Volume 18, Number 4 December 2013 – January 2014
Olympiad Corner
Below are the problems of the Dutch Team Selection Test for IMO 2013. Problem 1. Show that
2013 0 !(2013 )!2 ! 4026 n n nis the square of an integer.
Problem 2. Let P be the intersection of the diagonals of a convex quadrilateral ABCD. Let X, Y and Z be points on the interior of AB, BC and CD respectively such that . 2 ZD CZ YC BY XB AX
Suppose moreover that XY is tangent to the circumcircle of ΔBXY. Show that ∠APD=∠XYZ.
Problem 3. Fix a sequence a1, a2, a3, … of integers satisfying the following condition: for all prime numbers p and all positive integers k, we have
apk+1=pak–3ap+13.
Determine all possible values of a2013. (continued on page 4)
Andy Loo (Princeton University)
In the United States there are several annual math competitions organized by undergraduate students at different universities for high school enthusiasts, including the Harvard-MIT Math Tournament (HMMT), the Stanford Math Tournament (SMT), and, last but not least, the Princeton University Mathematics Competition (PUMaC). Started in 2006, PUMaC has grown into an international event in which high schoolers across America are joined by teams from as far away as Bulgaria and China on Princeton campus each year. PUMaC 2013 was held on November 16, engaging over 600 participants, and I was honored to serve as Problem Tsar (academic coordinator who heads the problem writing team). The responsibility of creating, grading and defending the problems and solutions of a competition of such scale and repute gave me an inspiring learning experience.
The competition is split into Division A (more challenging) and Division B (for less experienced contestants). Each team consists of eight students. In the morning, each contestant takes two out of four one-hour answer-only individual tests (Algebra, Geometry, Combinatorics and Number Theory, eight problems each) of his/her choice, followed by the one-hour Team Round, where members of the same team may discuss and work together (each team enjoying a separate room!).
The top 10 performers on each individual test (possibly with nonempty intersection) qualify for the Individual Finals, a one-hour proof-based test with three problems. I personally feel that an average Individual Finals problem lies somewhere near an IMO problem 1 or 4 in terms of difficulty. Remarkably, in PUMaC 2013, two contestants got a
perfect score on the Division A Individual Finals despite the time pressure! Also worth mentioning is the Power Round, which is a relatively long series of problems revolving around a central theme – knot theory in 2013 – released one week before the competition day for the teams to work on and turn in on competition day. (Teams may also enroll on a Power Round-only basis.) It usually takes frantic grading to determine the individual and team rankings in time for the award ceremony in the late afternoon, while mini-events such as Math (quiz) Bowl and Rubik’s cube as well as a lecture by a Princeton professor keep the participants entertained.
I would like to discuss a few problems in PUMaC 2013, not necessarily because they are the hardest, but mostly because they bring out certain lessons of problem solving we can learn.
Individual Finals B1.
Let a1 = 2013 and an+1 = 2013an for all
positive integers n. Let b1 = 1 and bn+1 =
20132012bn for all positive integers n. Prove that an > bn for all positive
integers n.
At first sight, one natural reaction to this problem would be to do induction. However, we would quickly realize that the assumption an > bn does not imply
an+1 > bn+1, as it does not imply
2013an 20132012bn. Many contestants performed pages of tedious calculations in vain. Are we doomed? It turns out that a clever little tweak to the induction idea would lead us to a crisp and compact solution:
Instead of an > bn, we shall prove an ≥
2013bn for all positive integers n. This is
clearly true for n = 1. If ak ≥ 2013bk for
(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 February 25, 2014.
For individual subscription for the next five issues for the 13-14 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. 4, Dec. 13 – Jan. 14 Page 2
some positive integer k, then
ak+1 = 2013ak
≥ 20132013bk = 2013bk20132012bk ≥ 2013bk+1.
There is something intriguing about this seemingly easy proof: if we cannot even prove just the original result, how come we can miraculously prove a stronger result? The answer to this paradox lies in the nature of mathematical induction: when we use induction, our task is essentially to prove the original statement about an arbitrary positive integer but equipped with an additional tool – the assumption that the statement is true for the preceding positive integer(s). If the statement is strengthened, what we need to prove becomes more demanding but the inductive hypothesis that we can use also gets more powerful. In the case of this problem, since the recurrence relations are exponential, the upgrade of the inductive hypothesis outweighs the increase in difficulty of the desired result.
Individual Finals A1.
