Q7 of SGS 1999 Trial (1 Viewer)

[scardizzle]

Could someone walk me through the solution to this question? Im having a little trouble understanding what the hell i'm supposed to show.

Thanks in advance

 

[xFusion]

you could say PA=a, PB=a +2r
so a Geometric Mean would be
the Arithmetic mean would be
So therefore it is obvious that

 

[untouchablecuz]

t²=a(a+2r) → a²+2ra-t²=0

a=(r²+t²)-r disregarding the negative root since a>0

(PA+PB)/2=(a+2r+a)/2=a+r=√(r²+t²)=t√(r²/t²+1) > t since r²/t²+1>1

√(PA x PB)=√(a(a+2r))=√(t²)=t

.'. (PA+PB)/2>√(PA x PB)

for b)

i'm not too sure, is it x²+y²=(AB)²

LOL

 

[scardizzle]

Thanks for the replies guys still confused about part 2 though..

Here's what the solutions say:

t^2 = a(a + 2x) so t is constant, so locus is the circle centre p radius t

anyone care to explain? I have no idea how they came to that conclusion from just looking at the above equation

also another q i dont understand (-_-)

sorry about constantly asking questions

 

[untouchablecuz]

i shall complete the rest later, english is calling