Pioeer Educatio The Best Way To Success INMO-018 proles ad solutios NTSE Olypiad AIPMT JEE - Mais & Advaced 1. Let ABC e a o-equilateral triagle with iteger sides. Let D ad E e respectively the id-poits BC ad CA : let G e the cetroid of triagle ABC. Suppose D, C, E, G are cocyclic. Fid the least possile perieter of triagle ABC. Solutio: BD.BC = BG.BE a. a. 3 a 3...(1) 4 By appoloius theore a + c = 4 a c () 4 Fro (1) ad () a c 3 a (Fro (1) ad ()) 4 4 c a (3) a + ust e eve a, ust e of sae partiy. Now, c = W. L. O. G let a a a a a a x N, y N (as a, of sae parity) c = x + y For y = 0, c = x a = ad c = a c = a = equilateral which is ot the case. y > 0 Pioeer Educatio SCO 30, Sector 40 D, Chadigarh +91-98155771, 017-461771 Page 1 of 6 www.pioeeratheatics.co
Pioeer Educatio The Best Way To Success NTSE Olypiad AIPMT JEE - Mais & Advaced Now c, x, y are sides of a right agle triagle ad sallest pythagorea triple is 5, 4, 3 ; secod sallest 5, 1, 13 For 5, 4, 3 we have c = 5, a 4, a 3 a = 7, = 1 Not possile For 13, 1, 5 we have c = 13, a 1, a 5 c = 13 ad a = 17, = 7 as 17 < 7 + 13 least perieter of ABC will e 7 + 13 + 17 = 37. For ay atural uer, cosider a 1 rectagular oard ade up of uit squares. This is covered y three types of tiles 1 1 red tile, 1 1 gree tile ad 1 lue doio. (For exaple, we ca have 5 types of tilig whe = : red-red; red-gree; gree-red; gree-gree ad lue.) Let t deote the uer of ways of coverig 1 rectagular oard y these three types of tiles. Prove that t divides t+1. Solutio: Let r, g, respectively e the uer of 1 tiles that ed with a red, gree ad lue tiles. Clearly, t = r + g +. To get a 1 ( + 1) tile edig i a red tile, we ca apped a 1 1 red tile to ay of the aove three. Hece r+1= r + g +. Siilarly, g+1 = r + g +. To get +1, we eed to apped a lue tile to a 1 ( 1) tile. Thus +1 = r 1 + g 1 + 1. Thus t+1 = r+1 + g+1 + +1 = (r + g + ) + (r + g + ) + (r 1 + g 1+ 1) = t + t 1 Thus we have recurrece relatio t+1 t. t 1= 0 whose characteristic equatio is Thus has characteristic roots 1±. Thus α 1 ad β1. Sice t1 = ad t = 5, we get A = t α 1 1 β t A 1 B 1 Aα ββ, where α β ad B. Thus λ y 1 0. Pioeer Educatio SCO 30, Sector 40 D, Chadigarh +91-98155771, 017-461771 Page of 6 www.pioeeratheatics.co
Pioeer Educatio The Best Way To Success Now, NTSE Olypiad AIPMT JEE - Mais & Advaced t 1 α β 1 1 1 1 α β α β = 1 1 α β 1 1 = α β = 1 t 1 α β Note that 1 1 1 1 1 1 α β 1 1 1..... is a iteger ad t + 1 is 4 divisile y t. 3. Let Γ1 ad Γe two circles with respective cetres O1 ad O itersectig i two distict poits A ad B such that O1AO is a otuse agle. Let the circucircle of triagle O1AO itersect Γ1 ad Γ respectively i poits C( A) ad D( A). Let the lie CB itersect i Γ i E; let the lie DB itersect Γ 1 i F. Prove that the poits C, D, E, F are cocyclic. Solutio: Clai : CB passes through O ad DB through O1 1 Proof : For circle Γ 1, AO1O AO1B ACB (1) Also for circle Γ3 AO1O ACO () Fro (1) ad () we set ACB = ACO CB CO CB passes through O Pioeer Educatio SCO 30, Sector 40 D, Chadigarh +91-98155771, 017-461771 Page 3 of 6 www.pioeeratheatics.co
