fternoon session ( 1 hours) 5. If we select at random three points on a given circle, nd the probability that these three points lie on a semicircle.

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1 The 8th Korean Mathematical Olympiad [ First round ] Morning session ( 1 hours) 1. Consider a nitely many points in a plane such that, if we choose any three points A; B; C among them, the area of 4ABC is always less than 1. Show that all of these nitely many points lie within the interior or on the boundary of a triangle with area less than 4.. For a given positive integer m, nd all pairs (n; x; y) of positive integers such that m; n are relatively prime and (x + y ) m =(xy) n, where n; x; y can be represented by functions of m. 3. Let A; B; C be three points lying on a circle, and let P; Q; R be midpoints of arcs BC; CA; AB; respectively. AP; BQ; CR intersect BC; CA; AB at L; M; N, respectively. Show that For which triangle ABC does equality hold? AL PL + BM QM + CN RN 9: 4. A partition of a positive integer n is a sequence ( 1 ; ; ; k ) of positive integers such that k = n and 1 k 1. Each i is called a summand. For example, (4; 3; 1) is a partition of 8 whose summands are distinct. Show that, for a positive integer m with n> 1 m(m+ 1), the number of all partitions of n into distinct m summands is equal to the number of all partitions of n, 1 m(m +1)into r summands (r m). 1

2 fternoon session ( 1 hours) 5. If we select at random three points on a given circle, nd the probability that these three points lie on a semicircle. 6. Show that any positive integer n(> 1) can be expressed by a nite sum of numbers satisfying the following conditions: (i) they do not have factors exept or 3. (ii) any two of them are neither a factor nor a multiple each other. That is, n = P N i=1 i 3 i, where i ; i (i=1;; ;N) nonnegativeintegers and ( i, j )( i, j ) < 0 whenever i 6= j. 7. Find all real valued functions f dened on real numbers except 0 such that 1 x f(,x)+f(1 )=x; x 6= 0: x 8. Two circles O 1 ;O of radii r 1 ;r (r 1 <r ), respectively, intersect at two points A and B. P is any point on a circle O 1. Lines PA;PB and a circle O intersect at Q and R, respectively. (1) Express y = QR in terms of r 1 ;r, and = \AP B. () Show that y =r is a necessary and sucient condition that a circle O 1 is orthogonal to a circle O.

3 [ Final round ] First day session (4 1 hours) 1. For any positive integer m, show that there exist integers a; b satisfying jaj m; jbj m; 0 <a+b p 1+p m+ :. Let A be the set of all non-negative integers. Find all functions f : A! A satisfying the following two conditions: (i) for any m; n A; f(m + n )=ff(m)g +ff(n)g (ii) for any m; n A with m n; f(m ) f(n ): 3. Let 4ABC be an equilateral triangle of side length 1;D a point on BC, and let r 1 ;r, be inradii of triangles ABD; ADC, respectively. Express r 1 r in terms of p = BD, and nd the maximum of r 1 r. 3

4 Second day session (4 1 hours) 4. Let O and R be the circumcenter and the circumradius of 4ABC, respectively, and let P be any point onthe plane ABC. Let perpendiculars PA 1 ;PB 1 ;PC 1, be dropped to the three sides BC; CA; AB. Express (4A 1B 1 C 1 ) (4ABC) of 4ABC. in terms of R and d = OP, where (4ABC) is the area 5. Let p be a prime number such that (i) p is the greatest common divisor of a and b; (ii) p is a divisor of a. Prove that the polynomial x n+ + ax n+1 + bx n + a + b cannot be decomposed into the product of two polynomials with integeral coecients, whose degrees are greater tham one. 6. Let m; n be positive integers with 1 n m, 1. Aboxislocked with several padlocks, all of which must be opened to open the box, and all of which have dierent keys. m people each have keys to some of the locks. No n people of them can open the box but any n + 1 people can open the box. Find the smallest number l of locks and then the numberofkeys for which this is possible. 4

