# Feyman's Triangle? How do you find the area of the inner triangle if the outside triangle is equilateral

If triangle $ABC$ is equilateral,$BD/BC=1/3, CE/CA=1/3,$ and $AF/AB=1/3$. What is the ratio of the area of triangle? I have problems analyzing this triangle I tried to use phythagorean, heron's formula, the formula $A$=$\frac{ab\sin(\theta)}{2}$ but I still can't figure the inside area. I've also done some research they said it's $1:7$? But what I want is to understand how they got $1:7$. • There's no D, E, and F in your diagram. (And this appears to be plane rather than solid geometry). Nov 28 '14 at 1:48
• See Routh's theorem. Nov 28 '14 at 2:41

There are many proofs of this result (see here for a generalized version), but perhaps the easiest is using the theorem of Menelaus, along with some area ratios. We use this figure. Consider collinear points $A, K, E$ on the sides of $\triangle DBG$. Menelaus's Theorem tells us:

\begin{align} \frac{DA}{AB} \frac{BK}{KG} \frac{GE}{ED} &= 1 \\ \frac{3}{1} \frac{BK}{KG} \frac{2}{1} &= 1\\ \implies \frac{BK}{KG} &= \frac{1}{6}\\ \implies \frac{KG}{BG} &= \frac{6}{7} \end{align}

Thus

\begin{align} [\triangle KAG] &= \frac{KG}{BG} [\triangle BAG] \\ &= \frac{6}{7} \frac{BA}{DA}[\triangle DAG]\\ &= \frac{6}{7} \frac{1}{3} [\triangle DAG]\\ &= \frac{6}{21} [\triangle DAG] \end{align}

By symmetry, we have that

$$[\triangle KAG] = [\triangle LDA] = [\triangle MGD] = \frac{6}{21}[\triangle DAG]$$

Thus,

\begin{align} [\triangle LKM] &= [\triangle DAG] - ([\triangle KAG] + [\triangle LDA] + [\triangle MGD])\\ &= ( 1 - \frac{6}{21} - \frac{6}{21} - \frac{6}{21})[\triangle DAG])\\ &= \frac{1}{7}[\triangle DAG] \end{align}

As desired.

• how did $\frac{GE}{ED}$ become $\frac{2}{1}$? Dec 3 '14 at 1:11
• @Mickey Remember that the cevians are trisecting the opposite sides. Dec 3 '14 at 2:44

Borrowing @extremeaxe5's diagram. If you don't know Menelaus' theorem, you can deduce $$\frac{KG}{BG}=\frac67$$ using vector geometry:

By hypothesis $$B=\frac23A+\frac13D$$ and $$E=\frac23D+\frac13G$$. Point $$K$$ lies on the line through $$AE$$, so it satisfies the equation $$\textstyle K=A+\alpha(E-A)=(1-\alpha)A +\frac23\alpha D+\frac13\alpha G\tag1$$ for some $$\alpha$$, and $$K$$ also lies on the line through $$GB$$, so $$K$$ also satisfies $$\textstyle K=G+\beta(B-G)=\frac23\beta A+\frac13\beta D+(1-\beta)G\tag2$$ for some $$\beta$$. Solve (1) and (2) simultaneously for $$\alpha=\frac37$$ and $$\beta=\frac67$$. In particular (2) says $$K = G+\frac 67(B-G)$$, which means $$K$$ is $$\frac67$$ of the way from $$G$$ to $$B$$.

From here you can continue as in @extremeaxe5's proof. An alternative is to observe the inner triangle is equilateral by symmetry, so it is enough to determine the length of any one of its sides. Plugging $$\beta=\frac67$$ into (2) yields $$K=\frac47 A + \frac27 D+\frac17 G$$. A similar argument gets $$L=\frac47 D+\frac27 G+\frac17 A$$. Subtract these: $$\textstyle K-L=\frac37A -\frac27 D-\frac17G=\frac37(A-D)-\frac17(G-D)$$ and compute the squared length: $$\textstyle |K-L|^2=(\frac37)^2|A-D|^2+(\frac17)^2|G-D|^2 -2\cdot\frac37|A-D|\cdot\frac17|G-D|\cdot\cos\theta$$ where $$\theta=60^\circ$$ is the angle between vectors $$A-D$$ and $$G-D$$. Simplify, using the fact that the outer triangle is equilateral, to obtain $$\textstyle|K-L|^2=\frac17|A-D|^2.$$