Pioeer Educatio The Best Way To Success NTSE Olypiad AIPMT JEE - Mais & Advaced Siilarly BD, passes through O1 Now BAE = 90 (as BF diaeter of Γ 1 ) ad BAF = 90 (as BF diaeter of Γ 1 ) FAE are colliear ad to O1O Let FEC = α O1OB = α (as O1O FE) or OOC 1 α O1D C α (o Γ 3) or FDC α FEC α FDC CDEF are cocyclic. 4. Fid all polyoials with real coefficiets P(x) such that P(x + x + 1) divides P(x 3 1). Solutio: Possiility (1) : P(x) is costat = c the P(x 3 1) = c ad P(x + x + 1) = c ad we are doe. Let P(x) e o cotat polyoial. As P(x + x +1) P(x 3 1) P(x 3 1) = P(x + x + 1) Q(x) where Q(x) i soe polyoial i x. P(x 1)(x + x + 1) = P(x + x + 1) Q(x) Wheever x + x + 1 i a root of P(x), (x 1) (x + x + 1) i also a root... (1) Let α e a root of P(x) such that α e axiu. Now tae x + x + 1 = α x = x1x = (say), roots with x1 + x = 1, Atleast oe root out of x1, x will have distace ore tha 1 (fro '1'). Let x1 1 x 1 x1 x 1 1 x 1 3 x 1 3 x 1 1 1 1 = x 1 1...() Fro oe we have (x 1) ( x x 1 ) = (x 1) α β(say) is aother root of P(x) = 0. Here B (x1)α x1 α α Which is a cotradictio α 0 α 0 All root of o cotat polyoial ust e '0'. P(x) = a. x, a R, N. A other solutio p(x) = c, c R. 5. There are 3 girls i a class sittig aroud a circular tale, each havig soe apples with her. Every tie the teacher otices a girl havig ore apples tha oth of her eighors coied, the Pioeer Educatio SCO 30, Sector 40 D, Chadigarh +91-98155771, 017-461771 Page 4 of 6 www.pioeeratheatics.co
Pioeer Educatio The Best Way To Success NTSE Olypiad AIPMT JEE - Mais & Advaced teacher taes away oe apple fro that girl ad gives oe apple each to her eighors. Prove that this process stops after a fiite uer of steps. (Assue that the teacher has a audat supply of apples.) Solutio: Let Iitially i th girl have ai apples Let as defie di = ai a i1 i = 1,, 3,,. here a+1 = a1 Cosider : Su d1 + d + d3 +...+ d = S1 (say) Let a > a 1 + a+1...(1) the after first step a a 1, a a 1, a a 1 1 1 1 1 ' d a 1 a 1 a a d...() 1 1 1 1 1 ' ad d a 1 a 1 a a d (3) 1 1 Equality i () ad (3) ca t hold siultaeously otherwise it will violate (1) ' ' 1 1 d d d d ' ' 1 1 S = d1 + d + d3. + d d... d S as Si 0ad Si is a decreasig sequece of o-egative itegers, it ca't e eep o decreasig forever. After soe fiite steps process will stop. 6. Let N deote the set of all atural uers ad let f : N N e a fuctio such that (a) f() = f()f() all, i N ; () + divides f() + f() for all, i N. Prove that there exists a odd atural uer such that f() = for all i N. Solutio: P(, ) : f() = f().f(); Q(, ) : + (f() + f()) P(1, 1) : f(11) = f(1). f(1) f(1) = 1 as f N Q(, ) : + (f() + f()) f() f() =.q, q soe odd uer N. If possile let q > 1 the there will exist a prie p such that p q p = odd prie. Also we set p f () p 1 p 1 p 1 P. : f p 1 f. f.f Pioeer Educatio SCO 30, Sector 40 D, Chadigarh +91-98155771, 017-461771 Page 5 of 6 www.pioeeratheatics.co
Pioeer Educatio The Best Way To Success P f p 1 NTSE Olypiad AIPMT JEE - Mais & Advaced Q(1, p 1) : 1 + (p 1) (f(1) + f (p 1) p (1 f p1 p 1 Which is a cotradictio q = 1 f() = (, 1) : ( + 1) (f() + f(1)) ( + 1) + 1 0 (od 3) or ( 1) + 1 0 (od 3) = odd. Also fro f() = f().f() f(...) f(). f()...f().... ties ties ties f( ) = () = Now Q(, ): ( + ) (f() f( )) i.e. f.(1) as (x + y) x y for = odd () Fro (1) ad () we get f N f f N f(). = 0 has ifiite divisors f() = for soe odd N Pioeer Educatio SCO 30, Sector 40 D, Chadigarh +91-98155771, 017-461771 Page 6 of 6 www.pioeeratheatics.co