Find (x+y+z) [Harvard-MIT] Guts contest

Рет қаралды 14,297

Prime Newtons

Күн бұрын

This problem is from the HMMT mathematics contest. It took me several days to figure this one out.

Пікірлер: 84

@Modo942000 3 ай бұрын

This is honestly such a beautiful solution that shows how areas of maths are all connected at the end of the day.

@childrenofkoris 2 ай бұрын

i was under the impression that it would be solved bu traditional algebra or equation

@quzpolkas 3 ай бұрын

I found another method for this, however I like yours more because it's more elegant :) First off, we can easily check that x, y, z must be all distinct numbers to satisfy the system (for example, from x=y follows x^2 = y^2 = 1/3, which leads to contradictions in equations 2 and 3 of out system, similar for y=z and z=x). Because of that, we can multiply both sides of equations 1, 2, 3 by (x-y), (y-z), (z-x) respectively (none of the brackets can be zero), use the cube difference formula, and add the 3 resulting equations together. The left side sums to zero, the right side sums to -4x + 3y + z, which gives us a new relation for x,y,z: z = 4x - 3y (mark it as equation 4). Then I noticed that original equations 1 and 2 can be summed, and the resulting right side sums to 5, which is the right side of equation 3. Why not equate them? After doing that and replacing z with 4x - 3y from equation 4, we get a second-order equation for x and y: 4x^2 - 8xy + y^2 = 0 (mark it as 5). None of the x,y,z can be zero (easy to check), so we can divide both sides by y^2, substitute x/y = t, use the quadratic formula and find two values for t: t = x/y = 1 + rad(3)/2, OR t = x/y = 1 - rad(3)/2 (rad(x) meaning square root of x). After that follows some gruelling algebra with radicals which I'm not gonna write here cuz this comment is already too long :P All that's left to do is substitute x = (1 + rad(3)/2)y into 4, get a formula for z in terms of y, substitute them both into, for example, original equation 1, and get the values of y, then find our needed sum x+ y + z = rad(5+2*rad(3)). However, I couldn't find a reason to discard the other root of equation 5: x/y = 1 - rad(3)/2 , which yields a different value for the sum: x+y+z = rad(5-2*rad(3)). They are both positive. Could this be a second possible value for answer, or did I miss something? Cheers!

@StaR-uw3dc 3 ай бұрын

Nice approach. We can get relation z=4x-3y also by subtracting equations in pairs: (2)-(1): (z-x)(x+y+z)=3 lets call it as (2_1) (3)-(2): (x-y)(x+y+z)=1 lets call it as (3_2) Comparing (x+y+z) from both above equations we get 3/(z-x)=1/(x-y) hence z=4x-3y From equation (3_2) we can notice that (x-y) should be positive (as x+y+z and 1 is positive) hence x>y and x/y>1. For this reason we should reject one solution for t=x/y=1-√3/2 as it is less than 1.

@quzpolkas 3 ай бұрын

@@StaR-uw3dc Yes, brilliant! I completely forgot that z>x>y, which I believe was mentioned in the video. Your argument brings some more clarity as to why that is though, in my opinion. I also didn't catch that for t = 1 - rad(3)/2 we get z = (1-2rad(3))y, which means y and z can't both be positive (since 1 - 2rad(3) in negative). Therefore the second value for t must be discarded. Thank you for the correction! 👍

@Nothingx303 3 ай бұрын

I was thinking 🤔 that good 👍 but the solution in the video is more juicy 😉 😊

@Attitude_boy_300 10 күн бұрын

Really appreciable 😊😊

@joseluishablutzelaceijas928 2 ай бұрын

That is a very beautiful solution, the decisive step was recognizing the cosine rule.

@sicaclerloxuty 3 ай бұрын

You're my favourite youtube math teacher

@anthonysnell4832 3 ай бұрын

This is an excellent example of how maths is so beautiful.

