Volume 18, Number 1 May-August 2013
Olympiad Corner
Below are the problems of the 2013 Asia Pacific Mathematical Olympiad.
Problem 1. Let ABC be an acute triangle with altitudes AD, BE and CF, and let O be the center of its circumcircle. Show that the segments
OA, OF, OB, OD, OC, OE dissect the
triangle ABC into three pairs of triangles that have equal areas.
Problem 2. Determine all positive integers n for which
2 ] [ 1 2 2 n n
is an integer. Here [r] denotes the greatest integer less than or equal to r. Problem 3. For 2k real numbers, a1,
a2, …, ak, b1, b2, …, bk define the sequence of numbers Xn by ). , 2 , 1 ( ] [ 1
n b n a X n i i i nIf the sequence Xn forms an arithmetic progression, show that
k
i 1aimust be
an integer. Here [r] denotes the greatest integer less than or equal to r.
(continued on page 4)
Ptolemy’s Inequality
Nguyen Ngoc Giang, M.Sc.
Ptolemy’s inequality is a beautifulinequality, but it is rather difficult to see when or how it can be applied to geometry problems. This inequality has often been used in gifted student selection exams in various places. In this article we will first look at how this inequality is derived.
Theorem (Ptolemy’s Inequality) Let
ABCD be a quadrilateral. We have ABCD+DABC ≥ ACBD with equality if and only if ABCD is a cyclic quadrilateral. C D X Y E
Proof. From A and from B draw two rays to cut diagonals BD and AC at X and at Y respectively such that ∠XAB =
∠DAC and ∠YBA =∠DCA.
Suppose AX cut BY at E. Then ∠BAC =∠EAD. We have ∆ABE ~ ∆ACD. So ) 1 ( BE AC CD AB AD AE CD BE AC AB
and we also have ∆AED ~ ∆ABC. It follows AD BC AC ED. (2) BC ED AC AD
Adding the equations (1) and (2), we have
ABCD+DABC=AC (BE+ED)
≥ ACBD.
Thus, for arbitrary quadrilateral ABCD, we have ABCD+DABC ≥ ACBD. Equality holds if and only if E belongs to BD. In that case ∠ABD =∠ACD or
ABCD is a cyclic quadrilateral.
This inequality can also be proved in a different way (cf vol. 2, no. 4 of Math
Excalibur).
Next we will look at applications of the theorem and Ptolemy’s inequality.
Example 1 An isosceles triangle ABC (with CA=CB) is inscribed in a circle with center O and M is an arbitrary point lying on the minor arc BC. Prove that
. AC AB MC MB MA C A M B
Solution. Applying Ptolemy’s theorem to AMBC, we have MABC + MBCA =
MC AB. From CA=CB, we have . AC AB MC MB MA
Example 2 (Pythagorean Theorem) For
right ∆ABC with ∠ACB=90°, we have
BC2 + AC2 = AB2. O A C B D
Solution. Draw a circle with midpoint O
of side AB as center and radius AB/2. Let ray CO intersect the circle at D. Applying Ptolemy’s theorem to ACBD, we have ADBC+BDAC = ABCD. Since AD = BC, BD=AC and CD=AB, we get
BC2 + AC2 = AB2.
Example 3 Let a and b be acute angles. Prove that
sin(a+b) = sin a cos b + sin b cos a. Solution. Let’s draw a circle with diameter AC = 1. Construct the rays AB and AD lying on opposite sides of the diameter AC such that ∠CAB = a and ∠CAD = b. Also draw diameter BE as shown in the next figure.
(continued on page 2)
Editors: 張 百 康(CHEUNG Pak-Hong), Munsang College, HK
高 子 眉 (KO Tsz-Mei)
梁 達 榮 (LEUNG Tat-Wing)
李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST
吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (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 October 1, 2013.
