SOLUTION THE ASSOCIATION OF MATHEMATICS TEACHERS OF INDIA BHASKARA CONTEST - FINAL - JUNIOR CLASS - IX & X

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Kathryn Shields
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1 Instructions:. Answer as many questions as possile.. Elegant and novel solutions will get extra credits.. Diagrams and explanations should e given wherever necessary. 4. Fill in FACE SLIP and your rough working should e in the answer ook. 5. Maximum time allowed is THREE hours. 6. All questions carry equal marks.. (a) Find all prime numers p such that 4p and 6p are also primes. () Determine real numers x, y, z, u such that xyz xy yz zx x y z 7 yzu yz zu uy y z u 9 zux zu ux xz z u x 9 uxy ux xy yu u x y 9 Sol. (a) First consider primes P,, 5 P, 4p 7 is prime ut 6P 5 is not P, 4p 7 is prime ut 6P 55 is not P 5, 4p 0 is prime & 6P 5 is also prime let P e prime greater than 5. 4P 5P (P ) º (P ) (mod 5) 6P 5(P P ) (P ) (P ) º (P ) (P ) (mod 5) & (P ) (P) (P ) (P ) (P ) º 0 (mod 5) \ P(4P ) (6P ) º 0 (mod 5) then (4P ) (6P ) º 0 (mod 5) ecause P & 5 are coprime. Hence 4P & 6P is a compost numer. Because each is greater than 5 and one of them is divisile y 5. \ only solution, P 5 () Add oth sides in each equations. (x )(y )(z ) 8...() (y )(z )(u ) 0...() (z )(u )(y ) 0...() (u )(x )(y ) 0...(4) [(x )(y )(z )(u )] 8000 (0) (x )(y )(z )(u ) 0...(5) (5) () (x )(y )(z )(u ) (x )(y )(z ) u 5 u 5, \ u

2 (5) () y 0 0 (5) () y 0 0 Þ y (5) (4) z 0 0 Þ z Þ x Þ x \ x, y, z, u is,,,. If x, y, z, p, q, r are distinct real numers such that Sol. x p y p z p p x q y q z q q x r y r z r r find the numerical value of æ ç ö èp q r ø Let p, q, r are roots of cuic in t \ x t y t z t t Þ - x t y t t z t Þ y t x t z t-t (x t)(y t) t(z t) Þ (x y t)(t)(z t) z(x t)(y t) Þ (tx ty t )(z t) z(x t)(y t) Þ tzx tzy t z t x t y t z(xy xt yt t ) t t (z x y z) (zx zy zx zy)t zxy 0 Þ t (x y z)t xyz 0 This equation is cuic in t whose roots are p, q, r. æ ç ö è t ø (x y z) æ ç ö xyz 0 è t ø

3 This equation is cuic in t whose roots are,, p q r. Þ x y z t t xyz 0 xyzt (x y z)t 0 xyzt (x y z)t 0 sum of zeros, i.e. p q r 0. ADC and ABC are triangles such that AD DC and CA AB. If ÐC AB 0 and ÐADC 00, without using trigonometry, prove that AB BC CD. Sol. Since ÐADC 00 ÐCAB 0 ÐABC ÐACB 80 ÐADC ÐABC 80 So ABCD is a cyclic quadrilateral. Produce BC to E such that CE CD Since ÐDCB \ ÐDCE 60 \ DCDE is an equilateral D. In DADE, ÐADE AD DE \ ÐAED ÐDAE 0 So ÐAEB 50 In DABE Ð BAE 50 \ AB BE AB BC CE AB BC CD A (a) a,, c, d are positive real numers such that acd. Prove that a c cd da ³ 4 a c d 0 a D () In a scalene triangle ABC, ÐBAC 0. The isectors of the angles A, B ande C meet the opposite sides in P, Q and R respectively. Prove that the circle on QR as diameter passes through the point P. Sol. (a) acd a 80 B C E a c a c a c d

4 Þ a a c - c a a ac c cd Þ ( a) é ê ë ú ( c) é a a ac ê ë c cd ú Þ ³ ( a) é 4 ê ë a a ac ú ( c) é 4 ê ë c cd ú (Q y Titu's formula ( ) 4 ³ ³ a a ac a a ac a a ac ) é a c ë a a ac c cd ú é a a ac ë a a ac a a ac acd ú é a a ac ë a a ac a a ac ú é a a ac ë a a ac ú Þ ³ 4 () Extend BA to X and CA to Y So ÐCAX 60 and ÐBAY 60 In DABP, AC is the exterior angle isector and BQ is the interior angle isector so Q is the excenter of DABP So QP isects ÐAPC. (In a triangle two external angle isectors and one internal isector are always concurrent) Similarly PR isects ÐBPA \ x y 80 x y 90 ÐQPR 90 Circle on QR as diameter passes through the point P. B R Y X Q y x y x P C 5. (a) Prove that x 4 x 6x 9x can not e expressed as a product of two polynomials of degree with integer coefficients. () n segments are marked on a line. Each of these segments intersects at least n other segments. Prove that one of these segments intersects all other segments. Sol. (a) Let x 4 x 6x 9x (x ax ) (x cx d) where a,, c, d are integers. x 4 cx dx ax acx adx x cx d x 4 (c a)x ( d ac)x (ad c)x d 4

