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Aha Problems
Приєднався 15 лип 2020
Can you see the hidden configuration? #geometry
In this video, I solve a British Maths Olympiad 2001 geometry problem from Round 2. The problem is simple if you spot the well-known configuration.
00:00 Drawing the setup
03:21 Pause and try.
03:33 Hint
03:46 Solution
00:00 Drawing the setup
03:21 Pause and try.
03:33 Hint
03:46 Solution
Переглядів: 143
Відео
What are Symmedians? How are they used?
Переглядів 92Місяць тому
In this video, I introduce symmedians and three characterisations: Angle, locus and length based criteria. We also identify a common configuration in which symmedians arise. 00:15 What is a 'symmedian'? 01:06 Angle condition for symmedian (useful for angle chasing) 01:33 What is an 'antiparallel' to a side in a triangle? How antiparallels produce cyclic quadrilaterals? 03:00 The locus of the mi...
IMO 2002: Incentre Excenter Lemma kills it!
Переглядів 75Місяць тому
A surprisingly easy solution to IMO 2002 P2 (G3). It mainly uses the Incentre-Excentre lemma. All the prereqs are explained. 00:24: Constructing the diagram 03:11 Basic Facts 07:06 Solution Credits to Viswanathan for the background music.
Reflections are beautiful!
Переглядів 47Місяць тому
In this video, I solve a cute problem from RMO 1998. Only reflections are needed to solve the problem. The solution starts at 1:00. Credits: Intro music at the start is due to Viswanathan.
Olympiad number theory: Suddenly a wild 7 appears!
Переглядів 3422 місяці тому
We discuss INMO 2013 number theory problem in this video. Its quite surprising that the only prime that solves the equation is 7. Lets see how. 00:08 The question 00:50 Factorisation 02:45 Bezout type argument 03:06 The wild 7 appears. 04:22 Case analysis 07:24 Arithmetic mistake (see comment by @aminzahedim.7548) 08:40 Final answer
Bertrand's Ballot Problem: Have you seen the gorgeous GEOMETRIC solution?
Переглядів 2382 місяці тому
Indian elections are right around the corner and we have to study a bit of election mathematics. So we will celebrate the election by learning a famous historical puzzle, Bertrand's Ballot Problem: "In an election where candidate A receives p votes and candidate B receives q votes with p bigger than q, what is the probability that A will be strictly ahead of B throughout the count?" This proble...
Can you solve this functional equation?
Переглядів 2913 місяці тому
We solve a functional equation from INMO 2005. 2:55 one-one and onto nature of functions from the properties of their composite.
Don't eliminate, do this instead!
Переглядів 1033 місяці тому
A system of linear equations, not solved by eliminating. We use high school algebra to solve it.
Pause and Solve: A single ROTATION solves the problem.
Переглядів 953 місяці тому
Given a triangle with ABC, erect squares ADEB and ACFG outwardly on the sides of triangle ABC. Let M be the midpoint of BC. Show that AM is perpendicular to GD. A general tip for IOQM, RMO, INMO, IMO, AIME, AMC: When you see a bunch of squares around a common point, Rotate by 90 degrees about that point! Edited by: Shreesha Shenoy Badthodi.
A STUNNING proof of Newton's theorem using parallelograms only!
Переглядів 1253 місяці тому
We present Roger Johnson's elegant proof of Newton-Gauss theorem. 00:00 Statement of Newton-Gauss theorem. 00:30 Fact 1 (about areas of parallelograms) 01:30 Fact 2 (converse of midpoint theorem) 01:52 Johnson's proof of Newton-Gauss theorem. This theorem is a classic olympiad side-dish.
The Complete Orthocentre Configuration: A path to a CELEBRATED circle!
Переглядів 1133 місяці тому
The simple altitudes of a triangle paves way to discover the complex terrain of circles buried underneath. An insane circle is waiting to be discovered. 00:00 Preview 00:20 Introduction to the configuration with orthocenter H (of triangle ABC) 01:43 Angles and Lengths 4:28 AH=2R cos A 4:39 Two circles with side and AH as diameter. 8:39 Proof that the circles are orthogonal. 10:00 An insane circ...
Aryabhata and Madhava's approximations of Pi!
Переглядів 464 місяці тому
Happy Pi day. Did you know about the Indian contribution to Pi? In fact, the Shatpata Brahmana, dated 6th century BCE accurately computes the value to 3.1.
4000 year history of Pi.
Переглядів 1634 місяці тому
On Pi day, we discuss the most famous constant of mathematics. We focus on Indian mathematicians, decimal expansions, experiments and some drama. 00:00 Introduction 00:20 What is Pi? 01:03 The name Pi was given in 1706. 01:16 Measure Pi. 03:06 Archimedes Circle Cutting Method 04:02 Aryabhata's shloka computing Pi. 06:28 Mnemonic for Pi. 09:18 Pi is irrational. 12:30 Lindemann proves Pi is not a...
