Mathematical Excalibur, Volume 6, Number 5

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Volume 6, Number 5

Volume 6, Number 5 January

January 200

January 2002222 –––– February 200

January 200

200 February 200

February 200

February 2002222

Vector Geometry

Kin Y. Li

Olympiad Corner

The 10th Winter Camp, Taipei, Taiwan, February 14, 2001.

Problem 1. Determine all integers a and b which satisfy that

2001 90 13 b b a + = .

Problem 2. Let an be sequence of real

numbers satisfying the recurrence relation

[ ]

2 , 1 ,2 ,....

, ₁

1=k a + = a n=

a _n _n

where [x] denotes the largest number which is less or equal than x. Find all positive integers k for which three exist three consecutive terms ai−1,ai ,ai+1 satisfy .2ai=ai−1+ai+1

Problem 3. A real number r is said to be attainable if there is a triple of positive real numbers (a, b, c) such that a, b, c are not the lengths of any triangle and satisfy the inequality

. 2 2 2_b _b _c _c _a a rabc> + +

(a) Determine whether or not 2 7

is attainable.

(b) Find all positive integer n such that n is attainable.

(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), Applied Math. Dept., HKPU

李健賢 (LI Kin-Yin), Math. Dept., 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 March

23, 2002.

For individual subscription for the next five issues for the 01-02 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]

A vector XY is an object having a magnitude (the length XY) and a direction (from X to Y). Vectors are very useful in solving certain types of geometry problems. First, we will mention some basic concepts related to vectors. Two vectors are considered the same if and only if they have the same magnitudes and directions. A vector

OX from the origin O to a point X is called a position vector. For convenience, often a position vector OX

will simply be denoted by X, when the position of the origin is understood, so that the vector XY = OY −OX will simply be Y – X. The length of the position vector OX_{= X will be denoted}

by X . We have the triangle inequality

Y X Y

X + ≤ + , with equality if and only if X = tY for some t ≥ 0. Also, cX

= c X for number c.

For a point P on the line XY, in terms of position vectors, P = tX + (1 – t)Y for some real number t. If P is on the segment XY, then t = PY/XY ∈ [0, 1].

Next, we will present some examples showing how vectors can be used to solve geometry problems.

Example 1. (1980 Leningrad High

School Math Olympiad) Call a segment in a convex quadrilateral a midline if it joins the midpoints of opposite sides. Show that if the sum of the midlines of a quadrilateral is equal to its semiperimeter, then the quadrilateral is a parallelogram.

Solution. Let ABCD be such a convex

quadrilateral. Set the origin at A. The sum of the lengths of the midlines is

2 B C D D C B+ − + + − and the semiperimeter is

2 B C D D C B + − + + − . So B C D D C B+ − + + − = B + C−D + D +C−B By triangle inequality, B +C−D≥ D C

B+ − , with equality if and only if ) (C D t B= − (or ABCD). Similarly, ≥ − +C B A A+C−B , with equality if and only if ADBC. For the equation to be true, both triangle inequalities must be equalities. In that case, ABCD is a parallelogram.

Example 2. (Crux Problem 2333) D

and E are points on sides AC and AB of triangle ABC, respectively. Also, DE is not parallel to CB. Suppose F and G are points of BC and ED, respectively, such that BF :FC = EG :GD = BE :CD . Show that GF is parallel to the angle bisector of ∠BAC.

Solution. Set the origin at A. Then E =

pB and D = qC for some p, q ∈ (0, 1). Let t = FC BF , then F = 1 + + t B tC and G = 1 + + t E tD = 1 + + t pB tqC . Since BE = tCD, so (1 – p)|B| = t(1 – q)|C|. Thus, B t p C t q t G F 1 1 1 ) 1 ( + − + + − = − =     ₊ + − | | | | 1 | | ) 1 ( B B C C t B p .

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Mathematical Excalibur Mathematical ExcaliburMathematical Excalibur

Mathematical Excalibur, Vol. 6, No. 5, Jan 02- Feb 02 Page 2

This is parallel to , | | | | B B C C + which is in the direction of the angle bisector of

. BAC

∠

* * * * * * * * * * * * * * * * * * * * * The dot product of two vectors X and Y is the number X⋅Y = |X||Y| cosθ , where θ is the angle between the vectors. Dot product has the following properties: (1) X⋅Y =Y⋅X ,(X+Y)⋅Z =X⋅Z + Y⋅Z and (cX)⋅Y =c(X⋅Y). (2) X2=X⋅X, X⋅Y ≤ X Y and OY OX ⊥ if and only if X⋅Y= 0.