Prove that 1 a2 2 1 b2 2 1 c2 2
1
6ab
c
2
1
6bc
a
2
1
6ca
b
2 for any positive real numbers a, b and c satisfying a2 + b2 + c2 = 1.The usual first step in proving such a symmetric inequality is to use the given condition to homogenize the inequality, i.e. to make the terms carry equal degrees. Afterwards, various inequality theorems can be applied. Here we first write the left-hand side as
2 2 2 2 2 2 2 2 2 3 2 2 1 2 2 3 1 2 2 3 1 b a c a c b c b a and note that by the AM-GM inequality, 3a2 + 3b2 ≥ 6ab and analogous inequalities hold. So 1 6ab c2 1 6bc a2 1 6ca b2 . 3 3 1 3 3 1 3 3 1 2 2 2 2 2 2 2 2 2 b c b c a c a b a
It suffices to prove the following inequality y x z x z y z y x 3 2 2 1 2 2 3 1 2 2 3 1 1 3x 3y z 1 3y 3z x 1 3z 3x y
where x, y and z are positive real numbers. At this stage, one may resort to passionate expansion and then apply Muirhead’s inequality and/or Schur’s inequality, or, alternatively, factorization and completing the square.
But I wish to share a solution using the majorization inequality (see Math Excalibur, vol. 5, no. 5, p.2): Without loss of generality we may assume x ≥ y ≥ z. Then
(3x + 3y + z, 3y + 3z + x, 3z + 3x + y) majorizes
(3x + 2y + 2z, 3y + 2z + 2x, 3z + 2x + 2y). Due to the convexity of the function f(t) = 1/t, the desired inequality follows by the majorization inequality.
Readers may also be interested in an alternative solution involving calculus: First, by Muirhead’s inequality (see Mathematical Excalibur, vol. 11, no. 1), we have
u3v3w + v3w3u + w3u3v ≥ u3v2w2 + v3w2u2 +w3u2v2 for any positive u, v, w. Letting u tx1/7, v ty1/7 and w tz1/7 where 0 < t < 1, we get
t 3x + 3y + z −1 + t 3y + 3z + x −1 + t 3z + 3x + y −1
≥ t 3x + 2y + 2z −1 + t 3y + 2z + 2x −1 + t 3z + 2x + 2y −1. Now, integrating both sides with respect to t from 0 to 1, we obtain nothing but the desired inequality!
Lastly I encourage all readers to try out the following problem which only one out of the 123 contestants attempting Combinatorics A got right. This is really
my favorite problem in PUMaC 2013 because I love eating sushi and find the setting very interesting:
Combinatorics A8.
Eight different pieces of sushi are placed evenly around a round table which can rotate about its center. Eight people sit evenly around the table. Each person has one favorite piece of sushi among the eight, and their favorites are all distinct. Sadly, they find that no matter how they rotate the table, there are never more than three people who have their favorite sushi in front of them simultaneously.
How many possible arrangements of the eight pieces of sushi are there? (Two arrangements that differ by a rotation are considered the same.) In 1908, a classic Chinese newspaper article famously raised three questions for the country: When can China first send an individual athlete to the Olympic Games? When can China first send a delegation to the Olympic Games? When can China first host the Olympic Games?
In closing, I would also like to ask three questions: When can Hong Kong first take part in the Power Round of PUmaC? When can Hong Kong first send a team to Princeton to join the main competition of PUMaC? When can a university in Hong Kong first host a math competition run by undergraduates for secondary school students?
As Dr. Kin Li (editor of Math Excalibur) observes, Hong Kong students need more opportunities to participate in different competitions and broaden their horizons. They will also be able to experience a beautiful university, make friends with some of the most brilliant brains from around the world, and learn team spirit especially through the Power Round and Team Round. With optimism, I hope my three questions will find answers before long.
For further information and past papers, please visit PUMaC’s website http://www.pumac.princeton.edu/
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 February 25, 2014. Problem 436. Prove that for every positive integer n, there exists a positive integer p(n) such that the interval [1, p(n)] can be divided into n pairwise disjoint intervals with each contains at least one integer and the sum of the integers in each of these intervals is the square of some integer.
Problem 437. Determine all real numbers x satisfying the condition that cos x, cos 2x, cos 4x, …, cos 2nx, … are
all negative.
Problem 438. Suppose P(x) is a polynomial with integer coefficients such that for every integer n, P(n) is divisible by at least one of the positive integers a1, a2,…, am. Prove that there
exists one of the ai such that for all
integer n, P(n) is divisible by that ai.