Parallelograms-3: Concurrency of Altitudes and the Euler line!
Переглядів 1064 місяці тому
Have you ever wondered why - altitudes of a triangle meet at a common point (orthocentre)? - Orthocentre, Circumcentre and Centroid of a triangle lie on a common straight line? In this video, we continue the parallelogram series. We only have a set of parallel lines feature in this video. And we will use this to get more parallelograms in the next video. The facts proved in this video are routi...
Only TWO facts needed! Problem 5 of IOQM '23
Переглядів 734 місяці тому
Only a basic fact about the medians and similar triangles is needed for this problem. Learn concepts in the videos for your IOQM/Olympiad preparation. Math Olympiad is very approachable if you learn concepts in small steps. Best of luck!
Three line solution to a RMO 2002 problem!
Переглядів 704 місяці тому
Three line solution to a RMO 2002 problem!
Pause and Solve: 2024 AIME II Problem 10 with hints.
Переглядів 964 місяці тому
Pause and Solve: 2024 AIME II Problem 10 with hints.
Parallelograms in Geometry - II: What is Varignon's parallelogram theorem?
Переглядів 874 місяці тому
Parallelograms in Geometry - II: What is Varignon's parallelogram theorem?
Three different methods for a geometry problem
Переглядів 1343 роки тому
Three different methods for a geometry problem
The Incenter-Excenter Configuration
Переглядів 2,5 тис.3 роки тому
The Incenter-Excenter Configuration
very good thank you
very nice video can you bring a series for inmo 2025
Very nice video. Bring more such topics
Thanks for the comment. I was not sure if people are interested in specialized configurations. You have encouraged me to do a few more videos like this :)
7:24 you made a slight mistake: if 4m+1=49, that means m=12 not 16. This would’ve already given you the solution you found not long after based on analyzing the last case; namely, m^2+3=3*49. Also, I think you brushed over the two key insights: a) the idea to take -3 to the LHS and factor the whole expression, and b) to use a very particular combination of 4m+1 and m^2+3 to arrive at a constant integer (independent of variable m), whence you could deduce something about their gcd. Eloquent explanation to a rather difficult problem, though. Thank you so much.
Oh... you are right about the mistake at 7:24. Looks like I am getting worse with arithmetic. Thank you for your comments.
Your English is difficult to understand. I'm interested in this problem but it is too difficult to decipher your thick accent. I don't know if talking slower would help. I know it would make the video longer. Please enunciate more clearly in the future and maybe I'll subscribe. Thank you.
I have been told I speak fast. It is especially true that I do not enunciate properly (especially the last syllables). Thanks for the feedback, I will work on it.
Beautiful
Thank you.
Please start taking classes for ioqm on youtube
Could u pls do projective geometry
Sure. I love that subject. Thanks for the suggestion.
Wonderful problem
Beautiful explanation
Thank you :)
Lovely observation
I love solving these problems! Consider the cubic polynomial ax^3 + bx^2 + cx + d=f(x) Now consider the biqaudratic polynoimal g(x) = x^4-f(x) Observe that as g(x) is monic and has roots at 3,-1,2 and -2 Now from Vietas relations a= 2+-2+-1+3 = 2 b =7 c = -8 d=-12 as usual thanks for the problem sir!
Beautiful solution sir!
Thanks Aditya!
noice, another approach wud be the show that M lies on the radical axis of circles (EA) and (AF) with centres being D and G respectively. motivation being:- noticing two nice circles and invoking the perp conditions of radical axis
@anudeep0129: Oh! I hadn't thought of that approach. But how do you show the M lies on the radical axis?
@@MudithaMath o lmao so i thought i had a proof but turns out i fakesolved, i'll try to continue with this approach and get back
@@MudithaMath so i asked one of my frnds abt this and he said he cudn't see any nice synthetic solution, but to trig bash and show that MD^2-DA^2=MG^2-GA^2 hence showing M lies on the radical axis.
cool solution man!
Thanks :)
even though you get less views i love how you keep uploading I love your videos and hope you keep going on like this
Thank you for support.
Great video! Just lacking audio quality at the start. Maybe try using something like "Dolby On" To record your audio!
Thanks!
Wow
i love his videos beacause I get to learn something more than school from him
Thank you :)
smooth entry
nice fountain pen
Thanks. It's a Pilot Custom 74 that was gifted to me.
wow that's cool@@MudithaMath
thank u very much sir , thank u very much , U HELPED TO UNDERSTAND THE INCENTER-EXCENTER configuration
we need more such videos sir!
Thanks Mrinal. I am adding more videos now.
Very nice video
Thank you so much
Woah I just explored ur channel! When will we have more vids on this channel! Btw I'm Anudeep
very helpful!
Excellent way of explanation! Keep up the good work!
Thank you brother, was really helpful for my Olympiad preparation.
thank you sir, this was very helpful. please continue posting videos
thank you sir for clearing the concept
Nice.
Nice