Example 3. (1975 USAMO) Let A,

B, C, D denote four points in space and AB the distance between A and B, and so on. Show that

. 2 2 2 2 2 2 CD AB BC AD BD AC + + + ≥ +

Solution. Set the origin at A. The

inequality to be proved is ) ( ) (B D B D C C⋅ + − ⋅ − +D ⋅D + (B − C)⋅(B − C) ≥B⋅B+(C−D)⋅(C−D).

After expansion and regrouping, this is the same as (B−C−D)⋅(B−C−D)

, 0

≥ with equality if and only if B – C = D = D – A, i.e. is BCAD is a parallelogram.

* * * * * * * * * * * * * * * * * * * * * For a triangle ABC, the position vectors of its centroid is

G = . 3

C B A+ +

If we take the circumcenter O as the origin, then the position of the orthocenter is H = A + B + C as OH

= 3OG. Now for the incenter I, let a, b, c be the side lengths and AI intersect BC at D. Since BD:CD = c:b and DI:AI = c b ca + :c = a:b + c, so D = c b cC bB + + and I = c b a cC bB aA + + + + .

Example 4. (2nd Balkan Math

Olympiad) Let O be the center of the

circle through the points A, B, C, and let D be the midpoint of AB. Let E be the centroid of triangle ACD. Prove that the line CD is perpendicular to line OE if and only if AB = AC.

Solution. Set the origin at O. Then , 2 B A D= + , 6 2 3A B C D C A E= + + = + + . 2 2C B A C D− = + −

Hence CD⊥OE if and only if (A + B – . 0 ) 2 3 ( ) 2C ⋅ A+B+ C = Since A⋅A = B B⋅ = C⋅C, this is equivalent to A (⋅ B , 0 )= ⋅ − ⋅ =

−C A B A C which is the same as ,OA⊥BC i.e. AB = AC.

Example 5. (1990 IMO Usused Problem,

Proposed by France) Given ABC∆ with no side equal to another side, let G, I and H be its centroid, incenter and orthocenter, respectively. Prove that ∠GIH>90$.

Solution. Set the origin at the

circumcenter. Then H = A + B + C, G = 3 C B A+ + , I = . c b a cC bB aA + + + + We need to show (G−I)⋅(H−I)= . 0 ) ( + < ⋅ − ⋅ + ⋅H I I I G H G Now A⋅A = B⋅B=C⋅C=R2 and 2B⋅C = B⋅B + C⋅C–(B−C)⋅(B−C)=2R2−a2, … . Hence, 3 ) ( ) (A B C A B C H G⋅ = + + ⋅ + + , 3 2 2 2 2 a b c R − + + = 2 ) ( ) ( ) ( c b a cC bB aA cC bB aA I I + + + + ⋅ + + = ⋅ , 2 c b a abc R + + − = ) ( 3 ) ( ) ( 4 ) ( c b a C B A cC bB aA H G I + + + + ⋅ + + = + ⋅ . ) ( 3 )] ( ) ( ) ( [ 2 4 2 2 2 2 c b a b a c a c b c b a R + + + + + + + − =

Thus, it is equivalent to proving (a + b + c)(a2+b2+c2) + 3abc > 2[a (b + 2 c) + b (c + a) + 2 c (a + b)], which 2 after expansion and regrouping will become a(a – b)(a – c) + b(b – c)(b – a) + c(c – a)(c – b) > 0. To obtain this inequality, without loss of generality, assume .a≥b≥c Then a(a−b)(a−

) )( (

) b a b b c

c ≥ − − so that the sum of the first two terms is nonnegative. As the third term is also nonnegative, the above inequality is true.

* * * * * * * * * * * * * * * * * * * * * The cross product of two vectors X and Y is a vector X×Y having magnitude |X||Y| sin θ , where θ is the angle between the vectors, and direction perpendicular to the plane of X and Y satisfying the right hand rule. Cross product has the following properties: (1) X×Y = −Y×X , (X + Y) ×Z = Z X× + Y ×Z and (cX) ×Y = c(X ×Y). (2) 2 | |X ×Y

is the area of triangle XOY. When X, Y ≠O, X×Y = 0 if and only if X, O, Y are collinear.

