Mathematical Excalibur, Volume 16, Number 2

Volume 16, Number 2

June - October 2011

Euler’s Planar Graph Formula

Kin Y. Li

Olympiad Corner

Below are the problems of the 28th Balkan Math Olympiad, which was held in May 6, 2011. Time allowed was 4½ hours.

Problem 1. Let ABCD be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at E. The midpoints of AB and CD are F and G respectively, and ℓ _{is the line through G} parallel to AB. The feet of the perpendiculars from E onto the lines _ℓ and CD are H and K, respectively. Prove that the lines EF and HK are perpendicular.

Problem 2.

Given real numbers x, y, z such that x+y+z = 0, show that

. 0 1 2 ) 2 ( 1 2 ) 2 ( 1 2 ) 2 ( 2 2 2 ₊ ≥ + + + + + + + z z z y y y x x x

When does equality hold?

Problem 3. Let S be a finite set of positive integers which has the following property: if x is a member of S, then so are all positive divisors of x. A non-empty subset T of S is good if whenever x, y _{∈T and x < y, the ration} y/x is a power of a prime number.

(continued on page 4)

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 November 10, 2011.

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

A graph G is consisted of a nonempty set V(G) (its elements are called vertices) and a set E(G) (its elements are called edges), where an edge is to be thought of as a continuous curve joining a vertex u in V(G) to a vertex v in V(G). A graph G is finite if and only if V(G) is a finite set. It is simple if and only if each edge in E(G) joins some pair of distinct vertices in V(G) and no other edge joins the same pair. In this article, all graphs are understood to be finite and simple. A graph is connected if and only if for every pair of distinct vertices a, b, there is a sequence of edges e1, e2, …, en

such that for i from 1 to n, edge ei joins

vi and vi+1 with v1 = a and vn+1 = b. A

graph is planar if and only if it can be drawn on a plane with no pair of edges intersect at any point other than a vertex of the graph. A planar graph divides the plane into regions (bounded by edges) called faces.

e

₂

e

₃

e

₆

e

₅

e

₄

e

₁

e

₇

e

₈

e

₉

v

₂

v

₃

v

₄

v

₅

v

₁

v

₆

v

₇

In the graph above, there are 7 vertices (labeled v1 to v7), 9 edges (labeled e1 to e9) and 4 faces (the 3 triangular regions and the outside region bounded by e1, e5, e7, e8, e9, e6, e3, e2, e1). The following theorem due to Euler relates the number of vertices, the number of edges and the number of regions for a connected planar graph and is the key tool in solving some interesting problems.

Euler’s Theorem on Planar Graphs

Let V, E, F denote the number of vertices, the number of edges, the number of faces respectively for a connected planar (finite simple) graph. Then V − E + F = 2, which we will called Euler’s formula.

We will sketch the usual mathematical induction proof on E. If E = 0, then since V(G) is nonempty and G is connected, we have V = 1 and F = 1. So V−E+F=2. Also, if E =1, then V = 2, F =1 and again the formula is true.

Suppose the cases E < k are true. For the case E=k, either there is a cycle (that is a sequence of edges e1, e2, …, en such

that for i from 1 to n, edge ei joins vi and

vi+1 with v1 = vn+1) or no cycle.

In the former case, removing en will

result in a connected graph with E decreases by 1, V stays the same and F decreases by 1 (since the two regions sharing en in their boundaries will

become one). The formula still holds. In the latter case, we call these graphs trees. It can be proved that they satisfy E=V−1 and F=1 (which implies Euler’s formula). Basically, removing any edge will split such a graph into two connected graphs with each having no cycle. This observation would allow us to do the induction on E.

Before presenting some examples, we remark that Euler’s formula also applies to convex polyhedrons. These are the boundary surfaces of three dimensional convex solids obtained by intersecting finitely many (half-spaces on certain sides of) planes. For example, take the surface of a cube, V=8, E=12, F=6 so that V − E + F = 2. For any convex polyhedron, we can obtain a connected planar graph by choosing a face as base, stretching the base sufficiently big and taking a top view projection onto the plane containing the base. The following is a cube and a planar graph for its boundary surface.

