Olympiad Corner
XII Asia Pacific Math Olympiad, March 2000:
Time allowed: 4 Hours
Problem 1. Compute the sum
∑
=−
+
=
101 0 2 33
3
1
i i i ix
x
x
S
for101
i
x
i=
.Problem 2. Given the following
triangular arrangement of circles:
Each of the numbers 1,2,...,9 is to be written into one of these circles, so that each circle contains exactly one of these numbers and
(i) the sums of the four numbers on each side of the triangle are equal;
(continued on page 4)
When we do a geometry problem, we should first look at the given facts and the conclusion. If all these involve intersection points, midpoints, feet of perpendiculars, parallel lines, then there is a good chance we can solve the problem by coordinate geometry. However, if they involve two or more circles, angle bisectors and areas of triangles, then sometimes it is still possible to solve the problem by choosing a good place to put the origin and the x-axis. Below we will give some examples. It is important to stay away from messy computations !
Example 1. (1995 IMO) Let A, B, C
and D be four distinct points on a line, in that order. The circles with diameters AC and BD intersect at the points X and Y. The line XY meets BC at the point Z. Let P be a point on the line XY different from Z. The line CP intersects the circle with diameter AC at the points C and M, and the line BP intersects the circle with diameter BD at the points B and N. Prove that the lines AM, DN, and XY are concurrent.
Z
X
Y
P
A
B
C
D
M
Q Q'
N
(Remarks. Quite obvious we should set the origin at Z. Although the figure is not symmetric with respect to line XY, there are pairs such as M, N and A, D and B, C that are symmetric in roles! So we work on the left half of the figure, the computations will be similar for the right half.)
Solution. (Due to Mok Tze Tao,
1995 Hong Kong Team Member) Set the origin at Z and the x-axis on line AD. Let the coordinates of the circumcenters of triangles AMC and BND be (x1, 0) and (x2,
0), and the circumradii be r1 and r2,
respectively. Then the coordinates of A and C are (x1−r1, 0) and (x1+ r1, 0),
respectively. Let the coordinates of P be (0, y0). Since AM ⊥ CP and the slope of
CP is −y0/(x1+ r1), the equation of AM
works out to be (x1+ r1)x−y0y=x −12 r12. Let
Q be the intersection of AM with XY, then Q has coordinates (0 ,( )/ 0) 2 1 2 1 x y r − .
Similarly, let Q' be the intersection of DN with XY, then Q' has coordinates
) / ) ( , 0 ( 0 2 2 2 2 x y r − . Since 12 2 2 1 x ZX r − = ' 2 2 2 2 x ,soQ Q r − = = .
Example 2. (1998 APMO) Let ABC
be a triangle and D the foot of the altitude from A. Let E and F be on a line passing through D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D. Let M and N be the midpoints of the line segments BC and EF, respectively. Prove that AN is perpendicular to NM.
A
B
C
D
M
E
N
F
(Remarks. We can set the origin at D and the x-axis on line BC. Then computing the coordinates of E and F will be a bit messy. A better choice is to set the line through D,E.F horizontal.)
Coordinate Geometry
Coordinate Geometry
Coordinate Geometry
Coordinate Geometry
Kin Y. Li
Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei)
梁 達 榮 (LEUNG Tat-Wing), Appl. 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 Janet Wong, 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 10, 2000.
For individual subscription for the next five issues for the 00-01 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin Li Department of Mathematics Hong Kong University of Science and Technology
Clear Water Bay, Kowloon, Hong Kong Fax: 2358-1643
Email: [email protected]
Mathematical Excalibur, Vol. 5, No.3, May 00 – Sept 00 Page 4
Solution. (Due to Cheung Pok Man,
1998 Hong Kong Team Member) Set the origin at A and the x-axis parallel to line EF. Let the coordinates of D, E, F be (d, b), (e, b), (f, b), respectively. The case b=0 leads to D=E, which is not allowed. So we may assume b≠0. Since BE ⊥ AE and the slope of AE is b/e, so the equation of line BE works out to be ex+by=e2+b2. Similarly, the equations of lines CF and BC are fx+by=f2+b2 and dx+by=d2+b2, respectively. Solving the equations for BE and BC, we find B has coordinates (d+e, b-(de/b)). Similarly, C has coordinates (d+f, b−(df/b)). Then M has coordinates (d+(e+f)/2, b−(de+df)/(2b)) and N has coordinates ((e+f)/2, b). So the slope of AN is 2b/(e+f) and the slope of MN is −(e+f)/(2b). Therefore, AN ⊥ MN.
