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Mitres on Pyramids

Date: 09/26/2002 at 09:57:07
From: Simon K
Subject: Pyramid internal angles

I am making a weather proofing construction for my home, and have to 
mitre the boards in a pyramid with a rectangular - not square - base, 
and an apex that is off to one side (though central in relation to the 
other sides) - say it is directly over the centre of one edge of the 

I have looked at two good solutions on your site, 

The first, 

   Miter for a pyramid  

appears to rely on an equal-sided base. Does it? Or is this 
irrelevant? What is the significance of bisecting the base lines to 
get m and M? Is it just to get a conveniently known length with a 
conveniently known angle? If mC and MC were not equal, would mJ and MJ 
both meet TC at right angles? The 45-degree angle of the line mM 
cutting across the base - is this significant? Must it be 45 degrees 
for "dropping a perpendicular" - or is it just an example? This is a 
beautiful and easy solution, just a bit of ignorance either side of 
the perfect square base solution makes it hard for me to use.

The second solution,

   Pyramid Construction 

gives me two problems also.

1. I cannot interpret the lines, the spine and the leaves of the book 
solution to the pyramid problem. Can you help? (Is it right to 
consider the lines the base, and the spine the pyramid edge/corner 

2. If you draw your first line from the center of the spine downward 
at an acute angle (10 degrees from the bottom half of the spine to the 
line) and the other line opposite it at 30 degrees from the spine, and 
then close the leaves of the book to 90 degrees, I think you will 
agree that the angle between the two lines is not 90 degrees, yet the 
solution suggests this - so I am confused. It looks very elegant and 
I hope it will solve my weatherproofing construction problem fast if
only I can interpret it correctly.

Simon K.

Date: 09/27/2002 at 11:15:19
From: Doctor Rick
Subject: Re: Pyramid internal angles

Hi, Simon.

   Miter for a pyramid  

does assume that the base is a square and the side triangles are 
congruent isosceles triangles. Because of the symmetry in the problem, 
the lines drawn perpendicular to a slanted edge from points equal 
distances out on the adjacent base edges will meet the slanted edge at 
the same point; the derivation depends on this.

In your case it sounds as if the base is a rectangle so its angles are 
right angles, but the side triangles are neither isosceles nor 
congruent. You need a more general solution.

   Pyramid Construction 

begins with the same special case (square base with congruent 
isosceles triangles for the sides). Dr. Mitteldorf, however, discusses 
the general case. He approaches it by describing a problem that is the 
reverse of yours: you know the angle between the sides of the pyramid 
(that's how wide the book is open) and you want to find the base angle 
(that's the angle between the diagonal lines on the pages). You are 
quite right in your interpretation: the spine is the slanted edge of 
the pyramid and the diagonal lines are the edges of the base.

Dr. Mitteldorf does not say that the angle between the lines is 90 
degrees when the book is open 90 degrees. The angle between the lines 
is found by his formula

  cos(x) = cos(a)*cos(b) + sin(a)*sin(b)*cos(c)

If you know that angle x is 90 degrees (as in your case), and you know 
angles a and b (the base angles of the side triangles at the vertex 
under consideration), then you can solve for cos(c) and then for c 
itself. The cosine of 90 degrees is 0, so in this case the formula 
reduces to

  0 = cos(a)*cos*(b) + sin(a)*sin(b)*cos(c)

which can be solved for cos(c) easily:

  cos(c) = -cos(a)*cos(b)/(sin(a)*sin(b))
         = -1/(tan(a)*tan(b))

  c = arccos(-1/tan(a)*tan(b))

If you know or you can find the base angles of the side triangles, 
then you can find the angle between the triangles using this formula. 
It will not be 90 degrees unless either tan(a) or tan(b) is infinite,  
or both are infinite (that is, a and/or b is 90 degrees).

As a bonus, here is a way to use geometry to derive the general 
formula. I'll look at one base vertex of the pyramid. I'll mark off a 
point on each of the three edges that meet at this vertex (exactly 
how they are marked off will be described in a moment, but for now, 
here's the figure so you can see what I'm referring to:


The three angles at this vertex are the angle in the base 
quadrilateral (x = BAC) and the angles in the two adjacent side 
triangles (a = BAD, and b = CAD). We want to find angle c = BDC. To 
do this, we can do essentially what Don Mohr (the writer of the 
second question) did, but for the general case. 

Mark off point B on one edge of the base, one unit from the vertex A. 
Drop a perpendicular from point B to the slanted edge, meeting the 
edge at D. From D draw another line perpendicular to AD and 
intersecting the second edge of the base at C. Thus angles ADB and 
ADC are right angles. Now we know enough about the triangles we have 
constructed to find the lengths of all the sides:

  AB = 1
  BD = sin(a)
  AD = cos(a)
  AC = AD/cos(b) = cos(a)/cos(b)
  CD = AD tan(b) = cos(a)*tan(b)

The triangle ABC is not a right triangle in the general case (though 
I think it is in your problem), so we need to use the Law of Cosines 
to find BC:

  BC^2 = AB^2 + AC^2 - 2*AB*AC*cos(x)
       = 1 + cos^2(a)/cos^2(b) - 2*cos(x)*(cos(a)/cos(b)

Now we turn our attention to triangle BCD, which again is not a right 
triangle, and apply the Law of Cosines:

  BC^2 = BD^2 + CD^2 - 2*BD*CD*cos(c)

  1 + cos^2(a)/cos^2(b) - 2*cos(x)*(cos(a)/cos(b) =
    sin^2(a) + cos^2(a)*tan^2(b) - 2*sin(a)*cos(a)*tan(b)*cos(c)

Multiply through by cos^2(b) to clear the denominators:

  cos^2(b) + cos^2(a) - 2*cos(x)*cos(a)*cos(b) =
    sin^2(a)*cos^2(b) + cos^2(a)*sin^2(b) 
    - 2*sin(a)*cos(a)*sin(b)*cos(b)*cos(c)

Isolate cos(x) on the left:

  2*cos(a)*cos(b)*cos(x) =
    2*sin(a)*cos(a)*sin(b)*cos(b)*cos(c) + cos^2(a) + cos^2(b)
    - sin^2(a)*cos^2(b) - cos^2(a)*sin^2(b)

A little trig simplification on the right side gives us

  2*cos(a)*cos(b)*cos(x) =
    2*sin(a)*cos(a)*sin(b)*cos(b)*cos(c) + 2*cos^2(a)*cos^2(b)

Now we can just divide through by (2*cos(a)*cos(b)):

  cos(x) = sin(a)*sin(b)*cos(c) + cos(a)*cos(b)

There's Dr. Mitteldorf's formula, derived by generalizing the method 
for a symmetrical pyramid.

- Doctor Rick, The Math Forum 

Date: 09/27/2002 at 19:36:17
From: Simon K
Subject: Thank you (Angles at an angle - or Mitres on pyramids with 
non-equilateral sides, and off-centre apexes.)

Dear Dr. Rick,
Thank you very much for replying so quickly. I really 
appreciate your site and your effort. It is VERY tough 
wrestling with problems like this on your own - to most 
people in ordinary life, this is EXTREMELY obscure stuff. 
Yet, absolutely essential to resolve a practical problem 
that is so small - just to join two pieces of wood cleanly!

I am replying now (very late at night) because I shall be 
away for 2 days and feel you deserve a response. I have not 
yet had time to digest and grasp, let alone apply your 
solution, so hope to write to you again later when I have. 
For now - thanks. 
Simon K.
Associated Topics:
College Polyhedra
College Trigonometry
High School Polyhedra
High School Trigonometry

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