Problem of the Week # 274 - Aug 01, 2017

[Ackbach]

Here is this week's POTW:

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Triangle $ABC$ has an area 1. Points $E,F,G$ lie, respectively, on sides $BC$, $CA$, $AB$ such that $AE$ bisects $BF$ at point $R$, $BF$ bisects $CG$ at point $S$, and $CG$ bisects $AE$ at point $T$. Find the area of the triangle $RST$.

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[Ackbach]

Re: Problem Of The Week # 274 - Aug 01, 2017

This was Problem A-4 in the 2001 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution (and extra kudos for a helpful TiKZ picture), which follows:

Spoiler

[TIKZ]\coordinate [label=above right: $A$] (A) at (4,3) ;
\coordinate [label=below left: $B$] (B) at (0,0) ;
\coordinate [label=below right: $C$] (C) at (6,0) ;
\coordinate [label=below: $E$] (E) at (2.29,0) ;
\coordinate [label=above right: $F$] (F) at (5.24,1.14) ;
\coordinate [label=above left: $G$] (G) at (2.47,1.85) ;
\draw (C) -- (A) -- (B) -- cycle ;
\draw (A) -- (E) ;
\draw (B) -- (F) ;
\draw (C) -- (G) ;
\node at (2.5,0.8) {$R$} ;
\node at (4.2,0.7) {$S$} ;
\node at (3.5,1.6) {$T$} ;
\node at (1.1,-0.2) {$1-\lambda$} ;
\node at (4.1,-0.2) {$\lambda$} ;
[/TIKZ]

Suppose that $E$ divides $BC$ into segments whose lengths are in the ratio $1-\lambda:\lambda$ (where $0<\lambda<1$). If the points $A,B,C$ are represented by vectors $\def\v{\mathbf} \v{a},\v{b},\v{c}$ then $E$ is given by $\lambda\v{b} + (1-\lambda)\v{c}.$

Suppose next (this was just a lucky guess, which turned out to be correct) that $F$ and $G$ divide $CA$ and $AB$ in that same ratio $1-\lambda:\lambda$. Then $F$, $G$ are given by the vectors $\lambda\v{c} + (1-\lambda)\v{a}$, $\lambda\v{a} + (1-\lambda)\v{b}$ respectively.

The midpoint $R$ of $BF$ is then given by $\frac12\bigl(\v{b} + \lambda\v{c} + (1-\lambda)\v{a}\bigr)$. If $R$ lies on $AE$ then it must also be given by $\mu\v{a} + (1-\mu)\bigl(\lambda\v{b} + (1-\lambda)\v{c}\bigr)$ for some $\mu$. Therefore $$\tfrac12\bigl(\v{b} + \lambda\v{c} + (1-\lambda)\v{a}\bigr) = \mu\v{a} + (1-\mu)\bigl(\lambda\v{b} + (1-\lambda)\v{c}\bigr).$$ Comparing coefficients on both sides of that equation, we get $\mu = \frac12(1-\lambda)$ (from the coefficients of $\v{a}$). Then the coefficients of $\v{b}$ and $\v{c}$ both agree provided that $\lambda^2 + \lambda = 1$.

By symmetry, that condition on $\lambda$ will also ensure that $S$ and $T$ are the midpoints of their respective segments. So the "lucky guess" does indeed describe the correct configuration for all the points on the diagram.

Now we need to think about areas. The area $|ABC|$ of the whole triangle is $1$. The triangle $EAC$ has the same height as $ABC$, but its base $EC$ satisfies $|EC| = \lambda |BC|$. Therefore $|EAC| = \lambda$. Since $T$ is the midpoint of $AE$, it follows that $|TAC| = \frac12|EAC| = \frac12\lambda.$

By symmetry, the triangles $RBA$ and $SCB$ have the same area $\frac12\lambda$ as $TAC$. But the whole triangle $ABC$ is the disjoint union of those three triangles together with $RST$. Therefore $\frac32\lambda + |RST| = 1$ and so $|RST| = 1 - \frac32\lambda.$

Finally, $\lambda$ is the positive solution of $\lambda^2 + \lambda = 1$, so $\lambda = \frac12(\sqrt5-1)$ and $$|RST| = 1 - \tfrac34(\sqrt5-1) = \boxed{\tfrac{7-3\sqrt5}4} \approx 0.072949.$$