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I am trying to solve the problem of constructing, with straightedge and compass, an equilateral triangle of given side length $a$ inscribed in a given triangle.

I found this post "Inscribe an equilateral triangle inside a triangle" and this other post "How to draw an equilateral triangle inscribed in another triangle?" but the construction must be made with straightedge and compass, using simple constructions such as arcs, parallel lines, perpendicular lines and that kind of thing.

I tried constructing the arcs capable of $120^{\circ}$ on the sides of the given triangles and noticed that the centers of the arcs form an equilateral triangle, but I don't know what to do after that.

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  • $\begingroup$ Did you see g.gov's solution in the 2nd linked post? That's constructible. $\endgroup$
    – Calvin Lin
    Apr 8 '20 at 1:39
  • $\begingroup$ That would be a general equilateral triangle inscribed, not with a given side lenght like is required in my case. $\endgroup$ Apr 8 '20 at 1:50
  • $\begingroup$ Something is wrong with your question. It seems that for a given triangle there is only one solution when a is not given. If a is different from the side of the equilateral triangle to be inscribed than there is no solution. $\endgroup$
    – Moti
    Apr 8 '20 at 4:32
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    $\begingroup$ @Moti For a given triangle there are infinitely many inscribed equilateral triangles geogebra.org/m/ey3aYYkK – $\endgroup$ Apr 8 '20 at 15:14
  • $\begingroup$ You are right! Will try it now:) $\endgroup$
    – Moti
    Apr 9 '20 at 1:52
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I believe the following diagrams and incorporated explanation will suffice. Let me know if it is not clear. Click on image to get a larger and clearer view.

enter image description here

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  • $\begingroup$ I just noticed that you accepted the first not proper response - this is not the way to solve math problems! $\endgroup$
    – Moti
    Jun 18 '20 at 4:04
  • $\begingroup$ I accepted it before you sent your solution, sorry $\endgroup$ Jun 18 '20 at 12:02
  • $\begingroup$ Thank you very much for posting the solution, it was really helpful $\endgroup$ Jun 18 '20 at 12:07
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$\mathbf{1}.$ Notations, definitions and classifications used in our answer TriangleTypes

The given scalene triangle is denoted by $ABC$. Its sides $a, b,$ and $c$ are sized according to $a > b > c$, and, hence its vertex angles $A, B,$ and $ C$ obey the inequality $\measuredangle A > \measuredangle B > \measuredangle C$, which implies that $\measuredangle A > 60^o$ as well. $\Delta$ stands for the area of $ABC$.

The sidelength of the inscribed equilateral triangle is denoted by $d$. The sidelength of the largest of the inscribable equilateral triangles is $d_{max}$, whereas that of the smallest is $d_{min}$. We denote the smallest and largest inscribed equilateral triangles as $DEF$ and $XYZ$ respectively. In a similar vein, $PQR$ and $STU$ are the sought pair of inscribable equilateral triangle with sidelength $d$.

For the ease of elucidating the construction, we discriminate between three types of triangles as depicted in $\mathrm{Fig.1}$. If the largest vertex angle of an obtuse triangles (i.e. $\measuredangle A$) is greater than or equal to $120^o$, we call it a triangle of Type-I. Type-II contains acute and obtuse triangles having just one angle (i.e. $\measuredangle A$), which is greater than $60^o$ and less than $120^o$. Acute and obtuse triangles having only one vertex angle (i.e. $\measuredangle C$) less than $60^o$ together with all equilateral triangles make up the group named Type-III.

$\mathbf{2}.$ Construction

The construction described below, in which we do a vertex-chasing, is, so-to-speak, a geometrical iteration, where the outcome at the end of each iteration is checked to see whether it has achieved the desired accuracy. This procedure makes sure that the points found in succession on the sides of $ABC$ converge very fast to the vertices of the coveted inscribed equilateral triangle. Because of its iterative nature, a pair of steady hands, a pair of sharp eyes, and a very sharp pencil are essential to achieve acceptably accurate outcome.

