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# Triangle with 60 degree show

2018-05-14 06:46:55

I do not understand how to do the following problem. I have attempted to use cos rule, sin rule and even constructions to make use of the 60 degree information but i still cant reach the answer.

By the Law of Cosines

$$c^2=a^2+b^2-ab$$

hence

\begin{align*}

&\frac{3}{a+b+c}-\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\[4pt]

=\;&\frac{3(a+c)(b+c)-(a+b+c)(b+c)-(a+b+c)(a+c)}{(a+b+c)(a+c)(b+c)}\\[4pt]

=\;&\frac{(3ab+3ac+3bc+3c^2)-(ab+ac+2bc+b^2+c^2)-(ab+bc+2ac+a^2+c^2)}{(a+b+c)(a+c)(b+c)}

\\[4pt]

=\;&\frac{c^2-(a^2+b^2-ab)}{(a+b+c)(a+c)(b+c)}\\[4pt]

=\;&\;0

\\[4pt]

\end{align*}

As to why the simplification above was destined to work, the idea is that the Law of Cosines, applied to the angle $C$, yields an if-and-only-if condition for $C=60^{\circ}$, hence, if $C=60^{\circ}$ implies some other identity relating $a,b,c$, we would expect it to be an algebraic consequence of the Law of Cosines.

• By the Law of Cosines

$$c^2=a^2+b^2-ab$$

hence

\begin{align*}

&\frac{3}{a+b+c}-\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\[4pt]

=\;&\frac{3(a+c)(b+c)-(a+b+c)(b+c)-(a+b+c)(a+c)}{(a+b+c)(a+c)(b+c)}\\[4pt]

=\;&\frac{(3ab+3ac+3bc+3c^2)-(ab+ac+2bc+b^2+c^2)-(ab+bc+2ac+a^2+c^2)}{(a+b+c)(a+c)(b+c)}

\\[4pt]

=\;&\frac{c^2-(a^2+b^2-ab)}{(a+b+c)(a+c)(b+c)}\\[4pt]

=\;&\;0

\\[4pt]

\end{align*}

As to why the simplification above was destined to work, the idea is that the Law of Cosines, applied to the angle $C$, yields an if-and-only-if condition for $C=60^{\circ}$, hence, if $C=60^{\circ}$ implies some other identity relating $a,b,c$, we would expect it to be an algebraic consequence of the Law of Cosines.

2018-05-14 08:49:32