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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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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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  • $\begingroup$ Thanks, good idea to do the reflection on all three sides. I still have one problem. What is the area of $AQCM$ ? $\endgroup$ – Alexander Mar 13 '13 at 21:39
  • $\begingroup$ Hold on, there is a flaw in my picture, correcting now $\endgroup$ – Valentin Mar 13 '13 at 21:49
  • $\begingroup$ see the edited answer $\endgroup$ – Valentin Mar 13 '13 at 22:00
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    $\begingroup$ $9+25\sqrt{3}/4$, because $50/4 = 25/2$. $\endgroup$ – Sungjin Kim Jan 18 '14 at 20:37
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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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  • $\begingroup$ Thank you, but what exactly is a,b,c,d,e,f,g,h,i in the drawing? $\endgroup$ – Alexander Mar 13 '13 at 21:22
  • $\begingroup$ Sorry, I was not referencing the drawing. $ag|AP|$, etc, are all right triangles which make up $ABC$. $\endgroup$ – Jonathan Rich Mar 13 '13 at 21:25
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    $\begingroup$ And how do you get a now? $\endgroup$ – Alexander Mar 13 '13 at 21:33
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A generic Solution for any lengths $d_1, d_2, d_3$:

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 $\triangle ABC$ and $E$ about $C$ through $60^\circ$, resulting in an equilateral triangle $\triangle BHC$ with $G$ congruent to $\triangle ABC$ with $E$. In particular we see that $CE$ is rotated through $60^\circ$ to $CG$, which shows that $\triangle CEG$ 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 $d1$:

enter image description here

Now we draw some simple conclusions:

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

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

  • BEG is the triangle with sides equal to the given $d_1, d_2, d_3$, 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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  • $\begingroup$ the cosine rule can be applied on the $\triangle {AEB}$ to find the side AB as the $\angle {AEB}$ IS $150^{\circ}$ $\endgroup$ – Nebo Alex Jan 28 '16 at 17:57
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SOLUTION $9 + \frac{25\sqrt{3}}{4}$.

Interestingly enough, this problem is Problem 13 of the American Invitational Mathematics Examination in 2012. As such, it was used to screen candidates for the American Mathematics Olympiad that year.

Art of Problem Solving details their solution to this problem in their Wiki and on their YouTube channel. The following points and figures were developed years ago based on their video.

We are asked here to find the area of the equilateral $\Delta{ABC}$ drawn in Figure (a) below and whose vertices lie 3, 4, and 5 units away, in that order, from a point $O$.

enter image description here

  • To begin our solution, define a pair of adjacent angles by rotating $\Delta{ABC}$ around $\color{blue}{A}$ by 60° clockwise (b) or counterclockwise (c). Alternatively, rotate around $\color{red}{B}$ (de). The rotation is indicated by the empty angle arcs in the figures.
  • The first adjacent angle in the pair involves $\color{blue}{\overline{AO}}$ (if the rotation was on $\color{blue}{A}$, or involves $\color{red}{\overline{BO}}$ if the rotation was on $\color{red}{B}$) in an equilateral triangle. This first adjacent angle thus measures 60°.
  • The second adjacent angle in the pair involves $\color{red}{\overline{BO}}$ (or $\color{blue}{\overline{AO}}$) and lies opposite the 5-unit-long rotated copy of $\color{green}{\overline{CO}}$ in a 3-4-5 triangle. This second adjacent angle thus measures 90°.
  • Thus, the shaded angles have total measure $m\angle{AOB} = 60° + 90° = 150°$.
  • By the law of cosines, $\Delta{ABC}$ has sides whose square is $AB^2 = AO^2 + BO^2 - 2(AO)(BO)\mathrm{cos}(m\angle{AOB}) = 3^2 + 4^2 - 2(3)(4)\mathrm{cos}150° = 25 + 12\sqrt{3}$.
  • Since $\Delta{ABC}$ is equilateral, its area is $\frac{\sqrt{3}{AB}^2}{4} = 9 + \frac{25\sqrt{3}}{4}$.
  • Just as an aside, note that rotations around $\color{green}{C}$ are unproductive despite resulting in equilateral and right triangles since not one of the resulting adjacent angles is right (fg). ■
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