For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
© Department of Mathematics, The Hong Kong University of Science and Technology
Mathematical Excalibur, Vol. 18, No. 1, May.-Aug. 13 Page 2 a cos a sin a cos b b p a+b sin b O A C B E D
Since AC and BE are diameters, ∠ABC, ∠ADC and ∠BDE are right angles, we see that AB = cos a, BC = sin a, CD = sin b, DA = cos b respectively. Also, ∠BED = ∠BAD = a+b and BD = p = sin(a+b).
Applying Ptolemy’s theorem to ABCD, we have ACBD = BCDA + CDAB, which is
sin(a+b) = sin a cos b + sin b cos a. Example 4 If an arbitrary point M lies on the circle circumscribed about square A1A2A3A4, then we have the relation . 2 4 2 2 2 3 2 1 MA MA MA MA A1 A2 A3 A4 M
Solution. Without loss of generality, assume that M lies on minor arc A1A4. Let MA1 = x1, MA2 = x2, MA3 = x3, MA4 = x4 and let the square have side a. Applying Ptolemy’s theorem to
MA1A2A3 and MA1A3A4, we have . 2 , 2 1 4 3 2 3 1a xa xa xa xa xa x
Cancelling a in both equations and rewriting the second equation as
, 2 4 1 3 x x x we get (x1+x3)2 + (x3−x1)2 = 2x22 + 2x42. Expanding the equation and cancelling 2 on both sides, we get
x12 + x32 = x22 + x42, which is the desired conclusion. Example 5 Let A1A2…An be a regular polygon that has an odd number of sides. Let M be a point on the minor arc
A1An of the circle circumscribed about it. Prove that the sum of the distances from the point M to the vertices Ai (i being odd) is equal to the sum of the distances from the point M to the vertices Ak (k being even).
A1 A2 A3 A4 A5 A6 A7 M
Solution. Let each side of the polygon have length a. Draw the diagonals A1A3,
A2A4, …, AnA2, which have a common length b. Next, draw the chords MA1,
MA2, …, MAn and let MAi have length di. Applying Ptolemy’s theorem to MA1A2A3,
MA2A3A4, …, MAn–1AnA1 and MAnA1A2, we have
ad1+ad3=bd2, bd3=ad2+ad4, ad3+ad5=bd4, bd5=ad4+ad6, … , adn–2+adn=bdn–1, bdn+ad1=adn–1, adn+bd1=ad2.
Adding these equations, we get (2a+b)(d1+d3+⋯+dn)
= (2a+b)(d2+d4+⋯+dn–1).
Cancelling 2a+b, this becomes
d1 + d3 + ⋯ + dn = d2 + d4 + ⋯ + dn–1, which is the desired conclusion.
Next we will generalize Ptolemy’s inequality to higher dimensions.
Example 6 Let ABCD be a tetrahedron in the three dimensional space. Prove that
ABCD+DABC > ACBD Solution. Draw a plane α which is parallel to lines AC and BD. Let A1, B1, C1, D1 be the feet of perpendiculars from A, B, C, D to plane α respectively. We have A1C1 =AC and B1D1 = BD. However, A1B1 ≤ AB, B1C1 ≤ BC, C1D1 ≤ CD, D1A1 ≤ DA and at least one of these is strict since A,B,C,D are not on the same plane.
A B C D A1 B1 C1 D1
Using the inequalities and equations above as well as Ptolemy’s inequality applied to quadrilateral A1B1C1D1 on the plane α, we have
ABCD + DABC > A1B1C1D1 + D1A1B1C1 ≥ A1C1B1D1
= ACBD.