Problem 439. In acute triangle ABC, T is a point on the altitude AD (with D on side BC). Lines BT and AC intersect at E, lines CT and AB intersect at F, lines EF and AD intersect at G. A line ℓ passing through G intersects side AB, side AC, line BT, line CT at M, N, P, Q respectively.
Prove that ∠MDQ =∠NDP.
Problem 440. There are n schools in a city. The i-th school will send Ci
students to watch a performance at a field. It is known that 0 ≤ Ci ≤ 39 for i=1,
2, …, n and C1+C2+⋯+Cn=1990. The
seats will be put in a rectangle arrangement with each row having 199 seats. Determine the least number of rows needed to satisfy the condition that all students from the same school must sit in the same row for all possibilities of the known conditions above.
*****************
Solutions
****************Problem 431. There are 100 people, composed of 2 people from 50 distinct nations, are seated in a round table. Two people sitting next to each other are neighbors.
Prove that it is possible to divide the 100 people in two groups of 50 people so that no 2 people from the same nation are in the same group and each person in a group has at most one neighbor in the group.
Solution. Jeffrey HUI Pak Nam (La
Salle College, Form 6), Math Group (Carmel Alison Lam Foundation Secondary School) and ZOLBAYAR Shagdar (Orchlon International School, Ulaanbaatar, Mongolia).
Suppose these 100 people V1, V2, …, V100 are seated in a round table in clockwise order. For n = 1,2,…, 50, call {V2n−1, V2n} a partner pair. We color V1 in black and color the person with the same nation as him, say Vr, in white. If Vr’s partner is not yet colored, then we color Vr’s partner, say Vs, in black (this completes the coloring of the partner pair {Vr, Vs}) and go on to color the person with the same nation as Vs in white. Repeat this process until we reach a Vr whose partner Vs was colored already, then Vr=V2 and Vs=V1 since the only partner pair not yet completing the coloring is {V1,V2} with V1 black and V2 waiting to be colored. This gives the first cycle. Then we start to form another cycle with a remaining partner pair. Since there are 100 people, we will eventually stop. At the end, there are two groups with 50 black’s and 50 white’s and the required conditions are satisfied.
Problem 432. Determine all prime numbers p such that there exist integers a,b,c satisfying a2 + b2 + c2 = p and a4+b4+c4 is divisible by p.
Solution. Ioan Viorel CODREANU
(Secondary School Satulung, Maramures, Romania), Jeffrey HUI Pak Nam (La Salle College, Form 6), KIM Minsuk Luke (The South Island School, Hong Kong, Year 13), Corneliu MĂNESCU- AVRAM (“Henri Mathias Berthelot” Secondary School, Ploieşti, Romania), Math Center (Carmel Alison Lam Foundation Secondary School) and O Kin Chit Alex (G.T. (Ellen Yeung) College). Without loss of generality, we may assume a ≥ b ≥ c ≥ 0. Then 0 ≡ (p − b2 − c2)2 + b4 + c4 ≡ (b2 + c2)2+ b4 + c4 = 2(b4 +b2c2 + c4) = 2(b2 −bc + c2) (b2 +bc + c2) (mod p). Next, 0 ≤ bc ≤ b2 −bc + c2 ≤ b2 +bc + c2 ≤ a2 + b2 + c2 = p.
Since a ≥ b ≥ c ≥ 0, if bc=a2, then a=b=c and p being prime implies a = 1 and p = 3. Otherwise bc < a2 leads to b2 +bc + c2= 0 or 1. If b = 0, then a2 = p contradicts p is prime. Then c = 0, b = 1 and a2 + 1 = p, which leads to 0 ≡ a4+b4+c4 = a4+1 ≡ 2 (mod p). Then p=2 and a=b=1, c=0. Therefore, the only solutions are p = 2 or 3. Problem 433. Let P1, P2 be two points inside ∆ABC. Let BC = a, CA = b and AB = c. For i = 1,2, let PiA = ai, PiB =
bi and PCi = ci. Prove that
aa1a2+bb1b2+cc1c2 ≥ abc.
Solution. Math Group (Carmel
Alison Lam Foundation Secondary School).
Let the complex numbers α, β, γ, μ, ν correspond to the points A, B, C, P1, P2 in the complex plane respectively. By expansion, we have . 1 ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( Then ba c c ca b b cb a a1 2 1 2 1 2 ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( = 1.
Multiplying both sides by abc, we get the desired result.
