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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?

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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: $$\frac12\begin{vmatrix}a_1+s_1&a_2&1\\ b_1+s_2&b_2&1\\ c_1+s_3&c_2&1\end{vmatrix}=3 \Rightarrow \underbrace{\frac12\begin{vmatrix}a_1&a_2&1\\ b_1&b_2&1\\ c_1&c_2&1\end{vmatrix}}_{=2}+\underbrace{\frac12\begin{vmatrix}s_1&a_2&1\\ s_2&b_2&1\\ s_3&c_2&1\end{vmatrix}}_{=1}=3\\ \frac12\begin{vmatrix}a_1+2s_1&a_2&1\\ b_1+2s_2&b_2&1\\ c_1+2s_3&c_2&1\end{vmatrix}=\underbrace{\frac12\begin{vmatrix}a_1&a_2&1\\ b_1&b_2&1\\ c_1&c_2&1\end{vmatrix}}_{=2}+\underbrace{\frac12\begin{vmatrix}2s_1&a_2&1\\ 2s_2&b_2&1\\ 2s_3&c_2&1\end{vmatrix}}_{=2}=4$$

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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 $A(10)=4$.

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  • $\begingroup$ could you please rename your A and B? Your solution is fine, but these letters are the names of runers so it becomes confusing. Thanks! $\endgroup$ – user376343 Jan 26 at 10:39

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