[music playing] Welcome to this brainteaser.

So there are two towns.

One we're going to call Mordor.

And one we're going to call Hogsmeade.

We want to build a pier on the shore between the two towns.

But the stingy city council wants to use as little material as possible.

So how do we minimize the path from Mordor to the pier and onto Hogsmeade?

So it's going to be somewhere along here.

And you want the shortest path that goes from Hogsmeade to the pier to Mordor.

Where does-- DR. DEREK MULLER: Hang on.

OK.

So the shortest total path?

DIANNA COWERN: Shortest total path.

JABRIL ASHE: So you're trying to minimize the complete distance.

DIANNA COWERN: Exactly.

You'll want it as close to a straight diagonal between the two towns as you can make it.

It's funny.

Some people answer questions like this intuitively, whereas I see the opportunity and set up some equations here.

I would put together an equation and then minimize the equation.

I just feel like I need to differentiate this expression here.

I'm stumped.

I feel like just throw it in the center.

DR. DEREK MULLER: Last step of my solution here is x equals ac on a plus b.

You're welcome.

I'm going to just offset it a bit to Hogsmeade, because it's further inland.

DIANNA COWERN: You actually got a really nice solution here.

I didn't know that I had the solution.

There's actually a pretty simple solution to this problem that involves a bit of geometry, but not much more than that.

I wonder if there's a really nice, beautiful, elegant solution.

I feel like triangles could be involved here.

Imagine I took Mordor and reflected it across the shoreline.

I'm going to fold the paper over at the shoreline and mark it so it goes through the paper, so that this distance from Mordor to the shoreline, and then this distance from the shoreline to reflected Mordor are the same distance.

Now, if we want to find the minimum distance between Hogsmeade and reflected Mordor, well that's easy.

The shortest path between two points is a straight line.

And now, we have a potential location for a pier.

But will this location minimize the distance from Hogsmeade to the pier to Mordor?

Well, we can show that the straight line distance is actually the same distance, because this side of the triangle is shared, and these two distances are the same.

Therefore, by similar triangles, this distance must be the same as this one.

So this location is looking promising.

And now we can see that by any other location of the pier, like over here, that would make the Hogsmeade-reflected Mordor distance longer.

And we can show again that it's the same distance as Hogsmeade-pier-Mordor, which means that any other location of the pier would make the path longer.

Therefore, this must be it.

This is the location of the pier.

So that is it.

That's the solution to our problem.

This is where our pier goes.

I love this problem because it's so elegant.

All you need is a little bit of geometry.

Now, you could solve it using more advanced geometry or even calculus by writing the two distances as a function of each other and minimizing that.

DR. DEREK MULLER: At a, b, and c, then the best place to put the pier is at a location where x is equal to a times c, which is the distance to there, divided by a plus b.

You're welcome.

But this is just so much more elegant.

So if I've got Mordor here and Hogsmeade there, what I'm going to do is reflect Hogsmeade.

DR. DEREK MULLER: Oh no.

Reflections, huh?

I should have expected this from a girl who did a video about mirrors.

It's also solving a problem with a different problem.

We were looking for one distance at the beginning, but we found a different distance that happened to be the same distance.

And by similar triangles and using these really simple arguments, we solved the first problem.

How do you like them apples?

This is getting weird.

But seriously, math is an essential tool for physics.

Algebra brought us constants.

And geometry brought us a way to draw our world with numbers.

And calculus is the only branch of math that has ever changed the way that I think.

But that's enough mental meandering.

Thank you for watching this video.

And happy physicsing.

---from the shoreline-- DIANNA COWERN: So you solved it.

I solved the problem.

DIANNA COWERN: Dude, that's awesome.

I can't believe this actually worked.

To be honest though, I had Dianna's assistance when I was doing some differentiation, because the chain rule is a long time in my past.

But it came back to me tonight.

DIANNA COWERN: I may not be able to resist putting that in the video.

That is OK.

This was a team effort.

I mean, can I use props for this easier way?

DIANNA COWERN: Yeah.

Props?

Sure.

So I would take a string, and I would put a pin here, put the string there.

And then I'd wrap the string around this, so that I could pull on this side.

I would anchor the string at different points here and see where I get the most coming out that side.

If I could use props, i would do it like that.