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- #1

- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

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- Feb 7, 2012

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\coordinate [label=left:{$B$}] (B) at (0,0) ;

\coordinate [label=right:{$C$}] (C) at (4,0) ;

\coordinate [label=left:{$A$}] (A) at (60:4) ;

\coordinate [label=left:{$D$}] (D) at (60:2) ;

\coordinate [label=right:$F$] (F) at (2.5,2.6) ;

\coordinate [label=left:{$D'$}] (H) at (300:2) ;

\coordinate [label=left:{$A'$}] (K) at (300:4) ;

\coordinate [label=below right:$E$] (E) at (intersection of B--C and F--H) ;

\draw [very thick] (A) -- (B) -- (C) -- cycle ;

\draw (E) -- (D) -- (F) -- (H) ;

\draw (B) -- (K) -- (C) ;

\draw[dashed] (A) -- (E) ;\node at (0.2,0.9) {$2$} ;

\node at (1.2,2.6) {$2$} ;

\node at (2.45,3.2) {$1$} ;

\node at (3.4,1.6) {$3$} ;

\node at (0.2,-0.9) {$2$} ;

\end{tikzpicture}

Let $A'BC$ be the reflection of $ABC$ in the line $BC$, with $D'$ the midpoint of $BA'$. The perimeter of $DEF$ is $DF + FE + ED = DF + FE + ED'$, and this is minimised when $FED'$ is a straight line (as in the diagram).

Now choose a coordinate system with $B$ as the origin and $C$ as the point $(4,0)$. Then $A = (4\cos60^\circ,4\sin60^\circ) = (2,2\sqrt3)$. Similarly, $F = \bigl(\frac52,\frac32\sqrt3)$ and $D' = (1,-\sqrt3)$. The line $FD'$ then has equation $y = \dfrac{5x-8}{\sqrt3}$. When $y=0$, $x=\frac85$. So $E = \bigl(\frac85,0\bigr)$.

Then $AE^2 = \bigl(2-\frac85\bigr)^2 + (2\sqrt3-0)^2 = \frac4{25} + 12 = \frac{304}{25}$, so $AE = \frac{\sqrt{304}}5 = \frac45\sqrt{19} \approx 3.487$.