Prove equilateral triangle in another equilateral triangle

Question

As shown above, given $AB=BC=AC$ and $BE=CF=AD$, to prove $DF=FE=ED$. (there's a typo on the image)

Furthermore, is such property true for any regular $n$-sided polygon？

Answer 1

ِTo draw lines AD, CF and BE one way is to draw a circle it's center at in center of triangle and draw tangents on it. Since this can be done in any regular polygon then there are always tangents from vertices which construct another regular polygon.In figure you can easily see that segments between vertices and other tangent are equal. Since the tangents from intersection of tangent lines on circle are equal then the sum of each two tangents construction each side of triangles are equal that means the sides of triangle DEF are equal.

Another method is that connecting the vertices to in-center of polygon. Now draw a circle with arbitrary radius. the intersection of this circle with lines connecting vertices to in-center creates three(for triangle) or n equal chords (for polygon)which construct a regular polygon.

Answer 2

Before we dive in, let us establish a small result: Let $\omega$ denote a cube root of unity, $$x, y\in \mathbb{R},\quad 0 < y < x < 1, \quad |1+x\omega+y\omega^2| \; \text{ always decreases on increasing } y$$ $$\text{ If the domain is further restricted to }2x-y-1< 0 \text{, it also always decreases on increasing } x$$ Proof:
The expression for squared magnitude of $z(x, y) = 1+x\omega+y\omega^2$ is: $$|z(x, y)|^2 = z(x, y)\cdot\overline{z(x, y)} = (1+x\omega+y\omega^2)(1+x\omega^2+y\omega) = 1+x^2+y^2-x-y-xy$$ This curve is really an elliptic paraboloid rotated by $45^{\circ}$ and with it's vertex shifted to (1, 1, 0). So there is little monotonicity along axes in general but it will do in our restricted domains. (We can also just take the easy road with partial derivatives here instead of these steps)

• In $z(a,b)$ if we add $\delta > 0$ in $b$, to get to another point in the domain, (i.e. $0\leq b+\delta<1$), the difference (using the above formula for $|z(x,y)|^2$): $$|z(a, b+\delta)|^2 - |z(a, b)|^2 = \delta^2 + 2\delta b - \delta - a\delta = \delta\cdot (\delta + b -1 + b - a) < (b-a) < 0$$
• If we are also given that we are restricted to $2x-y-1<0$, and in $z(a, b)$ if we increase $a$, by adding $\delta > 0$ to get to another point in the new restricted domain, i.e. $2(a+\delta) -b -1<0$ i.e. $2a-b-1<- 2\delta$, the difference: $$|z(a+\delta, b)|^2 - |z(a, b)|^2 = \delta^2 + 2\delta a - \delta - b\delta = \delta\cdot (\delta + 2a - b - 1) < \delta\cdot (\delta -2\delta)<0$$

Thus so long as we operate in this restricted domain of $(x, y)$, increasing either or both of them will cause a decrease in $|1+x\omega+y\omega^2|$.

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In the main proof we use complex numbers and without loss of generality assume that the triangle $\Delta ABC$ vertices can be placed on the cube roots of unity on the Argand plane. Let $A$ be on the complex number 1, $B$ on $\omega$ and $C$ on $\omega^2$. Denote the complex numbers for $D$, $E$ and $F$ by $z_0$, $z_1$ and $z_2$ respectively. It will be helpful to think of the vector representation of these complex number operations throughout.

