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It is straightforward to construct (straight-edge and compass) an isosceles triangle with area equal to a given triangle $\triangle ABC$, for instance as follows:

  1. Construct the line through $A$ parallel to $BC$ (demonstration of method);
  2. Construct the perpendicular bisector of $BC$ (demonstration of method);
  3. The perpendicular from (2) meets the line from (1) at $D$; draw $\triangle DBC$.

This triangle is isosceles, since $D$ is equidistant from $B$ and $C$, and has the same base and altitude as $\triangle ABC$ so has the same area.

Suppose we want to go further and construct an equilateral triangle with area equal to the given triangle — how might we go about that?


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  • $\begingroup$ I have posted this question partly because it has some poor answers elsewhere on the web, which I hope this will supplant in the search engines, and partly because I hope someone has a more elegant answer than my own construction. The fewer steps the better; I found I had to use the geometric mean construction and I wonder whether this step might be avoidable. $\endgroup$ – Silverfish Jun 8 '16 at 15:10
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    $\begingroup$ You may exploit: $$\Delta = \frac{\sqrt{3}}{4}\left(3R^2-3OG^2-2 \ell_N^2\right)$$ where $R$ is the circumradius, $OG$ the circumcenter-centroid distance and $\ell_N$ the side length of the inner Napoleon triangle :D $\endgroup$ – Jack D'Aurizio Jun 8 '16 at 15:57
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    $\begingroup$ You should post your own construction so that we don't duplicate your work. Linking to the "poor answers elsewhere on the web" would also be helpful. (It's no fun to put effort into an answer, only to have the questioner respond, "Yeah, I already did it that way." This happens more often than I like.) $\endgroup$ – Blue Jun 8 '16 at 16:00
  • $\begingroup$ @Blue The stuff I found elsewhere was simply not worth linking (e.g. the link might say "construct", but the "answer" might involve using trigonometry to calculate the appropriate length and then measuring it). I have managed to get my construction down to a less fiendishly long-winded form and posted it, though I suspect it could be simplified further. $\endgroup$ – Silverfish Jun 9 '16 at 0:46
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    $\begingroup$ No serious answer but only a comment. If you have unlimited patience :) draw a parallel to $a$ through A , perpendicularly bisect $a$ to find new vertex $A_1$. Go counterclockwise, do the same to $b$ creating new $B_1$...and so on ad infinitum .. Triangle $A_nB_nC_n$ quickly turns equilateral. $\endgroup$ – Narasimham Mar 23 '17 at 9:13
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The fewer steps the better; I found I had to use the geometric mean construction and I wonder whether this step might be avoidable.

Something like a geometric mean is unavoidable since the problem is a quadratic equation for the side length of the equilateral triangle.

Here is a relatively efficient construction using a geometric mean. The savings is in re-using one side of ABC as the base of the equilateral triangle, and in using a non-perpendicular line to measure altitude, and allowing the semicircle construction of geometric mean to be applied.

On one of the sides, say AB, build an equilateral triangle ABD. Extend line CD to intersect AB at P. Find a length $g$ equal to the geometric mean of PC and PD, and take a point H on PD with PH = $g$. The equilateral triangle whose "height" measured along line PCD (from P) is $g$ has the same area as the given triangle.

The construction of the geometric mean is made easier by extending line CD to E so that P is the midpoint of EC, then using ED as diameter of a circle and taking $g=|PG|$ for PG a perpendicular to CD with G on the circle. Then draw the parallel to AB through H and intersect it with AD (at point K) to cut off the correct side length (AK) of equilateral triangle.

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  • $\begingroup$ Is it necessary to construct E? I tried doing the construction in GeoGebra, under the assumption "E" was just a typo for "C" and it looks like it works fine. $\endgroup$ – Silverfish Jun 10 '16 at 13:06
  • $\begingroup$ You don't strictly need E but the construction of geometric mean comes out easier (I think) if the two segments do not overlap. $\endgroup$ – zyx Jun 10 '16 at 14:10
  • $\begingroup$ There is a choice of which side of AB on which to build the equilateral triangle. If D is on the opposite side from C, then PC and PD are in the right position to do the geometric mean construction (if you use a semicircle). I think, but am not completely sure, that the easiest thing to do when running the construction that way would still be to first find an E as before, on line CD so that P is midpoint of EC, so we again construct E as an auxiliary point (for finding the perpendicular to CD at P). Do you know an easier method for the mean that doesn't lead to using E? $\endgroup$ – zyx Jun 11 '16 at 16:29
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A possible construction.

enter image description here

  1. Draw $DE$ through $A$ in such a way that $DECB$ is a rectangle;
  2. Take $Q$ as the intersection of the angle bisectors of $\widehat{BDE}$ and $\widehat{DEC}$;
  3. Take $T$ on $CE$ such that $TE=EQ$;
  4. Take $F$ on $DT$ such that $DF=DE$;
  5. Let $\ell$ be the perpendicular to $BE$ through $E$;
  6. Take $G$ on $\ell$ such that $BG=BF$.

$EG$ is the side of an equilateral triangle with the same area of $ABC$.