Mathematical Excalibur, Vol. 16, No. 2, Jun.-Oct. 11 Page 2

Example 1. There are n > 3 points on a

circle. Each pair of them is connected by a chord such that no three of these chords intersect at the same point inside the circle. Find the number of regions formed inside the circle. Solution. Removing the n arcs on the circle, we get a simple connected planar graph, where the vertices are the n points on the circle and the intersection points inside the circle. For every 4 of the n points, we can draw two chords intersecting at a point inside the circle. So the number of vertices is V = n + nC4.

Since there are n−1 edges incident with each of the n points on the circle, 4 edges incident with every intersection point inside the circle and each edge is counted twice, so the number of edges is E = (n(n−1)+4 nC4)/2.

By Euler’s formula, the number of faces for this graph is F = 2 − V + E. Excluding the outside face and adding the n regions having the n arcs as boundary, the number of regions inside the circle is F − 1 + n = n + 1−V + E = 1+ nC4+ n(n−1)/2.

For the next few examples, we define the degree of a vertex v in a graph to be the number of edges meeting at v. Below d(v) will denote the degree of v. The sum of degrees of all vertices equals twice the number of edges since each edge is counted twice at its two endponts.

Example 2. A square region is

partitioned into n convex polygonal regions. Find the maximal number of edges in the figure.

Solution. Let V, E, F be the number of vertices, edges, faces respectively in the graph. Euler’s formula yields n+1 = F = 2−V+E or V= E + 1− n. Let A, B, C, D be the vertices of the square, then t = d(A) + d(B) + d(C) + d(D) ≥ 8 as each term is at least 2. Let W be the set of vertices inside the square. For any v in W, we have d(v) ≥ 3 since angles of convex polygons are less than 180_{°. Let s be the sum of d(v)} for all v in W. Since there are V−4 vertices in W, we have s ≥ 3(V−4). Now summing degree of all vertices, we get s + t = 2E. Then

2E−8 ≥ 2E− t = s ≥ 3(V−4) =3(E−3−n),

which simplifies to E ≤ 3n+1.

Finally, the case E = 3n+1 is possible by partitioning the square region into n rectangles using n − 1 line segments parallel to a side of the square. So the maximum possible value of E is 3n+1.

Example 3. (2000 Belarussian Math

Olympiad) In a convex polyhedron with m triangular faces (and possibly faces of other shapes), exactly four edges meet at each vertex. Find the minimum possible value of m.

Solution. Let V, E, F be the number of vertices, edges, faces respectively on such a polyhedron. Since each vertex is met by 4 distinct edges, summing all degrees, we have 2E = 4V.

Next, summing the number of edges in the F faces and observing that each edge is counted twice on the 2 faces sharing it, we get 2E ≥ 3m+4(F−m).

By Euler’s formula, we have

2 = V − E + F = (E/2) − E +F =F − E/2, which implies

4F − 8 = 2E ≥ 3m + 4(F−m). This simplifies to m ≥ 8. A regular octahedron is an example of the case m = 8. Sothe minimum possible m is 8.

Example 4. (1985 IMO proposal by

Federal Republic of Germany) Let M be the set of edge-lengths of an octahedron whose faces are congruent quadrilaterals. Prove that M has at most three elements. Solution. The octahedron has (4×8)/2=16 edges. By Euler’s formula, it has V = 2 + E − F = 2 + 16 − 8 = 10 vertices.

Next, let ni be the number of vertices v

with d(v) = i. Then, counting vertices and edges respectively in terms of ni’s, we

have V = n3 + n4 + n5 + ⋯ = 10 and 2E = 3n3 + 4n4 + 5n5 + ⋯ = 2×16. Eliminating n3, we get n4 + 2n5 + 3n6 + ⋯ = 2.

Hence, n4 ≤ 2, n5 ≤ 1 and ni = 0 for i ≥ 6.

Then n3 = 10 − n4− n5 > 0.

Let A be a vertex with degree 3. Assume M has 4 distinct elements a, b, c, d. Then the 3 faces about A are like the figure below, where we may take AB = a, BC = b, CD = c and DA = d. a b c d A B _C D F G E

Since ABCD and ABGF are congruent, so AF = b or d. Also, since ABCD and AFED are congruent, so AF = a or c. Hence, two of a, b, c, d must be equal, contradiction. Therefore, M has at most 3 elements.

