It's not from Olympiad, but from the national exams. Olympiad problems are much harder.

In the second problem, if you analyze the function inside the sine you can see that it is a semi-circle centred at (a/2,0) with a maximum at (a/2, a/2). So every pair of roots adds to a. You will get a pair of roots where the function has zeros (0 and a, which adds to a) as well as a pair of roots for every multiple of pi less than a/2. So, the sum of the roots will always be a*n where (n-1)*pi is less than a/2 and n*pi is greater than a/2. Since the sum of the roots is 100 = a*n, a = 100/n. Now, subbing into the inequality, (n^2 - n) < 50/pi < n^2. so n = 4 and hence a = 25.

Another way of solving #1 is to note that in the quadratic equation, that the square root part must be 0. Thus m^2-4n = 0 or n=m^2/4. Therefore substitute x=-3 and n=m^2/4 into the original quadratic, and you get another in m :- 9 - 3m + m^2/4 = 0 m^2 - 12m + 36 = 0 which gives a single root for m of 6 which yields 9 for n or m + n = 15.

First one doesn't seem so bad to me. If there's only one solution to a quadratic, the discriminant must be 0, and the rest must be -3. So -m/2 = -3, m = 6. m^2 - 4n = 0, 36 = 4n and n = 9. m + n = 15.

Second problem, quickly did some algebra and found our roots, x=(a(+/-)sqrt(a^2 - (2n pi)^2))/2 where n is a list of integers between 0 and... some upper bound. This is where the easy part ends. The upper bound has to do with (2n pi)^2 > a^2, since the square root won't allow n to go any higher, otherwise the root is no longer real valued. Through some number crunching, it becomes clear that the bound for n is floor(a/2pi)+1, I believe the +1 comes from n starting at 0 instead of 1. Then we just sum up all the roots. Two roots happen when n=0, another two when n=1, all the way up to our highest value, floor(a/2pi)+1. After preforming the summation and simplifying, it becomes clear our problem reduces down to a(floor(a/2pi)+1)=100. This makes it clear that a is an integer, a factor of 100. From here I checked each factor, eliminated all the factors of 100 except 25, as the floor(25/2pi)+1=4 and 4*25=100, so a=25 :D Never needed any help nor hints, although I did use a graphing calculator to understood exactly what the hell this equation was doing as you increased/decreased a (which looks very cool btw would recommend) , this was a really fun challenge. The first problem, in comparison to the second, was no difficulty and took me maybe a minute to solve, m=6, n=9, m+n=15.

Second one was a very cool problem, exactly my type, thank you for sharing ! Took me a good 10 minutes for the second one, first one was trivial though (don't know where you got it from but it looks like it could come out of a 14 year old's textbooks though lol)

That was a good one, it took me almost 15 minutes!

For problem 2, are we counting double roots twice? The question just says sum if the roots, doesn’t say to add up repeated roots twice and we do t know that all the roots we’re adding are distinct.

Well explained 🙌🏻

I really enjoyed the solution to the second problem. I could not get halfway there on my own. As well, needed to watch a few parts a couple times, and pause for a bit here and there to give myself opportunity to think it through, and to follow the math and the logic. Everything became crystal clear except for one thing at the very end. It felt like in the beginning, I understood it to mean we needed to find the maximum number for n. As such, when we find that maximum to be n = 3, I am lost as to why we need to check n = 0, n = 1, and n = 2. It feels as if we can accept that 3 is the value of n we need, and then from there solve for a. What am I not getting? Thanks.

You sound different. Have you changed mics?

I have another solution for question one. Usually when a quadratic only has one solution that means the slope is zero at that solution and the graph is touching the x axis. You have differentiate then equation and put value of x and you will get the value of m. You can then just put the values of m and x in the initial equation and you will get m and n separately as well.