5 The 8th Korean Mathemtical Olympiad s [First round] 1. Consider a nitely many points in a plane such that, if we choose any three points A; B; C among them, the area of 4ABC is always less than 1. Show that all of these nitely many points lie within the interior or on the boundary of a triangle with area less than 4. We can take 4ABC having the maximum area among triangles whose vertices are chosen from the given nitely many points. Then (4ABC) 1. Here (4ABC) denotes the area of 4ABC. Let 4LMN be the triangle whose medial triangle is 4ABC. Then (4LMN)=4(4ABC) 4: We may prove that the given nitely many points lie within the interior or on the boundary of 4LMN. Suppose a point P lies on the exterior of 4LMN. We may assume that a point P locates at the opposite side of N with respect to a line ML. Then the length of a perpendicular of P dropping to AB is larger than that of C dropping to AB. It follows that (4PAB) > (4CAB)=(4ABC) and P is never included in the given nitely many points. This completes the proof.. For a given positive integer m, nd all pairs (n; x; y) of positive integers such that m; n are relatively prime and (x + y ) m =(xy) n, where n; x; y can be represented by functions of m. If (n; x; y) is a solution of the given equation, then from x + y xy we have (xy) n = (x + y ) m > (xy) n. It follows that n>m. Let p be a common prime divisor of x and y, and let p a kx; p b ky. Here p a kx means that p a jx but p a+1 - y. Then p (a+b)n k(xy) n = (x + y ) m. Suppose b > a. Since p a kx ;p b ky, we see that p a kx + y and p am k(x + y ) m. It follows that am =(a+b)n>an and m>n. This is a contradiction. Similarly, a > b gives a contradiction. It concludes that a = b and x = y. Now we have x n = (x ) m = m x n and x (n,m) = m. Thus, x = t for some integer t. From 5

6 (n,m)t = m wehavet(n,m)=m. Since nt =(t+1)mand m; n are relatively prime, we obtain m =t; n =t+1. Answer. (n; x; y) =(m+1; m= ; m= ) (m: even) 3. Let A; B; C be three points lying on a circle, and let P; Q; R be midpoints of arcs BC; CA; AB; respectively. AP; BQ; CR intersect BC; CA; AB at L; M; N, respectively. Show that For which triangle ABC does equality hold? Since P is a midpoint of arc BC, wehave \BAP = \CAP = 1 \A. It follows that BL : CL = c : b, and BL = ca ba ; CL =. b+c b+c Now wehave AL PL + BM QM + CN RN 9: (1) AL PL = BL CL = a bc (b + c) Since (4ABL)+(4ALC) =(4ABC), we obtain 1 cal sin A = 1 bal sin A = 1 bc sin A and AL = bc sin A b + c sin A = bc b + c cos A : Form (1) we have Similarly, Finally, a bc PL = (b + c) b + c bc 1 a = cos A (b + c) 1 and cos A AL PL = 4bc A a cos = bc (1 + cos A) =bc 1+ b +c,a a a bc = = BM c + a QM =, 1 and b AL PL + BM QM + CN b + c RN = a b a, a c + b b, b c CN a + b RN =, 1: c c + a a + b + + b c b + c =, 1: a, 3 a + c a, c 1 +abc a b c 3 + a c a, c b a, a + c b b, b c n 1 +(ab + bc + ca) a, b b, c c a, 1 o +99; 6

7 equality holds for a = b = c. 4. A partition of a positive integer n is a sequence ( 1 ; ; ; k ) of positive integers such that k = n and 1 k 1. Each i is called a summand. For example, (4; 3; 1) is a partition of 8 whose summands are distinct. Show that, for a positive integer m with n> 1 m(m+ 1), the number of all partitions of n into distinct m summands is equal to the number of all partitions of n, 1 m(m +1)into r summands (r m). Since a partition of n into distinct m summands is of the form m = n; 1 > > > m 1, we have m 1; m,1 ; m, 3; ; 1 m. It follows that n = m m = m(m+1)=. This partition can be represented geometrically. Consider the array of points having 1 points in the top row, in next row, and so on down to m in the bottom row. 1 more than m points more than m, 1 points (1) m,1 more than points m more than 1 point In the rst row, we divide 1 points into m points and 1, m points, and in the second row, we divide points into m, 1 points and, (m, 1) points, etc. 1 : m points ( 1,m) points : m, 1 points (, (m, 1)) points () m,1 : points ( m,1, ) points m : 1 points ( m, 1) points total: m(m +1) points total: n, 1 m(m + 1) points P P m Since i, (m +1,i)0 and f n i=1 i, (m +1,i)g= i=1 i, m(m+1) = n, m(m+1), the right part of the array of points represents a partition of n, 1 m(m +1)into r summands (r n). This completes the proof. 7

