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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.

triangle

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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2 Answers 2

up vote 5 down vote accepted

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

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

enter image description here

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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
1  
$9+25\sqrt{3}/4$, because $50/4 = 25/2$. –  i707107 Jan 18 at 20:37
    
thanks, corrected –  Valentin Jan 18 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
1  
And how do you get a now? –  Alexander Mar 13 '13 at 21:33

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