Example 5. Let n be a positive integer.

A convex polyhedron has 10n faces. Prove that n of the faces have the same number of edges.

Solution. Let V be the number of vertices of this polyhedron. For the 10n faces, let these faces be polygons with a1, a2, …, a10n sides respectively, where the ai’s are arranged in ascending order .

Then the number of edges of the polyhedron is E = (a1 + a2 + ⋯ + a10n)/2. By Euler’s formula, we have

_{10 2.} 2 2 1+ + + _{+ =} −a a a n V L n (*)

Also, since the degree of every vertex is at least 3, we get

a1+a2+⋯+a10n ≥ 3V. (**)

Using (*) and (**), we can eliminate V and solve for a1+a2+⋯+a10n to get

a1+a2+⋯+a10n ≤ 60n −12. (***) Assume no n faces have equal number of edges. Then we have a1, a2, …, an−1

≥ 3, an, an+1, …, a2n−2 ≥ 4 and so on. This leads to

a1 + a2 + ⋯ + a10n

≥ (3 + 4 + _{⋯ + 12)(n−1) + 13×10} = 75n + 55.

Comparing with (***), we get 75n + 55 ≤ 60n −12, which is false for n.

Example 6. (1975 Kiev Math

Olympiad and 1987 East German Math Olympiad) An arrowhead is drawn on every edge of a convex polyhedron H such that at every vertex, there are at least one arrowhead pointing toward the vertex and another arrowhead pointing away from the vertex. Prove that there exist at least two faces of H, the arrowheads on each of its boundary form a (clockwise or counterclockwise) cycle.

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 November 10, 2011. Problem 376. A polynomial is monic if the coefficient of its greatest degree term is 1. Prove that there exists a monic polynomial f(x) with integer coefficients such that for every prime p, f(x) ≡ 0 (mod p) has solutions in integers, but f(x) = 0 has no solution in integers.

Problem 377. Let n be a positive integers. For i=1,2,…,n, let zi and wi be

complex numbers such that for all 2n

choices of ε1, ε2, …, εn equal to ±1, we have . 1 1

∑

∑

= = ≤ n i i i n i i iz εw ε Prove that

∑

∑

= = ≤ n i n i i i w z 1 1 2 2 _{| | .} | |

Problem 378. Prove that for every positive integers m and n, there exists a positive integer k such that 2k _{−m has at}

least n distinct positive prime divisors. Problem 379. Let ℓ be a line on the plane of _{∆ABC such that ℓ does not} intersect the triangle and none of the lines AB, BC, CA is perpendicular to ℓ. Let A’, B’ C’ be the feet of the perpendiculars from A, B, C to ℓ respectively. Let A’’, B”, C” be the feet of the perpendiculars from A’, B’, C’ to lines BC, CA, AB respectively. Prove that lines A’A”, B’B”, C’C” are concurrent.

Problem 380. Let S = {1,2,…,2000}. If A and B are subsets of S, then let |A| and |B| denote the number of elements in A and in B respectively. Suppose the product of |A| and |B| is at least 3999. Then prove that sets A−A and B−B contain at least one common element, where X−X denotes {s−t : s, t _{∈ X and} s ≠ t}.

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

Solutions

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

Problem 371. Let a1, a2, a3, … be a sequence of nonnegative rational numbers such that am+an= amn for all positive

integers m, n. Prove that there exists two terms that are equal.

Solution. U. BATZORIG (National

University of Mongolia), CHUNG Kwan (King’s College) and F7B Pure Math Group (Carmel Alison Lam Foundation Secondary School).

Let p and q be distinct primes. If ap and aq

are zeros, then we are done. Otherwise, consider

,

p q Na Na

q

n

and

p

m

=

=

where N is a positive integer that makes both Naq and Nap integers. Obviously, we

have m ≠ n and . ) ( ) ( _{q p p q n} m Na a Na a a a = = =

Other commended solvers: Samuel Liló ABDALLA (ITA-UNESP, São Paulo, Brazil).

