# A conjecture related to any triangle

Given one side $$AC$$ of any triangle $$\triangle ABC$$, we can draw the couple of circles with center in $$A$$ and passing through $$C$$ and with center in $$C$$ and passing through $$A$$, obtaining two points $$D,E$$ corresponding to the intersections of these two circles.

The same can be done with the other two sides, obtaining other $$2$$ couples of circles and other $$4$$ points $$F,G,H,I$$.

My conjecture is that

The sum of the areas of the triangles $$\triangle IDG$$ and $$\triangle FHE$$ is equal to the area of the triangle $$\triangle ABC$$.

(Notice that $$I,D,G$$ and $$F,H,E$$ are the intersection points of three distinct couples of circles, underlined by the different colors. I suspect that similar conjectures can be formulated starting from different choice of these points. E.g., if we consider the triangles $$\triangle EHG$$ and $$\triangle FID$$, the sum of the area of the first one and the area of the original triangle $$\triangle ABC$$ seems to give the area of the second one, as illustrated below).

This is probably a very obvious result, and I apologize in this case. However, I would need some hint to provide a compact proof of such conjecture. And I therefore thank you for any suggestion and comment!

• Your posts are always the best! – CaptainAmerica16 Nov 2 '18 at 18:01
• @CaptainAmerica16 Thank you! But, is all due to Geogebra! – user559615 Nov 2 '18 at 18:04
• Your questions are always so interesting - you should consider submitting some of them to math contests! – Carl Schildkraut Nov 2 '18 at 18:08
• Only suggestion I have for now is to consider a simpler restricted case maybe unit equilateral triangle to gain a foothold. – Karl Nov 2 '18 at 18:21
• I'd try to write the involved areas through vectors and exploit the fact that $D+E=A+C$ ($DE$ and $AC$ have the same midpoint) and $DE\perp AC$ ($\langle D-E,A-C\rangle=0$). – Jack D'Aurizio Nov 2 '18 at 18:29

This is a relative of a property of Napoleon Triangles, and/or what we might call "Petr-Douglas-Neuman" triangles: triangles determined by the apex vertices of isosceles triangles erected on the sides of a given triangle.

Let's coordinatize in the $$xy$$-plane of $$\mathbb{R}^3$$ (the third dimension will give us easy access to "signed area"), with $$A = (0,0,0) \qquad B = (c,0,0) \qquad C = (b \cos A, b \sin A,0)$$

Note that one traces $$A\to B\to C\to A$$ in a counterclockwise fashion.

Define $$(x,y,0)^\perp := (-y,x,0)$$, which performs a $$90^\circ$$ counterclockwise rotation of the vector $$(x,y,0)$$ about the origin. With this, we can define \begin{align} D_{\pm}&:=\frac12\left(\;(B+C)\mp t\;(C-B)^\perp\;\right) \\[4pt] E_{\pm}&:=\frac12\left(\;(C+A)\mp t\;(A-C)^\perp\;\right) \\[4pt] F_{\pm}&:=\frac12\left(\;(A+B)\mp t\;(B-A)^\perp\;\right) \end{align} where $$t := \tan\theta$$ for $$0<\theta<90^\circ$$ the base angle of an isosceles triangle. Sign choices guarantee that $$D_{+}$$ is the apex of such an isosceles triangle erected upon $$\overline{CA}$$ outside of $$\triangle ABC$$, while $$D_{-}$$ is the apex of the isosceles triangle that overlaps $$\triangle ABC$$. (So, for Napoleon Triangles, $$\theta = 30^\circ$$, while for the triangles in this question, $$\theta = 60^\circ$$.)

We choose $$+$$ to be outside the triangle, so that $$\triangle D_{+} E_{+} F_{+}$$ is traced in the same orientation as $$\triangle ABC$$. (Note that $$D_{-}E_{-}F_{-}$$ may-or-may-not have the opposite orientation; it depends upon whether $$\theta$$ is big enough to cause the apices of the isosceles triangles to move past each other.)