Example 6. (1984 Annual Greek

High School Competition) Let 1

A A2 A3 A4 A5 A be a convex 6 hexagon having its opposite sides parallel. Prove that triangles A1 A3 A 5 and A₂ A₄ A have equal areas. ₆

Solution. Set the origin at any point. As the opposite sides are parallel, ( A ₁

, 0 ) ( ) ₄ ₅ 2 × − = −A A A (A₃−A₂)×(A₅ ) 6 A − = 0 and (A3−A4)×(A6−A1) = 0. Expanding these equations and adding them, we get A₁×A₃+A₃×A₅

. 2 6 6 4 4 2 1 5 A A A A A A A A × = × + × + × + Now [A1 A3 A ] = 5 2 | ) ( ) ( | A₁−A₃ × A₁−A₅ = 2 | |A₁×A₃+A₃+A₅+A₅×A₁ . (continued on page 4)

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Mathematical Excalibur, Vol. 6, No. 5, Jan 02- Feb 02 Page 3

Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. 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. The deadline for submitting solutions is March 23, 2002.

Problem 141. Ninety-eight points are given on a circle. Maria and José take turns drawing a segment between two of the points which have not yet been joined by a segment. The game ends when each point has been used as the endpoint of a segment at least once. The winner is the player who draws the last segment. If José goes first, who has a winning strategy? (Source: 1998 Iberoamerican Math Olympiad)

Problem 142. ABCD is a quadrilateral with AB|| CD. P and Q are on sides AD and BC respectively such that ∠APB = ∠CPD and

AQB

∠ = ∠CQD. Prove that P and Q are equal distance from the intersection point of the diagonals of the quadrilateral. (Source: 1994 Russian Math Olympiad, Final Round)

Problem 143. Solve the equation cos cos cos cos x = sin sin sin sin x. (Source: 1994 Russian Math Olympiad, 4th Round)

Problem 144. (Proposed by José Luis Díaz-Barrero, Universitat Politècnica de Catalunya, Barcelona, Spain) Find all (non-degenerate) triangles ABC with consecutive integer sides a, b, c and such that C = 2A. Problem 145. Determine all natural numbers k>1 such that, for some distinct natural numbers m and n, the numbers 1km+ and kn+1 can be obtained from each other by reversing the order of the digits in their decimal representations. (Source: 1992 CIS Math Olympiad)

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

Solutions

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

Problem 136. For a triangle ABC, if sinA, sinB, sinC are rational, prove that cosA, cosB, cosC must also be rational. If cosA, cosB, cosC are rational, must at least one of sinA, sinB, sinC be rational?

Solution. CHAN Wai Hong (STFA

Leung Kau Kui College, Form 6), CHAO Khek Lun Harold (St. Paul’s College, Form 7), CHIU Yik Yin (St. Joseph’s Anglo-Chinese School, Form 6), LEUNG Wai Ying (Queen Elizabeth School, Form 7), LO Chi Fai (STFA Leung Kau Kui College, Form 6), WONG Tak Wai Alan (University of Toronto), WONG Tsz Wai (Hong Kong Chinese Women’s Club College, Form 6) and WONG Wing Hong (La Salle College, Form 4).

If sinA, sinB, sinC are rational, then by cosine law and sine law,

      ₊ ₋ = − + = c a b a b c c b bc a c b A 2 1 2 cos 2 2 2       ₊ ₋ = C A B A B C C B sin sin sin sin sin sin sin sin 2 1

is rational. Similarly, cosB and cosC are rational. In the case of an equilateral triangle, cosA = cosB = cosC = cos60 = $

2 1

is rational, but sinA = sinB = sinC =

2 3 60

sin $ = is irrational.

Other commended solvers: LEE Tsun Man Clement (St. Paul’s College, Form 3), LOONG King Pan Campion (STFA Leung Kau Kui College, Form 6), SIU Tsz Hang (STFA Leung Kau Kui College, Form 6) and TANG Chun Pong (La Salle College, Form 4).

Problem 137. Prove that for every positive integer n, n n 1/ / 1 ) 2 3 ( ) 2 3 ( + + − is irrational.

Solution. CHAO Khek Lun Harold (St.

Paul’s College, Form 7) and LEUNG Wai Ying (Queen Elizabeth School, Form 7). Let x=( 3+ 2)1 n/ . Since

(

3+ 2

)

(

3− 2

)

= 1, x−1=( 3− 2)1/n. If 1 − +x x is rational, then x2+x−2=(x+ 2 1 ) −

x – 2 is also rational. Since

(

k k

)

k k x x x x x x +1+ −( +1) =( + −1) + −

(

−1₊ −( −1)

)

− k k x x , by math induction, xn+x−n = 2 3 would be rational, a contradiction. Therefore, x + x−1 is irrational.

