How did Po-Shen Loh get 1/16 as the constant?

  • M2★ M3★ M4★ M5★

    Module 2 Week 4 Day 16 Bonus Explanation

    At 2:20, Po-Shen Loh explains that you need to multiply a constant (the yellow box at 2:20) with
    ( a + b + c )( -a + b + c)( a - b + c )( a + b - c ). How exactly did he get 1/16 from the 13 14 15 triangle?

  • MOD

    Hi @Jerry-Huang!
    Thanks for asking!
    We found out that for any triangle, the following is true:
    $$\text{Area}_\triangle^2={\color{orange}\boxed{\text{??}}}\times(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$ In addition to this, we found, using an absolutely different method, that \(\text{Area} _ {\triangle13-14-15}=84.\)
    So, on the one hand, we have$$\text{(Area} _ {\triangle13-14-15})^2=84^2=7056$$ And, on the other hand, we know that $$\text{(Area} _ {\triangle13-14-15})^2={\color{orange}\boxed{\text{??}}}\times(13+14+15)(-13+14+15)(13-14+15)(13+14-15)$$ $$\text{(Area} _ {\triangle13-14-15})^2={\color{orange}\boxed{\text{??}}}\times 42\times 16\times 14\times 12$$ $$84^2={\color{orange}\boxed{\text{??}}}\times 42\times 16\times 14\times 12$$
    So now we can find our \({\color{orange}"\text{yellow box}"}\): $${\color{orange}\boxed{\text{??}}}=\frac{84\times 84}{42\times 16\times 14\times 12}=\frac{42\times 2\times 84}{42\times 16\times 168}=\frac{42\times 168}{42\times 16\times 168}={\color{orange}\boxed{\frac{1}{16}}}.$$

  • M2★ M3★ M4★ M5★

    Ah. Thanks! I didn't think 1/16 would come from solving the equation, but apparently it does!

  • M2

    @Jerry-Huang I think so if you do all of the soling equations you get 42168/4216*168 which gets you 1/16 which is just only a matter of solving equations!!