Tough Geometry Problem

IMO A=1:3
V=SH/3=piR^2H/3
now, x(smaller radius)=R/2
V2(right one)=pi(R/2)^2H/3......(1)
V1(left side of cone)=V-V2 =piR^2H/4.....(2)
this gives, V2:V1=1:3

Hi earth@work,

I got the same result, but, it is apparantly wrong. OA is D, meaning, 1:7.

BTW, I followed the same steps as you; I used the formula for the area of a cone too.

Sorry, there was a miss calculation in solution above....Took height for V2 as H instead of H/2...have corrected below
V=SH/3=piR^2H/3
now, x(smaller radius)=R/2
V2(right one)=pi(R/2)^2(H/2)/3......(1)
V1(left side of cone)=V-V2 =1/3*piR^2H/4*7/8.....(2)
this gives, V2:V1=1:7
So answer is D=1:7

Thanks! However, could you please further elaborate on how you got x=R/2 and height of right cone= 1/2 of left cone.

Thanks for your help.

Given:

Let the radius of the smaller base be x
Let the radius of the larger base be y

Given AB/AC = 1/2, or AC = 2AB

Using the property of similar triangles,

x/y = AB/AC = 1/2
or y = 2x

Ratio of the volumes of the small (to the right and the large cone):
Vs/Vl = (pi * x^2 * AB/3)/(pi * y^2 * AC/3)
Vs/Vl = 1/8 (after simplification)

Vs/(Vl-Vs) = 1/7
Therefore, D is correct

smohsin wrote:Thanks! However, could you please further elaborate on how you got x=R/2 and height of right cone= 1/2 of left cone.

Thanks for your help.

Hi smohsin, attaching a figure with clarification