8 5. If we select at random three points on a given circle, nd the probability that these three points lie on a semicircle. Let A; B; C be selected points on a given circle, and let \AOB = x; \BOC = y;\coa = z. Then (1) x + y + z = x0; y0; z0: Three points A; B; C lie on a semicircle if and only if one of x + y z; y+z x or z + x y holds, that is, () x ; y or z : In space, the region of (1) is the union of the interior and the boundary of 4ABC and the region of () is the union of the interiors and the boundaries of 4A 0 B 0 C; 4AB 0 C 0 and 4BC 0 A 0. It follows that the required probability is 3 4. Answer cf. The required probability is equal to the probability that 4ABC becomes obtuse. 6. Show that any positive integer n(> 1) can be expressed by a nite sum of numbers satisfying the following conditions: (i) they do not have factors exept or 3. (ii) any two of them are neither a factor nor a multiple each other. That is, n = P N i=1 i 3 i, where i ; i (i=1;; ;N) nonnegativeintegers and ( i, j )( i, j ) < 0 whenever i 6= j. Note that = 1 ; 3=3 1 ;4= ;5= ;6= ;7= +3 1 ;8= 3 ;9=3 ;10= , etc. So if n 10, then the required representation is possible. We use induction on n. Suppose that, for every m( n), the required representation of m is possible. i) n is odd; 8

9 Since n +1 is even and n+1 <n,wehave n+1 = P N i=1 i 3 i, where (i, j )( i, j ) < 0 whenever i 6= j. Then n+1 = P N i=1 i+1 3 i and f( i +1),( j +1)g( i, j )=( i, j )( i, j ) < 0 whenever i 6= j. ii) n is even ; If n +1=3 k,wehave done. If n +16=3 k,we can take a positive integer k such that 3 k <n+1<3 k+1. Since 3 k is odd, n+1,3 k is even, and (n +1,3 n )=<n,wehave(n+1,3 k )== P N i=1 i 3 i, where ( i, j )( i, j ) < 0 whenever i 6= j. Then n +1=3 k + P N i=1 i+1 3 i. Since i +1>0; i <k (i=1;; ;N), we have3 k and i+1 3 i (i =1;; ;N) are neither a factor nor a multiple each other. From i) and ii) we know that the required representation of n + 1 is possible. This completes the proof. 7. Find all real valued functions f dened on real numbers except 0 such that Letting x =,y in the given equation, we have 1 x f(,x)+f(1 )=x; x 6= 0: x (1), 1 y f(y)+f(,1 y )=,y and letting x =1=y, wehave () yf(, 1 y )+f(y)= 1 y : Eliminating f(, 1 ) in (1) and (), we have y Conversely, iff(x)= 1 (x +1=x), then f(y)= 1 (y + 1 y ): 1 x f(,x)+f(1 x )= 1 +1 x,x3 +,x Answer. f(x) = 1 (x +1=x); x 6= 0: 1 +1 x 3 =x: 1 x 8. Two circles O 1 ;O of radii r 1 ;r (r 1 <r ), respectively, intersect at two points A and B. P is any point on a circle O 1. Lines PA;PB and a circle O intersect at Q and R, respectively. (1) Express y = QR in terms of r 1 ;r, and = \AP B. () Show that y =r is a necessary and sucient condition that a circle O 1 is orthogonal to a circle O. 9

10 (1) Case I. P lies in the exterior of a circle O ; Let \PAB = and AB = a. Then a =r 1 sin, PR PA = PQ PB = QR AB = y a ; PB sin = a sin = BQ sin =r ; PA sin( + ) =r 1: and In 4PBQ,wehave It follows that BQ = PB +PQ,PBPQcos(\BPQ) =PB h 1+ y a,y a cos(\bpq) i : 4r sin =4r1sin [f y a, cos(\bpq)g + sin (\BPQ)] and q y = a cos + a n q r n, r 1 sin = sin r 1 cos + r, r 1 sin Case II. P lies in the interior of a circle O ; Similarly, n q o y = sin r 1 cos(, )+ r,r 1 sin ; o : where \BPQ =, \AP B =, : () If a circle O 1 is orthogonal to a circle O, then r = r 1 tan (Case I); r = r 1 tan(, ) (Case II). It follows that y =r. Conversely, ify=r, then r = r 1 tan (Case I); r = r 1 tan(, ) (Case II). This completes the proof. Answer of (1); q y = sin fr 1 cos + r, r 1 sin g (P lies in the exterior of a circle O ); q y = sin fr 1 cos(, )+ r,r 1 sin g (P lies in the interior of a circle O ): 10

