Help me with this geometry problem

Factual Questions
1

… and don’t tell Ms. Flaherty, my ninth grade geometry teacher, that I couldn’t figure this out a mere 20 years later.

Imagine a pyramid with a square base and each face is at a 45 degree slope toward the top of the structure. Call the width of each side of the base a, and the line connecting the corner of the base to the top of the pyramid b.

Now each side of the pyramid is an isosceles triangle with a base of a and two sides of b. If you look at this triangle, what is the length of b in terms of a? Or, alternatively, what is the angle formed at ab?

I know it has to be between 45 and 90 degrees, and I want to say it’s 67.5, but that doesn’t seem to be right.

2

Let c be a line that connects the center of base a with the vertex b-b. As I work it out:

c = a / sqrt(2)
The Pythagorean theorm says that

b^2 = c^2 + (a / 2)^2
Combining these, we get

b = a * sqrt(3) / 2
Does that look right?

3

a=b, thus the triangles are equilateral thus all angles are 60°

This is how.

Lets call the centre of the base-square C, one corner A and the top B.

Angle CAB is 45°.
If base-sides are b, then line CA is b/(sqr rt 2).
If CAB is 45°, then CA=CB=b/(sqr rt 2), thus AB=b

Ergo

All sides are equal.

I may be wrong

4

Hmmm. How did you arrive at the length of c?

5

Sorry, that question was for Xema not Rodrigo.

6

I think Xema’s right. My own quick pass produced:

c as defined by Xema

b[sup]2[/sup]=c[sup]2[/sup] + (½*a)[sup]2[/sup]

7

The triangle consisting of the diagonal of the base has a side of length a*sqrt(2), two sides of length b, and two 45 degree angles which are not opposite the diagonal. So the angle opposite the diagonal is a right angle, and the Pythagorean theorem applies:

b[sup]2[/sup] + b[sup]2[/sup] = 2a[sup]2[/sup]

Therefore, a = b.

8

Sorry, I misunderstood Xema to mean c connected to the center of the base of a side, when he meant the center of the square base.

Rodrigo’s logic looks sound to me, though.

9

Thanks, all. When I was 14 I could have done that in my sleep. That’s what a career in Human Resources will do to you.

10

Um… Maybe I’m missing something here, but people seem to have arrived a two different answers by making two different assumptions. If the sloped faces make 45-degree angles with the plane of the base, then Xema’s analysis is correct and b = sqrt(3)*a/2. If the edges of the sloped faces make 45-degree angles with the base, then rodrigo and ultrafilter are right and a = b.

Ain’t solid geometry fun?

11

If I admitted I understood ultrafilter’s sig, would that make me even geekier than I realized? What if I said I was going to answer the angle question with trig because I thought it would be easier?

12

Since each face is sloping up at a 45 degree, that means the height of the pyramid is the same as the distance from the middle of a base to the centre, at 1/2 a. The diagonal of the base is asqrt(2), half of that is asqrt(2)/2.

Length of b is sqrt((1/2a)**2+(a*sqrt(2)/2)**2)

13

My answer agrees with Xema. The height of the pyramid is a/2 so the length of a face from tip to the center of the base is a/sqrt(2).

b is then sqrt(a[sup]2[/sup]/4 + a[sup]2[/sup]/2) = sqrt(3)*a/2

14

I knew I was guilty when I understood his NAME :smack:

15

I was a lot happier when I went to bed thinking that a=b. Now I’m going to have to figure out where sqrt(a[sup]2[/sup]/4 + a[sup]2[/sup]/2) = sqrt(3)*a/2 falls on my scale ruler.

16

I didn’t say it very well, but c is supposed to be a line that stats at the top and bisects one of the triangular faces.

According to my calculation, the angle sought in the OP is 54.7 degrees.

17

Here are two ways of visualizing the solution in your mind.

  1. Bisect the pyramid with a vertical plane through the middle of a face.
    The resultant cross-section is a triangle with a base of a and two sides c

  2. The pyramid sides slope at 45 degrees, so the apex is a right angle.
    c^2 + c^2= a^2 so a=sqrt(2)c

  3. c bisects each face into right triangles with sides a/2, c, and b
    (a/2)^2 + c^2 = b^2
    [a/2}^2 + [a/sqrt(2)]^2 = b^2

b= sqrt(3)/2 a

Theta= arcsin(c/b) = arcsin[asqrt(2) / asqrt(3)/2] = arcsin[2/sqrt(6)]
= 54.74 degrees

If you are more into 3D visualization, here’s a fun exercise:

  1. Make a cube from 6 pyramids (apices at the center, each base is a face)
    They fit exactly into a cube, since each edge is a, and each side forms a 45 degree angle to its perpendicular. (45+45=90)

  2. bisect the cube with the diagonal plane defined by any pyramid face.
    The resulting cross-section is a rectangle with sides a and sqrt(2)a.
    (The rectangle’s long side is a diagonal of a pyramid base/cube face.)

The pyramid edges of length b connecting the vertices to the center
(where the apices meet) can be shown to form diagonals by symmetry

From this figure, you can calculate the angles and the ratio between
b (the half-diagonal) and a (the basis of both cross-section sides)

  1. (2b)^2 = a^2 + (a sqrt(2))^2 = 3 a^2
    b = sqrt(3)/2 a

theta = arcsin[asqrt(2) / 2sqrt(3)/2] = arcsin(sqrt(2)/sqrt(3))
= 54.74 degrees