Problem 372. (Proposed by Terence ZHU) For all a,b,c > 0 and abc=1, prove that

) 1 ( ) 1 ( 1 ) 1 ( ) 1 ( 1 + + + + + + + abab bb bcbc a a . 4 3 ) 1 ( ) 1 ( 1 ≥ + + + + ca ca c c

Solution. V. ADIYASUREN (National

University of Mongolia) and B. SANCHIR (Mathematics Institute of the National University of Mongolia), F7B Pure Math Group (Carmel Alison Lam Foundation Secondary School) and Kipp JOHNSON (Valley Catholic School, Teacher, Beaverton, Oregon, USA).

Substituting a = z/y, b = x/z, c = y/x (say by choosing x=ab=1/c, y=1, z=a) into the inequality and simplifying, we get

∑

≥ cyc z y x f , 4 3 ) , , ( where ) ( ) ( ) , , ( 2 y x x y z z y z y x f + + + = and ). , , ( ) , , ( ) , , ( ) , , (xyz f xyz f yzx f zxy f cyc + + =

∑

Let g(x,y,z) = y2_(z2 _{+ zy + x}2 _{+ xy). By the}

Cauchy-Schwarz inequality, we have . ) , , ( ) , , ( 2 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≥

∑

∑

∑

cyc cyc cyc y z y x g z y x f So it is enough to prove (*) . 4 3 ) , , ( 2 2 _≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

∑

∑

cyc cyc z y x g y

Expanding and factorizing, we get ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

∑

∑

cyc cyc z y x g y 3 ( , , ) 4 2 2

∑

∑

∑

+ − + = cyc cyc cyc y x xy y x y 2 3 ( ) 4 4 2 2 2 2 ₌

_∑

_{− + +}

_∑

_{− ≥} cyc cyc y x y x y x ) ( ) ( ) 0. ( 3 2 2 2 2 2 2

This implies (*), which implies the desired inequality.

Other commended solvers: CHUNG Kwan (King’s College), NGUYEN Van Thien (Luong The Vinh High School, Dong Nai, Vietnam) and Paolo PERFETTI (Math Dept, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy). Problem 373. Let x and y be the sums of positive integers x1,x2,…,x99 and y1,y2,…,y99 respectively. Prove that there exists a 50 element subset S of {1,2,…,99} such that the sum of all xn

with n in S is at least x/2 and the sum of all yn with n in S is at least y/2.

Solution. William Peng and Jeff Peng. Arrange the numbers x1,x2,…,x99 in descending order, say xn(1) ≥ xn(2) ≥ ⋯≥

xn(99) so that {n(1), n(2),…, n(99)} = {1,2,…,99}. Let A = {n(2), n(4),…, n(98)} and B = {n(3), n(5), …, n(99)}. We have

.

) 1 (

∑

∑

∑

∈ ∈ ∈

≥

>

+

B j j A i i B j j n

x

x

x

x

If

∑

∑

,

∈ ∈

≥

B j j A i i

y

y

then let S =A_∪{n(1)}. Now S has 50 elements. Also,

∑

∑

∑

∈ ∈ ∈

≥

>

B j j A i i S i i

x

x

x

and

.

∑

∑

∑

∈ ∈ ∈

≥

>

B j j A i i S i i

y

y

y

Mathematical Excalibur, Vol. 16, No. 2, Jun.-Oct. 11 Page 4 So the sum of all xn with n in S is at

least x/2 and the sum of all yn with n in

S is at least y/2. If

∑

∑

,

∈ ∈

<

B j j A i i

y

y

then let S =B_∪{n(1)}. Again S has 50 elements. Now

∑

∑

∑

∑

∈ ∈ ∈ ∈

≥

>

+

=

B j j A i i B j j n S i i

x

x

x

x

x

₍₁₎ and

.

∑

∑

∑

∈ ∈ ∈

>

>

A i i B j j S i i

y

y

y

So the sum of all xn with n in S is at

least x/2 and the sum of all yn with n in

S is at least y/2.

Other commended solvers: U. BATZORIG (National University of Mongolia

)

and F7B Pure Math Group (Carmel Alison Lam Foundation Secondary School),

Problem 374. O is the circumcenter of acute _{∆ABC and T is the circumcenter} of _{∆AOC. Let M be the midpoint of} side AC. On sides AB and BC, there are points D and E respectively such that ∠BDM=∠BEM=∠ABC. Prove that BT⊥DE.