Problem 434. Let O and H be the circumcenter and orthocenter of ∆ABC respectively. Let D be the foot of perpendicular from C to side AB. Let E be a point on line BC such that ED⊥ OD. If the circumcircle of ∆BCH intersects line AB at F, then prove that points E, F, H are collinear.
Solution 1. Jeffrey HUI Pak Nam (La
Salle College, Form 6) and T. W. LEE (Alumni of New Method College). Let lines HE and AB intersect at F’. Let Γ be the circumcircle ofΔABC. Let H’
Mathematical Excalibur, Vol. 18, No. 4, Dec. 13 – Jan. 14 Page 4
be the intersection of line CD and Γ different from C. Let E’ be the intersection of lines DE and AH’.
O B A C H' H D E' E F'
Observe that since ∠H’DE’ = ∠HDE, AHD BAH ABC C H A 90
implies H’D=HD and the butterfly H’CBAH’ on Γ gives E’D=ED as OD⊥DE, we have ΔH’E’D≅ΔHED. Then
∠F’HD =∠EHD =∠E’H’D =∠AH’C =∠DBC. It follows ∠CHF’=∠CBF’. Then F’ is on the line AB and the circumcircle of ∆BCH. Therefore, F’=F and E, F, H are collinear.
Solution 2. Jerry AUMAN, Georgios
BATZOLIS (Mandoulides High School, Thessaloniki, Greece) and Jon GLIMMS (Vancouver, Canada).
O B A C H' H D E F I J
Let Π be the circle passing through C, H, B, F and let Γbe the circumcircle of ΔABC. Let line DE meet Γat I and J. Since OD⊥DE, D bisects chord IJ. Next, DCA BAC DBH DCF 90
implies D bisects AF. Hence AIFJ is a parallelogram. Then ∠IFJ =∠IAJ. Let H’ be the intersection point (different from C) of line CD and Γ. Then D bisects HH’ (see solution 1 -- Ed.) and IHJH’ is a parallelogram. So ∠IHJ =∠IH’J. Then
∠IFJ +∠IHJ =∠IAJ +∠IH’J = 180°. So I,F,J,H lies on a circle Σ.
Finally, the radical axis of Γ and Πis line BC, while the radical axis of Γ and Σis line IJ. So the radical center of Γ, Π, Σ is the intersection of lines BC and IJ, which is E. Therefore, E is also on the radical axis of Π and Σ, which is line HF.
Comments: One can also solve via coordinate geometry by assigning lines AB and CD as the x-axis and y-axis respectively.
Other commended solvers: Math Group (Carmel Alison Lam Foundation Secondary School), Vijaya Prasad NALLURI (Retired Principal, AP Educational Service, India) and Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Problem 435. Let n > 1 be an integer that is not a power of 2. Prove that there exists a permutation a1, a2,…, an of 1,2,…, n such that . 0 2 cos 1
n k k n k a Solution. Jeffrey HUI Pak Nam (La
Salle College, Form 6) and Math Center (Carmel Alison Lam Foundation Secondary School).
For integer n > 1, let ck = cos(2kπ/n) for k
= 1, 2, …, n. We have cn = 1, ck = cn−k and
n k n k n k k c 1 1 , 0 1 1 Re Re (*) where ω = e2πi/n. Suppose n = 2m+1, where m = 1,2,3,…. We have c1+c2+⋯+cm = −1/2 (using cn = 1 and ck = cn−k). Hence (2m+2)(c1+c2+⋯+cm)=−(m+1)c2m+1. Since ck = c2m+1−k, we have (2m+1)c1+2mc2+⋯+(m+2)cm = (m+2)cm+1+⋯+2mc2m−1+(2m+1)c2m. Subtracting the two displayed equations above and transposing all terms to the left, we get . 0 ) 1 ( ) 1 ( 2 1 2 1 1
k m m m k m k k c m c k kcThis solves the cases n = 3,5,7,…. Next, assuming the case n is true, we will show the case 2n is also true. Let dm =
cos(mπ/n) for m = 1,2,…,2n. The case n gives us an equation of the form
a1d2 + a2d4 + ⋯ + and2n= 0, (**) where a1, a2, …, a2n is a permutation of 1,2,…,n. Using (*), we have
n k n n k d d d 2 1 2 2 1 2 0 2 cos and
n k n n k d d d 1 2 4 2 0. 2 cos Subtracting these equations, we have d1+d3+⋯+d2n−1=0. For k=1,3,…,2n−1, we have
d2n−k = cos((2n−k)π/n) = cos(kπ/n) =dk.
