Kshitiz Mangal Bajracharya
Kshitiz Mangal Bajracharya
A place for unconstrained discussion of Mathematics!
I am a graduate (M.Sc.) on Mathematics from Central Department of Mathematics, Tribhuvan University, Nepal. I love Mathematics a lot and will try to discuss basics to more advanced Mathematics without being constrained under time and syllabus. My videos will be for the sake of Math only.
Пікірлер
its amazing that the integral does not depend on parameter "b"
Thank you
Thank you for this video which i searched many time , finally i found it in easy method
Thank you for this video!
topology ma video dinu rubber sheet
Nice elegant solution! 👍
Thank you! This is very clear and well done as I remembered it. Mr. Stengel was my high school geometry teacher. I remember this being extra credit on a exam. My only wish was that he named it Stengel's theorem. I am so happy that his work lives on.
I will edit the title now. You'll see Stengel's Theorem there. And if you know about some other works of Mr. Stengel, please let me know. Would love to go through them.
fngd
Keep it up kshitiz jee.....😍😍😍
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15:09 Nice video! & Nice problem!👍 I really enjoyed it! 👆 this moment won my ❤
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Good job #Kshitiz you are my source of inspiration.
Excellent.Keep it up
I like that ahhhaa at 4:33👌 Keep going bro👏
Kshitiz dai ma 11 ma pugesi math ko tution dinu na ghar ma bhayeni chalxa
Tyasko laagi timi ko ho vanera chinna paryo ni yaar. Real ID bata comment gareko vaye alli chinna sajilo hunthiyo ki ...
Ma pragun ho😁😁
If u are good at Maths make videos for B.Ed. Maths
I will make videos, but based on topics like Calculus, Analysis, Algebra, Geometry, etc. It will cover the course plus other topics. I just need more free time.
Of course if u have capacity please make videos from 1st year to 4 th year challenge...
@@netraluitel1999 If it's a "challenge" then sorry. I don't take challenges from anyone just to fulfill my or their ego. If that means I am "scared" or if that means I "don't know things properly", then so be it. If it's about wanting good content, then I need more time for making videos. A ten-minute video takes at least one hour of work. The lectures will be at least 60 minutes long. Just imagine the time and effort needed for such videos. I don't have free time, at least not now. I had planned to start Metric Space back in November of 2021. That's still on hold. There will be some short length random videos for a few months. I might get more time after that.
Math in higher level is very tough to understand .I challenge u to teach math by explanation
@@netraluitel1999 then sorry. I am not taking up any challenges. I will create content in my own terms. I don't have to prove myself to anyone.
Great! Nice! 🙂
Nicely explained!
Thank you! 😄
It's actually a worthy channel... Greatly and nicely explained!
It's really great to know that you liked my explanation and considered my channel worthy. That's very great of you. ❤️
♥️♥️♥️
Wow nice
très bien !!! Love from france
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Good explanation 🤟🤟
Thanks.
Video is very poor,even the formula can't be read.
I know that, sorry for inconvenience. I hope you find a better video where you can read the formula properly.
Very good explanation - best on KZread, but it's a pity the video is poor. Had to squint to see the symbols.
Thanks for appreciating the explanation. Regarding video quality, yeah .... it is kinda awful. There were many issues with my phone back then, and I was writing in the way I teach myself. I am planning to make a three part video for the three cases. I will use a better phone and won't squish a lot of things at the same time. That'll mark a temporary end for discussing Geometry. I will have to complete concurrency theorems and Morley's trisection theorem before that.
Good exposition, but you need to choose better tools for show it
Yes, I have figured that out now. Thanks for the suggestion.
I didn't make this video by keeping any competitive exams in mind. Here's a shorter video for the same question, and it's quite different from mine. But if you are looking for quicker ways to solve I recommend this video. If not, you can either choose to follow my solution or Amiya sir's solution or give your own solution. kzread.info/dash/bejne/gIaKj7B6nZfLhso.html
😍😍😍😍😍 Where I can get this type of more questions for practice vatiz sir 😁
You can search for these questions in older textbooks of Commercial Arithmetic for I.Com. level. Mathematicians like Dr. Bhanu Chandra Bajracharya, Krishna Bhadra Khanal and Ram Prasad Shrestha have written it elegantly some decades ago. You may check out R. S. Aggarwal's Arithmetic too. But avoid the so-called great practice books for school level. They include these as "challenging questions" without clarifying or discussing the concepts. A difficult question is always welcome but if the difficulty is created by not discussing the related concepts, then I don't approve them.
2:27 ... it's ((a_1+b_1)/2, (a_2+b_2)/2). Sorry about that.
I loved the way of solving it... very beautifully the usual geometry with Pythagorous rules and co-ordinate geometry are combined to solve the problem. Welldone 🤗🤗
Keep it up bro
СПАСИБО.ПРИВЕТ ИЗ БАКУ.
Very thorough explanation, thank you. But try to avoid standing in front of the whiteboard while adding things to the diagram. It blocks the view and one cannot be sure what is being written!
Thanks. About standing right in front of the board ... I will try not to do that again. Thanks for your suggestion.
See: harmonic mean.
Amazing! And clear! Thank you!
very nice
Thank you. 😁😁
Too small to see on a mobile phone.
Sorry for inconvenience. Will improve now.