Other commended solvers: CHAN Wai Hong (STFA Leung Kau Kui College, Form 6), SIU Tsz Hang (STFA Leung Kau Kui College, Form 6) and WONG Wing Hong (La Salle College, Form 4). Problem 138. (Proposed by José Luis Díaz-Barrero, Universitat Politècnica de Catalunya, Barcelona, Spain) If a + b and a – b are relatively prime integers, find the greatest common divisor (or the highest common factor) of 2a + (1 + 2a)(a2−b2) and 2a(a + 2a – 2

2

b )(a2−b2).

Solution. CHAO Khek Lun Harold

(St. Paul’s College, Form 7) and LEUNG Wai Ying (Queen Elizabeth School, Form 7).

Let (r, s) denote the greatest common divisor (or highest common factor) of r and s. If (r, s) = 1, then for any prime p dividing

rs

, either p divides r or p divides s, but not both. In particular p does not divide r + s. So (r + s, rs) = 1. Let x = a + b and y = a – b. Then

2a + (1 + 2a)(a2−b2) = x + y + (1 + x + y)xy = (x + y + xy) + (x + y)xy and 2a(a + 2a – 2 b )(2 a2−b2) = (x + y)(xy + x + y)xy. Now (x, y) = 1 implies (x + y, xy) = 1. Repeating this twice, we get

(x + y + xy, (x + y) xy) = 1 and

((x + y + xy + (x + y)xy, (x + y + xy)(x + y)xy) = 1. So the answer to the problem is 1. Other commended solvers: LEE Tsun Man Clement (St. Paul’s College, Form 3), POON Yiu Keung (HKUST, Math Major, Year 1), SIU Tsz Hang (STFA Leung Kau Kui College, Form 6), TANG Chun Pong (La Salle College, Form 4), WONG Chun Ho (STFA Leung Kau Kui College, Form 7) and WONG Wing Hong (La Salle College, Form 4).

Problem 139. Let a line intersect a pair of concentric circles at points A, B, C, D in that order. Let E be on the outer circle and F be on the inner circle such that chords AE and BF are parallel. Let

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Mathematical Excalibur, Vol. 6, No. 5, Jan 02- Feb 02 Page 4 G and H be points on chords BF and AE

that are the feet of perpendiculars from C to BF and from D to AE, respectively. Prove that EH = FG. (Source: 1958 Shanghai City Math Competition)

Solution. WONG Tsz Wai (Hong Kong

Chinese Women’s Club College, Form 6). Let M be the midpoint of BC (and AD). Since∠DHA=90$ ,∠ADH=∠DHM . Since BF|| AE, ∠BAE=∠FEA by symmetry with respect to the diameter perpendicular to BF and AE. Now

$ $ ₉₀ 90 −∠ = = ∠ =

∠FEA BAE ADH

– ∠DHM =∠AHG . So EF|| HG. Since EH ||FG also, EFGH is a parallelogram. Therefore, EH=FG. Other commended solvers: CHAO Khek Lun Harold (St. Paul’s College, Form 7), CHUNG Tat Chi (Queen Elizabeth School, Form 5), LEUNG Wai Ying (Queen Elizabeth School, Form 7), SIU Tsz Hang (STFA Leung Kau Kui College, Form 6) and WONG Chun Ho (STFA Leung Kau Kui College, Form 7). Problem 140. A convex pentagon has five equal sides. Prove that the interior of the five circles with the five sides as diameters do not cover the interior of the pentagon.

Solution. LEUNG Wai Ying

(Queen Elizabeth School, Form 7). Let the pentagon be A₁A₂A₃A₄A₅ and 2r be the common length of the sides. Let M be the midpoint of _ij A_iA_j and

i

C be the circle with diameter A_iA_i₊₁ for i = 1, 2, 3, 4, 5 (with A₆ = A₁). Since 180540−3⋅60=2⋅ and ∠A_i<

,

180$ there are at least 3 interior angles (in particular, two adjacent angles) greater than 60$. So we may suppose

. 60 , 2 1 ∠ > $ ∠A A Since A₃A₄=A₅A₄, we get A₄M₃₅ ⊥A₃A₅. Then M₃₅ is on C3, C4 and the points on the ray from A to ₄ M₃₅ lying beyond M₃₅ is outside C₃,C₄.

