[vineet.nitd]

Originally Posted by Aarav

Vineet, can you look at (3) again. LHS > 6 as (a+1/a) >= 2 for positive reals. Why 3 on RHS and that too with equality?

Aarav , yup u r bang on in saying that ...May be that is a mistake in RMO paper ..I called up and asked for the quest ..i was told this is exactly wot came there .I will be confirmed in 2 days ..as the question paper is with someone else .

btw ... keep this prob on hold for the time being

Will chip in wid my attempts tonight

[Aarav]

Originally Posted by vineet.nitd

Aarav , yup u r bang on in saying that ...May be that is a mistake in RMO paper ..I called up and asked for the quest ..i was told this is exactly wot came there .I will be confirmed in 2 days ..as the question paper is with someone else .

btw ... keep this prob on hold for the time being

Will chip in wid my attempts tonight

Problem 5 solved in the last 5 minutes of yet another boring marketing class :-) Answer is (4/3)^1/2. Was infact relatively easy, and I was not following a right approach.

[vineet.nitd]

Originally Posted by prade

2. Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.

a+13b =11x ; a+11b =13y

for minimum a and b ...we wud take minimum possible x and y ...


=>b = (11x -13 y)/2 ; a = (169y-121x)/2

a+b = 78y -55x ...

x and y also have to be +ve integers , so ..for min value of a+b and for +ve a and b,


and , (11/13) *x > y > (11/13)^2 * x

y=6 ,x =8 gives..


(a+b)min =28 with +ve a and b both

[vineet.nitd]

Originally Posted by vineet.nitd

5. Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD;AB=3CD and the area of the quadrilateral is 4. If a circle is drawn touching all the sides of the quadrilateral,find its radius.

Let CD=x
AD = 4 / {1/2 *4x} = 2/x
Radius = AD/2 = 1/x {sine 2 squares will be formed }
Using propertyof eual dist from a given point for two tangents to the same circle..
CB = 4x - 2/x
hence pythagorous , (AB-CD)^2 + AD ^2 = CB^2
=> 4x^2 + 4/x^2 = 4/x^2 + 16x^2 -16
=> 12x^2 =16 =>x= (4/3)^1/2
Hence radius = (3/4) ^1/2

[vineet.nitd]

repeat post ... edited

[vineet.nitd]

Originally Posted by Aarav

Problem 5 solved in the last 5 minutes of yet another boring marketing class :-) Answer is (4/3)^1/2. Was infact relatively easy, and I was not following a right approach.

Boss i guess u dint take the reciprocal ..u found out CD

[vineet.nitd]

Originally Posted by vineet.nitd

7. Let X be the set of all +ve integers greater than or equal to 8 and let f:X ®X be a function such that f(x+y)=f(xy) for all x>=4,y>=4.
If f(8 ) =9,determine f(9).

f(8 )=f(4+4)=f(16)= f(8+8 ) = f(64)=f(4*16)=f(20)=f(4+5)=f(9) = 9

hence f(9) = 9

[prade]

BETTER LATE THAN NEVER
Q2 a+10b+3b mod 11= a+2b mod 11
a+10b+b mod 13 = a-2b mod 13
=> a+2b = 11k
=> a-2b = 13m
a+b = (33k + 13m) /4
=> (k+m) mod 4 for least a+b k shud be more and m less.
take k = 3 and m = 1 => a+b = 28
Q5 (AB+CD).R = 4
CD.R = 1
(AB-CD)^2 = (2R)^2 + (4CD-2R)^2
EQUATING THE ABOVE EQNS R = ROOT(3)/2

[vineet.nitd]

Originally Posted by vineet.nitd

1. Let ABC be an acute-angled triangle and let D,E,F be the feet of perpendiculars from A,B,C respectively to BC,CA,AB. Let the perpendiculars from F to CB,CA,AD,BE meet them in P,Q,M,N respectively.Prove that P,Q,M,N are collinear.

Hey studs ..anyone started wid this one ... I am not getting how to start for this .....

Someone please throw light ...

[varun nakra1]

Originally Posted by vineet.nitd

Hey studs ..anyone started wid this one ... I am not getting how to start for this .....

Someone please throw light ...

hmmmmm....one can start by joining two of the intersection points nd then extending the line further to meet the others....then one needs to prove that the other angles are 90 each....for eg
draw FP perpendicular to BC and FN to BE...join PN and extend it to meet AD and AC at M and Q resp..then one needs to prove ki <FMD=90 and <FQC=90....
this is just another way of starting with this quesn...however I havn solved it yet...baad mein karoongaa...(procrastination )