source: #alcumus tags: #special-right-triangles #transformations #projection #decomposition course: [[Geometry]] [[Pre-Calculus]] status: #solved ### Problem An ant travels from the point $A(0,-63)$ to the point $B(0,74)$ as follows. It first crawls straight to $(x, 0)$ with $x \geq 0$, moving at a constant speed of $\sqrt{2}$ units per second. It is then instantly teleported to the point $(x, x)$. Finally, it heads directly to $B$ at 2 units per second. What value of $x$ should the ant choose to minimize the time it takes to travel from $A$ to $B ?$ #### Solution To minimize the time it takes will be challenging, because the function that determines the time taken is not fun to differentiate. This problem should be solved without [[Calculus]] anyway. The most important step is to turn the problem into one where distance is minimized instead. To do that, the ant needs to be perceived as moving at a constant speed. The two speeds that the ant travels at are $\sqrt{2}$ and then $2$. So the ant is traveling $\sqrt{2}$ times as fast after teleporting. Recognizing $\sqrt{2}$ as the hypotenuse of a [[Special Right Triangles#45-45-90]] with legs length 1, the path from portal to B can be broken down into 2 congruent and perpendicular components After shifting down $B$ to $B^\prime$ and breaking down the vector from the portal to $B^\prime$, we can find $B^{\prime\prime}$ as the right-angle vertex of the decomposition. ![[teleporting-ant.svg]] It turns out that $\forall x\in \mathbb{R}\implies B^{\prime\prime}(37,37)$ Since $B^{\prime\prime}(37,37)$ is a fixed point and we want to find the shortest path to it from another fixed point $A(0,-63)$, the path should be colinear because of the [[Triangle Inequality]], and must obviously pass through the portal entrance $(x,0)$. The equation of $\overleftrightarrow{AB^{\prime\prime}}$ is $\begin{aligned} y&=\frac{37-(-63)}{37}x-63\\ y&=\frac{100}{37}x-63 \end{aligned}$ And its x-intercept is $63\cdot 37 \div 100=\boxed{23.31}$