Example 3. (2000 IMO) Two circles Γ1 and Γ2 intersect at M and N. Let l be the common tangent to Γ1 and Γ2 so that M is closer to l than N is. Let l touch Γ1at A and Γ2 at B. Let the line through M parallel to l meet the circle Γ1 again at C and the circle Γ2 again at D. Lines CA and DB meet at E; lines AN and CD meet at P; lines BN and CD meet at Q. Show that EP=EQ. O1 O2 A B M C D N P Q E Γ Γ 1 2
(Remarks. Here if we set the x-axis on the line through the centers of the circles, then the equation of the line AB will be complicated. So it is better to have line AB on the x-axis.)
Solution. Set the origin at A and the
x-axis on line AB. Let B, M have coordinates (b,0), (s,t), respectively. Let the centers O1, O2 of Γ1, Γ2 be at (0, r1), (b,
r2), respectively. Then C, D have
coordinates (−s, t), (2b−s,t), respectively. Since AB, CD are parallel, CD=2b=2AB implies A, B are midpoints of CE, DE, respectively. So E is at (s, −t). We see EM ⊥ CD.
To get EP=EQ, it is now left to show M is the midpoint of segment PQ. Since O1
O2 ⊥ MN and the slope of O1 O2 is
(r2− r1)/b, the equation of line MN is
bx+(r2−r1)y=bs+(r2−r1)t. (This line
should pass through the midpoint of segment AB.) Since O2M=r2 and O1M=r1,
we get ) ( ) ( 2 2 2 2 2 r t r s b− + − = and 2 1 2 1 2 ) (r t r s + − = .
Subtracting these equations, we get b2/2=bs+(r2− r1)t, which implies (b/2, 0)
is on line MN. Since PQ, AB are parallel and line MN intersects AB at its midpoint, then M must be the midpoint of segment PQ. Together with EM ⊥ PQ, we get EP=EQ.
Example 4. (2000 APMO) Let ABC be a triangle. Let M and N be the points in which the median and the angle bisector, respectively, at A meet the side BC. Let Q and P be the points in which the perpendicular at N to NA meets MA and BA, respectively, and O the point in which the perpendicular at P to BA meets AN produced. Prove that QO is perpendicular to BC.
B
A
N
M
P
O
Q
(Remarks. Here the equation of the angle bisector is a bit tricky to obtain unless it is the x-axis. In that case, the two sides of the angle is symmetric with respect to the x-axis.)
Solution. (Due to Wong Chun Wai,
2000 Hong Kong Team Member) Set the origin at N and the x-axis on line NO. Let the equation of line AB be y=ax+b, then the equation of lines AC and PO are y=−ax−b and y=(−1/a)x+b, respectively. Let the equation of BC be y=cx. Then B has coordinates (b/(c−a), bc/(c−a)), C has coordinates (−b/(c+a), −bc(c+a)), M has coordinates (ab/(c2−a2), abc/(c2−a2)), A has coordinates (−b/a, 0), O has coordinates (ab, 0) and Q has coordinates (0, ab/c). Then BC has slope c and QO has slope −1/c. Therefore, QO ⊥ BC.
Example 5. (1998 IMO) In the convex
quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. Suppose that the point P, where the perpendicular
bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is a cyclic quadrilateral if and only if the triangles ABP and CDP have equal areas.
B
A
D
C
P
(Remarks. The area of a triangle can be computed by taking the half length of the cross product. A natural candidate for the origin is P and having the diagonals parallel to the axes will be helpful.)
