REGIONAL MATHEMATICAL OLYMPIAD, 2017 · REGIONAL MATHEMATICAL OLYMPIAD – 2017 | 08-10-2017...

RMO 2017

NATIONAL BOARD FOR HIGHER MATHEMATICSAND

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION TATA INSTITUTE OF FUNDAMENTAL RESEARCH

REGIONAL MATHEMATICAL OLYMPIAD, 2017

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REGIONAL MATHEMATICAL OLYMPIAD – 2017 | 08-10-2017

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1. Let AOB be a given angle less than 180° and let P be an interior point of the angular region determined by AOB. Show, with proof, how to construct, using only ruler and compasses, a line segment CD passing through P such that C lies on the ray OA and D lies on the ray OB, and CP : PD = 1 : 2.eku yhft, fd AOB ,d dks.k fn;k gqvk gS tks 180° ls de gS vkSj P, AOB }kjk fu/kkZfjr dks.kh; {ks=k esa]

,d Hkhrjh fcanq gSA izek.k ds lkFk fn[kkb, fd] dsoy :yj ,ao izdkj dk iz;ksx djrs gq,] P ls xqtjrs gq, ,d

,sls js[kk[kaM CD dh jpuk dSls djsaxs rkfd C v/kZ&js[kk OA ij fLFkr gks] D v/kZ&js[kk OB ij fLFkr gks] vkSj

CP : PD = 1 : 2 gksA Sol. Join OP and extend to Q such that OP : PQ = 2 : 1 Draw a line parallel to OA through Q. It cuts OB at point M (say) take point D on OB such that M is mid point of OD joint DQ and

Produce it to meet OA at N. Then by mid-point theorem, Q is mid point of DN. so OQ is median of ODN

As OP : PQ = 2 : 1, P is centroid DP : PC = 2 : 1

Construction of OP : PQ = 2 : 1

obtain mid-point L of OP.

with P as centre and radius PL, arc cuts OP produced at Q

QL PO

Construction of QM || OA : With O as centre, draw an arc to cut OQ at S1 and OA at SQ With Q as centre and same radius, draw an arc to cut OQ at S3

With S3 as centre and radius S1 S2, draw an arc to cut previous arc at S4 join QS4

S3

B

O

Q S4

S1

S2A





2. Show that the equation

a3 + (a+ 1)3 + (a + 2)3 + (a + 3)3 + (a + 4)3 + (a + 5)3 + (a + 6)3 = b4 + (b + 1)4

has no solutions in integers a, b.

fn[kkb, dh lehdj.k

a3 + (a+ 1)3 + (a + 2)3 + (a + 3)3 + (a + 4)3 + (a + 5)3 + (a + 6)3 = b4 + (b + 1)4

dk iw.kkZadksa a, b esa dksbZ gy ugha gSA

Sol. a3 + (a + 1)3 + (a + 2)3 + (a + 3)3 + (a + 4)3 + (a + 5)3 + (a + 6)3 = b4 + (b + 1)4

Let a + 3 = x

so LHS = (x – 3)3 + (x + 3)3 + (x – 2)3 + (x + 2)3 + (x – 1)3 + (x + 1)3 + x3

= 7x3 + 6x (12 + 22 + 32) = b4 + (b + 1)4

7x3 + 84x = b4 + (b + 1)4

7(x3 + 12x) = b4 + (b + 1)4

7x (x2 + 12) = b4 + (b + 1)4

Now 7 divides LHS but

b4 + (b + 1)4 1, 3, 6, 1, 6, 5, 1 (mod 7)

Hence the equation has no solutions in integers a & b.

3. Let P(x) = x2 + 1

2x + b and Q(x) = x2 + cx + d be two polynomials with real coefficients such that

P(x) Q(x) = Q(P(x)) for all real x. Find all the real roots of P(Q(x)) = 0.

eku yhft, fd P(x) = x2 + 1

2x + b ,oa Q(x) = x2 + cx + d nks cgqin gSa ftuds xq.kk¡d okLrfod gSa vkSj

okLrfod x ds fy, P(x) Q(x) = Q(P(x))A bl fLrfFk esa lehdj.k P(Q(x)) = 0 ds lHkh okLrfod gy dhft,A

Ans. 1

2, – 1

Sol. P(x) Q (x) = Q (P(x))

P(x) Q (x) = (P(x))2 + cP(x) + d

P(x) (Q (x) – P(x)) = cP(x) + d

2 xx b

2

1c x (d b)

2

= c 2 xx b

2

+ d

1c

2

x3 + (d – b) x2 + c 12 4

x2 + d b2 2

x + b

bc2

x + b (d – b) = cx2 + c

2x + bc + d

1c

2

x3 + c 1

d b c2 4

x2 + d b b c

bc2 2 2 2

x + 2bd b bc d = 0

1c

2

x3 + c 1

d b2 4

x2 + d c

b bc2 2

x + 2bd b bc d = 0

c = 1

2 , d – b =

1

2 ,

d

2– b +

b

2–

1

4 = 0

bd – b2 – bc – d = 0

c = 1

2, b = –

1

2, d = 0





Q(x) = x2 + x

2

P(x) = x2 + x2

– 12

= 0

P(Q(x)) = 0 let Q (x) = t

P(t) = 0 t2 + t

2–

1

2 = 0

t = 1

2, – 1

Q(x) = 1

2 and Q(x) = – 1

x2 + x

2 =

1

2 and x2 +

x

2 = – 1

2x2 + x – 1 = 0 and 2x2 + x + 2 = 0

x = 1

2 , – 1 and x

Real roots are 1

2, – 1

4. Consider n2 unit squares in the xy-plane centred at point (i, j) with integer coordinates, 1 i n,

1 j n. It is required to colour each unit square in such a way that when ever 1 i < j n and

