Minimizing a Path in the Coordinate plane

You start on point $A=(0,0)$ and you want to get to point $B=(10,1)$ . There is a circular object with radius $2$ blocking your way: it's equation is $(x-n)^2+y^2=4$ for some $n\in [2,8]$ . Let the shortest path from $A$ to $B$ such that you do not pass through the circular object have length $P$ . What should $n$ be such that $P$ is minimized?

Maybe surprisingly, the answer is not $n=8$ .

You can use Wolfram Alpha to bash it.

Diagram will be added ASAP.

#Geometry #Shortestpath #Bash #Minimize #CoordinatePlane

Easy Math Editor

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Comments

For n = 2, distance = 8 + pi, which I guess is the shortest distance.

Vineeth Chelur - 7 years, 1 month ago

Not quite! Good try.

Since nobody has replied, I will give the answer: the shortest distance is approximately $10.4583$ at $n\approx 6.56261$ .

Daniel Liu - 7 years, 1 month ago

After working out for nearly 4 hours, I found a smaller value. The answer is 10.3597 at n = 6.87425. I will post the solution tomorrow. The answer will be a bit smaller than this answer because of calculator limit I had to approximate.

Vineeth Chelur - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$