A quick geometry problem.

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We look at a solution to a nice and quick geometry problem.

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Do you ever worry about what might happen if Dr. Penn accidentally chose a *bad* place to stop? 🤔

ruathak
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Hey Michael! That final statement about geometry being taught poorly in the US and about how you’re still internalizing the rules used to solve simple problems such as these was relatable beyond belief. You make me (and surely many other students) confident about learning math and about how everyone learns at their own pace, with the means they have available. Thank you for being so open and honest!

diegosantoyo
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If you put a point G on AC such that CG=AD=BE, and construct the line BG, you get a triangle in the middle with 3 equal angles (because it's the exact same construction, just rotated 60 and 120 degrees), so it has to be 60 degrees

yoavshati
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Thank you for these videos. I find them professionally constructed and very instructive. I am 80 years old and retired. Every morning I try to solve one of the problems set in the videos and usually I do not succeed. But today I managed to solve the problem on my own. Hurrah.

levonnigogoosian
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Since lengths AD and BE are arbitrary, set them to zero meaning D coincides with A and E coincides with B, hence the angle in question coincides with base angle BAC which is known to be 60.

SanjayRay-dpvs
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exterior angle theorem is a short cut which could be used here

woodithwoodard
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I have another method that goes like this. Pick point G on CA such that CG = AD = BE. Then connect BG. The entire figure, by construction, has a 3-fold symmetry, so the small triangle in the center of the figure is equilateral, so all its angles, including the angle of interest, are 60 degrees.

eugenekim
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I’m glad you will build up geometry from the bottom because I’m new to geometry!

joeaverage
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Wording of the problem actually gives an interesting shortcut.

Since the length of AD/BE isn't given and the question still wants an absolute answer for theta then we can set AD/BE to what ever makes the math easiest. If you set them to length 0 then point D is exactly A and point E is exactly B => line AE is exactly AB and line CD is exactly CA => their intersection point (F) is exactly A and the angle they meet at is exactly BAC (60 degrees).

thebackyardmovies
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Honestly I like these easier problems - take me on your learning journey!

carsongbaker
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I like that you are building up from some easier problems. It makes it much easier to relate to. Thanks for expanding into geometry!

jjcadman
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consider a clockwise rotation of 120 degrees about the centre of the triangle. The transformation also rotates CD to AE with a clockwise 120 degrees. Then theta is 60.

quantalpha
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5:15 It’s okay Michael
5:45

Loading 93.8%... Have a great Sunday, everyone. Don’t forget to take care of yourself. Personally it’s gonna be a nice Sunday Football. Anyway...





Do there exist 5 points in the space such that for all n ∈ {1, 2, ..., 10} there exist two points such that distance between them is n?

goodplacetostop
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You're geometry rocks as far as I can tell..! Thanks for the video...!

tomgreg
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We can also note that the problem is independent to how long AD is as long as AD = BE due to the fact that it does not need to be specified. Then we can essentially choose D to be midpoint of AB.

s
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Some of these simpler geometry problems are totally fine. I have students from all levels and it's great to see a variety of problems that I can give them.

The book suggested here is also wonderful.

chuang
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I just thought of a super easy method, because it doesn't specify where d and e are, I just moved them really close to a and b, which basically just makes the 60 degree angle of the inside of the triangle.

Bruh-idqc
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Very humble Dr. Penn, thanks for the great videos.

Stev
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This is so awesome!! I’m not super good at problems like this and this just blew my mind

mvizzy
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Totally agree about Geometry and how it is taught. I only had about a half year of Geometry at a weak high school but still managed to go on.

doctorb