Halved Triangle

Let $\triangle ABC$ be the triangle, where $CA=b=7$ , $AB=c=8$ , and $BC=a=9$ . By Heron's formula . the area of $\triangle ABC$ , $A_\triangle = \sqrt{s(s-a)(s-b)(s-c)}$ , where semiperimeter $s = \frac 12 (a+b+c) = 12$ . Then $A_\triangle = \sqrt{12\cdot 3 \cdot 5 \cdot 4} = 12 \sqrt 5$ and the perimeter $2s= 24$ .

Let $EG$ be one of the lines that simultaneously bisect the area and the perimeter. Then the area of $\triangle BEG$ , $A_B = \frac 12 A_\triangle = 6\sqrt 5$ and $BE+BG = \frac {24}2 = 12$ . If $BG = x$ , then $BE=12-x$ .

We know that $A_\triangle = \frac 12 ac \sin B$ and $A_B = \frac 12 x(12-x) \sin B$ . Then we have

$\begin{aligned} \frac 12 x(12-x) \sin B & = \frac 14 ac \sin B \\ x(12-x) & = \frac {9\cdot 8}2 = 36 & \small \color{#3D99F6} \text{Rearranging} \\ x^2 - 12x + 36 & = 0 \\ (x-6)^2 & = 0 \\ \implies BG & = BE = 6 \end{aligned}$ .

Similarly, let $CF=y$ and $CD=12-y$ , then:

$\begin{aligned} y^2 - 12y + 31.5 & = 0 \\ \implies y & = \frac {12+\sqrt{144-126}}2 = \frac {12+3\sqrt 2}2 = CF \\ CD & = \frac {12-3\sqrt 2}2 \end{aligned}$

For the simultaneous bisecting line opposite $\angle A$ , we have:

$\begin{aligned} z^2 - 12z + 28 & = 0 \\ \implies z & = 6+2\sqrt 2 \approx 8.828 \end{aligned}$

Since $z$ is longer than the two sides (7 and 8), the line is outside $\triangle ABC$ . Therefore $\triangle ABC$ has only two such lines.

Let $\angle GHF=\theta = 180^\circ - \angle EGB - \angle DFC$ . Then:

$\begin{aligned} \tan \theta & = \tan (180^\circ - \angle EGB - \angle DFC) \\ & = - \tan (\angle EGB + \angle DFC) \\ & = \frac {\tan \angle EGB + \tan \angle DFC}{\tan \angle EGB \tan \angle DFC-1} & \small \color{#3D99F6} \text{See note: }\tan \angle EGB = \sqrt 5,\ \tan \angle DFC = 3\sqrt {10} - 4\sqrt 5 \\ & = \frac {\sqrt 5 + 3\sqrt {10} - 4\sqrt 5}{\sqrt 5(3\sqrt {10} - 4\sqrt 5)-1} \\ & = \frac {\sqrt {10} - \sqrt 5}{5\sqrt 2 -7} \\ & = \boxed{3\sqrt 5 + 2\sqrt{10}} \end{aligned}$

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Note:

$\begin{aligned} \tan \angle EGB & = \frac {BE\sin B}{BG-BE\cos B} & \small \color{#3D99F6} \text{Since }\frac 12\cdot 8\cdot 9\sin B = 12\sqrt 5 \implies \sin B = \frac {\sqrt 5}3 ,\ \cos B = \frac 23 \\ & = \frac {6 \cdot \frac {\sqrt 5}3}{6 - 6\cdot \frac 23} \\ & = \sqrt 5 \end{aligned}$

$\begin{aligned} \tan \angle DFC & = \frac {CD\sin C}{CF-CD\cos C} & \small \color{#3D99F6} \text{Since }\frac 12\cdot 7\cdot 9\sin C = 12\sqrt 5 \implies \sin C = \frac {8\sqrt 5}{21} ,\ \cos C = \frac {11}{21} \\ & = \frac {\frac {12-3\sqrt 2}2\cdot \frac {8\sqrt 5}{21}}{\frac {12+3\sqrt 2}2 - \frac {12-3\sqrt 2}2\cdot \frac {11}{21}} \\ & = 3\sqrt {10} - 4\sqrt 5 \end{aligned}$

The lines which bisects the area and the perimeter of a triangle are called Triangle Equalizers. Depending on the triangle, it can have 1, 2 or 3 Equalizers.

Click here for a detailed analysis of such lines

Here's what the halved triangle looks like: