Given a Golden Rectangle of length f and height 1, the inscribed ellipse would follow the equation:

The Golden Triangle, as depicted above, would set horizontally, symmetric to and bisected by the x-axis, touching the Golden Rectangle only at the leftmost point of the x-axis. I calculated that the line passing through the left tangent of the ellipse to the superscribed Golden Rectangle such that the angle it made with the x-axis would be 18° to result in a Golden Triangle would be:

Setting these two equations equivalent and solving for x gave the point (1/44)(5Ö5 + 9), with y = ±(1/11) Ö(25 - 2Ö5). Armed with this point, we can compute the three distances of interest in the inscribed Golden Triangle: the base, the length of each leg of the isosceles triangle, and the height. They are as follows:

Base = (2/11)Ö(25 - 2Ö5) (2.1)

Leg = (1/11)Ö[2(65 + 19Ö5)] (2.2)

Height = (1/11)(4Ö5 + 5) (2.3)

As the result was not elegant, I wondered about the OTHER Golden Triangle one can inscribe in the Golden Ellipse, the one set symmetric to and bisected by the y-axis, touching the superscribed Golden Rectangle only at the uppermost point of the y-axis, shown above. The line passing down and to the right from the y-axis at this point making an angle of 18° with the y-axis would be:

The Golden Ellipse equation would be as above. Setting the two equivalent and solving for x gives the position (1/62)Ö[2(145 + 19Ö5)]. Finding the corresponding y gives the three distances thus:

Base = (1/31)Ö[2(145 + 19Ö5)] (2.4)

Leg = (1/31)Ö[2(265 + 101Ö5)] (2.5)

Height = (1/62)(11Ö5 + 35) (2.6)

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