Cow-culus v Geometry (extra) - Numberphile 

If only the fence in the tree problem was a mirror. Run at the reflection of the tree.

It's the same problem -- the distance across the bridge is a constant, so it will differentiate out of the problem, and you can actually just ignore it and solve the problem as before. You just have to adjust the numbers so that the river has zero width (the villages will be closer together), solve, and then you can expand the width of the river to any arbitrary number (making the villages that much farther apart) without changing the solution. It will still be the minimal path.

You can make a lovely demonstration piece of it if you fold the paper along the river so the banks meet. Draw the straight line between the villages, then pull the paper out and draw the bridge in between the points where the route meets the banks.

Love her accent and the way she explains things.

These two videos remind me of a documentary that showed how medieval builders could figure out efficient construction and correct proportions by using geometry without any numbers. I wish they taught us the practical power of geometry in school.

Label the trees as tree1 and tree2 then reflect the trees across their closer sides and label them as reflection1 and reflection2. Finally draw the lines between reflection1 to tree2 and then reflection2 to tree1 That's what I would do. that's equivalent to the minimum path.

I like the second one because now that we know that we can reflect to make a "phantom" point, and just draw straight lines, this problem is trivial. That's what I love about math. Once you have a theorem, you can apply it to so many different things in both the world of hypotheticals and reality.

In the kids case the trees are the farmer and the cow, the fence is the river, just changing the fence the second time. If the angle of the fence were bigger than 90 degrees, the point of reflection could be a projection behind the other fence, requiring other method to solve.

In the first problem of Villages I feel it will be easier to assume the river length to near zero, then put a straight line from V1 to V2. Points interstcting river banks (near zero apart) can then be expanded to river width with straight bridge across. Which is shortest path

it appears that if you were to draw a straight line between the two villages, then where the line crosses the center of the river, that is where you would build the bridge.

I found a general solution this works for both the cow problem and the villages problem. Although complicated calculus is sooooo powerful: a= distance from village1 (or farmer) to river b= distance from village2 (or cow) to river c= length of the bridge (c=0 for the cow problem) d= distance projected on the river of the distance between village1 (farmer) and village2 (cow) y=sqrt(a^2+x^2)+c+sqrt(b^2+d^2+x^2-2dx) derivative of the function and set to 0 y'=x/sqrt(a^2+x^2)+(x-d)/sqrt(b^2+d^2+x^2-2dx) x/sqrt(a^2+x^2)+(x-d)/sqrt(b^2+d^2+x^2-2dx)=0 x*sqrt(b^2+d^2+x^2-2dx)+(x-d)sqrt(a^2+x^2)=0 x*sqrt(b^2+d^2+x^2-2dx)=(d-x)sqrt(a^2+x^2) b^2x^2+d^2x^2+x^4-2dx^3=(d^2+x^2-2dx)(a^2+x^2) b^2x^2+d^2x^2+x^4-2dx^3=a^2d^2+a^2x^2-2da^2x+d^2x^2+x^4-2dx^3 b^2x^2=a^2d^2+a^2x^2-2da^2x (a^2-b^2)x^2-2da^2x+a^2d^2=0 x=(da^2+-sqrt(a^2b^2d^2))/(a^2-b^2) x=(da^2+-abd)/(a^2-b^2) x=(a+-b)da/(a^2-b^2) x=da/(a-+b)

I believe that the Brady-Zvedelina (Bradelina) pair is more harmonic and fun for Mathematics than the Brad-Angelina (Brangelina) pair is or was for the movies. Anyway, the solution to the "Trees and Fences" problem can be easily solved if it is divided into 2 "Farmer and Cow" problems, like this: Part 1: consider 1.1 Tree-1 = Farmer; 1.2 Fence 1 (closest to Tree-1) = River; 1.3 Tree-2 = Cow; 1.4 Solve the first "Farmer and Cow" problem to find the point that touches Fence-1 (acute angle line); Part 2: consider 2.1 Tree-2 = Farmer; 2.2 Fence 2 (closest to Tree-2) = River; 2.3 Tree-1 = Cow; 2.4 Solve the second "Farmer and Cow" problem to find the point that touches Fence-2 (horizontal line); If the solution is by Calculus, I believe it will be the set of points from the results of two similar derivatives. Is that helpful?

