What is the formula for the distance between the centroid of an equilateral triangle and a point on, it at a given angle?

Suppose you have an equilateral triangle with each side of length $x$ , and its centroid $c$.
Given an angle $\theta$, what would the distance be of a line extending out from $c$ at angle $\theta$ to where it intersects with a line on the triangle?

Or in my horrible (not-to-scale) paint drawing, what is the length of the blue line for any given $\theta$?

I used your fig. Minimum distance $p = \dfrac{x}{2 \sqrt3}$ by $30^0, 60^0$ right triangle trig proportioning. The required distance

$$r = p \sec ( \theta - \pi/6)$$

which is polar coordinate equation of a straight line.

Subtract $120^0= 2 \pi/3,\,240^0 = 4 \pi/3$ for other sides if you wish to include other two sides ( clockwise).

Here's a rough picture ..

I have a way.....If the black line from centroid is parallel to one side then its length is $\frac{x}{3}$ and the angle between black line and side is $60^\circ$ So, if we are given an angle $\theta$ ((except angles at which $\sin(60^\circ+\theta)=0$ then the segment is simply $\frac x3$ ))

The segment length required $(y)$ is (By Law of Sines)

$$\frac{\sin(120^\circ-\theta)}{\frac{x}{3}}=\frac{\sin(60^\circ)}{y}$$ SO
$$y=\frac{x\sqrt3 }{6\sin(120^\circ-\theta)}$$

EDIT: This works as long as angle remains $60^\circ$..After that..One has to change reference line...But the same approach works

• Your answer makes sense to me, but when I try x=1, theta=14° I get y=1.19325 which is longer than the boundaries of my triangle. Mar 15, 2017 at 11:09
• @MarioY19 Please check your computation again...My calculator gives $y= 0.30030857966$ for the values you specified... Mar 15, 2017 at 11:18
• Sorry, I simplified incorrectly, I meant $\theta=-74$, also $\theta=110$ Mar 15, 2017 at 11:34
• Sorry..I forgot to mention it in answer...But if you try those values ..then $\theta$ is below black line..So angle between black line and side no longer remains 60..It becomes 120... Mar 15, 2017 at 11:40
• @MarioY19 I have given you a answer between a specified interval..You can keep changing triangles since the figure is symmetric..Same approach works Mar 15, 2017 at 11:49

Let $D=x\dfrac{\sqrt{3}}{6}$ be the distance from the center to any of the sides.

Here is a two-steps definition of what is called a "polar equation" of the equilateral triangle:

$$\tag{1}r(\theta):=\dfrac{1}{\max[\frac12, \cos(\theta)]}$$

$$\tag{2}y(\theta):= D*\min\left[r\left(\theta\right), r\left(\theta + 2 \tfrac{\pi}{3}\right), r\left(\theta + 4\tfrac{\pi}{3}\right)\right]$$

Here is a Mathematica script that generates the (polar) curves of $r$ and $y$:

r[t_] := 1/Max[1/2, Cos[t]]

y[t_] := Min[r[t], r[t + 2 Pi/3], r[t + 4 Pi/3]]

ParametricPlot[{r[t]*Cos[t], r[t]*Sin[t]}, {t, 0, 2 Pi}]

ParametricPlot[{y[t]*Cos[t], y[t]*Sin[t]}, {t, 0, 2 Pi}]


with output the graphics at the bottom.

Explanation : We have taken $D=1$. The first graphics is the border of a truncated disk with radius 2 ; when this figure undergoes a first rotation of 120°, then 240°, the intersection of the resulting 3 trucated disks, algebraicaly "rendered" by the "min" operation, is an equilateral triangle.

Remark: as your triangle is oriented, you will need of course a $\tfrac{\pi}{6}=30°$ supplementary rotation.

• I have completely written again my previous answer that was, I confess, erroneous. Mar 19, 2017 at 7:25