# How big is the equilateral triangle inscribed in a triangle?

Given a scalene triangle $ABC$ and an inscribed equilateral triangle whose vertices lie on different sides of $\triangle ABC$, what is the maximal ratio of the area of the equilateral triangle to that of the original triangle?

I would expect an answer as a ratio of polynomials in sines and cosines of the angles of $\triangle ABC$. I got such an expression via a clunky unsymmetrical method, but it but it was so messy that I gave up trying to simplify it. However, a more intelligent method might well yield a formula of reasonable length.

• Are you aware of (math.stackexchange.com/q/186432) ? Jan 18, 2017 at 23:22
• @JeanMarie: Yes, but that is a different question. Jan 19, 2017 at 7:15
• Though 4 years old - are you interested in a solution?
– Moti
Mar 22, 2021 at 6:53
• @Moti : Yes, certainly! Mar 22, 2021 at 7:13
• Variented question: docin.com/p-1294293529.html
– user1034536
Jan 5 at 9:53

We start with the process of solving the problem of "inserting" an equilateral triangle in the given triangle. We select the side of the triangle in front the largest angle to create the locus the smaller angles of the equilateral triangle.

When you have (the drawing on the right) the two circles with centers $$U$$ (radius $$r_1$$) and $$S$$ (radius $$r_2$$), you select the side of the equilateral triangle $$GH$$, in the larger circle (center $$S$$ in the drawing), you stretch it to intersect the smaller circle (not shown) and then you draw the line through this intersection and the third vertex of the equilateral triangle intersecting the other circle.

The triangle you created is the smallest triangle subscribing the equilateral triangle and is similar to the original triangle. You scale up the equilateral triangle by the similarity factor to get the largest equilateral triangle.

Note that the largest equilateral triangle is sharing a side with the triangle in which it is inscribed and that the $$60^\circ$$ angle of the equilateral triangle is included in the largest angle of the original triangle.

This means that to draw the largest equilateral triangle you do not need to go through the lengthy process; you draw a $$60^\circ$$ angle on the larger side of the largest angle of the triangle and from that vertex to the intersection with the opposite side is the largest equilateral triangle.

Let me know if you have a question.

• math.meta.stackexchange.com/q/33343/18398
– JRN
Mar 24, 2021 at 1:50
• @Joel I could not understand what issues you have with the question? What is the opinion? The answers were opinions - but not the question. The question is clear.
– Moti
Mar 24, 2021 at 5:36
• In your comment above, what "question" are you referring to? The question in this post on main? Or the question in the post on meta? Either way, I don't have "any issues with the question." But, if I'm not mistaken, it seems the OP does not understand your answer and doesn't want to ask you to explain it because they are "certain that any attempt at correspondence with this poster [you] to explain the present answer would be futile."
– JRN
Mar 24, 2021 at 7:38
• I did not vote to close the question on meta.
– JRN
Mar 24, 2021 at 7:43

There isn't a single inscribed equilateral triangle: by choosing a point on a side, rotating another side by $60^\circ$ around that point and marking the intersection point with the third side, we always get an inscribed equilateral triangle.

If you mean what is the maximum ratio $\frac{[ABC]}{[DEF]}$, the answer is given by a rescaled Napoleon triangle. We may notice that such ratio is also the ratio between the areas of the largest circumscribed equilateral triangle and the original triangle. On the ther hand, it is quite simple to describe the set of the circumscribed equilateral triangles: let we consider the Napoleon/Torricelli/Fermat configuration:

We have $O_B O_C\perp AV_A$ and so on. If we take a point $P$ on $\Gamma_A$ (the circumcircle of $BCV_A$), draw a line through $B$ till meeting $\Gamma_C$ at a new point $Q$, then draw a line through $A$ till meeting $\Gamma_B$ at a third point, the (grey) triangle defined this way is equilateral. It follows that we simply need the greatest $PQ$ segment among the segments built through the previous procedure. This segment has to be perpendicular to the radical axis of $\Gamma_A$ and $\Gamma_C$, so the Napoleon triangle of $ABC$ is the smallest inscribed equilateral triangle in $A'B'C'$. By rescaling, we get the solution. It is interesting to point out that the side length of $O_A O_B O_C$ is $\frac{1}{\sqrt{3}}AV_A$, and $AV_A=BV_B=CV_C$ is the length of the Steiner net of $ABC$, i.e. $FA+FB+FC$, with $F\in\Gamma_A\cap\Gamma_B\cap\Gamma_C$ being the Fermat point of $ABC$.

• Sorry, I meant the former. I hope that the revised wording of the question is definitive now. Jan 19, 2017 at 7:13
• The counterexample you have given looks to be in the case of an isosceles triangle, the equilateral inscribed triangles being symmetrical with respect to the axis of symmetry of the triangle, thus with the same area. Can such a phenomena arise in the case of non isosceles triangle ? Jan 19, 2017 at 8:03
• The counterexample is general, the depicted triangle is not isosceles. And: are you sure you want the largest inscribed equilateral triangle? To find the smallest is way more interesting. Jan 19, 2017 at 10:52
• Have you had a look at my "solution" in terms of fixed point(s) of a certain geometrical function (that should be expressed, I think, as a certain homography, which could explain that there are always two solutions) ? Jan 19, 2017 at 11:16
• I have fully understood now where was my misunderstanding. A very small remark in your very neat presentation. It is about your first paragraph, more precisely when you say "by choosing a point on a side," it should be added that the chosen point must be on a side such that the opposite angle is $\leq \pi/3$, otherwise it is not always possible. Jan 20, 2017 at 14:32

This is intended as a different view, but doesn't constitute a complete answer.

We are going to work on the following figure, which has a classical aspect (first part of Miquel's theorem, see (https://en.wikipedia.org/wiki/Miquel%27s_theorem))

Instead of having a fixed triangle with given angles $$\alpha, \beta, \gamma$$, and equilateral triangles inscribed in it, let us completely reverse the problem into: being given a fixed equilateral triangle $$UVW$$ with unit side, consider a circumscribed triangle $$ABC$$ with these angles.

This makes sense as the issue is size and rotation invariant.

Being given these angles $$\alpha, \beta, \gamma$$, where can $$A,B,C$$ be situated?

Evidently, $$A,B,C$$ are restricted to be on circles (or more precisely their external arcs) from which one can see resp. line segments $$VW$$,$$WU$$ and $$UV$$ under angles $$\alpha$$, $$\beta$$ and $$\gamma$$ resp..

Taking an arbitrary point $$A$$ on the $$\alpha$$ circle; line AW intersects circle $$\beta$$ in $$B$$; then, line $$BU$$ intersects circle $$\gamma$$ in $$C$$; at last, points $$C,V,A$$ are aligned, as can be established by angle chasing in triangles $$AWV$$, $$BUW$$ and $$CVU$$. This is our manner to prove the existence of an infinite number of ways to inscribe an equilateral triangle in a given triangle.

Remark: We have also a global view of the areas ratio $$R=[UVW]/[ABC]$$ whose maximum is looked for (this ratio has not been modified by the transformation); it is equivalent to look for the

$$\text{Minimization of } \ \ 1/R=[ABC]/[UVW]=1+\frac12(h_A+h_B+h_C)$$

where $$h_A,h_B,h_C$$ are the altitudes issued from $$A,B,C$$ onto $$VW$$, $$WU$$, $$UV$$, resp.

As the figure of this text is very similar to the figure given by @Jack d'Aurizio (though obtained in a different spirit), I do not complete (at least for the moment) my proof because it would paraphrase his proof. I am indebted to Jack d'Aurizio because a first draft of this text was on bad tracks : I hadn't perceived that there is a - continuous - infinity of inscribed equilateral triangles in a given triangle.)

See as well Johnson's triangles.