My friend finished making dodecahedrons, and her next project is to make regular icosahedrons (20-sided dice). Again she wants to know: what is the dihedral angle between any two adjacent faces?
Perhaps this can be solved without the use of spherical trig? ;P
It's time for some of my random and indiscriminate puzzle-solving!
Edit: I found out that I had made a mistake. F cannot be directly
above C, so I'm off by a degree. I'd fix it, but then I might as
well rewrite the whole thing. Oh well... It's not like,
say, English class, where it's mandatory for me to write this.
Speaking of which, I have an English paper to write.
If you
take a cross-section of an icosahedron, straight down the middle, you
can get a regular decagon. The angles are, of course 144 degrees
each. But these are not the dihedral angles! The cross
section is goes right through the middle of a row of triangles arranged
in a way similar to the below diagram:
____________
/\ /\ /
/ \ / \ /
/____\/____\/
The next diagram is merely for labeling points.
__________E
/\ /
/ \ /
A B C
/ \ /
D________\/
Now
let's look at the same triangle from a different point of view.
The triangle that includes A, B, and D is on the horizontal plane, the
other is bent slightly downward. For simplicity, let's say the
side of the triangle is 1 unit long.
Let's extend line AB
to point F, which is right above C. BCF is a right
triangle. Angle FBC is 36 degrees and BC is 1/2. CF must be
sin(36 deg.)/2.
Let's extend line DB to point G, right
above E. EG must be sin(36 deg.) because E is twice the distance
below the horizontal that C is. BGE is another right triangle,
and BE equals sqrt(3)/2. Using trig, angle GBE must be
asin(2*sin(36 deg.)/sqrt(3)).
The dihedral angle, DBE is
of course 180 deg.-asin(2*sin(36 deg.)/sqrt(3)) This is equal to
about 137.2566 degrees or 2.3956 radians.
Edited on January 14, 2005, 3:19 am
Edited on January 14, 2005, 4:11 am
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Posted by Tristan
on 2005-01-14 03:16:59 |
Solution