1 k < n, the three squares with centres at (i, k), (j,k), (j, ) have distinct colours. What is the

least possible number of colours needed ?

eku yhft, fd dkrhZ; ry esa n2bdkbZ oxZ (izR;sd dk {ks=kQy 1 gS) fn, x, gS ftuds dsanz (i, j) gSa] tgk¡

1 i n, 1 j n A izR;sd oxZ dks bl izdkj ls jax ls Hkjuk gS fd tc Hkh 1 i < j n o 1 k < n gS

mu rhuksa oxksZ ds jax fHkUu gksa ftuds dsanz (i, k), (j,k), (j, ) gSaA ,slh fLrfFk esa U;wure vko';d jaxks dh la[;k

dhft,A

Ans. 2n – 1

Sol. Squares centred at A, B, C are of distinct colors so any formed has distinct color squares.

No of colors required = number of diagonals = 2n – 1

as no two vertices on a diagonal will lie on

B (j, k)

C (j, )

A (i, k)





Color n

Color n –1

Color 2n–2

Color 2n–1

Color 1

5. Let be a circle with a chord AB which is not a diameter. Let 1 be a circle on one side of AB

such that it is tangent to AB at C and internally tangent to at D. Likewise, let 2 be a circle on the

other side of AB such that it is tangent to AB at E and internally tangent to at F. Suppose the line DC intersects at X D and the line FE intersects at Y F. Prove that XY is a diameter of . eku yhft, dh ,d oÙ̀k gS vkSj AB ,d thok gS tks fd O;kl ugha gSA eku dhft, fd 1 js[kk AB dh ,d

rjQ ,d o`Ùk gS tks js[kk AB dks C ij Li'kZ djrk gS vkSj oÙ̀k dks D ij Hkhrj ls Li'kZ djrk gSA mlh rjg

eku yhft, fd 2 js[kk AB dh nwljh rjQ ,d oÙ̀k gS tks js[kk AB dks E ij Li'kZ djrk gS vkSj oÙ̀k dks F

ij Hkhrj ls Li'kZ djrk gSA ekuk yhft, fd js[kk DC o`Ùk ls X D ij feyrh gS vkSj js[kk FE oÙ̀k

ls Y F ij feyrh gSA izekf.kr dfj, fd XY oÙ̀k dk O;kl gSA

Ans. 00 Sol.

–22

–2

2

O3

E

A B

D

O1

C

O2

M

X1

Y1

F

2

1

We have to prove that X1 coincides with X and Y1 coincides with Y Let O1 be the centre of . Also O2 and O3 be centre of 1 and 2 respectively. Let M be the

mid-point of AB. Further assume that diameter through M meets at X1 and Y1.





Now if O2 CD = CO2D = – 2 = X1 OD

(as O1X1 is parallel to O2 C )

Hence O1 X1 D = d X1 , C, D are collinear

i.e. X coincides with X1

Now assume FEO3 = EO3 F = – 2

Hence FO1Y1 = – 2 (as O3E & O1 Y1 are parallel)

so FY1O1 = F, E, Y1 are collinear

i.e. Y coincides with Y1

Hence XY is diameter of the circle.

6. Let x, y, z be real numbers, each greater than 1. Prove that

eku yhft, fd izR;sd okLrfod la[;k x, y, z la[;k 1 ls cM+h gSA izekf.kr dfj, fd %

x 1

y 1

+ y 1

z 1

+ z 1

x 1

x 1

y 1

+ y 1

z 1

+ z 1

x 1

Sol. Put x = 1 + A

y = 1 + B

z = 1 + C

2 A

2 B

+ 2 B

2 C

+ 2 C

2 A

A

B+

B

C +

C

A

A

B –

2 A2 B

+

B

C–

2 B2 C

+

C

A–

2 C2 A

0.

A B

B(2 B)

+ B C

C(2 C)

+ C A

A(2 A)

0

Now it is true for A = B = C ; without any loss of generality we may assume A > B > C.

Now A B B C C A

B(2 B) C(2 C) A(2 A)

A B B C C A

B(2 B) C(2 C) B(2 B)

1 1 (C A) (C A)

A(A 2) B(B 2) A(A 2) B(B 2)

C B

B(2 B)

+ B C

C(2 C)

C B

C(2 C)

+ B C

C(2 C)

0 Hence proved

REGIONAL MATHEMATICAL OLYMPIAD, 2017 · REGIONAL MATHEMATICAL OLYMPIAD – 2017 | 08-10-2017...

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