Hi everyone. I guess question 2 can b solved in an easy way.
If 11 divides a+13b , 11 divides a+2b
If a+11b is divisible by 13, a-2b is divisible by 13
Now, lets choose random values of a+2b and a-2b,
Let a+2b=22, a-2b=13, this does give integral soln
lets take a+2b=33 , a-2b=13 or a=23 and b=5
This is the least +ve value of a and b
Hence a+b=23

Great attempt Krish ..I dont have the solution for this question though.

Now tell me how can you say that PC is a straight line and eventually that DPCB has to be a quadrilatral ?I dont think u can zero in upon that ?
Awaiting ur reply :)

well Infact PC is a circular arc bulging outwards ... but am not sure how to calculate the area of the circular arc ...

No my dear ...I m afraid stiil not correct.......


Well , it will b tuff to prove that it is a unique value without th help of a compuer program , u dont need to do tht ..Just gimmeee that 1 value

This is certainly tedious i know . Get me the number .. anyone ???

Is the answer to this one:
/24??

Please post the full solution nd concept .I don have the final ans for this.

virtualvishal Says

is the answer 360...i think it is right:grab: :

No my dear ...I m afraid stiil not correct.......


Well , it will b tuff to prove that it is a unique value without th help of a compuer program , u dont need to do tht ..Just gimmeee that 1 value

krsh.vik Says

Is it 390?

is the answer 360...i think it is right:grab: :

Folks solve this prob :)

A point is randomly selected in a square. What is the probability that the angle from the closest vertice of the square to the random point to the next closest vertice is greater than or equal to 120 degrees

Is the answer to this one:
/24??

krsh.vik Says

Is it 390?

Krish this is not the ans .. try again ..:)

Editing : Posted on wrong forum

vineet.nitd Says

Not correct . Hint is there is only 1 such number and u will also have to prove this fact :)

Is it 390?

is the answer 130,258,514,1026?
i couldnt get a single value,and im not so sure bout this..if its correct il post my solution..

Not correct . Hint is there is only 1 such number and u will also have to prove this fact :)

This is not then correct ans ..ans is given in the above posts ..


New Problem
Lets say we have k lying in the range given by natural numbers greater than equal to 100 and less than equal to 1997 ,such that k divides 2^k+ 2.Find the value of K .

is the answer 130,258,514,1026?
i couldnt get a single value,and im not so sure bout this..if its correct il post my solution..

Is the ans

t+taninv(t) = x+c where t= root(y^2-1)

This is not then correct ans ..ans is given in the above posts ..


New Problem

Lets say we have k lying in the range given by natural numbers greater than equal to 100 and less than equal to 1997 ,such that k divides 2^k+ 2.Find the value of K .

vineet.nitd Says

If length of tangent at any point on the curve y =f(x) intercepted between the point and the x-axis is of length 1 . Find the equation of the curve

Is the ans

t+taninv(t) = x+c where t= root(y^2-1)

ok finally managed with a lot of help from a friend.

please see the figure
let angle CBF=x
so angle FCB= 90-x (as CBF is a right angled triangle at F)
and also PFB = 90-x (as BPF is aright angled triangle at P)

now angle FMB = angle FPB = 90 and they are on the same base FB. so the FMPB is a cyclic quadrilateral. thus angle contained by the arc on PB will be same.
therefore angle PFB=anglePMB=90-x

now in right angled triangle CDO angle COD will be = x
angle NOF is vertically opposite to angle COD thus they are equal to x
again NOMF is a cyclic quadrilateral(angle ONF+angle OMF, here OF is the diameter of the circle).the angle contained by the arc will be equal. so angle NOF = angle NMF=x

finally we have the angles as
NMF=x
angle FMB = 90
angle PMB = 90-x
adding these three angles we get 180 degrees, so the line PNQ is a straight line. similarly at N we get 180 degrees. thus finally proving that PMNQ are collinear.

I guess u have interchanged E and F in ur solution as it says something else in ur fig. edit ur post and correct it :)

You have convinced me fully dude .Thats a good proof .Kudos to you

Now this is a sexy thread!! Great going vineet, warrior (who would be leaving us soon to go to IIM A), varun, and Aarav.
With my pathetic interview at IIM Indore, I would be joining in this. Expect contributions from me after the 13th

The square is divided into four equal parts
OB is half part of a diagonal
for any point in the triangle OBD the nearest vertex is B and the second nearest vertex is A
angle APB = 120
angle ACB = 120
for all points in the quadrilateral DPCB the angle will be greater than 120
Hence the probability is equal to area of quadrilateral DPCB/One-eighth area of square :)

Great attempt Krish ..I dont have the solution for this question though.

Now tell me how can you say that PC is a straight line and eventually that DPCB has to be a quadrilatral ?I dont think u can zero in upon that ?
Awaiting ur reply

Folks solve this prob :)

A point is randomly selected in a square. What is the probability that the angle from the closest vertice of the square to the random point to the next closest vertice is greater than or equal to 120 degrees

The square is divided into four equal parts
OB is half part of a diagonal
for any point in the triangle OBD the nearest vertex is B and the second nearest vertex is A
angle APB = 120
angle ACB = 120
for all points in the quadrilateral DPCB the angle will be greater than 120
Hence the probability is equal to area of quadrilateral DPCB/One-eighth area of square

Folks solve this prob :)

A point is randomly selected in a square. What is the probability that the angle from the closest vertice of the square to the random point to the next closest vertice is greater than or equal to 120 degrees

please see the final solution in the next post

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

Someone please throw light ...

