Let the distances between the vertices of a unit equilateral triangle and a point on its incircle be a, b, and c. If a, b and c are in geometric progression, find the value of b.
In general, assuming a is the shortest distance, then
b=(2sqrt(2)+1)/sqrt(7)a.
In the unit equilateral triangle, the geometric series is quite pretty:
a= 10^(1/2)/5 -5^(1/2)/10
b= 35^(1/2)/10
c= 10^(1/2)/5+5^(1/2)/10
Edited on October 6, 2019, 10:02 am
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Posted by broll
on 2019-10-05 23:28:50 |
Possible solution
