It’s easy to show that the five acute angles in the vertices
of a
regular 5-pointed star total 180°.
Please show that the sum of these angles in an irregular 5-pointed star is also 180°.
Source: A. Korshkov, the Russian science magazine Kvant.
Let A, B, C, D, and E label (in that order) the five points
of the star.
Let I = AC /\ BE,
J = BD /\ CA,
K = CE /\ DB, /\ denotes intersection
L = DA /\ EC,
M = EB /\ AD.
5*180 is the sum (in degrees) of the interior angles of the
following five triangles: CIE, DJA, EKB, ALC, and
BMD.
3*180 is the sum (in degrees) of the interior angles of the
pentagon IJKLM.
Star is the sum (in degrees) of angles A, B, C, D, and E.
Therefore,
2*Star + 3*180 = 5*180
or
Star = 180 degrees.
QED
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Posted by Bractals
on 2016-12-14 13:15:45 |
Solution
