Without loss of generality we can assume that
the length of the sides of ABCDE is 1.
Applying x-y coordinates to the vertices of
ABCDE we have
B [0,0]
C [1,0]
A [cos(136),sin(136)]
D [1 + cos(76),sin(76)]
Point E must belong to the unit circles
with centers at A and D:
Circle-A: [x - cos(136)]^2 +
[y - sin(136)]^2 = 1
Circle-D: [x - {1 + cos(76)}]^2 +
[y - sin(76)]^2 = 1
There are two points that satisfy these
conditions. One makes ABCDE convex and
the other does not. The one that does -
E [cos(76),sin(76)]
Clearly it satisfies Circle-D. I will let
you confirm that it satisfies Circle-A.
Clearly |BE| = 1. Therefore, triangle ABE
is equilateral and quadrilateral BCDE is
a rhombus. Therefore,
angle(AEB) = 60
angle(BED) = angle(BCD) = 104
angle(AED) = angle(AEB) + angle(BED) = 164.
QED
You can't construct this pentagon with
straightedge and compass.
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Posted by Bractals
on 2018-04-12 12:38:48 |
Solution
