Cut it in half

Label the triangle as $ABC$ and the cut be $DE$ such that $AD=a$ and $AE=b$ . Then the area of $\triangle ADE$ is half of that of $\triangle ABC$ or

$\begin{aligned} \frac {ab \sin A}2 & = \frac 12 \cdot \frac {20\cdot 21}2 \\ \frac {ab}2\cdot \frac {20}{29} & = \frac 12 \cdot \frac {20\cdot 21}2 \\ \implies ab & = 304.5 \\ b & = \frac {304.5}a \end{aligned}$

And also the perimeter of $\triangle ADE$ is the same as quadrilateral $DECB$ or:

$\begin{aligned} a + b + DE & = 20+21-a + 29-b + DE \\ a+b & = 70 - (a+b) \\ \implies a+b = 35 \\ a + \frac {304.5}a & = 35 \\ a^2 - 35a + 304.5 & = 0 \\ \implies a, b & \approx 16.17712434, 18.82287566 \end{aligned}$

Since $21, 29 > a, b$ , there are two ways of "cutting off $\angle A$ " (the two red dash lines).

Now, using the same logic to cut off $\angle C$ , then $ab=290$ , $a+b=35$ , and $a,b \approx 13.46887113, 21.53112887$ . Since $20 < 21.53112887$ , there is only one way of cutting off $\angle C$ .

To cut off $\angle B$ , we have $ab= 210$ , $a+b=35$ , and $a,b \approx 7.689291565, 27.31070844$ . Since $27.31070844 > 21 > 20$ , there is no way to cut off $\angle B$ .

Therefore there are $\boxed 3$ ways to do the equal cut.

One straight line will pass through 2 sides of the triangle (or 1 vertex and 1 side, which will be included too). There are 3 cases: (i) $AC$ and $BC$ (ii) $AB$ and $BC$ (iii) $AB$ and $AC$

The case (i) is considered now: Let the side lengths of triangle (on the upper right) be as shown, where $0 <x \le 21$ and $0 <y \le 29$ . Then the side lengths of the quadrilateral (on the lower right) are as shown. Note that the side length $z$ is in fact unnecessary.

If both parts have the same perimeter, then $x+y+z = 20 +(29-y)+(z)+(21-x)$ , which means that $\textcolor{#D61F06}{x+y=35}$ . On the other hand, both parts have the same areas implies that $\frac{1}{2} xy \sin C = \frac{1}{2} \left( \frac{1}{2}(21)(29)\sin C \right)$ , that is $\textcolor{#D61F06}{xy=304.5}$ . From the sum and product, $x$ and $y$ are roots of the equation $u^2-35u+304.5=0$ . Solve the equation, we obtain $u \approx$ 16.18 or 18.82. Note that both answers less than 21, which means that $(x,y) = (16.18,18.82), (18.82, 16.18)$ are both possible answers. There are $\large \textcolor{#3D99F6} {2}$ answers in this case.

The case (ii) is considered now: Using the same argument, we have $\textcolor{#D61F06}{r+s=35}$ and $\textcolor{#D61F06}{rs=290}$ where $0<r\le29$ and $0<s\le 20$ . Both $r$ and $s$ are roots of equation $w^2-35w+290=0$ . Solve the equation, we obtain $w\approx$ 13.47 or 21.53. Hence $(r,s)=(21.53, 13.47)$ is an answer while $(r,s)=(13.47,21.53)$ is rejected as $s>20$ . There is $\large \textcolor{#3D99F6} {1}$ answer in this case.

The case (iii) is considered now: Using the same argument, we have $\textcolor{#D61F06}{p+q=35}$ and $\textcolor{#D61F06}{pq=210}$ where $0<p\le21$ and $0<q\le 20$ . Both $p$ and $q$ are roots of equation $v^2-35v+210=0$ . Solve the equation, we obtain $v\approx$ 7.69 or 27.31. Both answers rejected as either $p$ or $q$ > 21. There is $\large \textcolor{#3D99F6} {0}$ answer in this case.

So, there are $\large \boxed{\textcolor{#3D99F6}{3}}$ such lines.

Remarks: There is always a way to cut a convex shape in two so that they have equal area and equal perimeter.

Discussion can be found in this video .