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

[the_egonomist]

Aarav,

I seem to have assumed a lot of stuff in my approach.

Giving it a once over and will post.

Cheers.

[thebornattitude]

Originally Posted by Aarav

Treat A (0, 0), B (3, 0), C (3, 1), D (0, 1) and E is (3p, p) as it lies on AC. Let tanI denotes the Inverse of tan.

Slope of EB = tanI[p/(3p-3)] = 90 - µ (equal angle)
Slope of ED = tanI[(p-1)/(3p)] = 90 -2*µ
.....
.

Aarav, If thats the case, then BE is tangent to the circle which circumscribes the rectangle,,, and if so, then why am i getting a wrong answer.

[r_mondkar]

Originally Posted by Aarav

Please check your calculations, I'm confident that you have got it but for some mistake like Andrew Wiles made

yup there ws a mistake, E=(27/8,9/8 )

but i am getting ans=3:1

[r_mondkar]

double post

[mehrozkhn]

thanx for pointing out the mistake buddy

Originally Posted by arbit_rageur

Diagonals don't intersect at right angles for a rectangle...only true for a square and a rhombus.

[Aarav]

Geometrical solution to QQAD 046. Read this to open the circuits in your brain. In the exam, this question was expected to be solved using co-ordinate approach, the the solution for which is already shared.

Let F be the intersection of the lines BC and ED. The triangle BEF is isosceles, according to the given information. Let R be the point that is on the same side of line EB as F, and for which the triangle EBR is isosceles and right-angled, that is, BR=ER and < ERB=90 degrees.


Let, furthermore, T be the intersection of the circumscribed circle of rectangle ABCD with the line ED. BD is a diameter in the circle, therefore the segment BT is perpendicular to the line ED.

As ACTD is a cyclic quadrilateral, < ETC= < EAD = < ECF. The angles of the triangles ECF and ETC are pairwise equal, thus the two triangles are similar, and EC:EF=ET:EC, that is, EC^2=EF.ET.


As < ERB = < ETB = < 90 degrees, the quadrilateral ERTB is also cyclic, and < RTE =< RBE =< ERF. The triangles ERF and ETR are also similar, thus with the above reasoning we have ER^2=EF.ET.
We have obtained that EC^2=ER^2=EF.ET, that is, EC=ER, and hence our answer.

[ashima chopra]

i guess da ans. is 5
v cnt get da ans. by da help of da 2 conditions.......
plz let me knw if m rite

[thebornattitude]

Originally Posted by Aarav

As < ERB = < ETB = < 90 degrees, the quadrilateral ERTB is also cyclic, and < RTE =< RBE =< ERF. The triangles ERF and ETR are also similar, thus with the above reasoning we have ER^2=EF.ET.
We have obtained that EC^2=ER^2=EF.ET, that is, EC=ER, and hence our answer.

didnt get the part in bold........how come <RBE=<ERF

[sharathbt21]

Let R be the point that is on the same side of line EB as F

can you please give a diagram for this??

[shivam_01]

Originally Posted by Aarav

Treat A (0, 0), B (3, 0), C (3, 1), D (0, 1) and E is (3p, p) as it lies on AC. Let tanI denotes the Inverse of tan.

Slope of EB = tanI[p/(3p-3)] = 90 - µ (equal angle)
Slope of ED = tanI[(p-1)/(3p)] = 90 -2*µ
=> µ = tanI[(3p-3)/p], and 2*µ = tanI[(3p)/(p-1)].

Eliminiate µ using tan (2*µ) = 2tan(µ)/(1-tan^2(µ))
and we get p as [11 +/- root(11)]/10.

We need to find square root of [(3p-3)^2 + p^2]/[(3p-3)^2 + (p-1)^2] = 2.

You all will be stunned to note that the ratio AB/BC was irrelevant and infact our asked ratio is always √2:1. Check this in case of a square.

Those who understood what I wrote are suggested to prove the same result for any rectangle. This will help you grasp co-ordinate approach to geometry well.

Thnaks Aarav for serving as an eye opener and prompting all of us to think out of box which is good
thanks a lot mate