$D$ is on $BE$. This means $\vec{BD}$ is a scalar multiple (< 1) of $\vec{BE}$: $$z_0 - \omega = \lambda_0 \cdot (z_1 - \omega), \qquad 0 < \lambda_0 < 1$$ We ignore the degenerate cases of $\lambda_j = 0, 1$ Similarly, we obtain: $$z_1 - \omega^2 = \lambda_1 \cdot (z_2 - \omega^2),\qquad z_2 - 1 = \lambda_2 \cdot (z_0 - 1) \qquad 0 < \lambda_1, \lambda_2 < 1$$ If we make a change of variables: $v_k = z_k - \omega^k$, we note that these represent vectors aligned along arms $AD, BE, CF$, and thus they have equal magnitudes: $|v_0|=|v_1|=|v_2|$. We obtain: $$v_0+1-\omega = \lambda_0\cdot v_1, \qquad 0 < \lambda_0<1$$$$v_1+\omega-\omega^2 = \lambda_1\cdot v_2, \qquad 0 < \lambda_1<1$$$$v_2+\omega^2-1= \lambda_2\cdot v_0, \qquad 0 < \lambda_2<1$$ Just as an aside, adding up we get:$$(1-\lambda_0)\cdot v_1+(1-\lambda_1)\cdot v_2+(1-\lambda_2)\cdot v_0=0$$ But more importantly we have a complex system of equations parameterized by $\lambda_j$ which we hope to be able to solve and get $v_j$s of equal magnitude: $$\begin{bmatrix}1&-\lambda_0&0\\0&1&-\lambda_1\\-\lambda_2&0&1\end{bmatrix}\cdot\begin{bmatrix}v_0\\v_1\\v_2\end{bmatrix}=\begin{bmatrix}\omega-1\\\omega-\omega^2\\\omega^2-1\end{bmatrix}$$ $$\iff\begin{bmatrix}v_0\\v_1\\v_2\end{bmatrix}=\frac{1}{1-\lambda_0\lambda_1\lambda_2}\cdot\begin{bmatrix}1&\lambda_0&\lambda_0\lambda_1\\\lambda_1\lambda_2&1&\lambda_1\\\lambda_2&\lambda_2\lambda_0&1\end{bmatrix}\cdot\begin{bmatrix}\omega-1\\\omega-\omega^2\\\omega^2-1\end{bmatrix}$$ For us, the complex numbers $\omega-1$, $\omega-\omega^2$, $\omega^2-1$ are just representatives of three equal magnitude vectors aligned equally from each other. Each of the $v_is$ is the vector formed by this combination of these equally aligned equal magnitude vectors: $$[1\quad \lambda_i \quad \lambda_i\lambda_{i+1}]_{i=0,1,2} \qquad (\lambda_3\equiv\lambda_0) $$ and so far as the magnitude of $v_i$s is concerned it doesn't matter which weight is assigned to which.

So can we find distinct $\{\lambda_0, \lambda_1, \lambda_2\}$ such that these three combinations on equal magnitude and equally inclined vectors result in vectors of equal magnitude?

Assume without loss of generality $\lambda_0\leq\lambda_1\leq\lambda_2$. We will consider these combinations over the roots of unity, i.e. $1+\lambda_i\omega+\lambda_i\lambda_{i+1}\omega^2$. Firstly we note that except the last combination above i.e. $[1 \quad \lambda_2 \quad \lambda_2\lambda_0]$, the other combinations $[1 \quad \lambda_i \quad \lambda_i\lambda_{i+1}]$ are in the restricted domain we discussed in our little lemma. Because:

• $0<\lambda_i\lambda_{i+1}<\lambda_i<1$
• $-(1-\lambda_i)^2<0\iff 2\lambda_i - \lambda_i^2 -1<0 \implies 2\lambda_i - \lambda_i\lambda_{i+1} -1<0 \;\text{ since, } \lambda_i\leq\lambda_{i+1}$

Now consider these first two combinations over the roots of unity: $$1+\lambda_0\omega+\lambda_0\lambda_1\omega^2 \quad \text{ and } \quad 1+\lambda_1\omega+\lambda_1\lambda_2\omega ^2$$ If $\lambda_0<\lambda_1$, the second cannot have the same magnitude, since from the first to second we have increased both coefficients of $\omega^k$s and from our result, strictly decreased the magnitude. So clearly $\lambda_0=\lambda_1$. This will happen again if $\lambda_0=\lambda_1<\lambda_2$ as we would still be increasing the second coefficient. Thus $\lambda_0=\lambda_1=\lambda_2$. But this means each of the equal arms $AD, BE, CF$ is divided in the same ratio by $F, D, E$ respectively, and thus $FD, DE, EF$ are equal. $$\tag*{$\blacksquare$}$$

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Footnote: Even though the whole reason I took this algebraic complex number approach was to be able to generalize to the case of an $n$-regular polygon, as much as I have struggled, I have not yet succeeded. So posting this one solution I found.