# a pyramid

• Aug 24th 2008, 01:57 AM
perash
a pyramid
Consider a pyramid with a square base. The side length of the
base is 2 units and the height of the pyramid is 1 unit. Imagine
placing a cube inside this pyramid (resting on the base of the
pyramid) such that each of the four top corners of the cube is
touching each of the 4 slanted edges of the pyramid .
Find the dimensions of the cube. That is, find the value, in
units, of the length of one edge of the cube.
• Aug 24th 2008, 03:27 AM
ticbol
Quote:

Originally Posted by perash
Consider a pyramid with a square base. The side length of the
base is 2 units and the height of the pyramid is 1 unit. Imagine
placing a cube inside this pyramid (resting on the base of the
pyramid) such that each of the four top corners of the cube is
touching each of the 4 slanted edges of the pyramid .
Find the dimensions of the cube. That is, find the value, in
units, of the length of one edge of the cube.

So the the top four corners of the cube are along the slanted edges of the pyramid. Then the four edges of the base of the cube are parallel to the edges of the base of the pyramid each to each.

View the figure from the top, or on the top. We will get a vertical cross-section along one of the equal diagonals of the base of the pyramid.
The diagonal of the pyramid is 2sqrt(2) units long...by Pythagorean theorem.
The diagonal of the cube, whose edge is, say, x units long, is x*sqrt(2) units long.

Now view that said cross-section vertically, or from the side, or from one of the un-cut corner of the base of the pyramid.
The figure is that of an isosceles triangle whose base is 2sqrt(2) units long, and whose height is 1 unit long.
Inside it is a rectangle whose base is x*sqrt(2) units long, and whose height is x units long.
Above the rectangle is a smaller isosceles triangle whose base is x*sqrt(2) and whose height is (1-x) units long.
The two isosceles triangles are similar, and so, proportional.
By proportion,
2sqrt(2) /1 = x*sqrt(2) /(1-x)
2 = x /(1-x)
2(1-x) = x
2 -2x = x
2 = x +2x
x = 2/3 unit long ---------------answer
• Aug 24th 2008, 05:16 AM
Soroban
Hello, perash!

I used the same approach as ticbol, but got a different answer.

Quote:

Consider a pyramid with a square base. The side length of the base is 2 units
and the height of the pyramid is 1 unit. Imagine placing a cube inside this pyramid
(resting on the base of the pyramid) so that each of the four top corners of the cube
is touching each of the 4 slanted edges of the pyramid.

Find the dimensions of the cube.
That is, find the value, in units, of the length of one edge of the cube.

Looking down on the pyramid, we see:
Code:

      *-----------*       | *      * |       |  *  *  |       |    *    | 2       |  *  *  |       | *      * |       *-----------*             2
The diagonal of this square is $\displaystyle 2\sqrt{2}$

Slice the pyramid along a diagonal and we have this cross-section:
Code:

    -                *     :              * | *     :            *  |  *     :          *-----+-----*     1        * |    |    | *     :      *  |    |    |  *     :    *    |    |  2x|    *     :  *      |    |    |      *     - *---------*-----*-----*---------*       :      √2      :  x  :  √2-x  :
The lower-right right triangle is similar to the largest right triangle.

We have: .$\displaystyle \frac{2x}{\sqrt{2}-x} \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad 2\sqrt{2}x \:=\:\sqrt{2} - x$

. . $\displaystyle 2\sqrt{2}x + x \:=\:\sqrt{2} \quad\Rightarrow\quad (2\sqrt{2}+1)x \:=\:\sqrt{2} \quad\Rightarrow\quad x \:=\:\frac{\sqrt{2}}{2\sqrt{2}+1}$

The side of the cube is: .$\displaystyle 2x \;=\;\frac{2\sqrt{2}}{2\sqrt{2} + 1} \;=\;\frac{2(4-\sqrt{2})}{7}$

• Aug 24th 2008, 04:58 PM
ticbol
Quote:

Originally Posted by Soroban
Hello, perash!

I used the same approach as ticbol, but got a different answer.

Looking down on the pyramid, we see:
Code:

      *-----------*       | *      * |       |  *  *  |       |    *    | 2       |  *  *  |       | *      * |       *-----------*             2
The diagonal of this square is $\displaystyle 2\sqrt{2}$

Slice the pyramid along a diagonal and we have this cross-section:
Code:

    -                *     :              * | *     :            *  |  *     :          *-----+-----*     1        * |    |    | *     :      *  |    |    |  *     :    *    |    |  2x|    *     :  *      |    |    |      *     - *---------*-----*-----*---------*       :      √2      :  x  :  √2-x  :
The lower-right right triangle is similar to the largest right triangle.

We have: .$\displaystyle \frac{2x}{\sqrt{2}-x} \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad 2\sqrt{2}x \:=\:\sqrt{2} - x$

. . $\displaystyle 2\sqrt{2}x + x \:=\:\sqrt{2} \quad\Rightarrow\quad (2\sqrt{2}+1)x \:=\:\sqrt{2} \quad\Rightarrow\quad x \:=\:\frac{\sqrt{2}}{2\sqrt{2}+1}$

The side of the cube is: .$\displaystyle 2x \;=\;\frac{2\sqrt{2}}{2\sqrt{2} + 1} \;=\;\frac{2(4-\sqrt{2})}{7}$

I'm sorry to comment, since perash or anybody did not comment, but the cube should appear as a rectangle, not a square, in your diagram. The length of the base now of the rectangle should be (2x)sqrt(2). Not 2x as is in your diagram. You sliced along a diagonal of the base of the pyramid, remember.
• Aug 26th 2008, 03:17 AM
Soroban
Hello, ticbol!

Another blunder . . . *blush*

Quote:

The cube should appear as a rectangle, not a square, in your diagram.
The length of the base now of the rectangle should be (2x)sqrt(2).
Not 2x as is in your diagram.
You sliced along a diagonal of the base of the pyramid, remember.

Absolutely right!

I'll try to correct my work and get back soon . . .