Alternatively, consider the Cartesian coordinate system. Say A = (a1,a2,a3),
B = (b1,b2,b3), C = (c1,c2,c3) and D = (d1,d2,d3). Then Ptolemy’s inequality in 3-dimensional space is 3 1/2 1 2 3 1 2 ( ) ) (
i i i i i i b c d a 3 1/2 1 2 3 1 2 ( ) ) (
i i i i i i c b d a . ) ( ) ( 2 / 1 3 1 2 3 1 2
i i i i i i d b c a (*)To prove this, let xi = ai−bi, yi = ai−di,
zi = ai−ci. Also, let . , , 3 1 2 3 1 2 3 1 2
i i i i i i x z y If α or β or γ is 0, then (*) is obvious. Otherwise, none of them is 0. Dividing both sides by (αβγ)1/2, (*) becomes, ) ( 2 / 1 3 1 2 2 / 1 3 1 2 2 / 1 3 1 2
i i i i i i i q q p p where i i i z y p and . i i i z x q The last inequality is known as Minkowski’s inequality. By squaring and cancelling
3 1 2 3 1 2 i i i i q p from both sides, we arrive at the Cauchy-Schwarz inequality, which is a well-known inequality.Remark. In (*) and its proof, 3 can be replaced by any positive integer n and that gives Ptolemy’s inequality in
n-dimensional space.
The following are some exercises for the readers.
Exercise 1 A quadrilateral ABCD is inscribed in a circle with center O and ∠ABC =∠ADC=90°. Prove that BD =
AC sin∠BAD.
Exercise 2 Quadrilateral ABCD is convex. Its sides are AB=a, BC=b,
CD=c, DA=d and its diagonals are
AC=m, BD=n. Let φ=∠A+∠C. Prove
that m2n2 = a2c2 + b2d 2 − 2abcd cos φ. References
[1] Le Quoc Han (2007), Inside
Ptolemy’s Theorem (Vietnamese)
Education Publishing House.
[2] Vietnamese Mathematics and Youth Magazine.
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 sending
solutions is October 1, 2013.
Problem 421. For every acute triangle
ABC, prove that there exists a point P
inside the circumcircle ω of ∆ABC such that if rays AP, BP, CP intersect ω at D, E, F, then DE: EF: FD = 4:5:6.
Problem 422. Real numbers a1, a2,
a3, … satisfy the relations
an+1an + 3an+1 + an + 4 = 0 and a2013 ≤ an for all positive integer n. Determine (with proof) all the possible values of a1.
Problem 423. Determine (with proof) the largest positive integer m such that a mm square can be divided into seven rectangles with no two having any common interior point and the lengths and widths of these rectangles form the sequence 1,2,3,4,5,6,7,8,9,10, 11,12,13,14.
Problem 424. (Due to Prof. Marcel
Chirita, Bucuresti, Romania) In ∆ABC,
let a=BC, b=CA, c=AB and R be the circumradius of ∆ABC. Prove that
. 3 3 2 ) , , max( 2 2 2 R abc ab c ca b bc a
Problem 425. Let p be a prime number greater than 10. Prove that there exist distinct positive integers a1, a2, …, an such that n ≤ (p+1)/4 and
n n a a a a p a p a p 2 1 2 1)( ) ( ) (
is a positive integral power of 2. *****************
Solutions
****************Problem 416. If x1 = y1 =1 and for n>1,
xn = −3xn−1−4yn−1+n and yn = xn−1+yn−1−2,
then find xn and yn in terms of n only.
Solution. CHEUNG Wai Lam (Queen Elizabeth School, Form 3), F5D (Carmel Alison Lam Foundation Secondary School), Vijaya Prasad NALLURI (Retired Principal, AP Educational Service, Andhra Pradesh, India), Alex Kin-Chit O (G.T. (Ellen Yeung) College), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Writing out many terms, we may observe that {x2n−1}, {x2n}, {y2n−1}, {y2n} are arithmetic progressions. In fact, for n = 1, 2, 3, …, we can claim
x2n−1 = 17n−16, y2n−1 = −8n+9, x2n = −17n+12, y2n = 9n−9. We will prove these by induction. The case n = 1 follows from x1 = y1 = 1,
x2=−3−4+2=−5, y2=1+1−2=0, agreeing with the claim. If the case n is true, then by the definition of xn and yn,
x2n+1 = −3(−17n+12)−4(9n−9)+(2n+1) = 17(n+1)−16, y2n+1 = (−17n+12)+(9n−9)−2 = −8(n+1)+9, x2n+2 = −3(17(n+1)−16)−4(−8(n+1)+9) +(2n+2) = −17(n+1)+12, y2n+2 = (17(n+1)−16)+(−8(n+1)+9)−2 = 9(n+1)−9
and the induction is complete.
