# Equilateral triangle geometric problem

I have an Equilateral triangle with unknown side $a$. The next thing I do is to make a random point inside the triangle $P$. The distance $|AP|=3$ cm, $|BP|=4$ cm, $|CP|=5$ cm.

It is the red triangle in the picture. The exercise is to calculate the area of the Equilateral triangle (without using law of cosine and law of sine, just with simple elementary argumentation).

The first I did was to reflect point $A$ along the opposite side $a$, therefore I get $D$. Afterwards I constructed another Equilateral triangle $\triangle PP_1C$.

Now it is possible to say something about the angles, namely that $\angle ABD=120^{\circ}$, $\angle PBP_1=90^{\circ} \implies \angle APB=150^{\circ}$ and $\alpha+\beta=90^{\circ}$

Now I have no more ideas. Could you help me finishing the proof to get $a$ and therefore the area of the $\triangle ABC$. If you have some alternative ideas to get the area without reflecting the point $A$ it would be interesting.

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Well, since the distances form a Pythagorean triple the choice was not that random. You are on the right track and reflection is a great idea, but you need to take it a step further.

Check that in the (imperfect) drawing below $\triangle RBM$, $\triangle AMQ$, $\triangle MPC$ are equilateral, since they each have two equal sides enclosing angles of $\frac{\pi}{3}$. Furthermore, $S_{\triangle ARM}=S_{\triangle QMC}=S_{\triangle MBP}$ each having sides of length 3,4,5 respectively (sometimes known as the Egyptian triangle as the ancient Egyptians are said to have known the method of constructing a right angle by marking 12 equal segments on the rope and tying it on the poles to form a triangle; all this long before the Pythagoras' theorem was conceived)

By construction the area of the entire polygon $ARBPCQ$ is $2S_{\triangle ABC}$

On the other hand

$$ARBPCQ= S_{\triangle AMQ}+S_{\triangle MPC}+S_{\triangle RBM}+3S_{\triangle ARM}\\=\frac{3^2\sqrt{3}}{4}+\frac{4^2\sqrt{3}}{4}+\frac{5^2\sqrt{3}}{4}+3\frac{1}{2}\cdot 3\cdot 4 = 18+\frac{25}{2}\sqrt{3}$$

Hence

$$S_{\triangle ABC}= 9+\frac{25\sqrt{3}}{4}$$

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Thanks, good idea to do the reflection on all three sides. I still have one problem. What is the area of $AQCM$ ? – Alexander Mar 13 '13 at 21:39
Hold on, there is a flaw in my picture, correcting now – Valentin Mar 13 '13 at 21:49
see the edited answer – Valentin Mar 13 '13 at 22:00
$9+25\sqrt{3}/4$, because $50/4 = 25/2$. – i707107 Jan 18 '14 at 20:37
thanks, corrected – Valentin Jan 18 '14 at 23:40

You can solve this without any trig if you consider the properties of a equilateral triangle, and the fact that you've created six right triangles in which you know the length of the hypotenuse and the relationship:

$a+b = c+d = e+f$

$a^2 + g^2 = |AP|^2$

$b^2 + g^2 = |BP|^2$

$c^2 + h^2 = |BP|^2$

$d^2 + h^2 = |CP|^2$

$e^2 + i^2 = |CP|^2$

$f^2 + i^2 = |AP|^2$

Then:

$a+b = c+d = e+f$

$a^2 + g^2 = 9$

$b^2 + g^2 = 16$

$c^2 + h^2 = 16$

$d^2 + h^2 = 25$

$e^2 + i^2 = 25$

$f^2 + i^2 = 9$

Then:

$a+b = c+d = e+f = s$

$b^2 - a^2 = 7$

$d^2 - c^2 = 9$

$e^2 - f^2 = 16$

Then:

$b^2 - (s-b)^2 - 7 = s^2 + 2sb - 7 = 0$

$d^2 - (s-d)^2 - 9 = s^2 + 2sd - 9 = 0$

$e^2 - (s-e)^2 - 16 = s^2 + 2se - 16 = 0$

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Thank you, but what exactly is a,b,c,d,e,f,g,h,i in the drawing? – Alexander Mar 13 '13 at 21:22
Sorry, I was not referencing the drawing. $ag|AP|$, etc, are all right triangles which make up $ABC$. – Jonathan Rich Mar 13 '13 at 21:25
And how do you get a now? – Alexander Mar 13 '13 at 21:33

A generic Solution for any lengths d1, d2, d3: the distances d1, d2, and d3 are from points C, B, and A, respectively, to a fourth point E in the interior of the equilateral triangle ABC.

First, we rotate the figure of ABC and E about C through 60 degrees, resulting in an equilateral triangle BHC with G congruent to ABC with E. In particular we see that CE is rotated through 60 degrees to CG, which shows that CEG is an equilateral triangle, so that EG = CE = CG = d1. This triangle I will now refer to as EQUI(d1): an equilateral triangle with side d1:

Now we draw some simple conclusions:

• The point E divides the original triangle into three triangles, which I have made into a red one AEC, a yellow one AEB, and a white one BEC.

• The area of the white and red triangles together is the same as the area of BEG and CEG together.

• BEG is the triangle with sides equal to the given d1, d2, d3, which I will from now refer to such a triangle as T.

• We conclude that the white and red triangles together have an area equal to [T] + [EQUI(d1)] (where [x] is the area of x).

With similar reasoning we can draw similar conclusions about the areas of the white and yellow triangle together, and about the areas of the red and yellow triangle together.

Adding the areas of white+red, white+yellow, and red+yellow gives 2*[ABC]. It also gives 3*[T]+[EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)].

So, if we read in areas, we have:

[ABC] = 1.5*[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2.

We can also find the side length a by using Area of equilateral triangle => [ABC] = sqrt(3)/4 * a^2

√3/4 * a^2 = 1.5*[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2.

Which gives a = √((4/√3)(1.5[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2))

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the cosine rule can be applied on the $\triangle {AEB}$ to find the side AB as the $\angle {AEB}$ IS $150^{\circ}$ – Boris Jan 28 at 17:57