Next, since ∠A1>60$and A1A2=A1A5, 5

2A

A is the longest side of ∆A1A2A5. By the midpoint theorem, M₂₃M₃₅ =

r A A A A = > 2 2 2 1 5 2 _{so that} 35 M is outside 2 C . Similarly, M₃₅ is outside C . If ₅ 35

M is not outside C , then ₁ A₂M₃₅ 3 2 2 1A A A A = < and ∠A1M35A2 ≥ 90$. Since A₃M₃₅<A₃A₄ =A₂A₃ also, A₂A₃ must be the longest side of ∆A₂A₃M₃₅. Then ∠A2M35A3>60$. Similarly, . 60 5 35 1 > $ ∠AM A Then, we have , 60 2 35 1 < $ ∠AM A a contradiction. So 35 M is outsideC , too. 1 For i = 1, 2, 5 let d_i =M₃₅M_i_,_i₊₁−r>0. Let d be the distance from M₃₅ to the intersection point of the pentagon with the ray from A to ₄ M₃₅ lying beyond M₃₅. Choose a point X beyond M35 on the ray from A to ₄ M₃₅ with XM₃₅<d,d₁,d₂ and .d₅ Then X is inside the pentagon and is outside C₃, C₄. Also, for i = 1, 2, 5,

35 1 , 35 1 , M M XM XM_i_i₊ > _i_i₊ − = r + d_i−XM₃₅>r so that X is outsideC₁ ,C₂,C₅.

Comments: The point M₃₅ is enough for the solution as it is not in the interior of the 5 circles. The point X is better as it is not even on any of the circles.

Olympiad Corner

(continued from page 1)

Problem 4. Let O be the center of excircle of ABC∆ touching the side BC internally. Let M be the midpoint of AC, P the intersection point of MO and BC. Prove that AB = BP, if ∠BAC = 2∠ACB. Problem 5. Given that 21 regular pentagons P , 1 P , …, 2 P are such that 21 for any k∈{1, 2, 3, …, 20}, all the vertices of Pk+1 are the midpoints of the sides of P . Let S be the set of the _k vertices of P1,P2 ,...,P21. Determine the largest positive integer n for which there always exist four points A, B, C, D from S such that they are the vertices of an isosceles trapezoid and with the same color if we use n kinds of different colors to paint the element of S.

Vector Geometry

(continued from page 2)

Similarly, . 2 | | ] [ 2 4 6 2 4 4 6 6 2 A A A A A A A A A = × + × + × So [A1 A3 A ] = [5 A2 A4 A6].

Example 7. (1996 Balkan Math

Olympiad) Let ABCDE be a convex pentagon and let M, N, P, Q, R be the midpoints of sides AB, BC, CD, DE, EA, respectively. If the segments AP, BQ, CR, DM have a common point, show that this point also lies on EN.

Solution. Set the origin at the commom

point. Since, A, P and the origin are collinear, 0 = . 2 2 D A C A D C A P A = × + ×      + × = × So A×C = D×A. Similarly, B×D= B E× , C×E = A×C, D×A = B×D. Then .E×B = C×E So E×N = E×       + 2 C B

= 0, which implies E, N and the origin are collinear.

Example 8. (16th Austrian Math Olympiad) A line interesects the sides (or sides produced) BC, CA, AB of triangle ABC in the points A , ₁ B , ₁ C , respectively. ₁ The points A , 2 B , 2 C are symmetric 2 to A , ₁ B , ₁ C with respect to the ₁ midpoints of BC, CA, AB, respectively. Prove that A , ₂ B , ₂ C are collinear. ₂

Solution. Set the origin at a vertex,

say C. Then A₁=c₁B,B₁=c₂A ,C₁=A )

( 3 B A c −

+ for some constants c₁,c₂, 3

c . Since A , 1 B , 1 C , are collinear, 1 0 = (B1−A1)×(C1−A1) = (c₁−c₁c₂−c₁c₃+c₂c₃)A×B. Since , ) 1 ( 1 1 2 B A c B A = − = − A B₂= −B₁=(1−c₂)A and 2 C = (A + B) – C₁=c₃A + (1 –c )B, ₃ so A₂,B₂,C₂, are collinear if and only if 0 = (B₂−A₂)×(C₂−A₂)

= (c1−c1c2−c1c3+c2c3)A×B, which is true.