Using this, d1+ 3d3+ ⋯+ (2n−1)d2n−1 = d2n−1 + 3d2n−3 + ⋯ + (2n−1)d1. Adding the left and right sides, we get the equation 2n(d1+d3+⋯+d2n−1) = 0. So d1+3d3+⋯+(2n−1)d2n−1= 0. (***) Finally, taking twice the equation in (**) and adding it to the equation in (***), we solve the case 2n.
Comments: Titu ZVONARU
(Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania) pointed out that Problem 435 is the same as Problem 26753 in the Romanian Mathematical Gazette (G.M.-B) and a solution was appeared in G.M-B, No. 10, 2013, pp. 468-469.
Olympiad Corner
(Continued from page 1)
Problem 4. Determine all positive integers n ≥ 2 satisfying ) 2 (mod j n i n j i
for all i and j such that 0 ≤ i ≤ j ≤ n. Problem 5. Let ABCDEF be a cyclic hexagon satisfying AB⊥BD and BC=EF. Let P be the intersection of lines BC and AD and let Q be the intersection of lines EF and AD. Assume that P and Q are on the same side of D and that A is on the opposite side. Let S be the midpoint of AD. Let K and L be the centres of the incircles of ΔBPS and ΔEQS respectively. Prove that ∠KDL = 90°.
1. We prove the following more general statement. (5) We have (2m)l (m!)
2
(2m)! _(m)2. (2m)
(n!(m- n)!)2 = (n!(m- n)l)2 • (m!)2 - n m •Hence it suffices to show that
m
2: (':)
2=
e:).
n=O
We will do this combinatorially. Consider 21J1- balls, numbered from 1 up to 2m. Balls 1 up to m are coloured blue, and balls m
+
1 up to 2m are coloured red. We can choose m balls from these 2m balls in (2: ) ways.On the other hand, we ca.n also first choose n blue balls, with 0 ::::; n ::::; m, and then choose m - n red balls. Equivalently, we can. choose n blue balls . to include, and n red balls to not include. Hence the number of·ways in
which one can l!hoose m balls is also equal to
Hence this sum is equal to
C:).
This proves (5). 02. By the inscribed angle theorem, we have L.GZY
=
L.BYX and L.BXY=
LOY Z. It follows from this that
L.XYZ
=
180°-L.BYX- L.GYZ = 180°-L.BYX- L.BXY = L.ABG.Moreover, we see that 6.XBY rv 6.YGZ (AA). This implies that
~=l~gj.
From the assumptions, it immediately follows that IXBI = kiABj, IBYI =
iiBGidYGI
=
~jBGj, and jGZI = ~IGDj. Substituting this yields!IABI !IBGI
~lBO!
= i!GDI' . ·.?· 36 l •'· '·'L.XBY = L.YGZ = L.BGD. Hence !::lABG rv 6BGD (SAS). We
de-duce that LGAB
=
L.DBG. From this, it follows thatL.P AB
+
L.ABP=
LGAB+
L.ABD = L.DBC+
L.Al3D = L.ABG.We established before that L.ABG
=
L.XYZ, so by the ~xter~al' angle . theorem applied to 6.AB P, we haveL.BPC
=
L.PAB+
L.ABP=
L.ABC=
L.XYZ.As L.BPC
=
L.AP1J, it follows that L.APD = L.XYZ.3. Let q and t be primes. Substituting k = q and p = t yields
aqt+l = taiJ - 3at
+
13. Substituting k=
t and p=
q yieldsaqt+l = qat - 3aq
+
13.Hence the right hand sides are equal, so
taq- 3at =qat- 3aq,
or .equivalently,
(t
+
3)aq=
(q+
3)at·0
In particular, we have 5a3 = 6a2 and 5a7 = 10a2. Substituting k = 3 and
p = 2 now gives
13 .