ok finally managed with a lot of help from a friend.

please see the figure
let angle CBF=x
so angle FCB= 90-x (as CBF is a right angled triangle at F)
and also PFB = 90-x (as BPF is aright angled triangle at P)

now angle FMB = angle FPB = 90 and they are on the same base FB. so the FMPB is a cyclic quadrilateral. thus angle contained by the arc on PB will be same.
therefore angle PFB=anglePMB=90-x

now in right angled triangle CDO angle COD will be = x
angle NOF is vertically opposite to angle COD thus they are equal to x
again NOMF is a cyclic quadrilateral(angle ONF+angle OMF, here OF is the diameter of the circle).the angle contained by the arc will be equal. so angle NOF = angle NMF=x

finally we have the angles as
NMF=x
angle FMB = 90
angle PMB = 90-x
adding these three angles we get 180 degrees, so the line PNQ is a straight line. similarly at N we get 180 degrees. thus finally proving that PMNQ are collinear.

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

Someone please throw light ...

i think we will have to approach this problem using vector algebra. take one vertex say B as the origin then use the concept of position vectors and equation of line and finally we will have to use linear combination of vectors to prove that the points PQMN are collinear. i havent tried it but i had solved similar kind of problems while preparing for iitjee

thanks for the encouragement vineet
i took y = cosQ so it doesnt make a difference but i did not integrate properly
i get the integration part as
integrate cos^2y/siny dy = integrate x dx
then it is
log(tan y/2)+cosy = x+c
if still there is some error:( then please post the solution

take y=sinm to keep away from sign change ...

hence , integ{(cos^2m/sinm)} dm = integ{(cosec m - sinm)}dm = +/- x+ K
where K is a constant

lncosecm - cotm| + cosm = +/-x + K

Plug in the the values ,

(1-y^2)^1/2 + ln |{1- (1-y^2)^1/2}/y = +/- x + K

Excellent start deepak :)..but u deviated towards the end ...

Lemmee give u a hint ... take y = sinQ ..and u will get a genric eqn of a curve ...

Try again after that ...

thanks for the encouragement vineet
i took y = cosQ so it doesnt make a difference but i did not integrate properly
i get the integration part as
integrate cos^2y/siny dy = integrate x dx
then it is
log(tan y/2)+cosy = x+c
if still there is some error:( then please post the solution

the curve is y=f(x)
a point P on it be (x,f(x))
the slope of the tangent at this point is dy/dx (= f '(x) )at that point
so equation of tangent is
Y-y = f '(x)(X-x)
this meets the X axis so Y=0 to get the x co-ordinate
the point of intersection of the tangent with the X axis is (x- f(x)/f '(x), 0 )
now let this point be denoted by Q
given PQ=1
thus (x-x+f(x)/f '(x))^2 + (f(x)) ^2 = 1
put f(x)=y and f '(x)= dy/dx

we get y^2/(1-y^2)=(dy/dx)^2
solving this differential equation
(i dont know how to put the integration terms on the comp and other symbols :neutral:)
we will get the equation as x + (1/y) = constant
but in the calculation we simplify the term sec(cos^-1 y), this is 1/y only for y lies in -1<=y<=+1 - {0} ........(a)

thus the above curve holds x + (1/y) = constant is true for condition (a)
but for the other case it will be in inverse circular form

this is the basic approach of solving the problem.(what is the preposition to be used here? is it to)

Excellent start deepak :)..but u deviated towards the end ...

Lemmee give u a hint ... take y = sinQ ..and u will get a genric eqn of a curve ...

Try again after that ...

vineet.nitd Says

If length of tangent at any point on the curve y =f(x) intercepted between the point and the x-axis is of length 1 . Find the equation of the curve

the curve is y=f(x)
a point P on it be (x,f(x))
the slope of the tangent at this point is dy/dx (= f '(x) )at that point
so equation of tangent is
Y-y = f '(x)(X-x)
this meets the X axis so Y=0 to get the x co-ordinate
the point of intersection of the tangent with the X axis is (x- f(x)/f '(x), 0 )
now let this point be denoted by Q
given PQ=1
thus (x-x+f(x)/f '(x))^2 + (f(x)) ^2 = 1
put f(x)=y and f '(x)= dy/dx

we get y^2/(1-y^2)=(dy/dx)^2
solving this differential equation
(i dont know how to put the integration terms on the comp and other symbols :neutral:)
we will get the equation as x + (1/y) = constant
but in the calculation we simplify the term sec(cos^-1 y), this is 1/y only for y lies in -1<=y<=+1 - {0} ........(a)

thus the above curve holds x + (1/y) = constant is true for condition (a)
but for the other case it will be in inverse circular form

this is the basic approach of solving the problem.(what is the preposition to be used here? is it to)

Folks solve this prob :)

A point is randomly selected in a square. What is the probability that the angle from the closest vertice of the square to the random point to the next closest vertice is greater than or equal to 120 degrees

If length of tangent at any point on the curve y =f(x) intercepted between the point and the x-axis is of length 1 . Find the equation of the curve

vineet.nitd Says

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

Yes Vineet, I didn't take the reciprocal...the substitution variable was in the corner of the page that I forgot. Perhaps, age is catching up with me :-)

I will write the answers of the 2 problems that I didn't solve soon. I couldn't visit this thread on weekends.

I will love if anyone among you provide a better solution than mine to n(n+1) problem

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

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 ...

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

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

Aarav Says

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

repeat post ... edited

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

prade Says

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

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, 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

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.