@RomanOrekhov 3 ай бұрын

This geometric solution is much more beautiful than any algebraic for this problem I could come up with! Algebraic have intricacies about the second possible solution (sqrt(5-2*sqrt(3)) while in geometric this value just doesn't come up as line segments have positive length

@Jeremy-i1d 2 ай бұрын

Thank you for another wonderful video. Your solution is beautiful and it's hard to imagine another which could be as elegant. I love your focus on structure and ability to stand back to recognise familiar forms - like you did also in the video where you used the formula for tan (3 theta). I eventually managed to solve the problem by a brute force method but it was cumbersome. My approach was as follows: 1) Solve equation 1 as a quadratic in y to give y in terms of x, and similarly equation 3 as a quadratic in z to give z in terms of x 2) Add these to x to obtain R = x+y+z = 1/2*(R1+R2) where R1 = root(4-3x^2) and R2 = root (20-3x^2) 3) Calculate y^2, z^2 and yz from the expressions in 1) and sub them into equation 2 to obtain the equation R1R2 = -8 + 3x^2 + 6xR 4) Square the equation in 2) to obtain the second equation R1R2 = -12 + 3x^2 + 2R^2 5) Solve these two equations simultaneously to get x in terms of R as: x =(R^2-2)/3R 6) Square the equation in 3), subbing for R1^2 and R2^2 in terms of x, to obtain the cubic equation in x: 9Rx^3 + (6+9R^2)x^2-24Rx -4 = 0 7) Sub for x from 5) into this equation to obtain an equation in R only as: R^6-40R^4+52R^2=0 8) Factorise out R^2 (which cannot be 0) and complete the square on the other factor to obtain (R^2-5)^2 - 12 = 0, from which it follows that R = root (5+2root3), as you found. (positive square root must be taken at both R^2 and R stages because R^2 > 5 from the start of your solution and R>0.) Again, thank you for your lovely and blessed videos ❤

@hacerkayal1740 3 ай бұрын

Sir, I love how you explain mathematics to us in such beautiful ways, I'm glad to have you. ❤

@BartBuzz 3 ай бұрын

What an elegant solution. Math is so satisfying when we know how to apply the principles!

@hacerkayal1740 3 ай бұрын

Sir, The question is awesome but your solution is more perfect than the question❤ thank you for your teaching❤

@muhammadrizkyfebrian3566 Ай бұрын

watching you all night make me fallin in love with math again❤

@shmuelzehavi4940 Ай бұрын

Nice solution and explication ! I'll present a different (purely algebraic) approach: The original system of equations is: x^2+y^2+xy=1 (1) y^2+z^2+yz=4 (2) z^2+x^2+zx=5 (3) The goal is to evaluate: x+y+z , for x,y,z > 0 By subtracting eq. (2) from (3) we obtain: x^2-y^2+z(x-y)=1 Or: (x+y+z)(x-y)=1 (4) Similarly, by subtracting eq. (3) from (1) and (1) from (2) we obtain: (x+y+z)(y-z)=-4 (5) (x+y+z)(z-x)=3 (6) We denote: u=x+y+z and thus obtain: x-y=1/u ,y-z=-4/u ,z-x=3/u (7) By squaring these equations and adding the results we obtain: 2(x^2+y^2+z^2 )-2(xy+yz+zx)=26/u^2 (8) By adding eq. (1), (2), (3) we get: 2(x^2+y^2+z^2 )+(xy+yz+zx)=10 (9) By multiplying eq. (9) by 2, subtracting eq. (8) and dividing by 2 we get: (x+y+z)^2=10-13/u^2 , or: u^2=10-13/u^2 , which becomes: u^4-10u^2+13=0 (10) This is a quadratic equation in terms of u^2. The solution gives: u^2=5+2k√3 , (k=±1) (11) For x,y,z > 0 we must take the positive square root, which gives: u=√(5+2k√3) (12) By substituting expressions (11) and (12) in (7) , it's easy to show that the condition: x,y,z > 0 is satisfied only by k=1. Therefore, the single answer to this problem is: x+y+z=√(5+2√3) Remark: Without restricting to solutions which satisfy the condition: x,y,z > 0 , we have 4 distinct solutions, given by: : x+y+z=l√(5+2k√3) ,(k=±1,l=±1).

@annacerbara4257 3 ай бұрын

This is just a possible starting point: multiplying the first equation by (x-y), the second by (y-z), the third by (z-x) and adding the three equations found we obtain a linear equation in x,y,z which gives z=4x-3y and which allows you to eliminate z.

@Dr_piFrog Ай бұрын

You have an exciting and pleasing teaching method, not like other videos with a bit of Condescension.

@kinshuksinghania4289 3 ай бұрын

Finding connections between two seemingly unrelated things!! That’s the beauty of mathematics

@Ashleyleonard777 Ай бұрын

Awesomely done. A great show of a combination of algebra, geometry and trigonometry thinking solution. ❤

@ΑπόστολοςΠαπαοικονόμου-ν2θ 6 күн бұрын

Very nice combination of algebra, geometry and trigonometry. Well done!

