Concentric Equilateral Triangles

Question

I'm currently researching a particular dynamical system that is very geometric in nature. As part of this, I need to prove the following results (the second obviously implies the first). They are "obviously" true - but the proof doesn't seem so obvious.

1. Equilateral triangles $ABC$ and $DEF$ are concentric. Points $A,G$ and $H$ are collinear. Points $B,H$ and $I$ are collinear. Let $DG=FI=x$ and $EH=y$. Prove that $x=y$.

2. Now suppose the internal triangle $DEF$ is rotated about its centre by some small angle so that points $A$ and $G$ are still on opposite sides of line $DF$. Prove that the result is still true.

I have an algebraic solution to the first problem above. But I'm trying to find a clean geometric proof. I've made no progress on the second generalisation of the first result. I've tried all the usual tricks (congruence, rule of sines and cosines etc.) but to no avail. Does anyone have any approach to such problems that might work?

Answer 1

Proof of conjecture 2:

Let $M=AG\cap DF, N=BH\cap DE$ By Menelaus $\frac{FM}{MD}\frac{DG}{GE}\frac{EH}{HF}=1$ and $\frac{DN}{NE}\frac{EH}{HF}\frac{FI}{ID}=1$ but $IF=DG$ so $ID=GE$ and so $\frac{FM}{MD}=\frac{DN}{NE}$ but now since $DF=DE$ we have $MD=EN$. Now since $A,M,D$ are not collinear we have that $A\not=M$ and $B\not=N$ but from $DM=NE$ we have since $\angle ADM=180-\angle ADF=180-\angle BED=\angle BEN$ and $BE=AD$ that $BEN\cong ADM$ so $\angle BNE=\angle AMD$ so now $EN=DM,\angle HNE=\angle GMD,\angle NEH=120=\angle GDM$ so $ENH\cong DMG$ thus $y=x$. So we managed to obtain this thanks to $A,D,F$ not being collinear and in fact this is the only requirement. The proof is the same in other cases that have different placements of points because we can always get $AMD\cong BNE$. When $A,D,F$ are collinear we always have $DG=FI$ which we can easily see from Menelaus used in the proof.

Answer 2

let the two concentric equilateral triangles be $ABC, DEF$ are placed so that $DEF$ is inside $ABC.$ the three points $G, H, E$ are on the sides $DE, EF$ and $FD$ respectively. we also have the points $A, G, H$ collinear, a$B, H, I$ collinear and $DG = IF.$ let $O$ be the center of the two circumcircles.

claim 1: the triangle $GHI$ is an equilateral triangle.

proof: consider the triangles $DGO, FIO.$ they are congruent because $$\angle ODG = \angle OFI = 30^\circ, OD = OF = radius, DG = IF. \implies OG = OI, \angle OID + \angle OGD = 180^\circ$$

this also gives us $$\angle GOI = 120^\circ.$$ by looking at the isosceles triangle $OIG,$ and using the fact $\angle IOG = 120^\circ,$ we can see that $$\angle OIC = \angle OGI = 30^\circ,\, IG \parallel OF.$$

let $H'$ be the unique point on $EF$ such that $IGH'$is an equilateral triangle. the point $H'$ is unique because the base $GH$ of the equilateral triangle which $O$ as the circumcenter is fixed.

it seems clear, but i need to prove that this $H' = H.$

Answer 3

The generalized property is false (unfortunately). Here is a counterexample with simple coordinates :

$$ \begin{array}{llllll} A & (1,\frac{5}{\sqrt{3}}) & B & (2,-\frac{4}{\sqrt{3}}) & C & (-3,-\frac{1}{\sqrt{3}}) \\ D & (0,\frac{2}{\sqrt{3}}) & E & (1,-\frac{1}{\sqrt{3}}) & F & (-1,-\frac{1}{\sqrt{3}}) \\ G & (\frac{1}{3},\frac{1}{\sqrt{3}}) & H & (0,-\frac{1}{\sqrt{3}}) & I & (-\frac{2}{\sqrt{3}},0) \\ \end{array} $$

It is straightforward to check that with those values, $ABC$ and $DEF$ are both equilateral and centered at the origin, that $H$ is the middel of $[EF]$ and that

$$ \overrightarrow{HA}=3\overrightarrow{HG}, \overrightarrow{DE}=3\overrightarrow{DG}, \overrightarrow{BI}=\frac{4}{3}\overrightarrow{BH}, \overrightarrow{FD}=3\overrightarrow{FI}. $$

Here is the corresponding figure :

Answer 4

I'll approach this problem by relating the congruence of $\overline{DG}$, $\overline{EH}$, $\overline{FI}$ to how (or whether) $\overleftrightarrow{AG}$ contains $H$, $\overleftrightarrow{BH}$ contains $I$, and/or $\overleftrightarrow{CI}$ contains $G$.

