Find the radius

**Jenny20** · December 29th 2006, 10:27 PM

question

A bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by a Styroform sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used.

Thank you very much.

**anthmoo** · December 30th 2006, 01:26 AM

I'm not sure if I'm correct but here it goes...

Instead of calling the radius R, call it x as it's probably something we're both familiar with.

Lets only imagine the top left corner of the box and cut it out so we have box of sides of length of the radius x. Also in the box is top left quarter of our circle of radius x so it's not touching the top left corner.

Now we should find the length of the diagonal that goes from the centre of the big circle to the top left corner of the box:

$\sqrt x^2 + x^2=\sqrt 2x^2 = x\sqrt 2$

How do we get the radius of the smaller circle?

Well we get the length from the edge of the big circle to the top left corner leaving just the smaller circle within it. To do this it is just the diagonal length minus the radius of the big circle:

$x\sqrt{2} - x$

Which can not be canceled down further.

HOWEVER, this is the diagonal of the line from where the big circle touches the smaller one so really we have the diameter of the circle plus the bit in the corner. So in plain terms, we have the diameter plus top left corner minus the bottom right corner (PS. to visualise this, draw it all out with the diameter always drawn as the diagonal through the box).

So since the corners cancel out, we actually had the diameter all along! And in order to get the radius, just halve it. Which gives:

$\frac{x\sqrt{2} - x}{2}$

And since you want it in terms of R....

$SmallRadius=\frac{R\sqrt{2} - R}{2}$

I hope this is alright for you to understand! I was really working it out as I was going along so you have my thoughts here too and this is my second helping post I think too!

I think a good tip would be to draw out each stage as you read it!

Also you should note that you can keep working out the radiuses of each smaller circle like this an infinite number of times as it is impossible to reach the corner with circles!

Anthmoo

**CaptainBlack** · December 30th 2006, 01:44 AM

Originally Posted by anthmoo

I'm not sure if I'm correct but here it goes...

Instead of calling the radius R, call it x as it's probably something we're both familiar with.

Lets only imagine the top left corner of the box and cut it out so we have box of sides of length of the radius x. Also in the box is top left quarter of our circle of radius x so it's not touching the top left corner.

I'm having difficulty following your explanation, but to me it looks like you
are solving the 2-D analog of this problem rather than the 3-D problem
asked. Is this so?

RonL

**earboth** · December 30th 2006, 01:46 AM

Originally Posted by Jenny20

question

A bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by a Styroform sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used.

Thank you very much.

Hello Jenny,

I've attached a sketch of the described situation.

By Pythagorian rule you get:

$r^2+r^2=(r+x)^2$ . Thus $r+x=r \cdot \sqrt{2}$

You get the propoertion:

$\frac{R}{r}=\frac{R+2r+x}{r+x}$

$\frac{R}{r}=\frac{R+r+r+x}{r+x}=\frac{R+r}{r+x}+1$

$\frac{R-r}{r}=\frac{R+r}{r+x}$ . Now plug in the term for (r+x)

$\frac{R-r}{r}=\frac{R+r}{r \cdot \sqrt{2}}$ . Now solve for r. You'll get 2 solutions. The neagtive solution isn't very realistic with your problem. I've got:

$r=R\cdot \left(\frac{\sqrt{2}-1}{2} \right)$

EB

**CaptainBlack** · December 30th 2006, 02:02 AM

Originally Posted by earboth

Hello Jenny,

I've attached a sketch of the described situation.