5 on comparing coefficients of oth equations, we get a c...() d ac 6...() ad c 9...() d...(4) from equation (4), since & d oth are integers, we can have following possiilities. d Case, d d ac 6 ac 6 ac 7 also a c x x 7 0 D 9 4() ( 7) 7 D ¹ perfect square, so no integers a and c. Similarly & d will also not satisfy. Case 6, d 6 ac 6 ac & a c x x 0 D ( ) 4() ( ) D ¹ perfect square, So no integer a & c. Similarly & d 6 will also not satisfy. Case 4, d 4 ac 6 ac & a c x x 0 D ( ) 4()( ) 9 4 D ¹ perfect square, so no integer a & c. Similarly, d 4 will also not satisfy. Case 4, d ac 6 ac 9 & a c x x 9 0 D ( ) 4()(9) < 0 No real solution & hence no integer a & c. Similarly & d will also not satisfy. Case 5 6, d 6 ac 6 ac 4 & a c x x 4 0 D ( ) 4()(4) < 0 No real solution & hence no integer a & c. Similarly & d 6 will also not satisfy. Case 6 4, d 4 ac 6 ac & a c x x 0 D ( ) 4()() < 0 No real solution & hence no integer a & c. Similarly & d 4 will also not satisfy. So the given equation cannot e expressed as a product of two polynomials of degree with integer coefficients. 5

6 () Let us define each line segment y interval of type [a, ]. We will e referring a as left end and as right end. Let us consider a set X union of n left most (y picking left ends). The line segments and Y union of n right most (y picking right ends) line segments. Now in X, right most left end will e left of the left most right end otherwise there will e a line in X which will have less than n intersections. This implies each segment in X pairwise intersecting with each other which corresponds to n intersections of each line segments with in these n line segments in X. Similarly in Y left most right end will e right of the right most left end otherwise there will e a line in Y which will have less than n intersections. This implies each segment in Y is also pairwise intersecting with each other corresponds to n intersections of each line segments with in these n line segments in Y. Now if there is a any common line segment etween X and Y then this line segment will intersect with all line segments of X as well as all line segments of Y. So this line segment intersect with all other line segments and we are done. If there is no common line segment etween X and Y, then X and Y will e disjoint sets having n line segments. There will e one more line segment (as there are total n line segments) this line segment must intersect with each line segment contained in X as we need n intersection on each line segment similarly this line segment must intersect with each line segment in Y. So this is the line segment which intersects with other line segments and we are done. 6. If a,, c, d are positive real numers such that a c d and a d ad c c, find the value of a cd ad c. Sol. Let a c d k, k > 0 Let a a'k ; a' > 0 'k ; ' > 0 c c'k ; c' > 0 d d'k ; d' > 0 a c d k \ a' ' and c' d' Let a' sina, ' cosa c' sin, d' cos a d 4d c c k (sin a cos sina sin) k (cos a sin cosasin) sin a cos a cos sin cosasin sinacos sin(a ) sin(a ) sin(a ) sin(a ) sin (a ) sin (a ) sin (a ) Now, we have to find, a cd ad c 6

7 sin acos a sincos sinacos cos asin sinacos a sincos (sinacos cosasin ) sina sin sin( a) sin( a).cos( a-) sin( a) cos (a ) -sin ( a-) Alternate solution Let us consider a quadrilateral of side length a,, c, d such that a c d & a d ad c c...(i) Hence in DPQR. Þ cos0 c -PR c c -PR Þ c Þ PR c c...(ii) & in DPRS, a d -PR cos 60.ad P PR a d ad...(iii) By (i), (ii) & (iii), we can say that our construction of quadrilateral is correct. Also, when ÐP 90 Þ SQ a ìwhich are equal as ü í ý & ÐR 90 Þ SQ c d îgiven in questions þ a S 60 Now [PQRS] [PQR] [PRS]. sin 0.c.sin 60. ad d R c 0 Q [PQRS] (ad c)...(iv) 4 Also [PQRS] [PQS] [SRQ] [PQRS] (a cd)...(v) So, y (iv) & (v) a cd ad c 7

The 3rd Olympiad of Metropolises

The 3rd Olympiad of Metropolises Day 1. Solutions Problem 1. Solve the system of equations in real numbers: (x 1)(y 1)(z 1) xyz 1, Answer: x 1, y 1, z 1. (x )(y )(z ) xyz. (Vladimir Bragin) Solution 1.