Other commended solvers: Radouan
BOUKHARFANE, CHAN Long Tin (Diocesan Boys’ School), KWAN Chung Hang (Sir Ellis Kadoorie Secondary School (West Kowloon)), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM), SHUM Tsz Hin (alumni of City University of Hong Kong), Simon YAU C. K. and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).
Problem 417. Prove that there does not exist a sequence p0, p1, p2, … of prime numbers such that for all positive integer k,
pk is either 2pk−1+1 or 2pk−1−1.
Solution. Radouan BOUKHARFANE,
CHEUNG Wai Lam (Queen Elizabeth School, Form 3), F5D (Carmel Alison Lam Foundation Secondary School) and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).
Assume such sequence exists. (The cases
p0 = 2 or 3 can only yield no more than five terms, then contradiction arises.) For
k ≥ 2, we have pk > 3. Then either pk ≡ 1 or −1 (mod 6).
In the former case, 2pk+1≡3 (mod 6) cannot be prime. So pk+1=2pk−1≡ 1 (mod 6). Then repeating the same reason, we can only have pn=2pn−1−1 for all n > k. By induction, we have
pn=2n−k(pk−1)+1 for n > k. In the case
n=k+pk−1, we get ) (mod 1 2 1 k p p k
by Fermat’s little theorem. Then pn≡ 0 (mod pk) contradicting pn is prime.
In the latter case, 2pk−1 ≡ 3 (mod 6) cannot be prime. Similarly, this leads to
pn=2n−k(pk+1)−1 for n > k. Again, by Fermat’s little theorem, the case
n=k+pk−1 leads to contradiction. Problem 418. Point M is the midpoint of side AB of acute ΔABC. Points P and Q are the feet of perpendicular from A to side BC and from B to side
AC respectively. Line AC is tangent to
the circumcircle of ΔBMP. Prove that line BC is tangent to the circumcircle of ΔAMQ.
Solution 1. Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
A B P C M S O T Q N
Let O be the center of the circumcircle Γ of ΔBMP and S, T be the projections of
M, O onto line AC. Let OM intersect BP
at N. Now OM is the perpendicular bisector of BP. So ∠MNC=∠MSC=90° implies MNCS is cyclic. Since MS || OT,
γ=∠ACB=180°−∠OMS=∠MOT.
By the extended sine law, the chord MP of Γ satisfies . 2 sin 2 2 AB AP OM ABC OM MP AB
Hence OM=AB2/(4AP). Now, line AC is tangent to the circumcircle of ΔBMP if and only if OM=OT if and only if
. 2 1 4 2 4 cos cos 2 2 2 AB BQ AP AP AB BQ AP AB OM MS OT MOT ACB
Mathematical Excalibur, Vol. 18, No. 1, May.-Aug. 13 Page 4 Similarly, line BC is tangent to the
circumcircle of ΔAMQ if and only if . 2 1 cos 2 BA AP BQ BCA
The desired conclusion follows.
Solution 2. KWAN Chung Hang (Sir Ellis Kadoorie Secondary School (West Kowloon)).
Since AP⊥BC and BQ⊥AC, points
A,Q,P,B lie on the circle with M as center
and AB as diameter. Consider inversion with respect to this circle. Let X’ be the image of X under this inversion. We have
A’=A, Q’=Q, P’=P, B’=B. Since line AC is tangent to the circumcircle of Δ BMP, so the image of line AC and the
image of the circumcircle of ΔBMP are tangent. Since the circumcircle of Δ
BMP passes through M, the image of
this circumcircle is the line B’P’, which is line BC. Also, the image of line AC is the circle through A’, Q’, C’,
M, which is the circumcircle of Δ AMQ. So line BC is tangent to the
circumcircle of ΔAMQ.