By Pythagorian rule you get:

$r^2+r^2=(r+x)^2$ . Thus $r+x=r \cdot \sqrt{2}$

You get the propoertion:

$\frac{R}{r}=\frac{R+2r+x}{r+x}$

$\frac{R}{r}=\frac{R+r+r+x}{r+x}=\frac{R+r}{r+x}+1$

$\frac{R-r}{r}=\frac{R+r}{r+x}$ . Now plug in the term for (r+x)

$\frac{R-r}{r}=\frac{R+r}{r \cdot \sqrt{2}}$ . Now solve for r. You'll get 2 solutions. The neagtive solution isn't very realistic with your problem. I've got:

$r=R\cdot \left(\frac{\sqrt{2}-1}{2} \right)$

EB

2-D analog of the problem, not what was asked, which is for the maximum
sizes sphere that can be fitted into the corner of a box in the form of a
cube of side 2R containing a sphere of radius R.

(Probably wrong for the 2-D analog as well)

RonL

**Soroban** · December 30th 2006, 02:10 AM

Hello, Jenny!

A different approach . . . (and a different answer)

A bowling ball of radius $R$ is placed inside a box just large enough to hold it,
and it is secured for shipping by a Styroform sphere into each corner of the box.
Find the radius $r$ of the largest Styrofoam sphere that can be used.

Code:

                A    R    B
      *-------*-*-*-------*
      |   *     :     * / |
      | *       :     / * |
      |*       R:    /   *|R
      |         :   /     |
      *         : /       *
      *         * - - - - *C
      *         O    R    *
      |                   |
      |*                 *|
      | *               * |
      |   *           *   |
      *-------*-*-*-------*

In the upper-right, we have an R-by-R square: $OABC.$
The diagonal $OB$ has length $R\sqrt{2}.$

Diagonal $OB$ intersects the circle at $D$ (not shown).
Then: . $DB \:=\:R\sqrt{2} - R$

This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: . $\frac{R\sqrt{2} - R}{\sqrt{2}}$

The radius of the inscribed circle is: . $\frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$

Therefore: . $r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$

**anthmoo** · December 30th 2006, 02:12 AM

Originally Posted by CaptainBlack

I'm having difficulty following your explanation, but to me it looks like you
are solving the 2-D analog of this problem rather than the 3-D problem
asked. Is this so?

RonL

I know what you mean - would it be good enough to treat the situation as a 2-D one if we take the diagonal cross-section of the box?

**CaptainBlack** · December 30th 2006, 02:23 AM

Originally Posted by anthmoo

I know what you mean - would it be good enough to treat the situation as a 2-D one if we take the diagonal cross-section of the box?

No because the smaller sphere would not touch one side of the crossection.

RonL

**CaptainBlack** · December 30th 2006, 02:24 AM

Originally Posted by Soroban

Hello, Jenny!

A different approach . . . (and a different answer)

Code:

                A    R    B
      *-------*-*-*-------*
      |   *     :     * / |
      | *       :     / * |
      |*       R:    /   *|R
      |         :   /     |
      *         : /       *
      *         * - - - - *C
      *         O    R    *
      |                   |
      |*                 *|
      | *               * |
      |   *           *   |
      *-------*-*-*-------*

In the upper-right, we have an R-by-R square: $OABC.$
The diagonal $OB$ has length $R\sqrt{2}.$

Diagonal $OB$ intersects the circle at $D$ (not shown).
Then: . $DB \:=\:R\sqrt{2} - R$

This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: . $\frac{R\sqrt{2} - R}{\sqrt{2}}$

The radius of the inscribed circle is: . $\frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$

Therefore: . $r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$

Ho-Hum...

RonL - Too lazy to type the same thing a third time

**anthmoo** · December 30th 2006, 02:26 AM

Originally Posted by CaptainBlack

2-D analog of the problem, not what was asked, which is for the maximum
sizes sphere that can be fitted into the corner of a box in the form of a
cube of side 2R containing a sphere of radius R.

RonL

I tried to do it the diagonal way quickly and got something which I think is very incorrect...

$\frac{\sqrt{4x^2 + x^2\sqrt{8}} - x}{2}$

**anthmoo** · December 30th 2006, 02:29 AM

Originally Posted by Soroban

Hello, Jenny!