11 The 8th Korean Mathemtical Olympiad s [ Final round ] 1. For any positive integer m, show that there exist integers a; b satisfying jaj m; jbj m; 0 <a+b p 1+p m+ : Let f(x; y) = x + y p and S = ff(a; b)ja; b integers with 0 a m and 0 b mg. If f(a; b) = f(a 0 ;b 0 ), then a + b p = a 0 + b 0p and a = a 0 ;b = b 0. It follows that the number of elements in S is (m +1). The maximum of S is m + m p =(1+ p )m, and so if f(a; b) S, then 0 f(a; b) (1 + p )m. Devide the interval [0; (1 + p )m] into equal m +m small intervals. p p Then the lengths of small intervals are (1+ )m. Since the number of elements in S is = 1+ m +m m+ (m +1) =(m +m)+1, by peogion hole principle there exist f(a 1 ;b 1 );f(a ;b ) S such that f(a 1 ;b 1 );f(a ;b ) lie on the same small interval. We may assume that f(a 1 ;b 1 ) >f(a ;b ). Then p 0 <f(a 1 ;b 1 ),f(a ;b )< 1+. m+ Let a = a 1, a and b = b 1, b. Then 0 <f(a; b) 1+ p m+ f(a 1 ;b 1 ),f(a ;b )=f(a 1,a ;b 1,b )=f(a; b), and jaj m; jbj m: This completes the proof.. Let A be the set of all non-negative integers. Find all functions f : A! A satisfying the following two conditions: (i) for any m; n A; f(m + n )=ff(m)g +ff(n)g (ii) for any m; n A with m n; f(m ) f(n ): From (i) we have f(m )=ff(m)g +ff(0)g f(n )=ff(n)g +ff(0)g 11

12 and substrating two equations, we obtain ff(m)g,ff(n)g =ff(m ),f(n )g: It follows that if m n, then by (ii) we see that (1) f(m) f(n): Meauwhile, if m = n = 0, then we havef(0) = ff(0)g and, f(0) = 0 or f(0)=1. Case I. f(0) = 1. From f(m )=ff(m)g +1,wehave () f( n )= 1 )g +1]: [ff(n,1 From f(1) = ff(1)g +1wehaveff(1), 1g = 0 and f(1)=1. From () we have f() = 1; f( )=1; ; etc. In general, f() = f( )==f( n )=1 (n0). (3) For any positive integer m, there exists an integer n such that n,1 m< n. From (1) we have1=f( n,1 ) f(m) f( n ) = 1 and nally f(m) =1. Case II. f(0) = 0. From f(m )=ff(m)g wehave f(m ) Meanwhile, f() = ff(1)g. It follows that f( n ) n f( n,1 ) (4) = o = n f( n, ) = f f(m) g : o From f(1) = ff(1)g wehavef(1) = 0 or f(1)=. (i) f(1) = 0. = = From (4) we havef( n )=0 (n0). Similarly, f(n) =0 (n0). (ii) f(1) =. From (4) we havef( n )= n n f() o n = ff(1)gn+1 n (n0). By (3) we havef(m)iseven. ff(m +1)g = f((m +1) )f(m +1)=ff(m)g +ff(1)g > ff(m)g implies that f(m +1)>f(m): P It follows that f(m+1),f(m), 0, and n,1 ff(m+1),f(m),g = ),f(0), m=0 f(n n =0. Now wehavef(m+1),f(m), = 0 for all m =0;1; n,1. Since n is arbitrary, we obtain 1

13 f(m +1) = f(m)+ for all m =0;1;;. It follows that f(m) =m. Answer. 1; f(n) 0; f(n) n: f(n) 3. Let 4ABC be an equilateral triangle of side length 1;D a point on BC, and let r 1 ;r, be inradii of triangles ABD; ADC, respectively. Express r 1 r in terms of p = BD, and nd the maximum of r 1 r. Let l = AD. In 4ABD, wehave l =1+p,pcos 60 = p, p +1. Since the area of 4ABC is p 3=4, in 4ABD and 4ADC, we obtain Now wehave r 1 1+p+l = p 3 4 p; r, p + l = p 3 (1, p): 4 r 1 r = 3 p(1, p) 4(1 + p + l)(, p + l) = (1 + p)(, p)+l +3l = 3 p(1, p) 4, p, p + p, p +1+3l = 1 p(1, p) 4 1+l = 1 p(1, p)(1, l) = 1 p(1, p)(1, l) 4 4(1, l ) 4 1, p + p, 1 = 1 (1, l) 4 = p 1, p, p +1 p 3 1, = 1 4 =,p 3 : 8 1, p (p,1=) +3=4 Answer. r 1 r = 1 4 (1, p p, p +1); the maximum of r 1 r is,p 3 8 when p = Let O and R be the circumcenter and the circumradius of 4ABC, respectively, and let P be any point on the plane ABC. Let perpenticulars PA 1 ;PB 1 ;PC 1, be dropped to the three sides BC; CA; AB. Express (4A 1B 1 C 1 ) (4ABC) of 4ABC. in terms of R and d = OP, where (4ABC) is the area A quadrilateral PA 1 CB 1 is inscribed in a circle of a diameter CP. It follows that A 1 B 1 = PC sin C. Similarly, in a quadrilateral PB 1 AC 1 wehaveb 1 C 1 =AP sin A. Let D be the intersection point between a line CP and the circumcircle of 4ABC. 13