Other commended solvers: F5D
(Carmel Alison Lam Foundation Secondary School), MANOLOUDIS Apostolos (4° Lyk. Korydallos, Piraeus, Greece) and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).
Problem 419. Let n ≥ 4. M is a subset of {1,2,…,2n−1} with n elements. Prove that M has a nonempty subset, the sum of all its elements is divisible by 2n.
Solution 1. Juan G. ALONSO and Ángel PLAZA (Garoé Atlantic School & Universidad de Las Palmas de Gran Canaria, Spain), F5D (Carmel Alison Lam Foundation Secondary School), KWOK Man Yi (S2, Baptist Lui Ming Choi Secondary School), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM) and Aliaksei SEMCHANKAU (School 41 named after Serebryany, Minsk, Belarus).
If n is not in M, then by the pigeonhole principle, since M has n elements, it must contain at least one of the pairs {1,2n−1}, {2,2n−2}, …, {n−1, n+1}. This pair is a subset of M that we want. If n is in M, then let a1, a2, …, an be the elements of M. We may let an = n and
a1 be the minimum of a1, a2, …, an–1. Since M has at least 4 elements, we may assume a2 is not a1+n (otherwise
replace a2 by a3). Consider the n numbers
a2, S1, S2, S3, …, Sn–1, where Si is the sum of a1, a2, …, ai. By the pigeonhole principle and the fact a2≠a1+n, two of the
n numbers are congruent (mod n) and the
two numbers are not a1, a2. Hence, their difference (which is a sum of the ai’s) is equal to jn for some positive integer j. If j is even, then the set T of the ai’s in the sum is a desired subset of M. Otherwise j is odd. We can add an = n to T to get a desired subset of M.
Solution 2. CHEUNG Wai Lam (Queen Elizabeth School, Form 3) and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).
We will say {a,b,c} is bad if a,b,c ∈M and 2n divides a+b+c. Assume the contrary (which implies M has no bad subsets). Then M contains exactly one element in each of the sets {1,2n−1}, {2,2n−2}, …, {n−1, n+1}, {n}. In particular, n is in M. Now 1∈M ⇒ n−1∉M (otherwise {1, n−1, n} is bad) ⇒ n+1∈M ⇒ n−2∉M (otherwise {1, n−2, n+1} is bad) ⇒ n+2∈M ⇒ ⋯ ⇒ n+(n−1)=2n−1∈M ⇒ 1∉M, contradiction. Next 1∉M ⇒ 2n−1∈M ⇒ n+1∉M (otherwise {n, n+1, 2n−1} is bad) ⇒ n−1 ∈M ⇒ n+2∉M (otherwise {n−1, n+2, 2n−1} is bad) ⇒ n−2∈M ⇒ ⋯ ⇒ n − (n−1) = 1 ∈M, contradiction.
Other commended solvers: Radouan
BOUKHARFANE.
Problem 420. Find (with proof) all positive integers x and y such that 2x2y+xy2+8x is divisible by xy2+2y.
Solution. Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), KWAN Chung Hang (Sir Ellis Kadoorie Secondary School (West Kowloon)), LKL Excalibur (Madam Lau Kam Lung Secondary School of MFBM) and Aliaksei SEMCHANKAU (School 41 named after Serebryany, Minsk, Belarus).
Let a | b denote a divides b. Suppose
xy2+2y | 2x2y+xy2+8x. Then xy2+2y |
y(2x2y+xy2+8x) − (2x+y)( xy2+2y) = 4xy−2y2.
Since y > 0, we get xy+2 | 4x−2y. If 4x−2y < 0, then xy+2 | 2y−4x. Now 2y > 2y−4x ≥ xy+2> xy implies 2 > x. Hence
x = 1. Then y+2 | 2y−4 = 2(y+2)−8. So
y+2 | 8, which implies y = 2 or 6. We
can check (x,y) = (1,2) is a solution, but (1,6) is not.
If 4x−2y = 0, then (x,y) = (k,2k) for some positive integer k and we have
xy2+2y=4k3+4k | 2x2y+xy2+8x=8k3+8k for all positive integer k.