Solution. (Due to Leung Wing Chung,
1998 Hong Kong Team Member) Set the origin at P and the x-axis parallel to line AC. Then the equations of lines AC and BD are y=p and x=q, respectively. Let AP=BP=r and CP=DP=s. Then the coordinates of A, B, C, D are ) , (− r −2 p2 p , (q, r2−q2),( s −2 p2,p), ), ,
(q− s2−q2 respectively. Using the
determinant formula for finding the area of a triangle, we see that the areas of triangles ABP and CDP are equal if and only if . 2 2 2 2 2 2 2 2 pq q s p s pq q r p r − − − =− − − − − Since f(x) = − x2−p2 x2−q2−pq is
strictly decreasing when x ≥ |p| and |q|, equality of areas hold if and only if r=s, which is equivalent to A, B, C, D concyclic (since P being on the perpendicular bisectors of AB, CD is the only possible place for the center)
.
After seeing these examples, we would like to remind the readers that there are pure geometric proofs to each of the problems. For examples (1) and (3), there are proofs that only take a few lines. We encourage the readers to discover these simple proofs.
Although in the opinions of many people, a pure geometric proof is better and more beautiful than a coordinate geometric proof, we should point out that sometimes the coordinate geometric proofs may be preferred when there are many cases. For example (2), the different possible orderings of the points D, E, F on the line can all happen as some pictures will show. The coordinate geometric proofs above cover all cases.
Mathematical Excalibur, Vol. 5, No.3, May 00 – Sept 00 Page 4
Problem Corner
We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver’s name, home address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is October 10, 2000.Problem 106. Find all positive integer
ordered pairs (a,b) such that gcd(a,b)+lcm(a,b)=a+b+6, where gcd stands for greastest common divisor (or highest common factor) and lcm stands for least common multiple.
Problem 107. For a, b, c > 0, if abc=1,
then show that
. 3 + + + ≥ + + + + + c b a c b a b a c a c b
Problem 108. Circles C1 and C2 with
centers O1 and O2 (respectively) meet at
points A, B. The radii O1B and O2B
intersect C1 and C2 at F and E. The line
parallel to EF through B meets C1 and C2
at M and N, respectively. Prove that MN=AE+AF. (Source: 17th Iranian Mathematical Olympiad)
Problem 109. Show that there exists an
increasing sequence a1, a2, a3, ... of
positive integers such that for every nonnegative integer k,the sequence k+a1,
k+a2, k+a3 ... contains only finitely many
prime numbers. (Source: 1997 Math Olympiad of Czech and Slovak Republics)
Problem 110. In a park, 10000 trees
have been placed in a square lattice. Determine the maximum number of trees that can be cut down so that from any stump, you cannot see any other stump. (Assume the trees have negligible radius compared to the distance between adjacent trees.) (Source: 1997 German Mathematical Olympiad)
Comments. You may think of the trees being placed at (x,y), where x, y = 0, 1, 2, ..., 99.
*****************
Solutions
*****************
Problem 101. A triple of numbers (a1,
a2, a3)=(3, 4, 12) is given. We now
perform the following operation: choose two numbers ai and aj, (i ≠ j), and
exchange them by 0.6ai−0.8aj and
0.8ai+0.6aj. Is it possible to obtain after
several steps the (unordered) triple (2, 8, 10) ? (Source: 1999 National Math Competition in Croatia)
Solution. FAN Wai Tong (St. Mark's
School, Form 7), KO Man Ho (Wah Yan College, Kowloon, Form 6) and LAW
Hiu Fai (Wah Yan College, Kowloon,
Form 6).
Since (0.6ai−0.8aj)2 + (0.8ai+0.6aj)2
= a +i2 a2j, the sum of the squares of the
triple of numbers before and after an operation stays the same. Since 32 + 42 + 122 ≠ 22 + 82 + 102, so (2,8,10) cannot be obtained.