Solution. William Peng and Jeff Peng. X B Y T A C M O D E

By the exterior angle theorem, ∠ABC = ∠BDM > ∠BAM and also∠ABC = ∠BEM > ∠BCM. So ∠ABC is the largest angle in _{∆ABC. Then we have} 60_{° < ∠ABC < 90°. This implies O is} on the same side of line AC as B. Then T will be on the opposite side of line AC as O. Also, O, M, T are on the perpendicular bisector of line AC. Let X be the intersection of lines AB and ME. Let Y be the intersection of lines CB and MD. Now

∠DXE = 180_{°− ∠XBE −∠BEX} = 180_{°− 2∠ABC}

and similarly∠EYD =180_{°− 2∠ABC.} So ∠DXE =∠EYD, which implies D, X, Y, E are concyclic.

Next, since T is the circumcenter of ∆AOC, so

∠ATM =∠ATO = 2∠ACO = 2(90_°−∠BXE) = 180_{°− 2∠ABC} = ∠BXE =∠AXM. This implies A, M, T, X are concyclic. So ∠AXT = 180_{°− ∠AMT = 90°. Similarly,} ∠CYT = 90_{°. Then ∠BXT = ∠BYT,} which implies B, X, T, Y are concyclic. So ∠TBY =∠TXY = 90_{°−∠BXY. (*)} Since D, X, Y, E are concyclic,

∠BED +∠TBE =∠BXY +∠TBY = 90_{° by (*),} which implies BT⊥DE.

Other commended solvers: F7B Pure Math Group (Carmel Alison Lam Foundation Secondary School),

Problem 375. Find (with proof) all odd integers n > 1 such that if a, b are divisors of n and are relatively prime, then a+b−1 is also a divisor of n.

Solution. U. BATZORIG (National University of Mongolia), William Peng and Jeff Peng.

For such odd n, let p be its least prime divisor. Then n = pm_{a, where m is the}

exponent of p in the prime factorization of n. We will show a = 1.

Assume a > 1. Then every prime divisors of a is at least p+2. Also c = a+p−1 (> p) is a divisor of n. Since

gcd(c,a) = gcd(c−a,a) = gcd(p−1,a) = 1, this implies c=pr _{with r ≥ 2. Then d =}

a+p2_{−1 (> p}2_{) is also a divisor of n.} Similarly,

gcd(d,a) = gcd(d−a,a)= gcd(p2_{−1,a)= 1.} So d=ps _{with s ≥ 3. Finally, p}r_{−p = c−p =}

a−1= d−p2_{, which is divisible by p}2_{, while} pr_{−p is not. Therefore, a = 1.}

It is easy to check all n=pm _{with p an odd}

prime and m a positive integer indeed satisfy the condition.

Olympiad Corner

(continued from page 1) Problem 3. (Cont.) A non-empty subset T of S is bad if whenever x, y _{∈T and x < y,} the ration y/x is not a power of a prime

number. We agree that a singleton subset of S is both good and bad. Let k be the largest possible size of a good subset of S. Prove that k is also the smallest number of pairwise-disjoint bad subsets whose union is S.

Problem 4. Let ABCDEF be a convex hexagon of area 1, whose opposite sides are parallel. The lines AB, CD and EF meet in pairs to determine the vertices of a triangle. Similarly, the lines BC, DE and FA meet in pairs to determine the vertices of another triangle. Show that the area of one of these two triangles is at least 3/2.

Euler’s

Planar GraphFormula

(continued from page 2) Solution. Call {a,b} a hook if a, b are two consecutive edges on the boundary of some face of H. Call a hook {a,b} traversible if the arrowheads on a and b are both counterclockwise or both clockwise.

Note every hook is part of the boundary of a unique face. Let E be the number of edges on H and h be the number of hooks on H. As each edge on H is a part of 4 hooks, we get h = 2E. Next at every vertex v, d(v) ≥ 3. By the given condition on the vertices, there must be at least 2 traversible hooks through every vertex. Let V be the number of vertices on H, then there are at least 2V traversible hooks on H. Let h+ and h− be the number of

traversible and non-traversible hooks respectively on H. Then h+ ≥ 2V.