14 Since \PB 1 C 1 =\PCA 1 =\DAB and \PB 1 C 1 =\PAC,we see that \A 1 B 1 C 1 = \PB 1 A 1 +\PB 1 C 1 =\DAB + \PAC 1 =\PAD. From 4PAD wehaveap =sin B = DP =sin(\pad)=dp =sin \A 1 B 1 C 1 because \PDA =\B. It follow that AP sin(\a 1 B 1 C 1 )=DP sin B. Now wehave (4A 1 B 1 C 1 ) = 1 A 1B 1 B 1 C 1 sin(\a 1 B 1 C 1 ) = 1 (PC sin C) (AP sin A) sin(\a 1B 1 C 1 ) = 1 PC sin C sin A DP sin B = 1 PC DP sin A sin B sin C: Since (4ABC) =R sin A sin B sin C, we obtain (4A 1 B 1 C 1 ) (4ABC) = 1 4R PC DP: If P lies in the interior of a circle O, then PC DP = R, d, and if P lies in the exterior of a circle O, then PC DP = d, R. Finally, ifp lies on the circle O, then A 1 ;B 1 ;C 1 are collinear (Simson's line) and (4A 1 B 1 C 1 )=0. In summary, we obtain (4A 1 B 1 C 1 ) (4ABC) = jr, d j 4R : 5. Let p be a prime number such that (i) p is the greatest common divisor of a and b; (ii) p is a divisor of a. Prove that the polynomial x n+ + ax n+1 + bx n + a + b cannot be decomposed into the product of two polynomials with integeral coecients, whose degrees are greater tham one. (Eisenstein Theorem) Let f(x) =c 0 x n +c 1 x n,1 ++c n,1 x+c n be a polynomial with integral coecients. If there exists a prime number p so that p - c 0 ;pjc i (i=1;; ;n), and p - c n, then f(x) cannot be decomposed into the product of two polynomials with of lower degrees with integral coecients. (Proof) Let X n+1 X n+1 a i x i b j x j = c 0 x n + c 1 x n,1 + +c n,1 x+c n : i=0 j=0 From a 0 b 0 = c n ;p - c n wehavepja 0 ;p- b 0 or p - a 0 ;pjb 0. 14

15 Without loss of generality, wemay assume that pja 0 ; p - b 0. Comparing the coecients of x, we have a 0 b 1 + a 1 b 0 = c 1. It follows that pja 1 b 0 and pja 1. Suppose pja 0 ;pja 1 ; ;pja k,1. Then comparing the coecients of x k, we have pj P k i=0 a ib k,i and pja k b 0. It follows that pja k. Now we have pja i (i = 0; 1; ; ;n). If deg( P n+1 i=0 a i) = s, then deg( P n+1 i=0 b ix i )=n,s. Now wehavea s b n,s =c 0 and pjc 0. This is a contradiction, completing proof. holds. Since p - 1;pja; pjb; p - (a + b), by Eisenstein Theorem we see that the required proposition 6. Let m; n be positive integers with 1 n m, 1. Aboxislocked with several padlocks, all of which must be opened to open the box, and all of which have dierent keys. m people each have keys to some of the locks. No n people of them can open the box but any n + 1 people can open the box. Find the smallest number l of locks and then the numberofkeys for which this is possible. Let m people be denoted by A 1 ;A ; ;A m and write on each lock the symbols of people who do not hold a key for that lock. Since any n + 1 people can open the box, no lock has over n +1 symbols on it. For any n people, there is at least one lock with the corresponding n symbols. Thus the smallest number l of locks is m C n and any people must have the same number of keys (k = m,1 C n ). 15

IMO Training Camp Mock Olympiad #2

IMO Training Camp Mock Olympiad #2 July 3, 2008 1. Given an isosceles triangle ABC with AB = AC. The midpoint of side BC is denoted by M. Let X be a variable point on the shorter arc MA of the circumcircle