Optimal Ant Travel Path
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 $?
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- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
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- $\pi$
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- $\infty$
Solution
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 \[ 0 = \frac{100}{37}x-63 \rightarrow x = \frac{37\cdot 63}{100} = \boxed{23.31}.\]
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