I would solve the wide river problem like this: 1. Fold the paper at the south river bank. 2. Make this fold meet the north river bank. 3. Draw the straight line connecting the villages, crossing the now zero width river. 4. Unfold and build the bridge at the gap of the line.

I have another intuition. The path that light takes is always shortest distance. The path that the light travels in land will bend in river but will bend back and be parallel to previous path in land. However, constraint is that bridge is perpendicular so we say light doesn't bend in river. Still, the analogy is that paths in land are parallel.

These geometric solutions are very intuitive, I like it ❤

I'd say the bridge length isn't important as it's constant at any path you choose, so the path is just the same as in the cow problem

For the bridge problem, I'm a very simple man: I'd draw a straight line between the two villages, and where it intersects the middle of the width of the river, I draw the bridge (perpendicular to the middle of the river), and then connect each end of the bridge to the corresponding village. It's got the advantage that it works even if the river isn't straight or varies in width: you just need to find the middle of the width in a very tiny area, and never rely on assumption about the width of the river or orientation of the bridge. And over all, it just seems more logical to start from the goal, the shortest path, the straight line, and punctually adapt to local obstacles.

draw a line from v1 to v2. Now draw the bridge so the center of the bridge lies on the line. Now redraw v1 to bridge and bridge to v2. This will be the minimum path. The way you know intuitively that this works is if the river is 0 width you have the original line from v1 to v2 which is obviously minimal.

You just need to mirror the trees twice. Pick the closest tree to the fence in one of the 2 end space. link to the opposite tree in the first image and then link that to the image of the starting tree in the 2nd image.

Instead of the boy reflecting from the fence think about the garden reflecting so the boy will travel in a straight line

Easy (when you've seen the first video!) Just put a mirror on the blue lines and mark where the two trees are reflected "outside" the original shape (so to speak) and draw a straight line connecting the imaginary "reflection" trees. Then mark the two places on the blue lines where the new line crosses them.

Brady's suggestion was correct. If we made the houses on the same side of the river, then used our knowledge of the cow method, then it would give us the right path, and the point of reflection is the place where the bridge should be.

Despite what the video said, I ignore the width of the river (only considering the distances of each village from their nearer bank and from each other along the river). You get two sloping paths between villages and their nearer banks, each coming to points directly opposite each other across the river. As you widen and narrow the river, you change the length of the bridge, but not the lengths of the paths on land, so narrowing the river to 0 width and identifying the two ends of the bridge with each other means you get a path A->X->B made up of the two sloping paths, and it's obvious that that distance is minimised by the straight line A->B. Bring back the width of the river, and that only changes the total distance walked by the length of the bridge. If the bridge can go at angles up to some limit, you can do a similar trick when the straight line between the two villages is at a shallower angle than the bridge is allowed to go - instead of shrinking the river vertically, shrink it parallel to the bridge (which will be at the most extreme angle allowed) so that the two ends of the bridge (wherever it is) end up at the same point and take a straight line on that map. This is not very different from the video's solution of doing the bridge first - in both cases, you're finding a point so that the path from that point to B is the same length as the original path's sections on dry land.

Fold the paper in half along the river, then fold it in the opposite direction at the two river banks, so the river is folded out of the problem. Then draw a straight line. Then unfold the paper so the river is visible, and connect the gap in the two line segments you drew with the bridge.

4.25 is the potential distance shortest between the river problem if the towns are distance 1 from the river and the river is 0.5 in length and the distance horizontally is 2. this makes a very clear solution of triangles meeting in the center from the hypotenuses being 5. 4.25 is the shortest POTENTIAL distance in straight line calculation.