With your notations. The four triangles have the same base QR. So their area depends only on the height of M,N,O and P. In terms of coordinates, O is a convex linear combination of P,Q and R O = u*P + v*Q + w*R , u,v,w > 0 and u+v+w = 1 then M = (u/(1-w))*P + (v/(1-w))*Q and N = (u/(1-v))*P + (w/(1-v))*R Let h(X) be the height of X relative to QR , X in { M,N,O,P } h(O) = u*h(P) , h(M) = (u/(1-w))*h(P) , h(N) = (u/(1-v))*h(P) Now, (1-v)/u + (1-w)/u = (1+u)/u = 1 + 1/u Therefore, 1/h(P) + 1/h(O) = 1/h(M) + 1/h(N)
I will cover this one and will credit you in the video. That is, if I remember this comment after may be roughly eight months. But I will, if I remember. Been busy lately and will be busy for a long time.
@@kshitizmangalbajracharya5152 It's nice to keep busy. I wish i was more myself.
Please redo video draw the diagram 3 times its size shown here plus text size should be 3 time greater size .
Will do it, but not any time soon. Got to deal with exams, for a few months. I am planning on doing a three part video for the three cases.
Interesting. I saw it used and it was unfamiliar to me. I've like and bookmarked the video. I'll have a crack at proving it (my geometry is rusty, so I'll probaly fail), then come back and watch this either way.
If it's just a matter of being rusty, then it'll just take some time. You aren't going to fail. If I can do it, you can do it as well. 🙂🙂 Thanks for liking and bookmarking. 😁😁
Regarding being unfamiliar, I wasn't aware about it after so many years of teaching and studying Geometry, either. I happened to watch a number of videos regarding some Indian competitive exams (CAT exams, I believe), where they miraculously solved apparently impossible questions using the most unfamiliar results you can possibly imagine. Ladder Theorem was one of them. This 33 minutes of me rambling on like a lunatic was a culmination of three days of effort on focusing only on this particular thing. I searched everywhere and found almost nothing at all.
Great video! It went by way faster than I realized! One suggestion is that once you have the right angle case, it might be easier to let the proofs of the other two cases follow from the fact that if you have a triangle ABC and you move point A along a line through A parallel to line BC to some new location A', then the area of the new triangle A'BC is the same as the original. Further, if you had points D and E on sides AB and AC and you move each of *them* along lines through them parallel to BC until they land on lines A'B and A'C making points D' and E', the ratios of lengths of the relevant line segments will be the same as before (AD/DB = A'D'/D'B, A'E'/E'C). The same should hold for the other points in your diagram, in particular for the intersection point of DC and BE. In any case, since all the areas are equal in the new triangle to the original triangle, no matter where on the line through A parallel to the line BC you place A', you can take *any* starting triangle ABC and create triangle A'BC as a right triangle. Since the new areas are identical, any relationship between them in the new triangle must also hold for the original triangle ABC. It's pretty late, so I might be wrong, but I'm reasonably sure this idea would work to turn your 3 cases into one case and save you all the algebra and detailed diagrams those last two cases require. :) Great video! Thanks!
I will definitely try your way too!
Cool! Let me know how it works out :)
@@patrickpablo217 I will. But not too soon. It may take months. Exams are nearing, and I gotta prepare for the exams.
that seems very reasonable :) good luck with exams!
Can you again upload this video because it is ambiguous. It is a nice video. Thanks.
Sorry dear, I had to compromise the quality of the video because a year ago, I had my phone running out of space, and due to the pandemic, there weren't sufficient stationery goods.
Your proof is excellent. Just the film quality needs to be improved. The acute and obtuse triangles contain more notations, especially the obtuse one . It would be much better if the diagrams were drawn much bigger.
yes It will be much easer for him and us if he draw larger sketch and colour pens . but it is really excellent prove
Thanks gentlemen! About the quality .... I realized that it wasn't so good. I recorded the video almost a year ago, and to be honest, I started uploading the videos just to kill my time during the pandemic. I had many things on my phone (I still record using the phone), so I had to compromise the quality of the video. Regarding color pens .... I ran out of stock and in my locality, there weren't any. It was impossible for me to travel to the commercial area. Definitely, I am trying to improve the quality of my videos. I am just a student, and the university is running the classes now. So, it might take a while.
@Ken Fullman I am planning to make a three part video to cover Ladder Theorem in an unrushed manner. Thanks for your suggestion. I made videos just to kill my time during the pandemic last year. I will improve the quality a little bit.
Thanks so much, l too was looking for the proof of ladder theorem. Found it here. Well explained
You're welcome, and thanks a lot. It's great to know that this video is serving its purpose to the ones who are searching for a proof of the ladder theorem for areas.
Thanks a lot for this proof video sir. I can't explain how much you have helped me.......Keep it up sir.....we need more this kind of video.....Sir, don't quit uploading videos on KZread 🥺🥺🥺 Love from Bangladesh🇧🇩🇧🇩🇧🇩
Thanks for your kind words. And, do not worry. I won't quit uploading the videos. It is just that I won't be able to upload videos (on a regular basis) for the next six to eight months or so. But you will (probably) get three or four of these kind of random math videos. Thereafter, you will get more videos and tutorials ranging from school level math to graduate level math. Love from Nepal. 🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵🇳🇵
*at 4:25 I used beta again. The beta used in the previous similarity and this similarity are different.
really it is very interesting .❤️❤️❤️
Thank you sir.