No way the 1st question is a problem from Olympiad. I participate in many olympiads in my country (Kazakhstan) and I've seen so many crazy questions, 1st one was so easy but at the same time 2nd one was so hard

i was able to solve the first part : )

For the 2nd question From n=floor(a/2pi), we should get 2pi*n

If I only understood what "sum of all real valued roots" means I *might* be able to think of a solution, but I got nothing. Does that mean looking for the integral, or a limit? First q. is clearer to me and trivial to solve.

Problem 1: Not so difficult for me, I just differentiated the first equation and put x = -3 to get value of m, then I found value of n by putting m & x in equation 1. m + n = 15 Problem 2: I don't know if its correct, but my first step was to put everything that's inside sin to be equal to n(pi) n is some integer. Then squaring both sides, we get the quadratic equation, whose sum of roots = a and we are told that sum of roots of the quadratic = 100, so, a = 100??

Hi, could you solve this problem? I have a way of solution, but is very tedious, so maybe I forget something: Is 8th problem in Dudeney’s “536 curious problems and puzzles”: “Seven men engaged in play. Whenever a player won a game he doubled the money of each of the other players. That is, he gave each player just as much money as each had in his pocket. They played seven games and, strange to say, each won a game in turn in the order of their names, which began with the letters A, B, C, D, E, F, and G. When they had finished it was found that each man had exactly $1.28 in his pocket. How much had each man in his pocket before play?”

Thanks

While problem 2 is a nice problem, it has several flaws: 1. The problem says "real root", which could mean that the equation itself could be in complex domain. If that's the case the solution is not correct. 2. The x_1 and x_2 for different n *could* overlap. We need to prove that it couldn't.

The first is easy, year 10 level in France, may be 11 given the student needs to master quadratics, and it's in limited time. -3 is double root, so x2+mx+n factors as (x+3)^2, so m = 6 and n = 9, m+n=15.

first question was so easy it had me doubting whether I understood it correctly

Question 2: Trig equation = year 9 or 10. Easy. Quadratics = y10 or 11. Easy. But some details are hard to iron out! ----- 1) if a=0, ax-x^2=x(a-x) >= 0 only in the [0,a] interval, which is easily established by studying the sign of x(a-x). 3) sin()=0 if an only if its argument is of the k.Pi form, where k is an integer, so sqrt(x(x-a)) = k.Pi Which also means k >= 0. 4) so x^2-ax+k^2.Pi^2 = 0 5) delta = a^2-4k^2.Pi^2 >= 0. As k>=0, this is equivalent to a >= 2k.Pi. This is always possible if k = 0, because then a>=0, which we already know. 6) So the roots exist and their sum = a because a quadratic is always of the form x^2-Sx+P=0, where S is the root sum and P their product. 7) so a=100

Problem 1: I can take random values of m=2 and n=-3 to solve it, too.

I went quadratic equation for the first part. -3=-m/2 and (-m)^2-4n=0

I removed this comment as it sported a false asumption. a=25 is indeed the only answer.

Shout-out to all Georgians 🇬🇪

What is the reason you can't put -3 into the original equation and find solutions for 9-3m+n=0?

Interesting problems! In the second problem, we don't need to test n values one by one. From n π < a/2 < (n+1) π and a = 100/(n+1), we can get n(n+1) < 50/π < (n+1)². As 50/π = 15.xx..., which is slightly smaller than 4², we can get n = 3.

While explaining you sometimes jump the important thing of the solution ie the decision factor rather than calculation, which remains Inexplicit and lacks crystal clarity like in this 25/3pi still remains unclear

Tell me if I'm wrong, but I think the first one's solution comes from m>=3 and n = (m-3) * 3. Test 1: lets say m=3, n=0. Then: (-3)^2 + 3*(-3) + 0 = 0 => 9 + (-9) + 0 = 0 => 0 + 0 = 0 Test 2: lets say m=4, n=3. Then: (-3)^2 + 4*(-3) + 3 = 0 => 9 + (-12) + 3 = 0 => (-3) + 3 = 0 Test 3: lets say m=5, n=6. Then: (-3)^2 + 5*(-3) + 6 = 0 => 9 + (-15) + 6 = 0 => (-6) + 6 = 0 So m+n can be 3 (3+0) or 7 (4+3) or 11 (5+6) and so on.

