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

hariranjan

due to symmetry FE lies on BD: let points of intersection of AC & BD be O
then let FO=x then OE =1+x
let <OAE=z
in right triangles AOF & AEO
tan(45-z)=x/(4+x)
tanz=(x+1)/(4+x)
and tan(45-z)=(1-tanz)/(1+tanz)
solving x=2
hence FE=2+2+1=5

srikar2097

Quote:

Originally Posted by rajeev_hts

Let <EAD = p => <FAB = 45-p
Area(ECFA) = EF * AC = (a root(2) - 7)aroot(2)=a^2 - 7a/root(2)
ALso Area(ECFA) = FA * EA

In Tr(AED)

AE Sin p = 3

a = AE sin(45+p)

=> AE = Root((a root(2) -3)^2+9)

In Tr(AFB)

a=AF Cos p
AF Sin(45-p) = 4

=> AF=Root((a-4 root(2))^2 + a^2)

puting above

2 a^4 - 14 root(2) a^3 + 147 a^2 -336 Root(2) a + 576 =0

Still soving....

@Rajeev: How did you assume points E and F are on the diagonal?

srikar2097

Quote:

Originally Posted by hariranjan

due to symmetry FE lies on BD: let points of intersection of AC & BD be O
then let FO=x then OE =1+x
let <OAE=z
in right triangles AOF & AEO
tan(45-z)=x/(4+x)
tanz=(x+1)/(4+x)
and tan(45-z)=(1-tanz)/(1+tanz)
solving x=2
hence FE=2+2+1=5

What symmetry are you talking about? Can you elaborate on that? I don't symmetry. Points E and F are 3 and 4 units away from D and B. Also, they need not be on the diagonal.

rajeev_hts

Quote:

Originally Posted by srikar2097

@Rajeev: How did you assume points E and F are on the diagonal?

Just to make life easier but It still did not help to make my way any easier.

hariranjan

symmetry of A & C as both gives 45 deg angle to FE ; its not that F&E are symmetric to point of intersection O

srikar2097

Quote:

Originally Posted by rajeev_hts

Just to make life easier but It still did not help to make my way any easier.

That's true rajeev, I tried this method and ended up in a cul de sac. That's the reason I asked...

rajeev_hts

Probably you n me should have applied alternative approach.

I completely agree with hariranjan's sol.
ans(1) 5

srikar2097

Quote:

Originally Posted by hariranjan

symmetry of A & C as both gives 45 deg angle to FE ; its not that F&E are symmetric to point of intersection O

The points E and F are 3 and 4 units from D and B respectively. In some way a circle of radius 3 with centre D could be thought of and E is on the circumference of this circle inside the square. E need not lie on the diagonal of the square by default. Unless you can prove it. Similar with BF.

One constraint being EAF makes an angle of 45 so E is fixed somewhere. I'm just not convinced that it's on the diagonal...