In a high school physics class, we had a very similar bonus problem: You have a gecko that can crawl on the ground and on walls. It starts on the ground. The gecko wants to reach a fly on the wall. What's the shortest path? The trick in this problem was to "unfold" the floor and the wall, so that you could just draw a straight line between the two points. (Much easier to explain with a diagram. But this was a lovely little series!)

When you drew the river I thought about this problem and it's so easy!!!

Delightful! Thank you once again.

It's composed of two exact problems, minimzing the distance for a single trip between a tree and a fence, to do this we should minimize the length of the median of the distance between the two trees, we can do that by simply drawing the perpendicular median to the fence and voila, it's the median of minimum length, hence the median of the least triangle perimeter, we repeat this process for each tree.

Thanks so very much!!!

I've sketched out a solution to the homework problem but it looks like I can't post a link to it in the comment section. To sum up my sketch; Draw a reflection of the yard along one fence (I've chosen the fence closest to B, but the choice is arbitrary), labeling the new trees A' and B', with B' being the point closer to B. We draw a second reflection along the fence closer to A' (if you chose A first, you choose B' here), this time labeling the trees A'' and B'', with A'' being closest to A'. Now we draw a line from B to A' and from A' to B''. The total length of these lines is equal to the original of B-to-Fence-to-A-to-Fence-to-B.

My favorite numberphile presenter.

Would the bridge's location be the same if you drew a straight line between the 2 villages and took the mid-point of the line segment that overlaps the river and make it the mid-point of the bridge (making the bridge go N-S), would that have produced the same result?

The river is irrelevant in this case, since the time to transverse it is always the same. Just remove the river and draw a straight line. The second problem is the same as the cow problem. Use the equal angles property to find the rigth spot on both line segments.

Anyone interested in these geometrical problems should check out the game "Euclidea".

I guess its the same answer but Id do a straight line between villages, and then in the middle section of the river place the bridge. Then connect each end of the river to each village. However I start to see a difference if one village is exactly on the river shore

Sounds elliptical Take a piece of rope, tie it to the trees and treat them as focis. Trace out the two smallest possible ellipses , such that the first creates an intersection with the horisontal line with minimum amount of rope used. Repeat for the upward sloping line.

Trying to do this geometrically using points as seen at 5:05, let me know if anything is off. Been awhile since I've used geometry. What we would want to do is find the angle X-V1-A so that we could create the path V1-A that leads us to the start of the bridge so we could construct it. In order to do this, we would need to measure the distance of both villages to their side of the river, call them V1R and V2R. Then we would want to know the horizontal distance between the villages, call this D. We are going to add a new point directly to the right of point X which is D distance away, calling this point Y. This point will be directly below V2. We can now make a triangle of X-Y-V2. This is a right-angle triangle, with sides of XY=D and YV2=V1R+V2R. The key here is that the angle Y-V2-X = the angle X-V1-A. This angle can be calculated as tan^-1(D / (v1r + v2r) )

Idk how to actually do it, but I want to make an oval with the trees as the focal points; I think the points where the oval crosses the fence (closest to the corner) would be the best place to run to.

I think using the mid point between both the tress and then drawing perpendicular to both the sides would do the trick. I don’t have proof yet, but intuitively it makes sense. Because it would be the shortest distance if both the trees are the same.

Why am I getting homework here? 🤣

Simple and elegant =D Well... I dunno what's the cow problem, I'll look it up later. But I can tell the first 2 lines and the second 2 lines are independent so we just need to get the shortest distance for each pair. We get 2 triangles where 1 side is constant. And the sum of 2 others need to be the shortest... In practice it would be easy: just use a string. It would take the shortest path when you pull it. Sigh... I'm no mathematician. I'll go see the cow now.

It's all about conserving angle of incidence in the 3 problems

the second problem seems that it is the same with a first video before. but if it requires another solution, i would like to try to draw a circle whose centre is at a intersection point by using some arc.

I am a simple person. I see (cow)culus, click and like immediately.