A different approach . . . (and a different answer)

Code:

                A    R    B
      *-------*-*-*-------*
      |   *     :     * / |
      | *       :     / * |
      |*       R:    /   *|R
      |         :   /     |
      *         : /       *
      *         * - - - - *C
      *         O    R    *
      |                   |
      |*                 *|
      | *               * |
      |   *           *   |
      *-------*-*-*-------*

In the upper-right, we have an R-by-R square: $OABC.$
The diagonal $OB$ has length $R\sqrt{2}.$

Diagonal $OB$ intersects the circle at $D$ (not shown).
Then: . $DB \:=\:R\sqrt{2} - R$

This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: . $\frac{R\sqrt{2} - R}{\sqrt{2}}$

The radius of the inscribed circle is: . $\frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$

Therefore: . $r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$

Well I guess I'm wrong..

Worth a try though!

**Jenny20** · December 30th 2006, 02:54 AM

Hi all,

Thank you very much for your reply.

I have the answer for this question. The answer is :

r =(2-sqrt(3))R

**Jenny20** · December 30th 2006, 03:07 AM

And also the hint:

Take the origin of a Cartesian coordinate system at a corner of the box with the coordinate axes along the edges.

**CaptainBlack** · December 30th 2006, 05:15 AM

Consider one corner of the box, the ball of radius $R$ , and a smaller sphere
which is the largest that will fit between the large sphere (the ball) and the corner.
Let the radius of the small sphere be $\rho$

Take one corner of the box as origin with axes along the edges of the box.

Then the diagonal of the box is the line $\lambda (1,\ 1,\ 1)$ and the
centres of the small sphere in the corner choosen as origin, and the
large sphere both lie on this diagonal, and so does the kissing point
between the two spheres.

The equation of the small sphere is:

$(x-\rho)^2+(y-\rho)^2+(z-\rho)^2=\rho^2$

and the points at which the diagonal meets this satisfy:

$(\lambda-\rho)^2+(\lambda-\rho)^2+(\lambda-\rho)^2=\rho^2$ ,

or: $\lambda=\rho(1 \pm 1/\sqrt{3})$

Now doing the same for the large sphere gives:

$\lambda=R(1 \pm 1/\sqrt{3})$ ,

and as the spheres are kissing the larger of the roots for the small sphere
is equal to the smaller for the large sphere, so:

$\rho(1 + 1/\sqrt{3})=R(1 - 1/\sqrt{3})$

Hence:

$\rho=R\ \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}}$

Which simplifies down to:

$\rho=R\ (2-\sqrt{3})$

RonL

**CaptainBlack** · December 30th 2006, 06:11 AM

Originally Posted by Soroban

Hello, Jenny!

A different approach . . . (and a different answer)

Code:

                A    R    B
      *-------*-*-*-------*
      |   *     :     * / |
      | *       :     / * |
      |*       R:    /   *|R
      |         :   /     |
      *         : /       *
      *         * - - - - *C
      *         O    R    *
      |                   |
      |*                 *|
      | *               * |
      |   *           *   |
      *-------*-*-*-------*

In the upper-right, we have an R-by-R square: $OABC.$
The diagonal $OB$ has length $R\sqrt{2}.$

Diagonal $OB$ intersects the circle at $D$ (not shown).
Then: . $DB \:=\:R\sqrt{2} - R$

This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: . $\frac{R\sqrt{2} - R}{\sqrt{2}}$

The radius of the inscribed circle is: . $\frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$

Therefore: . $r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$

The 2-D analog of the full 3-D solution is:

$R(3-2\sqrt{2})\approx0.172\,R$ ,

while the above solution gives:

$R(\sqrt{2}-1)/(2\sqrt{2})\approx 0.146\,R$ ,

and construction (drawing a scal diagram and measuring) gives a factor more
like $\approx 0.175$ which allowing for errors in construction is consistent with the
first but not the second of these candidates.

RonL

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