REGIONAL MATHEMATICAL OLYMPIAD, 2017 ......MUMBAI & GOA REGION RMO – 2017 | 08-10-2017 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005

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RMO 2017

NATIONAL BOARD FOR HIGHER MATHEMATICS AND

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION TATA INSTITUTE OF FUNDAMENTAL RESEARCH

REGIONAL MATHEMATICAL OLYMPIAD, 2017 (Maharashtra and Goa Region)

TTEESSTT PPAAPPEERR WWIITTHH SSOOLLUUTTIIOONN && AANNSSWWEERR KKEEYY

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MUMBAI & GOA REGION RMO – 2017 | 08-10-2017




Time : 3 hours October 08, 2017 Total marks: 100

Instruction

1. Calculators (in any form) and protractors are not allowed.

2. Rulers and compasses are allowed.

3. Answer all the questions. Draw neat Geometry diagrams.

4. Maximum marks are mentioned next to each question.

5. Answerer to each question should start on a new page, clearly indicate the question number.

6. Mathematical reasoning will be taken into consideration while assessing the answers.

1. Consider a chessboard of size 8 units × 8 units (i.e. each small square on the board has a side length of 1 unit). Let S be the set of all the 81 vertices of all the squares on the board. What is number of line segments whose vertices are in S1 and whose length is a positive integer ? (The segments need not be parallel to the sides of the board.)

Sol. Number of line segment parallel to co-ordinate axis of 1 unit length equals to 8 × 9 × 2.

Number of line segment parallel to co-ordinate axis of 2 unit length equals to 7 × 9 × 2.

Number of line segment parallel to co-ordinate axis of 3 unit length equals to 6 × 9 × 2. and so on…..

Number of line segment parallel to co-ordinate axis of 8 unit length equals to 1 × 9 × 2.

Number of line segment not parallel to co-ordinate axis of 5 unit length equals to 5 × 8 × 2 × 2.

Number of line segment not parallel to co-ordinate axis of 10 unit length equals to 7 × 3 × 2 × 2.

Total line segment equals to 780.

2. For any positive integer n, let d(n) denotes the number of positive divisors of n; and let (n) denote the number of elements from the set {1, 2,….. n} that are coprime to n.

(For example, d(12) =6 and (12) = 4.)

Find the smallest positive integer n such that d((n)) = 2017.

Sol. d((n)) = 2017

As 2017 is a prime number

n) = n1 2 n

1 1 11 1 ...... 1p p p

where p1, P2 , ……Pn are the prime factors of n

Let (n) = t





d(t) = 2017 for minimum we have t is also minimum

t = 22016

(n) = 22016

n1 2 m

1 1 11 1 ...... 1p p p

= 22016

n= 22017

m = 1 , p1 = 2

3. Let P(x) and Q(x) be polynomials of degree 6 and degree 3 respectively, such that:

P(x) > Q (x)2 + Q(x) + x2 – 6, for all x R.

If all the roots of P(x) are real numbers, them prove that there exists two roots of P(x), any , such that < 1.

Sol. P(x) > (Q(x))2 + Q(x) + x2 – 6

P(x) > 21(Q(x)

2

+ x2 – 254

If all the roots of P(x) = 0 are real then P(x) < 0 between the roots. If P(x) < 0 then

21(Q(x)

2

+ x2 – 254

< 0

Now 21(Q(x)

2

is always positive

then 2 25x4

< 0 5 5x ,2 2

Now let the roots of P(x) be 1, 2, 3, 4, 5, 6 and they all belongs to 5 5x ,2 2

So six roots be between the length of interval 5 so by pigeonhole principal there will be atleast one pair of roots such that

1

Hence proved





4. Let 1, 2, 3,……40 be fourty parallel lines. As shown in the diagram, let m be another line that intersects the lines 1 to 40 in the points A1, A2, A3, ……. A40 respectively. Similarly, let n be another line that intersects the lines 1 to 40 in the points B1, B2, B3, ……. B40 respectively.

Given that A1B1 = 1, A40B40 = 14, and the areas of the 39 trapeziums A1B1 B2A2 , A2B2B3A3

,…….,A39B39B40 A40 are all equal ; then count the number of segments A, B whose length is a positive integer; where i {1, 2, …….40}

A1

A2

A3

A4

A5

I1

I2

I3

I39I39

I40

B1 B2

B3 B4

B39

B40

Sol. Area A1B1C is 21

1 sin sinx2 sin

Area A2B2C is 22

1 sin sinx2 sin

………………. & so on

A1 B1

C

B2x2 A2

x2

Now 2 2 2 2 2 21 2 1 1 3 2

1 sin sin 1 sin sinx x – x x x – x2 sin 2 sin

21x , 2

2x , 23x , ……………. are in A.P.

Now, 21x = 1. & 2

90x = 196 common difference = 5.

21x , 2

4x = 16 , 28x = 36, 2

17x = 81, 225x = 121, 2

40x = 196

x1 = 1, x4 = 4, x8 = 6, x17 = 9, x25 = 11, x40 = 14

5. If a, b, c, d R such that a > b > c > d > 0 and a + d = b + c; then prove that:

2 2 2 2(a b) (c d) a b – c d2





Sol. Let a = n + k, b = n, c = m + k, d = m, d = m , k > 0 , m > 0, n > 0

then bc – ad = mn + nk – mn – mk = k (n – m) 0

(bc – ad) > 0.

bc+ad> 2acbd

ac+bc+ a2d2 + b2d2 > a2 c2 + b2d2 + 2acbd

(a2 + b2 ) (c2 + d2) > (ac + bd)2

2 2 2 2(a + b )(c +d ) > (ac + bd)

–2ac – 2bd > – 2 2 2 2 2(a + b ) c d

a2+ b2 + c2 + d2 – 2ac – 2bd > (a2 + b2) + (c2 + d2) –2 2 2 2 2(a + b ) c d

(a – c)2 + (b – d)2 > 22 2 2 2(a + b ) c d

2(a – c)2 > 22 2 2 2(a + b ) c d

2 2 2 22(a c) (a + b ) c d

2 2 2 2(a b) (c d) (a + b ) c d2

Hence proved

6. Let ABC be acute - angled; and let be its circumcircle, let D be a point on minor arc BC of . Let E and F be points on lines AD and AC respectively, such that BE AD and DF AC. Prove that EF|| BC if and only if D is the midpoint arc BC.

Sol. Given: Let ABC be acute - angled; and let be its circum circle, let D be a point on minor arc BC of . Let E and F be points on lines AD and AC respectively, such that BE AD and DF AC.

To Prove: (i) EF || BC assuming D is mid point of arc BC

(ii) D is mid point of BC assuming EF || BC

proof: (i) ABAE EAF2

(since arc BD = arc DC)

AE = c cos A2

AP = 2bc Acosb c 2





B C B C2R 2sin cosAB b c 2 2AP 2b 2b

…………..(i)

Now ABCD BAD2

DC cos A2

= a2

since projection DC along BC is B2

Now angle ADFC C2

B CFDC2 2

OF = DC cos B C2 2

= a2

sec A2

cos B C2 2

Acos2Asin2

B CcosAF a 2 2

AAC 2b sin2

=

A B C2R 2cos cos2 2 22b

…………..(ii)

from (i) and (ii) AE AF EF || PCAP AC

EF || BC

Hence prove









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REGIONAL MATHEMATICAL OLYMPIAD, 2017 ......MUMBAI & GOA REGION RMO – 2017 | 08-10-2017 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005