i CB.
Since AACL is a right triangle with altitude Lh, .it follows that CE.Ck = CL!, and similarly in ABCK we have CF - CB = cfl. We therefore have
Furthermore, since LN is perpendicular to the diameter AC of kl, it also follows, that CL = CN must hold, andthe analogous argument for KM and Ire, yields CK 5 CM., _ .
In e&q, we see that 811 segments CK, CL, CM and CN are of equal length. The porn&
K, L, J4 and Nail therefore lie on a common circle with mid-point C, and KLMN is indeed a cyclic quadriiaterai as required.
Solution: For any i E {6,1,2,. . . ;‘2005} d’ . ’ rv+ble by 4,3 or 5 ‘we have f(v) = f($ + lj. The only vahres for y such that j’(y) and’ f(y i- 1) can be different are those reiativ&Iy. prime to 90.
She ~(30) = 8, the function f ‘can assume at most 8 d&rent vahres for y B {O; 1,2,. 1: 29},
‘and similarly for auy of the~foilowiug sets of thirtnmmtiw. integers. ; _.
Since iti e 30 * 66 + 25, f can assume d diierimf values in each of the 66 mnsectitive ae& of
30 comtive integers. For the last 25 ;t 1 = 26 elements of the.& - --.. - -... 10, 1,2,‘. . . ,2005], it &II
asame amoth& 8 values, since the largest integk ie& than 29 and reiativeiy prim& to 30 is 23, which is itself less than’25.
It foilows that the total maximum number of values the function f cau assume is 67 - 8 = 536.
a
6
Solution: First of ah, we note that all variables are equ81 to 4. times a fourth power, and can _ tberefore.not be negative. :
We note that the system is‘cyclic. If any two successive variabks’are not equal, without loss of generaiity say o i: b, comparing the Ihst two equations yiekls .
. -..- .._:L__.- _.._
b+c+d+e<c+d6e+f’ i .b<‘j”aud a<b<,f. .
Since a < 3,~coå the first and l&t equations’y&Ids
‘_ b+‘i+d&eka+b+c+d =+ :e<a and e’<a.<b<f. *_
Since e .< f, comparing the last two ,equations &Ids c
f+afbfc.<a+b+c+d =+ f<d and e<a<b<f<$. : ._
Since e <.d, 6oiriparing the fourth and fifth ‘&rations yields “’ - _-
f+a+b+c<eff+a+b .‘& ccc and c<e<ac.b<f<d. Fiially,%ii~e c < d,’ comparing the third and four@ equations yiehls ’ :
. ‘.
d.+e+f+a<e+j+a+b 3 d<b md:‘c<e-za<b<f<d<a,
. .
.
which is a ‘contradiction for b. It -follows that aI six variables must be equal, and since we ‘therefore have .
ha.= (40)“’ and o 2.0,
it folks that either 4u = land therefore a =‘f or 0 = 0 must bold. The only solutions .of the ~~,ofequationsaretb~f?~.(!,B,t,t,i,i) and(O,O,O,O,O,D). ’
Solution: Let C be thexf all &ita on’the surface of the cube and C’ the set of all poiirts obtained by reflecting the points of c in Q.’ For the sak of,sim&ity,~ we &II speak of the cube Cand the reflected cube c. - * . . .
Since Q is aninterior point of C; it must &. be an in&or point of C. The two cubes therefore have a common interior region.
‘Iv
e consider the set C n C’. This set must containan
infinite number of points. If this were not the case, C and C’ would only have a finite number of points in common (possibly none at aI1). Let us assume that this is the case. The cubes C and C’ ‘caamot intersect in common line seventy or have plane sections in common, but rithir one of the cubes (say ‘c) must then bejn the intesio~ of the other (say C’), with at most some orali of the wattice3 of C being points of C!. By reasons of symmetry; 0’ must then ai& be complete& in the interior of C, which is only possible XC = C’ (i.e. if Q is the mid-poiut of C). In thie casi C and C have an infinite number of points in common, contradicting the assumption that C II C’ only contains. a finite ntiber of points.
We see that C fi C’ must
contain
‘an infinite mm&r of points. Taking auy’of these points as a pint P on the surface of the cube C,. we see that P is aleo a point on: CI, and the symmetric point to P’ whh iespect to Q .must therefore r&o .lie’on the cube C, and can thus be chosen as a point R Side Q is certainly the mid-point of any such line segment’ PR, ti see: that aniniinite number of lies with the required properties exist.
