Consider triangle ABC. Point D is on AC, E is on BC and F is on AB. Given that triangle DEF is equilateral and that segments CD, BE and AF are equal in length, prove that triangle ABC must also be equilateral. I've played with this one for long enough using circles, sine law, cosine law, etc., and have almost reached the point of capitulation. Any hints would be appreciated. I'm not actually sure if the statement is true, but I haven't been able to find a counterexample, either, so I'm stuck both ways.


2 Answers 2


Let $a$, $b$, $c$ be the lengths of the sides opposite $A$, $B$, $C$, and let points $D$, $E$, $F$ lie on the sides opposite $A$, $B$, $C$. (The latter is slightly different than in the original question). Let $r = |DE| = |EF| = |FD|$ be the length of the sides of the equilateral triangle; let $s = |AF| = |BD| = |CE|$ be the lengths of the congruent sub-segments of the sides.

The Law of Cosines for each $r$-side in $\triangle AEF$, $\triangle BFD$, $\triangle CDE$ gives

$$\begin{align} r^2 = s^2 + (b-s)^2 - 2 s (b-s) \cos A \\ r^2 = s^2 + (c-s)^2 - 2 s (c-s) \cos B \\ r^2 = s^2 + (a-s)^2 - 2 s (a-s) \cos C \end{align}$$

Setting the right-hand sides equal and subtracting $2s^2$ yields: $$b^2-2bs-2s(b-s)\cos A = c^2-2cs-2s(c-s)\cos B = a^2 - 2 a s - 2 s(a-s) \cos C$$

Using the Law of Cosines (in $\triangle ABC$) to re-write each of $\cos A$, $\cos B$, $\cos C$, and then simplifying a bit, gives

$$\begin{align} &\phantom{=}\;a\left( s^2\left(-a^2+b^2+c^2\right)+ s b\left(a^2-\left(b+c\right)^2\right)+b^3 c\right) \\ &= b\left( s^2\left(\phantom{-}a^2-b^2+c^2\right)+ s c\left(b^2-\left(c+a\right)^2\right)+c^3 a\right) \\ &= c\left( s^2\left(\phantom{-}a^2+b^2-c^2\right)+ s a\left(c^2-\left(a+b\right)^2\right)+a^3 b\right) \end{align}$$

We can break the triple equality into a system of two quadratic equations in the parameter $s$. Eliminating $s$ gives a polynomial equation, $p(a,b,c)=0$; the factors of $p$ provide these equations: $$\begin{align} abc (a+b+c) &= 0 &(1) \\ a^2 b\left(a-b\right)+b^2c\left(b-c\right)+c^2 a\left(c-a\right) &= 0 & (2) \\ a b\left(a-b\right)\left(a\left(b-c\right)\left(a^2-b^2\right)-b^2c^2\right) \\ +b c\left(b-c\right)\left(b\left(c-a\right)\left(b^2-c^2\right)-c^2a^2\right) \\ +c a\left(c-a\right)\left(c\left(a-b\right)\left(c^2-a^2\right)-a^2b^2\right) &= 0 & (3) \end{align}$$

Solutions in each case correspond to ostensible solutions to the problem.

We can ignore (1), which has only trivial solutions. I believe (2)'s only non-trivial solutions (with the Triangle Inequality in play) require $a=b=c$, but I need to double-check this.

As for (3), clearly $a=b=c$ works; and, indeed, if (say) $a=b$, then $a=b=c$. So any non-equilateral solutions would need to be strictly scalene; it turns out that (3) admits scalene solutions.

Below is Mathematica's plot of (3), with $c=1$; the horizontal axis is $a$, and the vertical is $b$. The lines are $a+b=c$ and $b+c=a$ and $c+a=b$, which bound the region of feasibility according to the Triangle Inequality. The point $(1,1)$ seems to have been missed by the implicit plotter, but lots of other points appear in blue. (I believe there are even more. Zooms show an island near the $a+b=c$ border, for instance.)

enter image description here

Here's the figure for $a=8$, $b=8.58667$, $c=1$:

enter image description here

Equilateral $\triangle DEF$ extends beyond the confines of $\triangle ABC$. This may be why @Brian was having difficulty ruling-out non-equilateral $\triangle ABC$s.

I suspect equation (3) can be expressed in a way that's more illuminating; it's not clear to me, for instance, whether every solution to (3) corresponds to an "external" triangle. There's probably a whole other approach to this problem that makes everything obvious, but this is the best I have at the moment.