However, before attempting to construct an inscribed equilateral triangle with the given sidelength, we should make sure that such triangle or triangles actually exist. Otherwise, we could find ourselves chasing wild geese instead of vertices. For that matter, we need to carry out two additional constructions beforehand, one to determine the smallest inscribable equilateral triangle, while the other to find the largest. Nome of these constructions needs no iterations and, therefore, the exact location of the vertices of the sought equilateral triangles can be determined directly.

$\mathbf{3}.$ Construction of the smallest inscribable equilateral triangle of the given triangle $ABC$ SmallestIET

If you are dealing with a triangle of Type-I or Type-II, draw the angle bisector of the largest vertex angle (i.e. $\measuredangle A$) as shown in $\mathrm{Fig.3.1}$, so that it meets the longest side (i.e. $BC$) at $U$. Point $U$ is the vertex of the inscribed equilateral triangle that lies on the side $BC$ of $ABC$. If $ABC$ is a Type-III triangle, draw the angle bisector of the smallest vertex angle (i.e. $\measuredangle C$) to intersect the shortest side (i.e. $AB$) at $U$ (see $\mathrm{Fig.3.2}$). As in the previous case, point $U$ is one of the vertices of the inscribed equilateral triangle, but now it lies on the side $AB$ of $ABC$. Please note that, regardless of type of the triangle, if its second largest angle is equal to $60^o$ (i.e. $\measuredangle B = 60^o$), the angle to be bisected can be either $\measuredangle A$ or $\measuredangle C$ (see $\mathrm{Fig.3.3}$).

To complete the construction, draw two lines flanking the drawn angle bisector, so that each of them makes an angle of $30^0$ with it at $P$. Their internal intersection points with the nearest sides of $ABC$ mark the other two vertices of the inscribed equilateral triangle.

A triangle, whether it is scalene, isosceles, or equilateral, has only one smallest inscribable equilateral triangle. The two triangles share their incenter.

It is also possible to determine the value of $d_{min}$ numerically using the appropriate equation given below. $$d_{min}=\frac{2\Delta}{\left(b+c\right) \sin\left(30^o+\frac{A}{2}\right)} \tag{for Type-I & II triangles}$$ $$d_{min}=\frac{2\Delta}{\left(a+b\right) \sin\left(30^o+\frac{C}{2}\right)}\tag{for Type-III triangles}$$

$\mathbf{4}.$ Construction of the largest inscribable equilateral triangle of the given triangle $ABC$ LargestIET LargestIETSC

If $ABC$ is a Type-I triangle, its vertex $A$, which has the largest angle, coincides with one of the vertices (i.e. $Z$) of its largest inscribable equilateral triangle. One side of the inscribed triangle of this type of triangle (i.e. $YZ$) always lies on its side $CA$. Therefore, to obtain the vertex lying on the side $BC$, draw a line, which makes an angle $60^o$ with the side $CA$, through the vertex $A$ to meet the side $BC$ at $X$ (see $\mathrm{Fig.4.1}$). Since we now know two vertices of the sought inscribed equilateral triangle, its third vertex $Y$ on the side $CA$ can be easily found.

If $ABC$ is a Type-II triangle, as in the case of Type-I triangles, one of the vertices of the largest inscribable equilateral triangle $Y$ coincides with its vertex $A$, the vertex with the largest angle. However, this type of triangles have one of its sides (i.e. $YZ$) lying on the side $AB$ of $ABC$. The vertex lying on the side $BC$ can be pinpointed by drawing a line, which makes an angle $60^o$ with the side $AB$, through the vertex $A$ to meet the side $BC$ at $X$ (see $\mathrm{Fig.4.2}$).

If the triangle $ABC$ is of Type-III, its vertex $B$, where the second largest vertex angle is, harbours one of the vertices of the largest inscribable equilateral triangle, i.e. $Z$. One side of the inscribed triangle of this type of triangle (i.e. $ZX$) always lies on its side $BC$. To locate the vertex lying on the side $CA$, draw a line that makes an angle $60^o$ with the side $BC$ and goes through the vertex $B$ to meet the side $CA$ at $Y$ (see $\mathrm{Fig.4.3}$).

