Is it or not?

Geometry Level 3

Consider a triangle $ABC$ with side lengths $a,b$ and $c$ . Let the area of a triangle of this triangle be denoted as $\Delta$ . If $(a+b+c)abc = 16\Delta^2$ . Must the triangle $ABC$ be an equilateral triangle?

Yes There is insufficient information No

2 solutions

Sam Bealing
May 21, 2016

This is basically just a repeat of Rishabh Tiwari solution but with some LaTeX:

$\Delta=\dfrac{a+b+c}{2} r=\dfrac{abc}{4R} \Rightarrow a+b+c=\dfrac{2 \Delta}{r} \: abc=4 R \Delta$

$abc(a+b+c)=\dfrac{2 \Delta}{r} \times 4 R \Delta= \dfrac{8 \Delta^2 R}{r}$

$abc(a+b+c)=16 \Delta^2 \Rightarrow \dfrac{8 \Delta^2 R}{r} =16 \Delta^2 \Rightarrow 8R=16r \Rightarrow R=2r$

But by Euler's Inequality we have $R \geq 2r$ with equality if and only if the triangle is equilateral so the answer is:

$\boxed{Yes}$

Moderator note:

Rather than quoting Euler's inequality, it is better to explain why for a triangle, we have

$abc \geq (a+b-c)(a-b+c)(-a+b+c)$

with equality if and only if $a = b = c$ .

good approach to the solution...+1

Ayush G Rai - 5 years ago

Now it looks superb. Thanks. +1 !

Rishabh Tiwari - 5 years ago

Rishabh Tiwari
May 21, 2016

Using following formulae for circumradius & inradius :-

R = abc/4▲ ;

r = ▲/s

Now;

(4R▲)•(2▲)/r = abc •(a+b+c)

(8R)▲^2/r = 16▲^2

→ R = 2r

Now if it is an eq. ▲

Then R = a/sq.root3 &

r = a/2sq.root3

Hence R = 2r ,

& therefore

ABC must be an eq. ▲ !!!

Thank you.!

good one!! but use latex..+1

Ayush G Rai - 5 years ago

Is it or not?

2 solutions

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