Problem 102. Let a be a positive real
number and (xn)n≥1 be a sequence of real
numbers such that x1=a and
. 1 all for , ) 2 ( 1 1 1≥ + −
∑
≥ − = + n x kx n x n k k n nShow that there exists a positive integer n such that xn > 1999! (Source: 1999
Romanian Third Selection Examination)
Solution. FAN Wai Tong (St. Mark's
School, Form 7).
We will prove by induction that xj+1≥ 3xj for
every positive integer j. The case j=1 is true by the given inequality. Assume the cases j =1, ..., n−1 are true. Then xn ≥ 3xn−1 ≥ 9xn−2
≥ ... and
∑
− = + ≥ + − 1 1 1 ( 2) n k n k n n x kx n x x∑
− = − − − + ≥ 1 13 1 ) 2 ( n k n k n n ...) 9 1 3 1 )( 1 ( ) 2 ( + − − + + ≥ n n = 2 5 + n ≥ 3.So the case j = n is also true. Since a > 0, we can take
n > 1+ log3 (1999!/a).
Then xn ≥ 3n−1x1= 3n−1a >1999!.
Problem 103. Two circles intersect in
points A and B. A line l that contains the point A intersects again the circles in the points C, D, respectively. Let M, N be the midpoints of the arcs BC and BD, which do not contain the point A, and let K be the midpoint of the segment CD. Show that ∠ MKN=90°. (Source: 1999 Romanian Fourth Selection Examination)
α
B
A
C
K
D
M
N
M'
N'
Solution. FAN Wai Tong (St. Mark's
School, Form 7)
Let M' and N' be the midpoints of chords BC and BD respectively. From the midpoint theorem, we see that BM'KN' is a parallelogram. Now
.
90
90
M
M
K
B
M
K
B
N
K
N
N
K
′
∠
=
+
′
∠
=
+
′
∠
=
′
∠
o oLet α = ∠ NDB =∠ NAB. Then
α α tan 2 1 2 1 tan BD BC D N B M N N N K = ′ ′ = ′ ′ . Now ∠MCB=∠MCB= ∠CAB 2 1 ) 180 ( 2 1 DAB ∠ − = o = o−∠NAB 90 = o−α 90 .
Mathematical Excalibur, Vol. 5, No.3, May 00 – Sept 00 Page 4 So BD BC N B M C K M M M 2 1 cot 2 1 cotα α = ′ ′ = ′ ′ . Then KN′/N′N =MM′/M′K. So triangles N N K K M
M ′ , ′ are similar. Then ∠M ′KM NK N′ ∠ = and KN N KM M N K M MKN=∠ ′ ′−∠ ′ −∠ ′ ∠ =∠KN′D−(∠N′NK+∠N′KN) =90o
Other commended solvers: WONG
Chun Wai (Choi Hung Estate Catholic
Secondary School, Form 7).
Problem 104. Find all positive integers
n such that 2n−1 is a multiple of 3 and (2n−1)/3 is a divisor of 4m2+1 for some integer m. (Source: 1999 Korean Mathematical Olympiad)
Solution. (Official Solution)
(Some checkings should suggest n is a power of 2.) Now 2n−1 is a multiple of 3 if and only if (−1)n≡2n≡(mod 3), that is n is even. Suppose for some even n, (2n−1)/3 is a divisor of 4m2+1 for some m. Assume n has an odd prime divisor d. Now 2d−1≡3 (mod 4) implies one of its prime divisor p is of the form 4k+3. Then p divides 2d−1, which divides 2n−1, which divides 4m2+1. Then p and 2m are relatively prime and so
1≡(2m)p−1=(4m2)2k+1≡−1 (mod p), a contradiction. So n cannot have any odd prime divisor. Hence n=2j for some positive integer j.
Conversely, suppose n =2j. Let Fi=
i
2 2 + 1. Using the factorization 22b−1= (2b−1)× (2b+1) repeatedly on the numerator, we get 1 2 1 3 1 2 − = − j n F F F L .
