North Macedonian Math Olympiad | 2019 Q4
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- Опубліковано 21 жов 2020
- We solve a nice problem that combines number theory with a functional relationship.
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No matter what the functional equation is, My answer will be f(x)=x😄😆
f(2x) = f(x)
The total number of divisors of N=(p_1^r_1)...(p_k^r_k) is not (r_1+1)!...(r_k+1)! but rather (r_1+1)...(r_k+1).
Yeh those factorals shouldn't be there.
Yep
Exactly.
Eg : 24=2^2*3^2
It obviously can't have (2+1)!(2+1)!=36 divisors
well numbers can have more than its value in divisors, i propose an even simpler counterexample can be enumerated directly: 8 = 2^3, it has 4 (not 4! or even 3!) divisors 1,2,4,8
This was yet another great problem. Loving these
20:19
Hello, what’s up? Here’s the daily problem. It’s always hard to explain a geometry problem without a diagram but I think this one is clear enough.
PQRS is a rectangle with PQ = SR and PS = QR. The points A, B, C, and D are the midpoints of sides PQ, QR, RS, and SP, respectively. The point E is the midpoint of line segment AD.
Show that it is always true that the area of rectangle PQRS is four times the area of triangle BCE.
Since area of triangle in a parallelogram is half of the parallelogram, the problem is equivalent to show that 2*Ar.(ABCD) = Ar.(PQRS). Join AC and BD to meet at O. We can easily show that ∆OAD is congruent to ∆PDA. Hence 2*Ar.(OAD) = Ar.(OAPD) similarly we do for other 3. Hence adding these 4 equations we will have 2*Ar.(ABCD) = Ar.(PQRS).
Since PQRS is A QUADRILATERAL thus PQRS is a parallelogram of area 1/2(PQRS)
Since BCE has height and base same as ABCD thus area(BCE)=1/2•area(ABCD).........
................=1/4•area(PQRS)
I'm sure there's a more elegant way to do this, and I look forward to seeing that elegance in action :-D
I can set S to the origin of a Cartesian plane and orient PS along the y-axis and RS along the x-axis without loss of generality. I define |PQ| = a and |QR| = b, so the area of PQRS is equal to a*b.
By the definition of a midpoint, the following points live at the following coordinates:
A: (a/2, b)
B: (a, b/2)
C: (a/2, 0)
D: (0, b/2)
E: (a/4, 3b/4)
So, |BC| = sqrt(a^2 + b^2)/2, and line segment BC lies along the line L: ab/2 = bx - ay, the slope of which being b/a.
Next, I define a point F that lives at (x0, y0) such that EF and BF are perpendicular, and BF lies along the same line as BC. So then the slope of the line along which EF lies must be the negative reciprocal of the slope of the line along which BF lies. Using the definition of a slope and cross multiplying gets the following equation:
(a*2 + 3b^2)/4 = ax0 + by0
Plugging (x0, y0) into the equation for line L gets:
ab/2 = bx0 - ay0
That system of equations can be solved for:
x0 = (a/4) * (a^2 + 5b^2)/(a^2 + b^2)
y0 = (b/4) * (3b^2 - a^2)/(a^2 + b^2)
So, it can be shown that |EF| = ab/sqrt(a^2 + b^2)
Using BC as the base of triangle BCE and EF as the height,
A_BCE = 0.5 * |BC| * |EF| = 0.5 * (0.5 * sqrt(a^2 + b^2)) * (ab/sqrt(a^2 + b^2)) = ab / 4 = A_PQRS / 4, so A_PQRS = 4 * A_BCE
Answer..........
Let the length of PQ be 2x and the length of PS be 2y. It follows that PA=AQ=SC=CR=x and PD=DS=QB=BR=y. The area of rectangle PQRS is PQ × PS = (2x)(2y) = 4xy.
Therefore, we need to show that the area of BCE is (4xy)/4=xy. Since PQRS is a rectangle, each of the corner angles is 90◦ and each of the four corner triangles are right triangles. Also, each of the four corner triangles, APD, BQA, CRB, and DSC, has base x and height y so their areas are the same. The total area of these four triangles is4×(xy/2)= 2xy.
The length of the hypotenuse of each of the four right triangles is√(x²+y²), so AB=BC=CD=DA=√(x²+y²) and ABCD is a rhombus. Then, it follows that AD is parallel to CB.
