# Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.

Given that $$G$$ is the centroid of $$\Delta ABC$$, $$GA = 2\sqrt{3}$$ , $$GB = 2\sqrt{2}$$, $$GC = 2$$ . Find $$[\Delta ABC]$$.

What I Tried: Here is a picture :-

I know the centroid divides each of the medians in the ratio $$2:1$$ . So $$AD = 3\sqrt{3}$$ , $$BE = 3\sqrt{2}$$ , $$CF = 3$$ .
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$ Where $$m_1,m_2,m_3$$ are the medians of the triangle and $$s_m = \frac{m_1 + m_2 + m_3}{2}$$ .

After putting the respective values for the medians I get that $$[\Delta ABC]$$ is :- $$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$ $$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$

I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?

Can anyone help me?

• We have $2AB^2+2AC^2 = BC^2+2AD^2$ and two similar relations Oct 27 '20 at 7:50

(This is not likely to be what you're looking for.)

I think in this problem you can use a simpler solution.

Construct point $$H$$ outside $$\overline{AC}$$ such that $$AGCH$$ forms a Parallelogram. We have

1. $$\overline{AH}=\overline{GC}=2$$
2. $$\overline{AG}=2\sqrt3$$
3. $$\overline{GE}=\overline{EH}\Longrightarrow \overline{GH}=\overline{GB}=2\sqrt2$$

Since $$\overline{AG}^2=\overline{AH}^2+\overline{GH}^2$$, we know that $$\angle AHG=90^{\circ}$$.

Note that $$\triangle AGE=\frac{1}2\triangle AGH=\frac{1}2\cdot\frac{1}2\cdot2\cdot2\sqrt2=\sqrt2$$.

Therefore $$\triangle ABC=6\triangle AGE=6\sqrt2$$.

• Nice solution! But it's not a general way. Are you agree? Oct 27 '20 at 8:10
• Yes I agree. This is not a general way. I just want to share this neat solution in this special case.
– user808951
Oct 27 '20 at 8:12
• I think your solution is a best here. +1 Oct 27 '20 at 8:15

From where you left,

$$A = \displaystyle 3 \sqrt{{(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}}$$

Take the first two terms, it is of the form $$(a-b)(a+b)$$ so we have,

$$(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3}) = 2\sqrt2$$

Next two terms can be taken as (a+b-c)(a-b+c)

$$((\sqrt{3} + 1) - \sqrt{2})((\sqrt{3} -1) + \sqrt{2}) = 3 - 1 - 2 + \sqrt 2 (\sqrt3 + 1) - \sqrt2 (\sqrt3 - 1) = 2 \sqrt2$$

So $$A = 6 \sqrt2$$

• Got it, thanks for completing my working. Oct 27 '20 at 8:13
• @Anonymous you are welcome. Oct 27 '20 at 8:14

I think, it's better to use $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$ $$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ because from the given easy to get $$a^2$$, $$b^2$$ and $$c^2$$.

Indeed, $$\frac{1}{3}\sqrt{2b^2+2c^2-a^2}=2\sqrt3,$$ $$\frac{1}{3}\sqrt{2a^2+2c^2-b^2}=2\sqrt2$$ and $$\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=2,$$ which gives $$\frac{1}{3}(a^2+b^2+c^2)=4(3+2+1)$$ or $$a^2+b^2+c^2=72,$$ which gives $$2(72-a^2)-a^2=108$$ or $$a^2=12.$$ By the similar way we obtain: $$b^2=24$$ and $$c^2=36$$, which gives $$S_{\Delta ABC}=6\sqrt2.$$

• Ok so is that a new formula to get the area of the triangle? Oct 27 '20 at 7:50
• @Anonymous No! It's just the Heron's formula. I added something. See now. Oct 27 '20 at 7:50
• Oh I see, ok fine. Oct 27 '20 at 7:55
• Can you tell me how you got $a^2,b^2,c^2$ individually, or am I missing something? . Ok I got it. Oct 27 '20 at 8:00

It is indeed a theorem that can be generalized, although Mathworlddoes n't say so explicitly.

The area of a triangle formed by medians ( computed for example from Brahmagupta/Heron formula ) is three-fourths the area formed by the corresponding sides of the given triangle.

It can also be proved by projective geometry.

Linear scale $$k=\sin \frac{\pi}{3}$$ can be established from relative proportions of the simplest equilateral triangle.

In our case hypothetical medians after scaling up full sides from centroid by $$\text{50%} : 3(\sqrt 3, \sqrt 2,1)$$ calculates to $$\dfrac{9}{\sqrt 2};$$

So the circumscribed triangle area would be:

$$\dfrac{{\dfrac{9}{\sqrt 2}}} {\sin^2\dfrac{\pi}{3}} = 6 \sqrt 2.$$