Minimum perimeter

Relevant wiki: Reflection

Reflect the triangle $ABC$ and the point $D$ along the side $AB$ . The reflection of $D$ marked as $D'$ . Note that $DE=D'E$ .

Reflect the triangle $ABC$ and the point $D$ along the side $AC$ . The reflection of $D$ marked as $D'_1$ . Note that $FD=FD'_1$ .

Now $DE+EF+FD=D'E+EF+FD'_1$ , which the minimum length is clearly the same as $D'D'_1$ .

It is not difficult to calculate that $GD'=8+5-2\cos 60^{\circ} = 12$ and that $GD'_1=12\sin 60 ^{\circ}=6\sqrt{3}$ . Hence $s=D'D'_1=6\sqrt{7}$ and thus $\lfloor 100s\rfloor =\lfloor 600\sqrt{7}\rfloor =1587$

The only point pivoted for triangle DEF is D. Think of the triangle as being composed of a single thread curled across bars AB and AC. As you pull the thread harder, the perimeter reduces. As a consequence something else happens as well. The angle of incidence of thread on the bar will be equal to the angle of reflection, when the perimeter is minimum. The angles have been labled as $\theta$ and $\phi$ .

Now using the property that sum of all angles of a triangle is $180^o$ , we have, $\theta + \phi = 120^o$ . The remining angles are then found in terms of $\theta$ and $\phi$ as shown in figure.

It can be seen that there are 3 similar triangles in the picture, in the following order,

CDF ~ BED ~ AEF

The sides of a normalised triangle, also similar to these triangles are assumed to be in the ratio 1 : a : b.

All sides in the figure can be found in terms of the assumed ratios as demonstrated,

we also have, $AF/AE = a$ from the properties of similar triangles. Substituting values of AF and AE from figure and solving, we get,

a = 2/3

To find b, cosine law is used,

$b^2 = 1^2 + (2/3)^2-2.(2/3).1.cos(60^o)$

on solving,

$b = \sqrt7 / 3$

The three sides of the triangle are found to be, $3*\sqrt7 / 3, 7*\sqrt7 / 3, 8*\sqrt7 / 3$ .

Thus, perimeter = $6. \sqrt7$

I used the coordinate system, despite it is quite brutal and a lot of "tough" equations! Assume that the coordinate of $A$ is $\left( 0,0 \right)$ , then the corresponding coordinates of $B$ and $C$ are $\left( 10,0 \right)$ and $\left( 5,5\sqrt { 3 } \right)$ , respectively.

It is obvious that $DH=\sqrt { 3 }$ and $AH=9$ (because $AB=10$ and $BH=1$ , therefore, the coordinate of $D$ is $(9,\sqrt { 3 } )$ . Next, because $F$ lies on side $AC$ of equilateral triangle $ABC$ , so $F$ have a coordinate of $(m,\sqrt { 3 } m)$ , for real number $m$ satisfies $0<m<5$ . Beside that, the coordinate of $E$ is $(n,0)$ for real number $n$ satisfies $0<n<10$ .

By the distance formula:

$DE=\sqrt { (9-n)^{ 2 }+(\sqrt { 3 } -0)^{ 2 } }$

$DF=\sqrt { (9-m)^{ 2 }+(\sqrt { 3 } -\sqrt { 3 } m)^{ 2 } }$

$EF=\sqrt { (n-m)^{ 2 }+(0-\sqrt { 3 } m)^{ 2 } }$

So

$s\quad =\quad DE+DF+EF\quad =\quad \sqrt { (9-n)^{ 2 }+(\sqrt { 3 } -0)^{ 2 } } +\sqrt { (9-m)^{ 2 }+(\sqrt { 3 } -\sqrt { 3 } m)^{ 2 } } +\sqrt { (n-m)^{ 2 }+(0-\sqrt { 3 } m)^{ 2 } }$

We use caluclus here: Define the function $f(m,n)=\sqrt { (9-n)^{ 2 }+(\sqrt { 3 } -0)^{ 2 } } +\sqrt { (9-m)^{ 2 }+(\sqrt { 3 } -\sqrt { 3 } m)^{ 2 } } +\sqrt { (n-m)^{ 2 }+(0-\sqrt { 3 } m)^{ 2 } }$ .

In order to minimize the value of the function, then the partial derivatives of the function must equal zero, simultaneously:

$f_{ m }(m,n)=\quad \frac { 4m-n }{ \sqrt { 4{ m }^{ 2 }-2mn+n^{ 2 } } } +\frac { 2m-6 }{ \sqrt { { m }^{ 2 }-6m+21 } }=0$

$f_{ n }(m,n)=\quad \frac { n-m }{ \sqrt { 4{ m }^{ 2 }-2mn+{ n }^{ 2 } } } +\frac { n-9 }{ \sqrt { { n }^{ 2 }-18n+84 } }=0$

Solving the above system of equations yields $m=\frac { 7 }{ 3 }$ , $n=7$ , satisfy the conditions above.

Plus the recently-found values into the formula for $s$ , then $s=6\sqrt { 7 }$ . Therefore $\left\lfloor 100s \right\rfloor =\left\lfloor 600\sqrt { 7 } \right\rfloor =\left\lfloor 1587.450787... \right\rfloor =1587$ .