The area of rhombus ABCD is equal to the area of PQRS minus the area of the four corner right triangles. So, the area of rhombus ABCD = 4xy − 2xy = 2xy.
Let h be the perpendicular distance between AD and CB, two opposite parallelsides of the rhombus. The area of rhombus ABCD is h×BC, but the area of rhombus ABCD is also 2xy, so h×BC = 2xy. Then the area of BCE=(h×BC)/2=2xy/2=xy.
Hi,
14:08 : so the condition f(1)=1 becomes redondant with f(p-1)=p-1 : let p=2
For fun:
2 "so on and so forth", including 1 "and so on and so forth",
1 "the next thing I want to say",
1 "no what I want to do",
2 "all the way up to",
10 "let's go ahead and", including 1 "let's go ahead and point that out", 1 "let's go ahead and clean this up" and 1 "let's go ahead and do that",
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12:42 : whouah! whouah!
Wow ,such a beautiful solution involving Primes and the Wilson theorem ,Plz sir continue the problems from different mathematical Olympiad s from around the world.
Great video bro
Can you Do a backflip or something in the intro, I saw you do one in the differential equation series
I am in class 8 and seeing your videos for past few months, thanx, for your videos
How about a video demostrating that wilson theorem. Never heard of it before. Any cool theorem demonstration will make cool content for the channel!
🇲🇰Finally something from our neighbours!
(Конечно нешто од нашите комшии!)
Greetings from Belgrade, Serbia!
Here's another one, from the MMO (Macedonian Mathematical Olympiad) 2016, Q1.
Solve the equation in N: 1+x^z+y^z=LCM(x^z, y^z).
Solution:
We have two cases, GCD(x^z, y^z)=1 and GCD(x^z, y^z)=d, where d!=1.
1) We get the following equation, 1+x^z+y^z=x^z*y^z, or (x^z-1)(y^z-1)=2, which can be easily solved in cases.
2) So d | x^z and d | y^z. And from the given equation, because d | GCD(x^z, y^z) (d is GCD(x^z, y^z) and d divides both x^z and y^z, d must divide 1, so d=1, which contradicts the assumption that d!=1.
@@aleksijtasic2605 Why do you get 1+x^z+y^z=x^z*y^z? Shouldn't the righthand side be 1?
@@JM-us3fr Sorry. I meant LCM instead of GCD in the text of the equation (in the text of the problem). 😆
If a|b and a|c and b-c isn’t 0 then a
This has got me thinking. It's quite easy to get to the point that n!+1 must divide f(n)!-n!. Getting beyond this point should be simple but it's annoyingly difficult. Keep in mind that f(n)! is of the form L! with L>=n. So L!-n! is divisible by n! Now the point is that this should imply that L!-n! is then not divisible by n!+1 unless L=n. This looks like it should be true but I lack a proof if one does in fact exist.
So...I was watching a presh talwaker video (i hate-watch them) and in the belgium mystery one, he solves it whatever. it was you have a 2 digit number, you add 3 in front of it and double the whole thing, that new number is 27 times the original number, find the original number. my question to you is, if you wanted to find all bases such that you could solve that question, what bases would they be? and what solutions would you get? i dont know if i said it right, but im talking about a general solution for that question over all bases. in base-10, its 24. what is it in base-11? base-12? base-13?...base-n?
But WAS it a good place to stop?
What i did was to set m=n. this will give f(n)! | n! which means f(n)!
8:19 can we just compare with mod f(1)
if f(1) > 1
f(1)+f(1)! = 0 mod f(1)
1 + f(1)! = 1 mod f(1)
and we came to a contradiction faster
Lol this channel went from 3k subs to 80k so fast
17:03 Didn't we say as long as the difference isn't 0, but then it turned out to be 0 so doesn't that stop the argument?
Yes and we also said at 2:42 that zero has infinitely many divisors, but at 5:25 that there is no "good notion of a number dividing zero". Hmmm, this proof leaves me perplex...
@@jacquesbabaud3610 Yeah, his argument was kind of garbled there. Certainly 0 is divisible by every integer n, because 0=0n. He should have just said that the inequality he wants at 6:30 holds in the case that f(1)-1 is not zero (since then it equals (1+f(1)!)k for a positive integer k).
If k|a and k|b then k|a-b
But then, likewise, if k|b and k|a then k|b-a
That implies k=
But the function is from natural number to natural number,but i don't know if its correct
Yeah it should be k | |a-b|, cuz only the function f is from natural numbers to natural numbers not the whole equation.