@ion5964 3 ай бұрын

You can never unsee the geometry in this problem, but I would never have guessed it

@ScreamAnonyme 2 ай бұрын

The solution is poetic ! I think that the person who composed this problem had derived it from a right triangle ABC (AB=1; AC= 2 and BC = rad 5) and a point O inside the triangle is positioned in such a way so that OA, OB and OC are equal to x, y and z and angles AOB=BOC=AOC= 120 degrees

@BurningShipFractal 3 ай бұрын

I didn’t expect there would be geometry to solve the problem

@joaocarvalho5832 2 ай бұрын

Is just a beauty, the way how the resolution was done!

@epsilonxyzt 7 күн бұрын

Never Stop Teaching!

@rakeshsrivastava1122 3 ай бұрын

Geometry solving algebra! Great synergy in math.

@childrenofkoris 2 ай бұрын

YOU ARE MY FAVOURITE TEACHER

@clemensetz Ай бұрын

What a fantastic channel!

@panagiotisvlachos6114 2 ай бұрын

Three boards filled with the solution: That's why, I love Mathematics!! Greetings from Greece!!

@andrasferencz7948 25 күн бұрын

Brilliant! Just brilliant!

@avinashpokhriyal3657 3 ай бұрын

One elegant solution is also there. I have solved it without using any cosine function. Simply Algebra..!

@reniervanrooyen3417 3 ай бұрын

hes name is "the chmistry teacher" he used the gauss method to solve for this type of question, he first lugged the initial equations into matrix form, where he then manipulated it to the point some variables = 0 then was able to answer the question, cause this is almost like big simultaneous equation

@rui_santos 3 ай бұрын

Beautiful problem. Greetings from Portugal.

@kinshuksinghania4289 3 ай бұрын

The beauty of mathematics!!

@Ranoake 23 күн бұрын

It would help to know the explicit domain of x, y, and z, not just positive, positive reals? positive integers? You get different solutions depending on that answer. I assumed integers and got x + y + z = 1, but back substitution shows no solution.

@dougaugustine4075 3 ай бұрын

Wow (again)!!! The trig formulas are so easy to treat lightly, but dang, it's really amazing how you are able to see their application in a problem. This isn't the first time I've seen you do this.

@xyz9250 3 ай бұрын

Don’t have access to pen and paper atm, but I think may able to make use of the formula x^3 - y^3 = (x-y)(x^2 + xy + y^2) = (x-y) * 1, do this for the other two combinations, and add the three resulting equations, LHS will be 0, and RHS be first degree. Still additional work to get the final solution

@arampak 2 ай бұрын

I think there is a typo in the thumbnail: y instead of x in the last line.

@tessfra7695 3 ай бұрын

I'm sorry if this is a silly question, but at ~ 12.50 when sir said the longest side of triangle PQR is 2, I got confused...isn't the longest side sqrt 5 ?

@javadkhusro 3 ай бұрын

He was pointing at sqrt 5 when he said longest, so that is the hypotenuse, so 1 and 2 are the base and height to get the area.

@Grecks75 3 ай бұрын

I solved the system algebraically and finally arrived at explicit formulas for all of x, y, z, and their sum s:=x+y+z. I found four different solutions of the system and also 4 different answers s to the problem corresponding to those 4 solutions. That means, you missed three of the four solutions! Here is a complete list: Let w denote the positive, real root of 3*w^2 - 4 = 0. 1) x1 = (1+w)/√(5+3w), y1 = w/√(5+3w), z1 = (4+w)/√(5+3w), s1 = +√(5+3w) 2) x2 = -(1+w)/√(5+3w), y2 = -w/√(5+3w), z2 = -(4+w)/√(5+3w), s2 = -√(5+3w) 3) x3 = (1-w)/√(5-3w), y3 = -w/√(5-3w), z3 = (4-w)/√(5-3w), s3 = +√(5-3w) 4) x4 = -(1-w)/√(5-3w), y4 = w/√(5-3w), z4 = -(4-w)/√(5-3w), s4 = -√(5-3w) This is just the complete solution of the problem. You can check them by "simply" plugging in those values in the original equation system. Getting there was rather elaborate and ugly, you don't wanna know. 😂

@Grecks75 3 ай бұрын

Oh, I just saw we were only supposed to find solutions for *positive* x, y, z; I solved it without that restriction, nevermind. When we restrict x, y, z to positive values, then I think only one solution (no. 1 above) remains.