If 4x−2y > 0, then xy+2 | 4x−2y. Now 4x > 4x−2y ≥ xy+2 > xy. So y < 4. The case y = 1 leads to x+2 | 4x−2 = 4(x+2)−10. Hence, x+2 | 10. Then (x,y) = (3,1) or (8,1). Both can easily be checked to be solutions. The case y = 2 leads to 2x+2 | 4x−4 = 2(2x+2)−8. Hence, x+1 | 4. Then (x,y) = (1,2) or (3,2), which are not solutions. The cases y = 3 leads to 3x+2 | 4x−6. Then 3x+2 | 4(3x+2) − 3(4x−6) = 26, which leads to (x,y) = (8,3), but it is not a solution.
So the solutions are (x,y) = (3,1), (8,1) and (k, 2k) for all positive integer k.
Other commended solvers: Radouan
BOUKHARFANE, F5D (Carmel Alison Lam Foundation Secondary School), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania), ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).
Olympiad Corner
(continued from page 1)
Problem 4. Let a and b be positive integers, and let A and B be finite sets of positive integers satisfying:
(i) A and B are disjoint;
(ii) if an integer i belongs either to A or to B, then either i+a belongs to A or i–b belongs to B.
Prove that a|A|=b|B|. (Here |X| denotes the number of elements in the set X.) Problem 5. Let ABCD be a quadrilateral inscribed in a circle ω, and let P be a point on the extension of
AC such that PB and PD are tangent to ω. The tangent at C intersects PD at Q
and the line AD at R. Let E be the second point of intersection between
AQ and ω. Prove that B, E, R are
be the cente~ of its circumcircle. Show that the segments OA,-OF, 00, dissect the triangle ABO into three pairs of triangle~ that have equal areas.
Solution. Let M and N be midpoints of sides. BC and AG, respectively. Notice that
L.MOG = lL.BOG = LEAB L.OMG = 90° = L.AEB, so triangles OMG and AEB are
· 2 ' OA
similar and we get
~~
=
~~.
For triangles ONA
and BDA we also have~~
=
BA. Then~~ = ~~ or BD· OM= AE · ON. ·
Denote by S(<I>) the area of the figure <I>. So, we see that S(OBD) = ~BD· OM =
lAE. ON= S(OAE). Analogously, S(OGD) = S(OAF) and S(OGE) =8(0BF).
2 . .
Probl~m 2. Determine all positive integers n for which r~t~
2
is an integer. Here[r]
denotes the greatest integer less than or equal to
r.
·
Solution. We will show that there are no positive integers n satisfying the condition of the problem.
Let m=
[Vn)
and a= n-m2• We have m~ 1 since n ~ 1. From n2+1 = (m2+a)2+1
=
(a-
2)2 + 1 (mod (m2+
2))', it follows that the 'condition of the· problem is equivalent to the fact that(a-
2)2+
1 is divisible by m2+
2. Since we have0
<
(a-
2?+
1 ::; max{22, (2m- 2)2}+
1 ::; 4m2+
1<
4(m2 + 2),we see that (a:_ 2)2
+
1 = k(m2+
2) must hold with k = 1, 2 or 3. We will show that noneof these can occur.
Case 1. When k
=
1. We get (a- 2)2- m2 = 1, and this impli~ that a- 2=
±1, m= 0must hold, but this contradicts w.ith fact m ~ 1.
Case 2. When k
=
2. We have (a- 2)2+
l
= 2(m2+
2) in this case, but any perfectsquf!Ie is congruent to
0,
1, 4 mod8,
~nd therefore, we have(a- 2)
2+ 1
=
1, 2,5
(mod8),.
while 2(m2
+
2)
=
4,6
(mod8).