A nice problem from somewhere in the world you probably never heard of 😂
These "natural numbers" seemed to make heavy use of 0. Maybe we should have included it from the beginning
Haha, "whole numbers" yes?
EDIT: to be honest I haven't actually watched the video haha, just scrolling down the comments... I saved the video to my Watch Later playlist
Mickael said at the beginning that 0 is excluded of the solutions, that does not mean that it is excluded from results of calculus.
Good
Let x,y be positive integers, solve the equation.
x(x+y)=y^2+1
Any idea? );
Good exercise!!!!
x(x+y)=y^2+1
If x=y then wehave that x=y=1
If x is bigger than y then LHS(Left side) is bigger than RHS(Right side) there no solution
If y=x+a then we have that
x^2=a^2+ax+1
If a=x or a bigger than x there are no solution so x=a+b and we have that
b(a+b)=a^2+1
For a=b=1 we have x=2 and y=3
for a=2 and b=3 we have x=5 and y=8
And so on!!!!!
@@dionisis1917 thanks!
Best books for math olympiad beginners.(i am in class 9th and interested in math olympiad.)
Depends on what you mean by beginner, but check out Mathematical Circles and The USSR Olympiad Book (former is highly recommended)
Edit: Just a FYI, Mathematical Circles has nothing to do with circles
Arthur Engel Problem solving strategies
@@physicsnovice656 umm.. not for beginners now is it
Assuming you're Indian, check out Challenge and Thrill of Pre College Mathematics. YT Course by Prashant Jain is also pretty good in case you're interested in that. For problems you can also see NMTC's Gems book series.. it was written for the NMTC math competitions. Past AMC problems are good for complete beginners otherwise see problems from Problems 1-6(or so) of AIME, Singapore Math Olympiad, Purple Comet Math Competition, HMMT November round, and of course, lastly, past PRMO papers. OK, that's all I could remember.
For RMO you should see Pathfinder for Math Olympiads (Pearson Publication) and Art and Craft of Problem Solving by Paul Zeigt along with previous questions.
If you were overwhelmed, just stick to Mathematical Circles as your first book. It starts from the basic. Then cover Geometry from Challenge and Thrill of Pre College Mathematics. Lastly, attempt previous PRMO papers.
@@l1mbo69 I'm not sure if it's worse suggestion in terms of difficulty compared to The USSR Olympiad Book. It has 14 chapters:
The Invariance Principle
Coloring Proofs
The Extremal Principle
Enumerative Combinatorics
Number Theory
Inequalities
The Induction Principle
Sequences
Polynomials
Functional Equations
Geometry
Games
Further Strategies
Each chapter has introduction with theorems/proofs and example problems and then list of problems with solutions. Someone who mastered normal hs material should be able to follow most of the book if he puts some effort.
Also "102 combinatorial problems from the training of the USA IMO team "
is good book for combinatorics going from hs level to harder problems, with solutions.
Looks like that title has already angered some nationalists
Why prove for both any n and any m that f(n!) = f(n)! and f(m!) = f(m)! Inst that redundant ?
If you hadn't done that for n=p-1 first then how would you substitute f(p-1!)=f(p-1)!=(p-1)! in the last one where m=p-1
I love your channel so far, but Michael, the name of the country is not Macedonia. Macedonia is where I was born and well, I know very weel where I was born and raised. In Greece. At least you could have tried to say North Macedonia. This name is still pitiful for the Greeks, but at least that's the official name now.
Macedonia? It is not a seperate country, why does it have seperate problems from the rest of Greece?
That was quite a vicious and violent comment.
Χαχα έλεγα κ γω κάποιος δε θα το πει
@@jimjim3979 πάντως δεν είπα κάτι που δεν ισχύει
North Macedonian*
@@brankojangelovski3105 bruh moment
@@brankojangelovski3105
1) No you.
2) That's the official name of your country according to the Prespes treaty.
3) You guys steal names and history from others to validate your fantasies. You're slavs speaking a slavic language and other from your STOLEN name you have no relations whatsoever with Macedonia or Macedonian history or Alexander the great.
Jesus Christ is that all you people have to care about
It's actually pronounced South-west Bulgaria.
It's North Macedonia (aka Skopje) , not Macedonia. Macedonia is a state of Greece. The birthplace of Alexander the Great. And that's a good place to stop!
It is North Macedonia
Μacedonia is a cerain part of Greece