Oxford Mathematics Interview Question - Solved in 2 Ways!

Поділитися
Вставка
  • Опубліковано 24 гру 2024

КОМЕНТАРІ • 18

  • @Theproofistrivial
    @Theproofistrivial 28 днів тому +4

    Multiplying by abc on 3=1/a + 1/b + 1/c implies ab+ac+bc=3.
    expanding (a-1)(b-1)(c-1) and subbing in the different values we know shows (a-1)(b-1)(c-1)=0 so at least one of a,b,c=0. wlog we say a=1. b=1/c so sub into b+c=2 and find (c-1)^2=0 so c=1 and hence b=1. thus (a,b,c)=(1,1,1) is the only sol.

    • @JPiMaths
      @JPiMaths  28 днів тому

      Nice solution! Thanks for sharing!

  • @trueblaze8930
    @trueblaze8930 16 днів тому +1

    my pure algebra approach- I hadn't heard of any of the formulae or techniques mentioned in the video so here is how I solved it.
    ab+bc+ac=3
    ac+bc+ab=3 (1.) (rearranged)
    a+b+c=3
    multiply by c
    ac+bc+c^2=3c (2.)
    c^2-ab=3c-3 (2. -1.)
    abc=1, ab=1/c
    c^2-1/c=3c-3 (multiply by c)
    c^3-1=3c^2-3c
    c^3-3c^2+3c-1=0
    (c-1)^3=0 so c=1 only
    then c=1 so ab=1, using a+b+c=3
    a+b=2 (3.)
    a^2+2ab+b^2=4
    a^2+b^2=2
    1/a+1/b=2 (multiply by ab)
    b+a=2ab
    so by (3.), a+b=b+a, so 2ab=2,
    b=1/a
    sub 1/a for b in a^2+b^2=2
    a^2-2+1/a^2=0
    a^4-2a^2+1=0
    solve for (a^2-1)^2=0, a=+-1 but +ve so =1
    so b=1 also therefore a,b,c=1,1,1

  • @nonroutineganit7682
    @nonroutineganit7682 21 день тому

    Given a , b , c > 0
    a + b + c = 3
    1/a + 1/b + 1/c = 3
    ab + bc + ca = 3
    abc = 1
    let a,b,c be the roots of the polynomial
    f(x) = x³ - 3x² + 3x - 1 = 0
    f'(x) = 3x² - 6x + 4 > 0 for all x € R
    Then f(x) is increasing function for all x > 0
    => a = b = c
    Therefore , a = b = c = 1

    • @JPiMaths
      @JPiMaths  20 днів тому

      @@nonroutineganit7682 ah I like this approach, I hadn't thought of it this way!

  • @Mo-zj6hz
    @Mo-zj6hz 20 днів тому

    What is the blackboard software and pen you use to present online ?

    • @JPiMaths
      @JPiMaths  20 днів тому

      @@Mo-zj6hz I use bitpaper at the moment, but I'm sure there may be better ones! And I just use a basic writing tablet

    • @Mo-zj6hz
      @Mo-zj6hz 18 днів тому

      @JPiMaths thanks and to record your screen ?

  • @MrConverse
    @MrConverse 28 днів тому

    Do you need f(x) or can we start that part by just considering (x-a)(x-b)(x-c)? Also, the part at 4:56 seemed a little hand-wavy to me. Maybe you could show a proof of that in another video.

    • @dontspam7186
      @dontspam7186 28 днів тому

      whats hand-wavy about it, if we define a polynomial with roots a, b and c, and we work out that afterwards the polynomials roots are purely 1, we know a,b and c are all 1?

    • @MrConverse
      @MrConverse 28 днів тому

      @ because polynomials factor uniquely just like natural numbers do?

    • @JPiMaths
      @JPiMaths  28 днів тому +3

      I was being a bit hand-wavy to be fair. To clarify, we show that f(x) is a cubic with roots 1, 1 and 1, and since there is 'no more room' for any other roots (as f is a cubic), it must follow that the roots coincide with those as defined in f, namely a, b and c.

  • @aliwelchoo
    @aliwelchoo 27 днів тому

    Can't you just use 1/n < 1 for all n>1 and 1/n = 1 only for n=1 therefore all the numbers must be 1 else 1/a + 1/b + 1/c < 3

    • @artemis477
      @artemis477 26 днів тому

      No because what about 1/n when n

    • @aliwelchoo
      @aliwelchoo 26 днів тому

      @artemis477 for positive integers?

    • @JPiMaths
      @JPiMaths  26 днів тому

      Nice try, but n needn't be an integer.

    • @aliwelchoo
      @aliwelchoo 26 днів тому

      Oh yeah lol helps if I read the question

    • @shiveshpratapsingh3501
      @shiveshpratapsingh3501 25 днів тому

      ​@@JPiMathsdoesn't matter if n is an integer what matters is n>1 in order for 1/n