How to draw an equilateral triangle inscribed in another triangle?

Question

Given a triangle $\Delta$ABC, how to draw any inscribed equilateral triangle whose vertices lie on different sides of $\Delta$ABC?

Answer 1

(See figure and Matlab program below). We are going to show how to capture the issue in all its generality, and thus build a program giving as many inscribed equilateral triangles $UVW$ as desired, the key observation being an affine relationship between the evolution speeds of vertices $U$ and $V$ (and $W$).

One can assume WLOG, using an adequate translation, that $A$ has coordinates $\binom{0}{0}$. Let us name $\binom{a}{b}$ and $\binom{c}{d}$ the resp. coord. of $B$ and $C$. In the example given below, $B$ is in the 3rd quadrant and $C$ in the 4th quadrant, but this is unimportant as long as the names of the vertices $A,B,C$ are chosen such that triangle $ABC$ is positively oriented.

Consider an inscribed equilateral triangle $UVW$. Let $s$ and $t$ be defined by :

$$\vec{AU}=s \vec{AB} \ \ \ \ \& \ \ \ \ \vec{AV}=t \vec{AC} \ \ \ \ \text{with} \ 0 \leq s,t \leq 1.$$

Let $\binom{x}{y}$ be the coordinates of $W$.

It is not difficult to see that $UVW$ is an equilateral triangle if and only if:

$$\tag{1}\vec{AW}=\underbrace{\tfrac12(\vec{AU}+\vec{AV})}_{\text{midpoint} \ I \text{of} \ UV}+\underbrace{\tfrac{\sqrt{3}}{2}Rot_{-\tfrac{\pi}{2}}(\vec{UV})}_{\text{altitude} \ \vec{IW}} \ \iff \ \begin{cases}x=\tfrac12(as + ct + \sqrt{3}(dt-bs))\\ y=\tfrac12(bs + dt + \sqrt{3}(as - ct))\end{cases}$$

Now we must constrain $A,B,W$ to be aligned. This will be done in this classical way (see formula (2) in (http://mathworld.wolfram.com/Collinear.html)).

$$\tag{2}\begin{vmatrix}a&c&x\\b&d&y\\1&1&1\end{vmatrix}=0.$$

By plugging expressions (1) in (2), one gets a formula that allows $t$ to be expressed as an affine function $\alpha s + \beta$ of $s$ (I don't reproduce this formula : it is expressed in the program). This dependency of $t$ on $s$ explains that the loop in the program is controlled by values of $s$ as we can see:

clear all,close all;hold on;axis equal
xA=0;yA=0;% assumed
a=-1;b=-3; % B's coord.
c=5;d=-4; % C's coord.
%a=-rand;b=-rand;c=rand;d=-rand;
Tr=[0,a,c,0;
    0,b,d,0];
plot(Tr(1,:),Tr(2,:),'b');
r=sqrt(3);
for s=0:0.05:1
   N = s*((b+a*r)*(a-c)+(b*r-a)*(b-d))+2*(c*b-a*d);
   D = (c+d*r)*(b-d) + (c*r-d)*(a-c) ;
   t = N/D;
   x=(a*s + c*t - b*r*s + d*r*t)/2;
   y=(b*s + d*t + a*r*s - c*r*t)/2;
   W=[x;y];
   U=s*[a;b];V=t*[c;d];
   Tequ=[U,V,W,U];
   if t>0 && t<1 && inpolygon(W(1),W(2),Tr(1,:),Tr(2,:))
      plot(Tequ(1,:),Tequ(2,:),'color',rand(1,3))
   end;
end;

Remark: one can establish that if $U$ has a rectilinear motion on $AB$, then the center of triangle $UVW$ (little red circles on the figure) undergoes as well a rectilinear uniform motion.

Answer 2

Construction

1. Use Morley's theorem to construct an equilateral triangle $\triangle XYZ$ that lies strictly inside $\triangle ABC$.

2. Through the vertices of $\triangle XYZ$, draw three lines parallel to the sides of your original $\triangle ABC$. These lines form a new triangle $\triangle A'B'C'$ that is similar to $\triangle ABC$ and lies strictly inside it. (Thus the equilateral triangle $\triangle XYZ$ is now inscribed in the newly constructed $\triangle A'B'C'$.)

3. Enlarge the figure so that $\triangle A'B'C'$ becomes the size of $\triangle ABC$. The enlarged image of $\triangle XYZ$ is the sought-for inscribed triangle.

Answer 3

Assume $A=(0,0)$, $B=(b,0)$, $C=(c,h)$ with $b>0$, $0<c<b$, and $h>0$. We shall construct an equilateral triangle with one side horizontal as follows: Draw a horizontal line $y=h'$, where $h'$ is determined by the condition $$h'={\sqrt{3}\over2}\>{h-h'\over h}\>b\ .\tag{1}$$ This line intersects the two legs of the triangle in two points $P$ and $Q$. Let $M=(m, h')$ be the midpoint of $PQ$. Then $P$, $Q$, and $R:=(m,0)$ form an equilateral triangle.

The condition $(1)$ ensures that $h'={\sqrt{3}\over2}\>|PQ|$. Solving for $h'$ one obtains $$h'={h\, b\over{2\over\sqrt{3}}h+b}\ .$$

Answer 4

Another way is to draw a bisector $AD$ of $\angle CAB$, $D\in BC$, find the intersection of the sides $AC$ and $AB$ with the line $AD$, rotated $\pm30^\circ$ around $D$:

Edit

\begin{align} D&=\frac{b\cdot B+c\cdot C}{b+c} ,\\ |AF|=|AE|&=\frac{|AD|}{2\,\sin(30^\circ+\tfrac\alpha2)} . \end{align}