There are a few noteworthy special cases. All triangles, which has a vertex angle equal to $120^o$ (i.e. $\measuredangle A = 120^o$), have two identical largest inscribed equilateral triangles, which do not overlap as shown in $\mathrm{Fig.4.4}a$. If the second largest angle of the given triangle is equal to $60^o$ (i.e. $\measuredangle B = 60^o$), the given triangle and its largest inscribable equilateral triangle share the shortest side (i.e. $AB$) as depicted in $\mathrm{Fig.4.4}b$. All isosceles triangles have two partially overlapping identical largest inscribed equilateral triangles (see $\mathrm{Fig.4.4}c$). An equilateral triangle and its largest inscribed equilateral triangle are one and the selfsame (see $\mathrm{Fig.4.4}d$). All triangles other than isosceles triangles have a unique largest inscribed equilateral triangle.

Following equations can be used to calculate the value of $d_{max}$. $$d_{max}=\frac{2\Delta}{a \sin\left(60^o+C\right)} \tag{ for Type-I triangles }$$ $$d_{max}=\frac{2\Delta}{a \sin\left(60^o+B\right)} \tag{ for Type-II triangles }$$ $$d_{max}=\frac{2\Delta}{b \sin\left(60^o+C\right)} \tag{ for Type-III triangles}$$

$\mathbf{5}.$ Construction of inscribed equilateral triangles with a given sidelength $d$ IET

Once you know for sure that there are inscribed equilateral triangles with a given sidelength, you can follow the steps outlined below to construct them. We hope that the series of diagrams from $\mathrm{Fig.5.1}$ to $\mathrm{Fig.5.4}$ would help you to comprehend the description.

Draw the angle bisector of the largest angle $\measuredangle A$ of the given triangle $ABC$ to meet its largest side $BC$ at $D$. As shown in $\mathrm{Fig.5.1}$, draw a circle or an arc with $D$ as the center and $d$ as the radius to cut the sides $CA$ and $AB$ at $Q$ and $U$ respectively, each of which serves as the educated guess to start the geometrical iteration leading us to one of the sought pair of inscribable equilateral triangles with sidelength $d$, i.e. either $PQR$ or $STU$.

Obviously, to construct $PQR$, we need to consider the point $Q$. As shown in $\mathrm{Fig.5.2}$, we draw a circle with $Q$ as the center and $d$ as the radius to cut the side $AB$ at $R$. Next, draw a circle with $R$ as the center and $d$ as the radius to cut the side $BC$ at $P$. If you measure the sides of the triangle $PQR$ after the end of this first iteration, you will find that $QR = RP = d$, but $PQ ≠ d$. As a consequence, we have to perform further iterations as follows. Draw a circle with $P$ as the center and $d$ as the radius to intersect the side $CA$ and move the point $Q$ to this point of intersection. Now, you may find that $QR ≠ d$. Therefore, we proceed along by drawing a circle with $Q$ as the center and $d$ as the radius to intersect the side $AB$. This point of intersection is the new location of $R$. Now, you have to measure $RP$ to check whether it is exactly equal or almost equal to $d$. If you are satisfied with the length of $RP$, you can stop the iteration, because you have found one of the two inscribable equilateral triangles to a certain degree of accuracy. However, if you want to increase the accuracy of the construction, you have to iterate further to improve the positions of the three vertices $P$, $Q$, and $R$ (e.g. $\mathrm{Fig.5.3}$). To find the other inscribable equilateral triangle $STU$ (e.g. $\mathrm{Fig.5.4}$), a similar series of iterations starting from the point $U$ in $\mathrm{Fig.5.1}$ should be carried out.

$\mathbf{6}.$ Points to ponder

You may have already noticed that we have not provided any proof of what we have stated in our answer. All our deductions set out above are evidence-based, meaning our inferences came only through the observations made during a thorough analysis of the problem. If you find mistakes, errors, or counter-evidence, please post them. If we cannot rectify errors or are unable to argue against counter-evidence, we are ready to take this post down immediately.

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