Part 1: Solve for x the equation to get (x+m/2)^2=m^2/4-n => x=-m/2+/-sqrt(m^2/4-n). If you substitute x with -3 now you get two equations: 1) -6=m+sqrt(m^2-4n) 2) -6=m-sqrt(m^2-4n) This means that m=-6 and therefore n=9.

Well problem #1 has multiple valid solutions for m and n, for example 6 and 9 but also 4 and 3 ...

For the first problem If x is only equal to -3 ie both roots are equal to -3 Thus -3-3=-m Thus m=6 (-3)×(-3)=n thus n=9 m+n=9+6=15 It didn't even require paper and pen

Either SAT is too easy or I'm missing something.

4:28 - teehee. That audio edit, a loud "pi squared".

Personally I would guess multiples of 100 until one works for the second one. I got problem 1 properly though :)

"the only possible" what ? 😂

If this the hardest probkem, most likely I missed something obvious. If (-m +/- sqrt(m^2-4n))/2 = -3, then m = 6 and m^2-4n = 0, so n = 9. Indeed for x^2 + 6x + 9= 0 the only solution is x = -3, so m+n = 15. What have I missed?

25/2pi =3?? I dont understand

As a former math major... Huh?

I am missing something, 25/(2 x pi)=3.98.....🤔🤔

i was able to solve first part i must be genesis

First 15

Why can't m=4 and n=3?

Why n = |_a/2pi_| and not n

Solved the first problem in my head in less than 10 seconds. The qudratic equation whose only root in -3 is: (x+3)(x+3)=0 Expandi it x^2+6x+9=0 Therefore m=6 n=9 m+n=15

Hardest NA math problem

Sorry, i have objections. To let "x == -3" be the root to any quadratic expression, ANY curve fits that request that goes through x == -3 with it's bottom tip.There are infinitely many curves that fit that request. What you stipulate with your approach is that the Y-VALUE (!) of that curve shall be that x squared. The y value IS ALWAYS ZERO as is written in the problem description in the first place. Correction: We have no factor at the x^2, that's why we have only one principal curve that gets shifted by m and n. Had to do it myself to become a believer. For the second problem: I couldn't figure out how a sum of pi's should in any kind equal any integer. That those guys wanted the sum of the "x" at the roots for the sin (the root of the sin and the x are two completely different things, you know?!) was far off my abilities for "creative interpretation". The sin curve has roots. Those are the Pi's. The argument for the sin curve with its roots is provided by another function. That function IS a root. That root gets its own argument provided by a quadratic expression. THAT ONE has its own roots. Those are the x that make the value of that last quadratic function zero. Bot those roots have nothing, but completely nothing at all to do with what was requested by the author of that question. I think the problem here is that mathematicians use terms like "root" in a way that is evading normal mortal ones. Or it has to do with language barriers. In german, when we speak of "Roots" alias "Wurzel", we really mean "Wurzel". Else we speak of "Nullstellen" - which is clearly and unambiguously differentiated from "Wurzel" of any other kind in german. Maybe german mathematicians have less problems with those terms. I was completely confused. After clearification of that part of the problem, i was able to solve it by myself, but i needed a lot more intermediate steps, which got omitted by the explanations in the video. For example: To get the clue that the root expression for the x is completely out of any meaning, you must get the visualization that you have a quadratic curve (despite being rooted) that always generates symmetric curvature to both sides of a symmetry point (which shows as that ominous "a") for the whole span of what is requested for the value range by the problem. Both for the curve as well as for the value range, you need a "clue-full" understanding. In hindsight, the problem as such was extremely interesting. But maybe a little more explanation before you get to the solution in such cases would benefit the egos of non studied math lovers :D

Lol.... hardest questions! 🤣😂

Who are indians here

What can’t m=0 and n=-9? And m+n = -9

Why you not taking -π,-2π.....-nπ Please give the reason