CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[sanchit_states]

If u have issue with similar triangle u can do it wit co-ordinate geo funda also.
ground and poi of 10cm line as origing.co -od of pocontact is (0,10) whcih dvd line b/w (-8, and (r,r) in ratio of 8:r.thus (8r-8r)=0 and y-co-od((8r+8r)/8+6)=10-->r=40/3.

[IIM maniac]

s + p - 2√sp = 1 => s+p = 1 + 2√qr

Aarav, firat of all thanks a ton for such a great effort !!
i didnt get this above quoted step, sorry for asking too many question but hope you help me improve.

Ameya

[Aarav]

Originally Posted by IIM maniac

s + p - 2√sp = 1 => s+p = 1 + 2√qr

Aarav, firat of all thanks a ton for such a great effort !!
i didnt get this above quoted step, sorry for asking too many question but hope you help me improve.

Ameya

It is given in the problem that ps = qr.

[IIM maniac]

Originally Posted by Aarav

It is given in the problem that ps = qr.

Got it !!! Got it !! Got it !! ha...finally..
Thanks Aarav, i started with the problem right from the scratch on the paper with writing down each and every step and eventually it taught me how was it solved, great solution. I had solved this question in morning assuming P & S are squares, but the GCD thing is just rocking..Thanks for teaching me new concept.

Ameya

[srikar2097]

Originally Posted by sanchit_states

If u have issue with similar triangle u can do it wit co-ordinate geo funda also.
ground and poi of 10cm line as origing.co -od of pocontact is (0,10) whcih dvd line b/w (-8, and (r,r) in ratio of 8:r.thus (8r-8r)=0 and y-co-od((8r+8r)/8+6)=10-->r=40/3.

Is this a reply to some question? If yes then what question? There is a thread going on here. Please be careful enough to post such things... Thanks.

[Aarav]

Originally Posted by srikar2097

Is this a reply to some question? If yes then what question? There is a thread going on here. Please be careful enough to post such things... Thanks.

He was answering QQAD # 018, the problem put on Sunday.

[srikar2097]

Originally Posted by Aarav

Given that s-p > r -q as p < q <= r < s
=> (s-p)^2 > (r-q)^2 => (s-p)^2 + 4sp > (r-q)^2 + 4rq [becuase sp = rq, we add 4sp in LHS and 4rq in RHS]
=> (s+p)^2 > (r+q)^2
=> s+p >= r+q+1 [as all numbers are integers] -> (1)

Suppose √s - √p = 1 (the other possibility is √s - √p < 1 that we will see later)
=> s + p - 2√sp = 1 => s+p = 1 + 2√qr [becuase sp = rq] but (1) tells that √qr >= q + r => r = q [By AM-GM rule on positive numbers] and p+s = 2q + 1.

Now ps = q^2 => gcd(p, s) = 1 (the explanation is below)
Because if x divides gcd (p, s) and x is prime (or it will be product of two or more primes, but we assume the base case which covers other case as well), then x would divide q [because p = ax, q = bx => ps = abx^2 = q^2 and gcd(a, b) = 1) and thus x dividing 2q+1 [= p+s = x(a+b)] is a contradiction => each of p and s is a perfect square [gcd(p, s) = 1]. [if x divides q then x divides 2q => it can not divide 2q+1]

If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

=> In all s and p are perfect squares. Now take X -> only possible s is 81 => p = 64
Now take Y, p > 50 => p can be 64 or 81 but if p = 81 then s = 100 (not possible as s < 100) => p can only be 64. The information on r is required to cross-check if our data in hand is correct and it indeed is as √64.81 = 72.

Please let me know if it's still unclear.

Aarav, first of all thanks for this beautiful solution. I have got a couple of clarifications (I have marked them in bold in your reply above)-
1) s+p >= r+q+1 [as all numbers are integers] -> (1) .
Why did we add 1 to the RHS. I mean it's given that they are integers, but by removing the squares why do we need to add one?

2) If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

Here we could directly say that p+s < p+s (since p+s=1+2√sp) and hence a contradiction? Right?

P.S: That GCD way to solve we really something.

[implex]

Originally Posted by srikar2097

Aarav, first of all thanks for this beautiful solution. I have got a couple of clarifications (I have marked them in bold in your reply above)-
1) s+p >= r+q+1 [as all numbers are integers] -> (1) .
Why did we add 1 to the RHS. I mean it's given that they are integers, but by removing the squares why do we need to add one?

2) If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

Here we could directly say that p+s < p+s (since p+s=1+2√sp) and hence a contradiction? Right?

P.S: That GCD way to solve we really something.

(s+p)^2>(r+q)^2 and each of p,q,r,s are integers
this is only possible if p+s>=r+q+1

the point is we can write it as p+s>r+q but if we add 1 to the right side, the inequality changes to one with equality as well!

[Aarav]

Originally Posted by srikar2097

Aarav, first of all thanks for this beautiful solution. I have got a couple of clarifications (I have marked them in bold in your reply above)-
1) s+p >= r+q+1 [as all numbers are integers] -> (1) .
Why did we add 1 to the RHS. I mean it's given that they are integers, but by removing the squares why do we need to add one?

2) If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

Here we could directly say that p+s < p+s (since p+s=1+2√sp) and hence a contradiction? Right?

P.S: That GCD way to solve we really something.

1) has been answered by Implex

2) can be seen as we have proved that s+p >= p+q+1. To write in detail

√s - √p < 1 then p + s < 1 + 2√ps = 1 + 2√qr <= 1 + q + r (AM-GM) <= p + s

[srikar2097]

Originally Posted by implex

(s+p)^2>(r+q)^2 and each of p,q,r,s are integers
this is only possible if p+s>=r+q+1

the point is we can write it as p+s>r+q but if we add 1 to the right side, the inequality changes to one with equality as well!

Thanks Impex. But yaar, still not clear. I had seen the equality sign being added, but how and why exactly does addition of 1 leads to this? Can you elaborate on this point?