Now, we calculate the signed area of $$\triangle XYZ$$ via $$|\triangle XYZ| := \left(\;(Y-X)\times(Z-X)\;\right)\cdot\left(0,0,\frac12\right)$$ Effectively, we take half the third coordinate of the cross product; the other two coordinates are $$0$$, anyway. (That's how the third dimension comes in handy!) With this definition, we have, for signs $$d$$, $$e$$, $$f$$, \begin{align} |\triangle ABC| &= \frac12b c \sin A \\[6pt] |\triangle D_d E_e F_f| &= \frac14 |\triangle ABC|\left(1+t^2(de+ef+fd)\right) + \frac18 \left(a^2 d + b^2 e + c^2 f\right) \\[6pt] |\triangle D_{-d} E_{-e} F_{-f}| &= \frac14 |\triangle ABC|\left(1+t^2(de+ef+fd)\right) - \frac18 \left(a^2 d + b^2 e + c^2 f\right) \end{align} Thus,

$$|\triangle D_{d}E_{e}F_{f}| + |\triangle D_{-d}E_{-e}F_{-f}| = \frac12 |\triangle ABC| \left(\; 1 +t^2 (de+ef+fd) \;\right) \tag{\star}$$

For $$\theta = 30^\circ$$ (Napoleon), $$t^2=1/3$$, and $$(\star)$$ becomes $$\frac16|\triangle ABC|\left(\;3+de+ef+fd\;\right) = |\triangle ABC| \qquad\text{when d, e, f match }$$ And, for $$\theta = 60^\circ$$ (OP), $$t^2 = 3$$, so $$\frac12|\triangle ABC|\left(\;1+3(de+ef+fd)\;\right) = -|\triangle ABC|\qquad\text{when exactly two of d, e, f match}$$

As a side note, this recent question asks about the concurrency of segments through the apex vertices of the isosceles triangles in general.

• Thank you, wonderful! I would have never guessed that this was related to Napoleon triangles. – user559615 Nov 3 '18 at 4:06
• I found a similar conjecture, but using, for each side, the couple of ellipses with foci in $A,C$ and passing through $D,E$ and with foci in $D,E$ and passing through $A,C$. In this case, for each side we obtain $4$ points (where the couple of ellipses intersects), which will lead to couples of triangles to sum/subtract to obtain the area of the starting triangle. Do you think that your method can also be applied in that case? In your view, is it worth to post such problem? – user559615 Nov 3 '18 at 6:58
• @AndreaPrunotto: You can, of course, post whatever you like. My methodology can be adapted to the idea of your double-ellipse construction, but all the sign options become just that much more complicated to track. It would be very helpful to have a specific goal. Have you verified exactly which possible triangle pairs actually give add/subtract to get the area (and/or the negative of the area) of the starting triangle? And/or, do you need actually need four triangles so that all the constructed vertices get used? Etc, etc, etc. The more information, the better. – Blue Nov 3 '18 at 8:40
• Thanks for your comment. Sure, I see your point. It takes a while to check all the relations among these triangles. But I think I understand that the solution to the problem with the ellipses would be an extension of your proof. I try to make a sketch and maybe I edit the post with such observation. Thanks again for your work, as elegant as always! – user559615 Nov 3 '18 at 9:35

I guess you refer to oriented areas for the triangles.

I have built a geogebra model and moving B I see some cases that can match your conjecture only using this definition of area. So this is a case similar to your first one. If I move point B to reach segment AC, the triangle ABC degenerates to a segment (Area of ABC equals 0) but the other triangles do not. Though, if B is any point of the straight line AB, both triangles IDG and EFH are symmetrical, so if we consider an oriented area, one has positive area, the other one has the same negative area and only in this case the sum is correct.

If I understand well this is a counter example to your conjecture in the original form.

• I believe the problem given was of the form "if $x$, $y$, and $z$ are the areas of the triangles, then one is the sum of the other two" -- i.e. it allowed for signed areas. Also, nowhere in the question is it noted which of the intersection points to call $D$ and which to call $E$, for example. – Carl Schildkraut Nov 2 '18 at 20:59
• The suggested construction consists in 6 additional points besides ABC so we have 6! / 3! possible combinations in general. Much less if we suppose that each point must be the intersection of circles of different colors. – Francesco Iovine Nov 2 '18 at 21:09
• I do think it is an interesting observation. But I have to think about it. Thanks for your efforts! – user559615 Nov 2 '18 at 21:11