C of C Math meet discussions

[varun nakra1]

The link posted as CofC math meet in QQAD by Aarav sir is gud.
The three level problems of all the years are very much CAT like..So we can attempt some of them and start our discussions here..The link to the same is
C of C Math Meet : samples
I hope fruitful discussions will be carried out here solving the purpose of this space.

[kekambus]

I have compiled some problems which are giveing me a headache.

Plz find the attachment and provide ur inputs

[varun nakra1]

Originally Posted by kekambus

I have compiled some problems which are giveing me a headache.

Plz find the attachment and provide ur inputs

Hat(White)

Bag(White,Black)

Now the cases could be Hat(white,white) or Black(white,black)

It is saying that the probability that the ball which is left is White provided that the ball already been taken out is white is what...That means our sample space is the probability of the ball being taken out to be White,so we need to find as to what percentage chance is out of this probabilty that the ball which is left is also white..

NOw the prob that the ball being taken out is white =
(Prob that a white ball is taken out of the bag)*(Prob that a white ball is taken out of the hat) = (1/2)*1 plus
(Prob that a black ball is taken out of the bag)*(Prob that a white ball is taken out of the bag) = (1/2)*(1/2) = 1/4

Now the the probabilty that the leftover ball is also white is in jus this case :
Hat(white,white)..so prob that this case happens and the white ball is taken out is (1/2)(1)
Hence the reqd prob is (1/2)/(1/4 + 1/2) = 2/3..

[varun nakra1]

The length of the angle bisector of a triangle has a standard result which could be proved using Stewart's Theorem...Ill use that first

we have BD^2 = AB.BC - AD.CD

After that using ABC to be similar to BCD and using AB/BC = AD/CD ,one can establish the following results
AD=1 ; AB.CD=1 ; BD = AC.CD

using all this , (AB - 1/AB) = BD^2 Now since triangle is isoceles so we have AB=AC
using that we find BD=1
hence AB - 1/AB = 1 so this gives AB = (1+root5)/2..btw this is also known as Golden Ratio..nd one should remember the value of it also to be 1.618..

[varun nakra1]

It should be 4*(12-6root2) which is approx 14 feet..isn this the answer?

next waley ka 1/16 aa raha hai mera...ill solve the next two by tmrw..

[varun nakra1]

Let the side of the longer base be x and that of the shorter one be y..and the length of the equal sides be a..Now using Ptolemy's theorem

x^2 = xy + a^2 =>x = [y +/- root(y^2 + 4a^2)] / 2
Since x is an integer, y^2 + 4a^2 = k^2 and x = (y +/- k)/2
Also x>y => k>y and x= (y+k)/2
Also both y and k should be of the same parity,i.e. both even or both odd...

Considering, y=2m and k=2n we get x=m+n and m^2 + a^2 = n^2...Now the smallest possible such triplet with integer values is (3,4,5)..THis gives the smaller value of x = 8 and y=6 but a=4 which is <y..so nt possible

That is we need to find Pythagorean triplets such that a/m>2

Considering, y=2m+1 and k=2n+1 we get x=m+n+1 and a^2 = (n-m)(n+m+1)...Now the smallest possible such triplet with integer values is n=2 and m=1..THis gives the smaller value of x = 4 and y=3 but a=2 which is less than y which is not possible..so lemme think further....here a>2m+1