My tool of choice is my Extended Ceva's Theorem, introduced in this answer, but reformulated here:

Extended Ceva's Theorem. Lines $\overleftrightarrow{A^+B^{-}}$, $\overleftrightarrow{B^+C^-}$, $\overleftrightarrow{C^+A^-}$ through the edges of $\triangle ABC$ concur if and only if $$\alpha^+ \beta^+ \gamma^+ + \alpha^-\beta^-\gamma^- + \alpha^+\alpha^- + \beta^+\beta^- +\gamma^+\gamma^- = 1 \tag{$\star$}$$

where $$\alpha^+ := \frac{|BA^+|}{|A^+ C|} \qquad \beta^+ := \frac{|CB^+|}{|B^+A|} \qquad \gamma^+ := \frac{|AC^+|}{|C^+B|}$$ $$\alpha^- := \frac{|CA^-|}{|A^- B|} \qquad \beta^- := \frac{|AB^-|}{|B^-C|} \qquad \gamma^- := \frac{|BC^-|}{|C^-A|}$$

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Now for the problem at hand.

Extend the edges of $\triangle DEF$ to meet the sides of $\triangle ABC$ at points $P_i$ and $Q_i$; by symmetry, there are $p$ and $q$ such that $$ \frac{|BP_a|}{|P_aC|} = \frac{|CP_b|}{|P_bA|} = \frac{|AP_c|}{|P_cB|} = p \qquad \frac{|CQ_a|}{|Q_aB|} = \frac{|AQ_b|}{|Q_bC|} = \frac{|BQ_c|}{|Q_cA|} = q $$

Moreover, extend $\overline{AG}$, $\overline{BH}$, $\overline{CI}$ to meet the edges of $\triangle ABC$ at $A^\prime$, $B^\prime$, $C^\prime$, and define $$\alpha := \frac{|CA^\prime|}{|A^\prime B|} \qquad \beta := \frac{|AB^\prime|}{|B^\prime C|} \qquad \gamma := \frac{|BC^\prime|}{|C^\prime A|}$$

By the ECT with substitutions

$$A^+ \to P_a \qquad A^-\to A^\prime \qquad B^+\to C \qquad B^-\to Q_b \qquad C^+\to A \qquad C^-\to C^\prime$$ so that $$\alpha^+ = p \qquad \beta^+ = \gamma^+ = 0 \qquad \alpha^- = \alpha \qquad \beta^- = q \qquad \gamma^- = \gamma$$ we have that $\overleftrightarrow{CI}$ contains $G$ (which is to say that $\overleftrightarrow{AG}$, $\overleftrightarrow{CI}$, $\overleftrightarrow{DE}$ concur at $G$) if and only if $$1 = \alpha\;(\; p + q \gamma \;) \tag{$1$}$$

Likewise, $\overleftrightarrow{AG}$ contains $H$ if and only if $$1 = \beta\;(\; p + q \alpha \;) \tag{$2$}$$ and $\overleftrightarrow{BH}$ contains $I$ if and only if $$1 = \gamma\;(\;p + q \beta \;) \tag{$3$}$$

Now, the problem asserts that $\overleftrightarrow{AG}$ and $\overleftrightarrow{BH}$ contain $H$ and $I$, respectively; therefore, $(2)$ and $(3)$ hold. The problem also asserts that $\overline{DG}\cong\overline{FI}$, so that $\overline{AC^\prime}\cong\overline{CB^\prime}$, which implies $\alpha = \gamma$. Provided that $p\neq 0$, we may conclude that $\alpha = \beta = \gamma$, that all three concurrencies hold, and that $\overline{DG}\cong\overline{EH}\cong\overline{FI}$.

Note. When $p=0$, the figure is such that $\overleftrightarrow{DF}$ contains $A$ (and so forth for other edge-lines and vertices). Here, we need not assume $\overline{DG}\cong\overline{FI}$, as the fact that $\alpha = \gamma$ (and therefore that congruence) follows from assuming $(1)$ and $(2)$ (with $\beta \neq 0$). There's no guarantee, though, that $(1)$ holds. @Ewan's answer gives a case in which it doesn't.

Note. When $q=0$, the figure is such that $\overleftrightarrow{DE}$ contains $A$ (and so forth). By the problem's construction, any point $G$ on $\overline{DE}$ causes $H$ and $E$ to coincide, whereupon $I$ and $F$ also coincide; to satisfy the condition $\overline{DG}\cong\overline{FI}$, we must have that $G$ and $D$ coincide, as well. This, obviously, satisfies all the concurrencies and gives that $\alpha = \beta = \gamma$.