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 $x \ge 0$, 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 $B$?
The abrupt jump in the middle of the ant's trip makes the problem difficult to analyze geometrically.  (A solution using calculus is possible, but the algebra becomes a little intense.)  The speeds of $\sqrt{2}$ and 2 units per second are also suggestive, as is the fact that the teleportation begins on the $x$-axis and ends on the line $y=x$, which makes an angle of $45^\circ$ with the $x$-axis.    Therefore we transform the entire latter portion of the ant's trip by rotating it clockwise by $45^\circ$ and scaling it down by $\sqrt{2}$.  This has the effect of removing the teleportation all together, reduces the ant's speed to $\sqrt{2}$ for the second part of the journey, and moves the destination point to $(37,37)$.


In other words, an equivalent problem is to ask where the ant should cross the $x$-axis if it wishes to crawl from $(0,-63)$ to $(37,37)$ at a uniform rate of $\sqrt{2}$ units per second in the least amount of time.  Of course, it is now clear that the ant should crawl in a straight line.  The equation of this line is $y=\frac{100}{37}x-63$, and it crosses the $x$-axis when $y=0$, so \[ 0 = \frac{100}{37}x-63 \rightarrow x = \frac{37\cdot 63}{100} = 23.31. \]