Proving Pick's Theorem | Infinite Series

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You could spend hours doing trigonometry to determine the area of a complex shape or you could simply plug in Pick’s Theorem.

Previous Episode - 5 Unusual Proofs

Written and Hosted by Kelsey Houston-Edwards
Produced by Rusty Ward
Graphics by Ray Lux

Sources and further references:
David Eppstein’s Math Fun

Proofs from the Book by Martin Aigner and Günter M. Ziegler

Comments answered by Kelsey:

J.H.

M.K.D

SleepyGuyy

[buchweiz]

So I was hesitating to write how much I like this channel since everybody is telling you that anyway, but other creators keep persuading me that it is still important for you to hear these words. And you deserve all the praise you can get. I have studied a very math-heavy field in college (physics), so most of the educational channels on math on youtube don't really cut it for me in the sense that I either already know the topics they cover, or they don't go into enough detail or it's plain boring, but not these series though. Every video is well written, deep and on an interesting topic, and you explain everything in such a way that even schoolchildren would understand it perfectly. And then you also engage with the audience in the most thought and curiosity provoking way possible! You have talent for this stuff and please continue making these videos! This channel is fantastic!

[George]

I had the privilege of being one of a dozen students proving conjectures with Paul Erdos. That man saw around corners.

[tannisbhee]

Dear Dr. Houston-Edwards, we need more people like you on Earth, or perhaps we need more visibility for educators like yourself. Thank you for putting forth the time and effort to share your love of mathematics, as well as to the the rest of the crew.

Thank you.

[iankrasnow]

A new episode of PBS infinite series is now one of the highlights of my week. Keep em coming!

[Twisol]

In your proof that planar graphs have Euler characteristic 2, I think you missed a step. Since we're allowing loops on a single vertex, we need to consider whether the edge we pick is a loop or not. If it's not, the logic works fine -- contraction of that edge removes one edge and one node, and we're fine. But if it's a loop, contraction of that edge removes one edge *and one face*! This is still fine (faces and edges can cancel out too), but the argument is subtly different between the two cases.

Personally, I would have focused strictly on straight-line planar graphs, which can't have loops to begin with. The base case is just as clean if not cleaner. Barring that, we could show that every planar graph with loops can be reduced to one with the same Euler characteristic but without loops; the proof would proceed exactly as before.

[gonzalowaszczuk]

This channel is awesome. I hope these don't get defunded

[amaarquadri]

3:25 "When we contacted the edge we removed one edge and one vertex so the Euler characteristic didn't change."

This assumes there is only 1 edge between those two vertices. In the case where there are k edges between the vertices, then you remove k edges, 1 vertex, and k-1 faces. However, since 1-k+(k-1)=0 the Euler characteristic is still unchanged.

[docthorium]

6:14 The answer is trivial and is left as an exercise for the reader.

[appsenence]

I saw this popup on my chrome app and I had my stoner friends over and I was acted like it was no biggo but you know, I love this show and lööve it when you upload, couldn't wait till they left, checked their busses and everything 10 times just to see this, now I'm stoned too so here we go

[MATHguide]

This is not a simple proof, but the video makes it digestible. Thank you!

[damon]

at 3:40, if you contract an edge that is part of a triangle, it removes the face but also merges two edges leaving the Euler characteristic the same but you missed that in the video. minor detail but it needed pointing out

[raykent]

If maths fails you, you could try archimedes. Extend the surface in the third dimension to give a volume. Submerge it in a bath. Measure the volume of water displaced and divide by the thickness. Run naked down the street shouting eureka (optional). Works for any shape.

[shubhamshinde]

PBS Digital Studios is a gift to the world!

[notEphim]

1:43 *connected

Also, I enjoy more the other proof. We prove the formula for rectangles, then for right triangles, then for any triangles, and then for any polygon using triangulation

[avethakur]

Amazing illustration...even after so many years it's still fascinating, useful and intriguing 🥰

[nikhilpatel]

6:14 Assume that two of the points of the triangle lie on a line of slope m. Then they are (m^2)+1 units apart. The third point can go either one of the two parallel lines closest to the line with the original two points. A little geometry shows that both of these lines are exactly 1/((m^2)+1) units away from the original, so we have a triangle with base (m^2)+1 and height 1/((m^2)+1), which just has area 1/((m^2)+1) * ((m^2)+1) / 2 = 1/2.

Also, awesome video thanks for sharing!

[jaimeduncan]

Excellent series, no pun intended, I hope the cuts proposed by the new administration don't hurt this and Space time channels. Keep the good work.

[benjaminpedersen]

I am really glad that you are doing this channel. People need to have a better understanding of what mathematics is about.
That said, you forgot to argue/mention that the graph obtained by contracting an edge is still planar.

[Hedshodd]

Showing that the triangles in 6:14 all have the same area is quite simple, because you can turn them into one another using transformations that don't change the area of the object.
These transformations are moving, turning and shearing. First, you move and turn the triangle on the top left so that one it's sides exactly matches one of the other two. Let's call this side a, the one you just used to align the triangles. After that, shear the point that is NOT a vertex connected to a so that it matches the triangle we are looking for (shearing is done by moving the vertex or corner parallel to an edge that it is not connected to).
The fact that turning and moving doesn't change the area of the triangle is easy, because you can either just shift your 'camera' again to match the original triangle, or show that the formula for the area of a triangle still holds true.
Shearing is a bit more trickier, but can be done graphically. It's probably easier to show it using a square. Take one side of the square, call it a, for example. The side opposite of a is the one we want to shear, let's call it c. Now, shearing would be done by moving c parallel to a. Draw a new c' that is the sheared version of the edge c you want to have, and draw the square using c' over the one using c, so you have to overlapping squares with one shared edge, a. Notice something? The difference between the two squares is visible as two triangles. If you were precise enough with your drawing, the two triangles are the exact same, they're just flipped. This means that you can actually visualise shearing as taking one triangle away on one side and adding the exact same triangle back on on the other side, meaning that you didn't change the net area; you just subtracted and added the exact same area on the original area.

[sameetakumavat]

The way you explain is awesome and you explained each and every term so clearly it helped me a lot so thanks for the video 😘