Two choices are made, so the first and second legs of the journey are independent. Each leg of the journey is exactly the farmer/cow problem.

Draw a straight line between the two villages.. then take the centre of river where the straight line crosses and take that as the centre of the bridge

Or you could fold the paper along the on of the banks and align the edge of the fold along the other bank of the river - therby making the river disappear in a concertina - then draw the straight line between the two villages - then unfold the paper to reveal the river again ‐ and then draw the bridge.

Could you achieve the same result by drawing a straight line between the two points and setting the center point of the bridge at the point where the line intersects the center of the river?

The width of the river is completely irrelevant, you can ignore it and shrink it to width 0 but you should keep the bank on the side of V1 in the same place.

2:43

Different solution for the villagers ?? first a straight line from v1 to v2, then take the middle of the line (in the middle of the river) and turn it perpendicular to the river making it "the bridge" then erasing the lines outside the river, then connect the villages to the bridge....... ??

Based on fist video its a minor variation that seamed obvious .. Draw a line from Village 1 to Village 2 .. The point on that line at which the line crosses the midpoint of the river should also be the midpoint of the bridge ..

My solution was, cut the river from the paper, draw straight line between the villages and then paste the river back.

I love math.

how unimaginative is that? you're doing it on paper, so fold it and thereby spacetime until there is no river, draw the straight line and boom.

So she's implying that you approach the fence in such and angle that the angle to either tree is the same? So the angle between the trees would be 2a if the angle to one tree was a.

very nice

the 2 angles when the lines meet the river on both sides of the river are the same. This is the same as the last problem.

Make folds in the paper to hide the bridge, draw the straight line, then unfold to reveal the bridge.

the first one seems so trivial, the river is a constant in the optimization problem and can be ignored completely

2:44 What is that red rectangle doign there?

Fold the paper so that the river is entirely hidden. Now draw the line directly between the two villages.

Replace the fences by mirrors and run to the image of the trees !

i was thinking about taking the river as a line with no width, same solution

Folding top of river to bottom of river turns it into first problem

I feel like the kids should run Tree->Fence->Fence->Tree instead to make the Problem more interesting.

Just fold the paper to hide the river...

It's light going through a glass pane. The ray of light on both sides will be parallel.

I saw the first solution immediately. I have to think about the second one.

These are very simple problems after seeing the farmer with the cow. That's a weird sentence, but I stand by it.

Sounds like fin. Enjoy!

cow-culus and geo-moo-try

opportunity missed to fold a 1/2 m long paper in such a way that the river is gone^^

I want more homework!! :)

Ha I saw that one straight away , 8⁠-⁠)

Well that was trivial

Yeah. Cow-culus vs Meow-metry

Disappointed there was no paper folding to make the width of the river zero.

The homework is just two separate farmer/cow problems - each is it's own path minimisation.

i smell an elipse. Draw the smallest elipses so there is a point (or two) which touch the fence, then run from whichever tree, that point, the other tree, that point again, then the first tree.

I sense an oval around the trees;)

hmm stoped visoe in mid stream .. second problem same as farmer except twice .. once for each (tree wall tree) leg ..

The first problem with the cow was trivial. This one was easier 😛

~~ I see the letter "i" in your future. ~~ (If you're an engineer I see the letter "j" in your future instead.)

The width of the river actually *isn't* relevant. Just pretend it has zero width, connect the villages with a straight line, and then the river can be whatever length you want. Since the river has constant width, it doesn't actually matter where the bridge is, you just need it to connect the points on the straight line.

Very sly... telling us that the children cannot leave the play area to make using the phantom path solution less obvious :P

Same as the other problems, just make the same angles.

Visually the best solution seems to be as drawn - but I guess with the mirror hint you have to draw a line from each tree perpendicular to each fence (4 lines in total) and then run towards the point midway between where these lines intersect each fence.

Balance

you mean "Geomootry"

Sorry, but this is way too easy to be interesting