# The parallelogram area, determinant and runners

Three runners $$A$$, $$B$$ and $$C$$ run along parallel tracks with constant speed. On start the area of the triangle $$ABC$$ is equal to $$2$$. $$5$$ seconds later it is equal to $$3$$. What could it be $$5$$ more seconds later?

I understand that the area of a triangle is half the area of a parallelogram spanned by vectors corresponding to the position of the runners. So, we need to look at the sign of the determinant and its value. But I don’t understand how to solve the problem. I know that there are two possible answers: 8 and 4, but I don’t understand how to get them.

It is clear that the speed of the runners is different, and the area can either decrease from the start or increase. But what after?

Let $$A(a_1,a_2), B(b_1,b_2),C(c_1,c_2)$$ be the initial positions of the runners.
The area of the triangle is: $$\frac12\begin{vmatrix}a_1&a_2&1\\ b_1&b_2&1\\ c_1&c_2&1\end{vmatrix}=2.$$
Without loss of generality, we can assume the runners run horizontally with the constant speeds $$s_1,s_2,s_3$$, respectively. Then: $$\frac12\begin{vmatrix}a_1+s_1&a_2&1\\ b_1+s_2&b_2&1\\ c_1+s_3&c_2&1\end{vmatrix}=3 \Rightarrow \underbrace{\frac12\begin{vmatrix}a_1&a_2&1\\ b_1&b_2&1\\ c_1&c_2&1\end{vmatrix}}_{=2}+\underbrace{\frac12\begin{vmatrix}s_1&a_2&1\\ s_2&b_2&1\\ s_3&c_2&1\end{vmatrix}}_{=1}=3\\ \frac12\begin{vmatrix}a_1+2s_1&a_2&1\\ b_1+2s_2&b_2&1\\ c_1+2s_3&c_2&1\end{vmatrix}=\underbrace{\frac12\begin{vmatrix}a_1&a_2&1\\ b_1&b_2&1\\ c_1&c_2&1\end{vmatrix}}_{=2}+\underbrace{\frac12\begin{vmatrix}2s_1&a_2&1\\ 2s_2&b_2&1\\ 2s_3&c_2&1\end{vmatrix}}_{=2}=4$$
Let us consider the positions of the runners as functions of time, i.e., $$s_0(t),s_1(t),s_2(t)\in\Bbb R^2$$. Using a coordinate system following the first runner, we may assume $$s_0(t)=0$$ for all $$t$$. Hence, the (signed) area $$A(t)$$ of the triangle is, as you said, given by $$2A(t) = \det(s_1(t), s_2(t)).$$ Let the tracks point in direction $$v\in\Bbb R^2$$ so that the constant speed trajectories may be described as \begin{align*} s_1(t) &= s_1(0)+t\lambda_1 v,\\ s_2(t) &= s_2(0)+t\lambda_2 v, \end{align*} where $$\lambda_1,\lambda_2\in\Bbb R$$ are the speeds of the two runners.
Hence, using the properties of the determinant, we get \begin{align*} 2A(t) &= \det(s_1(t), s_2(t)) \\ &= \det(s_1(0)+t\lambda_1v, s_2(0)+t\lambda_2v) \\ &= \det(s_1(0),s_2(0)) + t \det(s_1(0), \lambda_2 v) + t\det(\lambda_1 v,s_2(0)) + t^2\det(\lambda_1 v,\lambda_2 v) \\ &= \det(s_1(0),s_2(0)) + t \big[\det(s_1(0), \lambda_2 v) + \det(\lambda_1 v,s_2(0))\big]. \end{align*} Simply put, we have $$A(t) = A_0 + t B$$ is a linear function of time. From $$A(0) = 2$$ and $$A(5)=3$$ you can determine $$A_0=2$$ and $$B=\frac 1 5$$ and calculate $$A(10)=4$$.