5
a2=
13,hence a2
=
5. We deduce that for all primes p, we have ap = (v+;)a2 =p+3.
Substituting k
=
4 and p=
3 givesa13 = 3a4 - 3a3
+
13.As a13
=
16 and aa=
6, it follows that 3a4 :== 21, or equivalently, a4=
7.Finally, substitute k = 4 and p = 503 to get
a2013
=
a4-503+1=
503 · a4-3a5o3 + 13 = 503 · 7-3 · (503 + 3) + 13 = 2016. Hence a2013 = 2016, so this is the only possible value.It remains to check that this value is attained for· some sequence of integers, i.e. there exists a sequence satisfying the co~dition. Define the sequence at, a2, a3, ... by an
=
n + 3 for all n 2 1. Then for all primes p and all positive integers k, we haveapk+l = pk+4 = pk+3p-3p+4 = (pk+3p)-3p-9+9+4 = pak-3ap+13,
as desired. D
4. We first show tlJ_at n satisfies the condition if and only if (~)
=
i + 1 mod 2 for all i such that 0::; i::; n. Suppose that(7)
=
i+1 mod 2 for all i, then we have (~) + ('"~)=
i + 1 + j + 1=
i + j mod 2 for all i and j. Hence nsatisfies the con~lition. Conversely, suppose that n satisfies the condition.
As (~)
=
1, we have i=
1 +(7)
mod 2 for all i, so(7)
=
i - 1=
i + 1 mod 2 for all i..Write n as n = 2k + m with 0 ::; m
<
2k. Since n2
2, we may assu~e that k 2 1. Consider(
n ) _ (2k+m) _ (2k+m)(2k+m-1)· .. (m+4)(m+3)
2k - 2 - 2k - 2 - (2k-2)(2k- 3) ... 2' 1 . .
The product in the denominator has
L
2"2
2J
factors divisible by 2,L
2"4
2J
·factors divisible by 4, ... ,
L
;~:tJ
factors divisible by 2k-l and no factors divisible by 2k. The product in the numerator consists of 2k- 2 consecutive factors, hence has at leastL
2"2
2J
factors divisible by 2, at leastL
2"4
2J
. k
factors qivisible by
4, ... ,
at leastL
;k:~J
factors divisible by 2k-l. We ded1+ce that the number of factors 2 of the product in the numerator is at least that of the one in the denominator. If 2k occurs as factor of the numerator, then the number of factors 2 of the product in the numerator :is greater than that of the one in the denominator. This is the case ifm + 3 :<?··2k, i.e. if m ::; 2k - 3. So if m ::; 2k - 3, then (
2
~o~2
) is even, whereas 2k - 2 is even as well. Hence n=
2k + m does not satisfy the condition.We deduce that n can only satisfy the condition if there exists a k 2 2 such that 2k - 2 ::; n ::; 2k - 1. If n is odd, then · (~) + (~) = 1 + n
=
0 mod 2, so n does not satisfy the condition. This implies that n can only'HI
l
l
satisfy the condition if n is of the form n = 2"' - 2 with k 2 2 is. Suppose that
n
is of that form. We show thatn
satisfies the condition. We have the following.(
2k-
2) _
(2k-2)(2k - 3) · · · (2"'- c)(2k - c-1).c - c·(c-1) .. ·2·1 ·
Note that the number of factors 2 in 2k-i is equal to the number of factors 2 in i for 1 ::; i ::; 2k - 1. BenGe the number of factors 2 in the numer~.tor is equal to the number offactors 2 in the product (c+ 1) · c· (c-1) · .. 2, or eqQivalently, the number of factors 2 in the denominator plus the number of factors 2 in c + 1. We deduce that
e" ;
2) is even if and only if c is odd. Hence n = 2k - 2 satisfies the condition.We conclude that n satisfies the condition if and only if n is of the form
2k-2with k·2 2. · D
5. The configuration is fixed by the order in which the points A up to F occur
on the circle, and by the condition on the points P and Q on AD. So there
is no need· to distinguish between different configurations. (The positions
of P and Q with respect to one another are irrelevant.)
First note that Sis the centre of the circumcircle of ABOD EF, as AD is its
diameter, since LABD
=
90°, and S is the midpoint of AD. We will showthat
"i
..
KDS = LKBS. AsKS is the angle bisector of LBSD, we haveLBSK = LKSD. Furthermore, ISDj = ISBI, as the segments SD and
SB are both radii of the circumcircle of ABODEF. Since
I
SKI=
!SKI,we have 6.DSK ~ .6.BSK .(SAS). Hence LKDS = LKBS.
As BK is the angle bisector of LOBS, we have LKBS
=
LOBK=
!LOBS. Hence LKDS
=
!LOBS..Analogously, we find LLDS = !LQES, which implies that LLDS
=
~·
(180°...:LFES) = 90°-.~LFES. Hence LLDS+LKDS = 90°-~LFES+
!LOBS. Now note that 6.SBO ~ 6.SEF (SSS), so LFES =LOBS. We