Since Fi divides Fj−2 for i < j, the Fi’s
are pairwise relatively prime. By the Chinese remainder theorem, there is a positive integer x satisfying the simultaneous equations x ≡ 0 (mod 2) and
1
2
2
−≡
ix
(mod Fi) for i=1, 2, …, j−1.Then x=2m for some positive integer m and 4m2+1= x2+1≡ 0 (mod Fi) for i=1,
2,…j−1. So 4m2+1 is divisible by F1F2…Fj-1=(2n−1)/3.
Problem 105. A rectangular
parallelepiped (box) is given, such that its intersection with a plane is a regular hexagon. Prove that the rectangular parallelepiped is a cube. (Source: 1999 National Math Olympiad in Slovenia)
Solution. (Official Solution)
D'
A'
B'
C'
Α
Β
C
D
X
Z
Y
O
N
P
K
M
L
As in the figure, an equilateral triangle XYZ is formed by extending three alternate sides of the regular hexagon. The right triangles XBZ and YBZ are congruent as they have a common side BZ and the hypotenuses have equal length. So BX=BY and similarly BX=BZ. As the pyramids XBYZ and OB′ NZ are similar and ON XY 3 1 = , it follows B' Z BZ 3 1 = . Thus we have BB BZ 3 2 =′ and similarly AB BX 3 2 = and CB BY 3 2 = .
Since BX=BY=BZ, we get AB=BC=BB′. Other commended solvers: FAN Wai
Tong (St. Mark’s School, Form 7).
Olympiad Corner
(continued from page 1) (ii) the sums of the squares of the four numbers on each side of the triangle are equal.
Find all ways in which this can be done.
Problem 3. Let ABC be a triangle. Let
M and N be the points in which the median and the angle bisector, respectively, at A meet the side BC. Let Q and P be the points in which the perpendicular at N to NA meets MA and BA, respectively, and O the point in which the perpendicular at P to BA meets AN produced. Proved that QO is perpendicular to BC.
Problem 4. Let n, k be given positive
integers with n > k. Prove that
⋅ − < − < − ⋅ + − k n−k n k n k n k n k n k n k n k n k n n !( )! ( ) ! ) ( 1 1
Problem 5. Given a permutation (a0, a1,
..., an) of the sequence 0,1,..., n. A
transposition of ai with aj is called legal
if i > 0, ai = 0 and ai−1 + 1 = aj. The
permutation (a0, a1, ..., an) is called
regular if after a number of legal transpositions it becomes (1,2, ...,n,0). For which numbers n is the permutation (1, n, n−1, ..., 3, 2, 0) regular ?
2000 APMO and IMO
In April this year, Hong Kong IMO trainees participated in the XII Asia Pacific Mathematical Olympiad. The winners were
Gold Award
Fan Wai Tong (Form 7, St Mark’s School)
Silver Award
Wong Chun Wai (Form 7, Choi Hung Estate Catholic Secondary School) Chao Khek Lun (Form 5, St. Paul’s College)
Bronze Award
Law Ka Ho (Form 7, Queen Elizabeth School)
Ng Ka Chun (Form 5, Queen Elizabeth School)
Yu Hok Pun (Form 4, SKH Bishop Baker Secondary School)
Chan Kin Hang (Form 6, Bishop Hall Jubilee School)
Honorable Mention
Ng Ka Wing (Form 7, STFA Leung Kau Kui College)
Chau Suk Ling (Form 5, Queen Elizabeth School)
Choy Ting Pong (Form 7, Ming Kei College)
Based on the APMO and previous test results, the following trainees were selected to be the Hong Kong team members to the 2000 International Mathematical Olympiad, which was held in July in South Korea.
Wong Chun Wai (Form 7, Choi Hung Estate Catholic Secondary School) Ng Ka Wing (Form 7, STFA Leung Kau Kui College)
Law Ka Ho (Form 7, Queen Elizabeth School)
Chan Kin Hang (Form 6, Bishop Hall Jubilee School)
Yu Hok Pun (Form 4, SKH Bishop Baker Secondary School)
Fan Wai Tong (Form 7, St. Mark’s School)
