First of all. The bigger circle on the drawing is a cylinder. Three smaller circles are spheres. And they are situated as you can see on the drawing.
I must find radius of cylinder (r). R is radius of the sphere. So my solution is the following:
X=H-R => R*sqrt(3)-R => R*[sqrt(3)-1]
r=(2*H)+X => 2*[R*sqrt(3)]+X => 2*[R*sqrt(3)]+R*[sqrt(3)-1] => R*[3*sqrt(3)-1]
August 4th 2005, 10:35 AM
Sorry mistakes were made before! I realised after reading the following posting from tibol.
Working out r=R[3*sqrt(3)-1] gives r=4.196152423*R so the radius of the cylinder would be four times greater than that of the smaller sphere. This seems too large.
The triangle that is inscribed in the larger circle is equilateral - a fact that can be used to make progress. Each angle is 60 degrees
Try drawing the radius so that it bisects one of these angles.
The hypotenuse of the triangle formed is of length y, the adjacent side is of length R. Angle is half of 60 or 30.
| \ y
C R B
Now cos(30)=(sqrt(3)) / 2) = Adj / Hyp = y / R
sqrt(3)/2 = R/ y
y = 2/(sqrt(3) * R
y = 1.1547 * R
r=1.1547R + R
Adding on the the radius of the smaller circle R gives r=2.1
There are probably a few ways of doing this problem. I'll be the first to admit that I've assumed a lot about the radius of the cylinder bisects the angles of the triangle neatly!
August 4th 2005, 11:37 AM
Why did you say r = 2H +X ?
Even if you assumed r = 2H +X, you should explain why you assumed that is true.
(I corrected my mistake before. I put 2H now instead of the 2R before.)
Here is one way to solve for r.
On your diagram, the lefthand end of the green radius r is the center of the cylinder. It is also the center of the "concave, curved triangular" area enclosed by the 3 smaller circle. It is also the intersection point of the angle bisectors of the 3 angles of the equilateral/equiangular triangle whose 3 sides are 2R each.
These 3 angle bisectors divide the equilateral triangle into 3 congruent isosceles triangles, anyone of which has:
>>>base = 2R
>>>2 sides that are same in length = y long each
>>>2 base angles = 30 degrees each
>>>angle included by the two equal sides = 120 deg
If we draw a line segment from the apex to the midpoint of the base, two congruent right triangles are formed, each of which has these:
>>hypotenuse = y
>>one leg = R
>>the other leg = "line from apex to midpoint of base"---this is perpendicular to the base.
cos(30deg) = R/y
y = R / cos(30deg)
y = (1.1547)R
Back to your diagram,
r = y +R
r = (1.1547)R +R
r = (2.1547)R -------------answer.
I forgot, you want exact answers.
y = R / cos(30deg) = R / sqrt(3)/2 = 2R / sqrt(3) = 2R[sqrt(3) /3]
y = (2/3)R*sqrt(3)
r = [1 +(2/3)sqrt(3)]R
r = [(1/3)(3 +2sqrt(3))]R --------answer.
August 4th 2005, 01:05 PM
I just wonder about the way of finding the radius.
By the way, I thought something like that: "It is also the intersection point of the angle bisectors of the 3 angles of the equilateral/equiangular triangle whose one side is 2R." But I didn't try this way. As I see, this is the right way to be progress. However I admit that using the sine of the acute angle (cos 30deg = sub leg / opposite leg) surprised me. I have used this way but on another excercises.
"Why did you say r = 2R +X ?"
This was r=2H+X. I just looked at my drawing and thought that it could be right but was wrong.
August 4th 2005, 11:05 PM
Oops, yes, it was r = 2H +X, not r = 2R +X.
"....(cos 30deg = sub leg / opposite leg) ..."
No, it is not like that. (Besides, what is sub leg?)
It is cos(30deg) = (adjacent side)/(hypotenuse).
"If we draw a line segment from the apex to the midpoint of the base, two congruent right triangles are formed, each of which has these:
>>hypotenuse = y
>>one leg = R
>>the other leg = "line from apex to midpoint of base"---this is perpendicular to the base."
The acute angle 30 degrees is included by the hypotenuse y and the leg R.
So, adjacent side = R
hypotenuse = y
And, cos(30deg) = adj / hyp = R/y
The opposite side here is the "line from apex to the midpoint of the base"
"But I didn't try this way. As I see, this is the right way to be progress.
.......... I have used this way but on another excercises....."
I have been saying in Math, there are many different ways/solutions to the same one answer. So, try as many ways as you know until you come to the correct answer. If you cannot get the answer in one way only, don't stop there.