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

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

[Modo]

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

[quzpolkas]

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!

[loxuty]

You're my favourite youtube math teacher

[joseluishablutzelaceijas]

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

[anthonysnell]

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

[Jeremy-id]

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 ❤

[hacerkayal]

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

[ΑπόστολοςΠαπαοικονόμου-νθ]

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

[BartBuzz]

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

[RomanOrekhov]

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

[shmuelzehavi]

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).

[hacerkayal]

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

[Ashleyleonard]

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

[muhammadrizkyfebrian]

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

[ScreamAnonyme]

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

[kinshuksinghania]

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

[Dr_piFrog]

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

[annacerbara]

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.

[panagiotisvlachos]

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

[rakeshsrivastava]

Geometry solving algebra! Great synergy in math.