@Grecks75 3 ай бұрын

Here is a sketch of what I did to get those solutions: A) Since we're mainly interested in s=x+y+z, let's first introduce s into equations (2) and (3) and exchange it for z (i.e. eliminate z). We then get: (4) s^2 - 2xs - ys = 3 (5) s^2 - xs - 2ys = 4 (B) Subtract (4) from (5) to get: (6) xs - ys = 1 (C) Combine (4) and (5) again to eliminate either xs or ys and express s^2 in terms of the other: (7) s^2 = 2 + 3xs = 5 + 3ys (D) Multiply (1) with s^2 and eliminate either xs or ys using (6) and get: (8) s^2 = 3(xs)^2 - 3(xs) + 1 = 3(ys)^2 + 3(ys) + 1 (E) Combine (7) and (8) to eliminate s^2 and determine w^2 := (ys)^2: (9) 3(ys)^2 = 4 (F) Now we're nearly done. Determine s^2 from the two possible ys=+/-w using (7). Take the square root out of s^2 to arrive at the four possible values of s. (G) If you want you can now also calculate x and y from xs, ys, and s. Subtract x+y from s to get z. All in all this is much more elegant than I said at first. 😄

@pinoes31416 8 күн бұрын

(x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+yz+zx) let C=(x+y+z) A=(x^2+y^2+z^2) B=(xy+yz+zx) Adding the 3 equations we get 2A+B=10 so A=5-B/2 C = sqrt(A+2B)=sqrt(5+(3/2)B) ... this is what you did. sorry. I will give it a try

@Abby-hi4sf 2 ай бұрын

Wonderful!

@williamspostoronnim9845 3 ай бұрын

Очень красивое и непростое решение. А у меня симплекс-методом по Нелдеру-Миду получилось Х=0.8270349, Y=0.2843326, Z=1.842614, а сумма = 2.95398. Немного отличается от sqr((5+2*sqr(3))=2.90931. Однако в чем дело, не пойму.

@makehimobsessedwithyou6412 3 ай бұрын

you are so clever!

@sergiosereno1489 3 ай бұрын

Amazing!

@justliberty4072 2 ай бұрын

Third equation on the initial screen is wrong (y^2 instead of x^2).

@khetamrout4849 3 ай бұрын

Yay another math problem

@rajeshsaha6734 12 күн бұрын

I have discovered another simple method.

@PrimeNewtons 12 күн бұрын

Please share

@Attitude_boy_300 10 күн бұрын

Plz share algebraic soln 😢 to this

@kapishkumar4252 3 ай бұрын

similar question came in ioqm 2021

@mrl9418 3 ай бұрын

Bravo

@chathuragalkandage3437 3 ай бұрын

Can you please solve for x,y and z separately using the triangle, I tried with theta between side with size=1 and x. I used the x+y+z finding with sine law on each mini triangle to formulate an equation for theta as a*sin(theta)+b*cos(theta)=c. This involves tedious numerical calculations, nice if there is an elegant way to find x,y and z.

@chathuragalkandage3437 3 ай бұрын

x^2 = (7+4q(3))/(15+6q(3)), y^2= 4/(15+6q(3), z^2=4(13+4q(3))/(15+6q(3)), xy=(4+2q(3))/(15+6q(3)), yz = (4+8q(3))/(15+6q(3)), zx = (16+10q(3))/(15+6q(3))

@Attitude_boy_300 25 күн бұрын

Can anyone suggest me a cool algebric soln to this question.

@Ron_DeForest 3 ай бұрын

How does someone get your email? It’s not in the description.

@ThePayner11 3 ай бұрын

It's stated at the start of the vid :)

@shikharjaiswal8525 3 ай бұрын

Hello respected sir It can also be solved using simple algebra And thank you for your beautiful solution 😮😊

@lukion27 3 ай бұрын

Please share your algebraic solution. I would like to see it.

@MyNameMeansGiftFromGod 3 ай бұрын

Mathematics is the language of God.

@sunil.shegaonkar1 3 ай бұрын

At 10:45, you shift from 3D to 2D and that is little hard to digest and apply the same geometric relations.

@PrimeNewtons 3 ай бұрын

Everything was 2D. Flat on the board.

@lukion27 3 ай бұрын

How is the average math student to think to use a trig formula to solve an algebra problem? When I looked at this, it never would have occurred to me to use the LOC formula. Where does such inspiration come from? Is there something in the problem to look for that points to the LOC? I would have totally missed it. Learning how to recognize this would greatly improve how people approach various systems of equations that look absolutely messy to solve algebraically.