rmo mumbai 2014 paper & solution

RMO_2014(MUMBAI) PAPER & SOLUTION PAGE # 1

Regional Mathematical Olympiad-2014(Mumbai Region)

Time : 3 hours December 07, 2014

Instructions :

� There are six questions in this question paper. Answer all questions.

� Each question carries 10 points

� Use of protractors, calculators, mobile phone is forbidden.

� Time allotted : 3 Hour

1. Three positive real numbers a, b, c, are such that a2 + 5b2 + 4c2 � 4ab� 4 bc = 0, can a, b, c, be the lengths

of the sides of a triangle ? Justify your answer.Sol. Given : a2 + 5b2 + 4c2 � 4ab� 4 bc = 0

a2 + 4b2 + 4c2 + b2 � 4ab � 4bc = 0

(a�2b)2 + (2c � b)2 = 0a = 2b , 2c = ba = 2b = 4cNow,a + b = 4c + 2c = 6cb + c = 2c + c = 3c but a = 4cSo, a, b, c can not be the side of traingle

2. The roots of the equation x3 � 3ax2 + bx + 18c = 0 form a non-constant arithmetic progression and the rootsof the equation x3 + bx2 + x � c3 = 0 form and a non-constant geometric progression. Given that a, b, c are realnumbers, find all the positive integral values a and b.

Sol. Given : The roots of the equationx3 � 3ax2 + bx + 18c = 0 ......(i)from an AP. Let roots are� d, and + dSo, � d + + + d = 3a= aPut x = = a in euqation (i)a3 � 3a3 + ab + 18c = 0ab + 18c = 2a3 ......(ii)Now, euqation x3 + bx2 +x � c3 = 0 ......(iii) from a non -constant G.P.

Let roots are r,,r

So, product of roots 3 = c3 = cput x = = c in euqation (iii)3 + b2 + � c3 = 0c3 + bc2 + c � c3 = 0bc2 + c = 0c (bc +1) = 0c = 0 or bc + 1 = 0c = 0 (not possible), bc + 1 = 0

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bc = �1

c = b1

Put c = b1

in equation (ii)

ab2 � 18 = 2a3b

= b = a18

aa 42

Hence, a = 1, 2, 3, 6 ,9, 18but only a = 2 given positive integres value of b = 9So, (a, b) = (2, 9) Ans.

3. Let ABC be an acute-Angled triangle in which ABC is the largest angle of the triangle. Let O be itscircumcentre.The perpendicular bisectors of BC and AB meet AC at X and Y respectively. The internalbisectors of AXB and BYC meet AB and BC at D and E respectively. Prove that BO is perpendicular toAC if DE is parallel to AC.

Sol. Given : that XM is perpendicular bisectors of BC. YN is perpendicular bisector of AB.XD and YE are angle bisectors

Now, DE || AC BDAD

= EBCE

..... (1)

XD is angle bisectorsXBAX

DBAD

..... (2)

YE is angle bisectorsYBCY

EBCE

..... (3)

From euqation (1), (2) and (3)

YBCY

XBAX

AYCY

XCAX

AYYB

XCXB

XC = AY

Now, in ABCY

M

A

CB

X

E

DN

ABCsin

BCAsin

ABBC

CsinAsin ..... (4)

also cosC = cos XCM = XC2

BCXC

2/BCXCCM

and cosA = cos YAN = AY2AB

AY2/AB

AYAN

ABBC

AcosCcos ..... (5)

from equation (4) and (5)

AcosCcos

CsinAsin sin2A = sin2C A = C


4. A person mvoes in the x�y plane moving along points with integer co-ordinates x and y only. When she is at

point (x, y), she takes a step based on the following rules :

(a) if x + y is even she moves to either (x+ 1, y) or (x + 1, y + 1);

(b) if x + y is odd she move to either (x, y + 1) or (x + 1 , Y + 1) .

How many distnct paths can she take to go from (0, 0) to (8, 8) given taht she took exactly three steps to the

right ((x, y) to (x + 1, y)) ?

Sol. According the question.

She takes exactly 3 step to the right and she need to go finally at (8,8) So she must have taken 5

diagonal steps. Further she can a diagonal step from any point on the plane.

Note that choosing the 5 instance when she takes a diagonal step actually determines the path she

takes. This is because at any instant when she does n�t take the diagonal step she has only one choice

(right or up depending an x+y) . therefore the number of paths is 511

5335 CC .

5. Let a, b ,c be positive numbers such taht

1c1

1b1

1a1

1

Prove that (1 + a2) (1 + b2) (1 + c2) 125. When does the equality hold ?

Sol. Let p = a + b + c , q = ab + bc + ca

and r = abc by AM GM

3cba 3 abc p 3 3 r

Now, abc a + b + c + z

r p + 2

r 2 + 3 3 r

r � 2 3 3 r

r3 � 8 � 6r2 + 12 r 27r

r3 � 6r2 + 12r � 8 27r

r3 � 6r2 �15 r� 8 0

r2 (r+1) � 7r (r+1) �8(r+1) 0

(r+1) (r2�7r�8) 0

(r+1) (r+1) (r�8) 0

so r 8 since a,b,c are positive

P 6 3 r3p red no.

(1+a2) (1+b2)(1+c2) = (1+a2) (1+c2+b2+b2c2)

= 1 + c2 + b2 + b2c2 + a2 + a2c2 + a2b2 + a2b2c2

= r2 + (a2b2+b2c2+c2a2) + (a2 + b2 + c2) + 1

82 + 6 × 8 + 3 3 28 + 1

= 64 + 48 + 12 + 1

= 125

since a2 + b2+c2 3 3 8

a2b2 + b2c2 + c2a2 6x

r2 82


6. Let D, E, F be the points fo contact of the incircle of an acute-angled traingle ABC with BC, CA, ABrespectively.Let l

1, l

2, l

3 bethe incentres of the triangles AFE, BDF, CED, respectively. Prove the lines l

1 D,

l2 E, l

3 F are concurrent

Sol. To proveI,D,I

2 E and I

2F

are concurrent for this we will prove thatI,D, I

2E and I

2F

are angle bisectors of DEF.In AFE

AEF = AFE = 90 �2A

AEAFcesin so AEF=AFE)

Now AE is tangent and EF is a chord

E

CDB

F

A

I1

I3I2

so by alternate segment theorem

AEF = EDF = 90 � 2A

....(1)

Since I1 is the incentre of AFE

so EI1F = 90 +

2A

....(1)

from eq. (1) and (2)EDF + EI

1F = 180º

so DEI1F is a cyclic quadrilateral

so, I1 lie on the cirlce DEF

Now I1E = I

1F so EDI

1=FDI

1

hence DI1 is angle bisector of EDF

similarly EI2 and FI

3 are angle bisector of DEF and DFE respectvely.

Hence DI1,

EI2 and FI

3 are concurrent.

rmo mumbai 2014 paper & solution

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