  • $\begingroup$ This looks promising. I agree that with the Triangle Inequality in play that the only applicable solution to (2) is a = b = c. Judging from your graph of (3) it would seem that every solution other than a = b = c corresponds to an external triangle, but I suppose that would have to be proved. Thanks so much. $\endgroup$
    – Brian
    Aug 25, 2012 at 17:04
  • $\begingroup$ @Brian: It's dangerous to judge much of anything from that particular graph. Equation (3) is, after all, a degree-7 polynomial, so its general behavior isn't nearly as tidy or predictable as, say, a quadratic. Mathematica's implicit plotter may have missed a great deal more of the structure than the point $(1,1)$ and the oval island of solutions I happened to notice near the line $a+b=c$. Caveat graphor. $\endgroup$
    – Blue
    Aug 25, 2012 at 17:23
  • $\begingroup$ Yes, of course, caveat graphor. At least I have a new phrase to add to my math vocabulary. :) I did notice one typo near the beginning; the third of the cosine equations has a "c" where there should be an "a". It didn't continue, but I just thought I'd mention it. $\endgroup$
    – Brian
    Aug 25, 2012 at 17:46
  • $\begingroup$ Blue, OP states: Point D is on AC, E is on BC and F is on AB. Why doesn't $\Delta$ABC enclose $\Delta$DEF? $\endgroup$
    – Fred Kline
    Aug 25, 2012 at 17:56
  • $\begingroup$ @FredDanielKline: What I've shown is that vertices of an equilateral $\triangle DEF$ could lie on extended sides of non-equilateral $\triangle ABC$. This wasn't something I'd really anticipated when I started looking at the problem; it happened to fall out of my analysis. (I love when that happens!) So, I've actually addressed a neat extension of OP's question. I'm still curious (though dubious) about possible scalene $\triangle ABC$s with "enclosed" equilateral $\triangle DEF$s, but I need a bit of a breather before continuing analysis of equations (2) and especially (3). $\endgroup$
    – Blue
    Aug 25, 2012 at 18:29

Draw triangle DEF. Draw triangle ABC over it, with A=D, B=E, and C=F. The triangles are identical and have the same center point. Rotate triangle DEF counter-clockwise about the center point. Because the points D, E and F take a circular route, we need to shrink the triangle DEF to keep its points on the sides of triangle ABC.

Rotation and shrinking do not change the shape of triangle DEF and we did not alter triangle ABC, which remains equilateral.

In addition to rotation and dilation, we need to shift the center point of the inner triangle by half its original distance from center to vertex.

The two triangles are congruent at the start and then the inner triangle becomes smaller but remains similar (the angles don't change.)

Other triangles don't work Trying this by crafting a right triangle ABC and putting a small offset from each vertex for the points D, E, and F, then connecting the dots, we will find that triangle DEF is no longer a right triangle.

Example Example

The three outer triangles are congruent by Side-Angle-Side theorem. This means that the three sides of the inner triangle are also equal, thus making the inner triangle an equilateral triangle.

Animated Example

  • 2
    $\begingroup$ How do you know that $DEF$ is obtained from $ABC$ by rotation and shrinking? That seems to be begging the question. $\endgroup$
    – joriki
    Aug 24, 2012 at 21:14
  • $\begingroup$ @joriki When visualizing this, I found it easier to fix $DEF$, rotate $ABC$ and expand that to $A'B'C'$ so that the vertices of $DEF$ lie on the correct edges of $A'B'C'$. The fact that $A'B'C'$ is still an equilateral triangle follows from looking at similar triangles. Nice intuition, Fred. Too bad I can only upvote once. $\endgroup$ Aug 24, 2012 at 21:47
  • $\begingroup$ Fred Daniel Kline. The image of a rotating triangle is intuitively helpful, although I find it easier to visualize it the way Rick Decker mentions. However, I tend to agree with @joriki that this approach assumes ABC is equilateral from the start and then is rotated and shrunk/expanded. The only information my question provides is that the segments CD, BE and AF are equal in length; if it can be concluded from this that DEF and ABC share a center then your observations would be applicable and the proof would be complete. Or have I misinterpreted? Thank you for your interest in my question. $\endgroup$
    – Brian
    Aug 24, 2012 at 23:52
  • $\begingroup$ @Brian, we can rotate one of the triangles 120 degrees, which causes the segments CD, BE and AF to range in size from zero to the length of a side of ABC. $\endgroup$
    – Fred Kline
    Aug 25, 2012 at 1:57
  • $\begingroup$ Moebius Transformations shows the stuff we can do in this area. $\endgroup$
    – Fred Kline
    Aug 25, 2012 at 2:11

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