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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!)

circle, triangle, square

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. :)

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  • $\begingroup$ yes! equilateral triangle! i will add atop in the description! $\endgroup$ – Astrophysics Math Apr 29 '16 at 21:18
  • $\begingroup$ Perhaps the easiest construction approach would be to start with the circle, and construct a point on it that makes a $30^\circ$ angle with the horizontal. (GeoGebra will let you create an angle of any size you like; alternatively, you can rotate a point about the circle's center.) Then construct the tangent to that point, which gives a line containing a side of the triangle. Use tangents at the top and bottom of the circle to determine where to cut off that side, then go on to complete the triangle and square. Anyway ... I get that the target ratio is $1.6233\dots$. No gold here. $\endgroup$ – Blue Apr 29 '16 at 21:24
  • $\begingroup$ Thanks @Blue! I will try that. Did you get the 1.6233 using geogebra? :) $\endgroup$ – Astrophysics Math Apr 29 '16 at 21:27
  • $\begingroup$ Since there are three tangents to the circle, including the imaginary top and bottom lines, you can draw the angle bisectors at the two intersections of the tangent lines, and use the intersections of the angle bisectors as the centre of the circle. $\endgroup$ – peterwhy Apr 29 '16 at 21:31
  • $\begingroup$ In the Cartesian plane the distance from the point $(u,v)$ to the line $A x+ B y+C=0$ is $|A x+ B y+ C|/\sqrt {A^2+B^2}.$... If $(0,0)$ is the foot of the perpendicular of the triangle, and $(0,\sqrt 3)$ is the top vertex of the triangle, the center of the circle is $(u,\sqrt 3\;/2).$...Then the distance from $(u,\sqrt 3\;/2)$ to the left leg of the triangle is equal to $\sqrt 3\;/2. $ This enables us to find $u$ because the left leg of the triangle is part of the line $ -x+y/\sqrt 3-1=0.$ $\endgroup$ – DanielWainfleet Apr 29 '16 at 23:42
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Here is a diagram of the situation.

enter image description here

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.

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enter image description here

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

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 ... $$\frac{4}{11}\;\left(\;1 + 2 \sqrt{3}\;\right) = 1.6233\dots \neq 1.618\dots$$

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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 $$\frac{4r/\sqrt3}{(2\sqrt3-1)r/\sqrt3} = \frac{4}{2\sqrt3-1} = \frac{8\sqrt3+4}{11} \ne\phi$$

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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.

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This is not the proper description of the Golden Ratio. The definition of Golden Ratio involves a rectangle which gives a similar rectangle after a square is cut off from it.

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  • $\begingroup$ This does not answer the question, so it should be left as a comment (either to the question or another answer). ... That said, one may certainly discuss the Golden Ratio without ever mentioning the standard rectangle construction; indeed, the fun of $\phi$-hunting comes from discovering unexpected (and typically non-rectangular) configurations of elements exhibiting the ratio. (Granted, the configuration in question here is only a near miss, but it's not an uninteresting one. See OP's other questions for more attempts, including successes.) $\endgroup$ – Blue Sep 21 '19 at 19:49
  • $\begingroup$ This is not the proper description of the Golden Ratio. The definition of Golden Ratio involves a rectangle which gives a similar rectangle after a square is cut off from it. The Golden Ratio is the ratio of lengths of sides of such a rectangle. $\endgroup$ – H. Tomasz Grzybowski Sep 23 '19 at 9:49

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