In every face where the boundary arrowheads do not form a cycle, there are at least two changes in directions on the boundary, which result in at least two non-traversible hooks. Let F be the number of faces on H. Let f+ be the

number of faces the boundary arrowheads form cycles. Let f−=F−f+.

Then h− ≥ 2f−.

By Euler’s formula, V− E + F = 2. Then 2f+ = 2F − 2f−

= (4 + 2E − 2V) − 2f− ≥ 4 + h − h+ − 2f− = 4 + h− − 2f− ≥ 4,

which implies f+ ≥ 2. This gives the desired conclusion.

~o ltv-t'OYl.s 井的叫叫 ~Y\e.rJ 叫 l~

)

仰，之

PROBLEM 1

Let ABCD be a cyclic quadrilateral which is not a trapezoid and whose diagonais et at E.The midpoints of AB and CD are F and G respectively

,

and t is the line through G parallel bo AB.The fedof theperpendicularsborn E onto the lines t and CD are H and k,respectively-Prove that the lines EF and HK are perpendicular.

Solution. The points E

,

K

,

H

,

G are on the circle of diameter GE

,

80

LEHI< = LEGI<.

Al帥，from LDCA= LDBA and g~ =劈 itfollow8 CE 2CE 2BE BE Eδ= 石百=五X=BF'

therefore 6CGE fV 6BFE. In particular

,

LEGC = LBFE

,

so by

(•)

LEHI< = LBFE.

But HEiFB and so

,

since FE and HK are obtained by rotations of these iines by the same(directed) angle

,

FEl.HI< ..

B

A

PROBLEM 2

Given real numbers x

,

y

,

z such that x

+

y

+

z = 0

,

show that x(x

+

2) . y(y

+

2) . z(z

+

2)

一一一一一+一一一一一+一一一句> O.

2x2

+

1 ' 2y2

+

1 ' 2z2_{十1 一}

When does equality hold?

)

, 71. ''， 1 、

'. Solution. The inequality is clear if xyz = 0

,

in which case equality holds if and only if x = y = z 早 O.

Hénceforth assume xyz 手 oand rewrite the"inequality 聞 (2x

+

1)2 . (2ν +1)2. (2z+1)2

+一一一一+一一一一> 3.

2X2

+

1 ' 2y2

+

1 ' 2z2

+

1 一

Notice that (exactly) one of the products 旬，仰， zx is positi珊， s叮 yz> 0

,

to get

n4--llh 、巴巴'，一句 i-可」 叫一+戶，一+:~ l 可 --z 啥』 -ZHr--z-d 二 ME----l +-41: 一舟，二十

upmfjh-d

2-j-z2->一=>­ no-、 1 ， JM4 自 •• 41 日 一+ +一 -a" , u z-z nd-nd

+

nA-1--ti 咱 i--sT

+-Z4

U-u

n4-nd (by Jensen) (for x

+

y

+

z = 0) (for yz

>

0) Here equality holds if and only if x =: 1 and y

=

z

=

-1/2. Finally

,

since

(2x

+

1)2 自 2(x- 1)2 .., 2x2(x 一 1)2

一一一一一一+一一一一一一 -3= ，.~

,-

-'.>0.

2X2

+

1 . x2_{十 1 (2x2}

+

_1)(x2

+

_{1) 一，} Z 巴lR，

the conclusion follows. Clearly

,

equality holds if and only if x

= 1 ，日o

y

=

Z.= -1/2. Therefore

,

jf xyz 手 0 ，equality holds if and only íf one of the numbers is 1

,

and the other two are -1/2

PROBLE

lV{

3

Let S be a finite s~t of positive integers which has 七he following property: if x is a memberof S

,

then so are all positive divisors of x. A non-empty subset T of S is good if whenever x

,

y E T and x

<

y

,

the ratio y/x is a power of a prime number. A non-empty subset T of S is bad if whenever x

,

y E T and x < 仇 the ratio y/x is not a power of a prime number. We agree that a singleton subset of S is both good and bad. Let k be the largest possible size of a good subs的 ofS. Prove that k is also the smallest number 'of pairwise-disjoint bad subsets whose union is S.