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  • $\begingroup$ (+1) This is very neat, what was the intuition that led to this approach? $\endgroup$ – Silverfish Jun 9 '16 at 9:47
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    $\begingroup$ @Silverfish: to construct $2ab\cos\theta$ as the difference between $a^2+b^2-2ab(\cos\pi/2+\theta)$ and $a^2+b^2$. $\theta$ here is $\cos^{-1}\frac{1}{\sqrt{3}}$. $\endgroup$ – Jack D'Aurizio Jun 9 '16 at 13:06
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Thank you to all you brilliant mathematicians for your ingenious methods on this topic. I do not understand trigonometry very well because I am not very well educated, but this is Euclid's approach to the problem which is general for generating similar figures of any kind of a given area, which, although perhaps cumbersome, I found convenient and readily intelligible. An equilateral triangle is made, a parallelogram constructed equal to it, another equal to the desired area applied alongside it in the parallels, then a mean proportional taken between the lengths of the two parallelograms. Because as the first line is to the third, so is the figure on the first to the similarly described figure on the second, the equilateral described on the mean proportional contains the desired area. The other methods are simpler I just thought I'd post this because it's purely geometrical. Below is a link to the proposition and my construction. http://aleph0.clarku.edu/~djoyce/elements/bookVI/propVI25.html construction

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Let the required equilateral triangle have side $s$, and the given triangle have base $b$ and altitude $h$, so:

$$ \frac{1}{2}bh = \frac{1}{2}s^2 \sin 60^\circ \implies s^2 = b \left( \frac{h}{\sin 60^\circ} \right)$$

Therefore we can construct $s$ as the geometric mean of $b$ and $\frac{h}{\sin 60^\circ}$.

enter image description here

  1. Draw the perpendicular from $A$ to $BC$ (produced if necessary); let the foot of the perpendicular be $F$.
  2. Draw an arc center $A$ through $F$ and another centre $F$ through $A$; these intersect at $P$ so that $\angle FAP = 60^\circ$.
  3. Draw the angle bisector of $\angle FAP$; let $Q$ be its intersection with $BC$ (produced if necessary), so that $\angle FAQ = 30^\circ$.

Considering $\triangle AFQ$ we have $\sin 60^\circ = \cos 30^\circ = \frac{AF}{\color{red}{AQ}}$, so that $\color{red}{AQ} = \frac{AF}{\sin 60^\circ}$. We need to take the geometric mean of this length and the base.

enter image description here

  1. Draw an arc, center $C$ and radius $\color{red}{AQ}$, to construct $R$ on $BC$ produced such that $\color{red}{AQ} = \color{red}{CR}$.
  2. Construct the midpoint $O$ of $BR$, and hence draw a circle with diameter $BR$.
  3. Construct the perpendicular to $BR$ through $C$; let $S$ be its intersection with the circle from step (5).

Now by considering similar triangles $\triangle BCS$ and $\triangle SCR$, or bearing in mind that $\color{green}{CS}$ is one half of a chord bisected by the diameter $BR$ and then applying the intersecting chords theorem, we have $\color{green}{CS}^2 = BC \cdot \color{red}{CR}$. Hence $\color{green}{CS}$ is the required geometric mean, and is the side of an equilateral triangle with area equal to that of the given triangle.

enter image description here

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I wondered whether it was possible to do this by "morphing" an isosceles triangle into an equilateral one — here's an approach that creates an equilateral triangle by moving the apex of the isosceles triangle, then scales its sides to restore the original area.

Without loss of generality, assume $\triangle ABC$ is isosceles with $AB=AC$. (Otherwise, construct an isosceles triangle with area equal to the given triangle, e.g. by the method given in the question.)

enter image description here

  1. Draw a circle center $B$ through $C$, and another center $C$ through $B$.
  2. Take $D$ as the intersection of these circles that lies the opposite side of $BC$ to $A$; $\triangle BCD$ is equilateral.
  3. Draw $AD$ and take $M$ to be its intersection with $BC$; $M$ is the midpoint of $BC$.

Although $\triangle BCD$ is equilateral, it isn't the triangle we need because it does not have the same area as $\triangle ABC$. They share base $BC$, so the ratio of their areas is simply the ratio of their altitudes:

$$ \frac{\text{Area } \triangle ABC}{\text{Area } \triangle BCD} = \frac{AM}{MD}$$

So to obtain the desired equilateral triangle we must scale the lengths of $\triangle BCD$ by $\sqrt \frac{AM}{MD}$.

enter image description here

  1. Construct $O$ as the midpoint of $AD$ and hence draw a circle diameter $AD$.
  2. Produce $BC$ to meet the circle from (5) at $E$, then by the intersecting chords theorem we have $EM^2 = AM \cdot MD$ and hence $\tan \angle EDM = \frac{EM}{MD} = \sqrt \frac{AM}{MD}$.
  3. Take $P$ on $AD$ such that $PD = BC$.
  4. Construct the perpendicular to $AD$ through $P$, and take $Q$ as its intersection with $DE$.

Then $PQ$ is the side of an equilateral triangle with area equal to the given triangle, since $$\frac{PQ}{BC} = \frac{PQ}{PD} = \tan \angle EDM = \sqrt \frac{AM}{MD} = \sqrt \frac{\text{Area } \triangle ABC}{\text{Area } \triangle BCD}$$

Equilateral triangle with same area as given isosceles triangle

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