Thus, this case cannot occur either. .Case 3. When k
=
3. We have (a- 2)2 + 1 = 3{m2+
2) in this case. Since any perfectsquare is congruent to
0
or 1 mod 3, we have (a-2)2+1
=
1,2 (mod3),
while3(m
2+2)
=
0
.(mod 3), which shows that this case cannot occur either. ·
Problem 3. For 2k real numbers a1, a2, ... , ak; b1, b2, ... , bk define the seque~ce of numbers Xn by
k
Xn =
I)O-i~ +
bi] (n=
1, 2, ... ) ..i=l
If
the sequence Xn forms an arithmetic progression, show that2:~.;
1
ai must be an integer. Here[r]
denotes the greatest integer less than or equaltor.
Solution. Let us write A = :Z:::~=l ai and B =
:z=:=l
bi; Summing the corresponding terms · of the following inequalities over i,ain
+
bi - 1<
[ain+
bi] ::; ain+
bi,we obtain An+ B- k
<
Xn<
An+ B. Now suppose that { Xn} is an arithmetic progre.!!sion. with the common difference d, then we have nd = Xn+l-X1 and A+ B- k<
X1 ::; A+ BCombining with the i~equalities obtained above, we get
A(n+ 1) +B-k
<
nd+X1<
A(n+1) +B,or
An- k_::; An+ .(A+ B-~1)-k
<
nd <An+ (_A+ B-::-X1) <An+ k,from which we conclude that
lA - dl
<.
~ must :P.old. Since this inequality holds for any positive integer n, we must have A= d. Since {Xn} is a sequence of integers, d must be an integer also, ·and thus we conclude that A is ·also an integer.4 • • • '
Problem 4. Let a and b be ·positive integers, and let A and B be finite sets of integers satisfying:
(i)
A and B are disjoint; ·(ii) if an integer i belongs either to A or to B, then either i
+
a belongs to A or i - bbelongs to B.
Prove that aiAI = biBI. (Here lXI denotes the number of-elements in the set X.)
Solution. Let A* = {n- a : n E A} and B* = {n
+
b : n E B}. Then, by (ii),A
u
B ~A*u
B* and by (i),lA
u
Bl::; lA*u
B*l::; IA*I+
IB*I = IAI+
IBI =lAu
Bl. (1)Thus, A U B.= A* U B* and A* and B* have no element in common. For each finite set X
of integers, let 2:(X) = 2:xex x. Then
l:)A)
+
l:(B)=
l:CAu
B)=
l:(A*u
B*) = l:(A*)+
l:(B*)= l:(A)- aiAI
+
L(B)+
biBI,(2)
which implies aiAI =
bjBj.
Problem 5. Let ABCD be a quadrilateral' inscribed in & circle
w,
and let P be a point on the exten~lion of AG such that P B and P D are tangent tow. The tangent at C intersectsP D at Q and the line AD at R. Let E be the second point of .int~rsection between AQ and
w. Prove that B, E, Rare collinear. ·
Solution. To show B, E, R are collinear, it is equivalent to show the lines AD, BE, CQ
are concurrent. Let OQ intersect AD at R and BE intersect AD ?-t R'. ·We shall- show
RD/!tA = R'D/R'A so that R = R'.
Since .6PAD is similar to .6P DC and b.P AB is similar to 6P BC, we have AD/ DC =
PAjPD = PA/PB =AB/BC. Hence, AB· DC= BC· AD. By Ptolemy,s theorem,
AB_:pp =BC· AD= ~CA· DB. Similarly CA· ED = CE ·AD= ~AE ·DC.
.,_ DB .2DC
AB=
CA.)DC
2ED CA- AE. (3)(4)
Since the triangles RDC and RCA are similar, we have ~g = g~ = ~~, Thus using
(4)
p RD
=
RD ·RA=(RC):l
=(DC)
2 =(2ED)
2RA RA2 RA CA AE . (5)
Using the similar triangles ABR' and E;R' . we have
R'
D/R! B = E .. D/AB .U. •t~..
· 'l · , . smg ue
srm1 ar tnangles DBR' and EAR' we have R'A/R'B =EA/DB. Thus using (3) and (4), .
R'D = ED·DB _
(2ED)
2R'A EA·AB- AE (6)