Solution. Notice fìrst that a bad subset of S contains at most one element from a good one

,

to deduce that a parti七ion of S into bad subsets has at least as many members

間 ama.ximal good subset.

Notice further that the elements of a goòd subset of S must be among the terms of a geometric sequence whose ratio is a prime: if

x

<

y

<

z

are elements of a good subset of S

,

then y 口 xp自 andz = 的β = xp白qß for some primes p and q and some positive integers αand 戶，so p = q for

z/x

to be a power of a prime.

Next

,

let P 口 {2 ， 3 ， 5 ， 7， 11γ ﹒.

}

denote the set of all primes

,

let m = max{exppx: x ε S andp ε P} ，

where expp x is the exponent of the prime p in the canonical decomposition of x

,

and

notice that à maximal good subset of S must be of the form 柄， α仇 αpm}for some prime p and some positive integer αwhich is not divisible by p. Consequently

,

a ma.ximal good subset of S has m

+

1elements

,

so a partition of S into bad subsets has at least m

+

1 members.

Finally

,

notice by maximality of m that the sets

此話 {x: x ξ S and 2二師p x

==

k (mod m

+

1)}

,

k = 0

,

1

, . .. ,

m

,

pEP

form a paJ:tition of S into m 十 1bad subsets. The conclusion follows.

PROBLEM 4

Let ABGDEF be a convex hexagon of area 1

,

whose opposite sides are parallel. The lines AB

,

GD and EF meet in pairs to determine the vertices of a triangle. Similarly

,

the lines BG

,

DE and FA meet in pairs to determine the vertices of another triangle: Show that the area of at least one of these two triangl巴s is at le且t3/2.

Solution. Unless otherwise stated

,

throughou七 theproof indices take on values from

o

to 5 and are reduced modulo 6. Label the vertices of the hexagon in circular order

,

Ao

,

Al' . . " A5

,

and let the lines of support of the alternate sides AiAi+l and Ai+2Aï+3 meet at Bi' To show-that the area of at least one of the triangles BoB2B4

,

BIB3B5 is greater than or equal to 3月， it is suffic加nt to prove that the total area of the six triangles Aï+lBiAï+2 is at least 1:

5

LareaAi+l或Ai+2 主 L i=O

To begin with

,

reflec七 each Bi through the midpoint of 七hesegment Ai+lAi+2 to get the point!> B~. We shall prove that the six triangles Ai+l 月Ai+2 cover the hexagon. To this end

,

refiect A2i+l through the midpoint of the segment A2iAzi十Zto get the points A~i+l' i

=

0

,

1

,

2. The hexagon splits into three parallelograms

,

AZiA況+lA2i+2A~ï+ l'i

=

0

,

1

,

2

,

and a (possibly degenerate) triangle

,

Ai A~A~. Notice first 七hat each parallelogram A2iA2ï+lA2ï+2A~i+l is covered by the pair of triangles (A2iB~i+5 A2i+l， A2i+lB~iA創刊),

, i = 0

,

1

,

2. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle AiA~A~ ， To 出is end

,

it is su的cient to prove 出的 A2iB~ï+5 主 A2iA~i+5 and A2j+2到j 主 A2j+iA~j+3 for some indices i

,

j E {O

,

1

,

2}. To establish the first inequality

,

notice that

tö get

Similarly

,

A2iB~i+5 = A2i+lB2i

+5,

A2iA~i+5 = A2i+4A2i

+5,

i = 0

,

1

,

2

,

AIB5 AoB5 , A3Bl AZA3

一一一一=-一一:- anα -一一=一一一­

A4A 5 A5 B 3 ---- AoAl AoB5 '

台全iB~ï+5 一 1

1;1AML+5-4

12r

h

j+Z月

j

_

1

月何+2A~

_j+3

叫

whence the conclusion.

8₃_ A.4 內 、-.---~

\、/\艾

5 / A'伊:心一一一一-

-三，

A2 Aj失……唸…--持自 f\\/口 4 ~3 .---_.巴 84 Ao ~