An isosceles triangle has two congruent legs of length 1. The
Steiner circumellipse of the triangle is then drawn, and the foci of the ellipse are marked.
Question 1: Is it possible for the foci to lie upon the pair of congruent legs of the triangle? If so, how long is the triangle's base?
Question 2: Is it possible for one focus to lie upon the base of the triangle? If so, how long is the triangle's base?
The quick answer is YES to both questions.
I made a Desmos graph to play with.
Let f=half the base length.
The height of the triangle h=sqrt(1-f^2)
The medians of the triangle are the center of the ellipse (by definition) and in an isosceles triangle this is 1/3 of the way up the height so b=2h/3 (the b in the ellipse formula = half the vertical axis)
For simplicity the ellipse center is (0,0)
The rest you can see in the graph
https://www.desmos.com/calculator/t8tjzn3t78
you can either drag the red points or the slider with f.
For question 1:
The focus length is c=2*sqrt(4f^2-1)/3
and the sides have x intercepts 2f/3
Setting these equal f=sqrt(3)/3 so the base is 2*sqrt(3)/3
For question 2:
The focus length is the same (but vertical)
the base is on the line y= -sqrt(1-f^2)/2
Setting these equal f=sqrt(5)/5 so the base is 2*sqrt(5)/5
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Posted by Jer
on 2023-02-17 13:05:58 |
Solution
