Golden Ratio Conjecture in three simple Geogebra shapes--circle, triangle, and square.

Question

A circle, equilateral triangle, and square of equal heights are all placed on the same horizontal line as shown below. The circle is tangent to the triangle which is centered upon the left edge of the square. A line is drawn from the center of the circle to the right edge of the square, passing through the center of the square. The line is cut into two segments by the right side of the triangle, as shown.

Show that the ratio of the length of the blue segment to the green segment is the golden ratio 1.618. (Is it? It seems so!)

I have been playing around in geogebra, but I was unable to get the circle tangent to the triangle as shown in the figure, which I drew in Adobe Illustrator. Any geogebra assistance would be appreciated! How do I move/translate a simple object in such a manner? I am used to dragging and dropping it in Adobe Illustrator, but geogebra is much better suited to these golden ratio conjectures. Thanks for all your help! I have been successful with geogebra with a couple other constructions which I will share soon. :)

Answer 1

Here is a diagram of the situation.

Let us calculate the relevant lengths, assuming that the radius of the circle is $1$ so that we don't have to deal with scaling. One should note that $\angle CAB$ is $\pi/6$, equal to $\angle CDE$. Thus, the length of $AC$ is $\frac{1}{\cos(\pi/6)}$. The length of $CE$ and $EF$ separately is $\tan(\pi/6)$. Noting that $\cos(\pi/6)$ is $\sqrt{3}/2$ and $\tan(\pi/6)=\frac{1}{\sqrt{3}}$, we find that the length from $A$ to $F$ is $\frac{4}{\sqrt{3}}$. The length from $F$ to $G$ is $2-\frac{1}{\sqrt{3}}$. The ratio of these $$\frac{\frac{4}{\sqrt{3}}}{2-\frac{1}{\sqrt{3}}}=\frac{4}{2\sqrt{3}-1}$$ which is not the golden ratio, but is $1.623$ instead.

Answer 2

Let the radius of the circle be $r$. The length of the blue segment outside the triangle is: $$\frac r{\sin 60^\circ} = \frac{2r}{\sqrt 3}$$ The length of the blue segment inside the triange is half of that: $$2\cdot r\tan 30^\circ = \frac{2r}{\sqrt 3}$$ The length of the green segment is: $$2r - r\tan 30^\circ = \left(2-\frac1{\sqrt3}\right)r$$ The ratio of blue : green is $$\frac{4r/\sqrt3}{(2\sqrt3-1)r/\sqrt3} = \frac{4}{2\sqrt3-1} = \frac{8\sqrt3+4}{11} \ne\phi$$

Answer 3

Let the side of the equilateral triangle be $s$. Then the altitude of the triangle (which is also equal to the side of the square and the diameter of the circle) is $\frac{\sqrt3}{2}s$.

If you draw the radius of the circle to the point where the circle is tangent to the triangle, it is one leg of a $30$-$60$-$90$ triangle with part of the the blue line as the hypotenuse. That triangle is congruent to one of the small right triangles you can see in the upper part of the equilateral triangle. So the portion of the blue line between the center of the circle and the near side of the equilateral triangle is equal to the hypotenuse of one of those right triangles, which is $\frac12 s$.

The other part of the blue line (between the sides of the equilateral triangle) also has length $\frac12 s$, so the whole blue line has length $s$.

Another way to see this is to circumscribe a hexagon around the circle, using part of the tangent side of the equilateral triangle as one side of the hexagon. The draw all three main diagonals of the hexagon, cutting it into six equal equilateral triangles. We can then trace a parallelogram with the blue line as one side and the base of the equilateral triangle as the opposite side, so those two line segments have the same length.

(That figure should also show you how to construct the figure in Geogebra: from the lower left corner of the equilateral triangle, construct a line parallel to the opposite side of the triangle; the blue/green line of course is parallel to two sides of the square and halfway between them; and the center of the circle is at the intersection of those two lines.)

The length of the green line is the side of the square minus the part of the blue line between the midline of the triangle and the side; that is, $\frac{\sqrt3}{2}s - \frac14 s$, which is $\frac{-1 + 2\sqrt3}{4}s$.

The ratio of the blue segment to the green is therefore $\frac{4}{-1 + 2\sqrt3} \approx 1.62330968$, which is not equal to the golden ratio.

Answer 4

$$\frac{2t+s}{2r-t} = \frac{2r\tan 30^\circ+r\sec 30^\circ}{2r-r\tan 30^\circ}$$

So far as the conjecture goes, we can stop here, since there's clearly no chance of introducing $\sqrt 5$, and thus no appearance of the golden ratio. For completeness, though, we can evaluate the ratio and get ... $$\frac{4}{11}\;\left(\;1 + 2 \sqrt{3}\;\right) = 1.6233\dots \neq 1.618\dots$$