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$\Delta ABC$ is an equilateral triangle and $AD = BE = CF$. Prove that $\Delta DEF$ is an equilateral triangle.

  • $\begingroup$ Has anyone considered either of these approaches? $$1. \text{Drawing "auxiliary" lines}$$ $$2. \text{Assuming that DEF is not equilateral and reaching a contradiction}$$ $\endgroup$ Apr 18, 2012 at 16:24
  • $\begingroup$ approach 2 is OK , But how to draw the auxiliary lines?I have no ideas. $\endgroup$
    – yibotg
    Apr 21, 2012 at 12:48
  • $\begingroup$ I have racked my brain for a solution that doesn't involve circles, with no success. Good luck to you! $\endgroup$ Apr 21, 2012 at 14:02
  • $\begingroup$ OK,thanks a lot. $\endgroup$
    – yibotg
    Apr 23, 2012 at 10:38

2 Answers 2


Draw circles centered at A, B, and C, with radius AD=BE=CF.

Triangle $\triangle$DEF must lie in the region between these 3 circles, touching each circle with a vertex.

Draw the rotationally symmetric solution as $\triangle$D'E'F'. (Side exercise: Show it must exist.)

Now suppose $\triangle$DEF $\ne$ $\triangle$D'E'F'.

Up to here, everything I've said is very straightforward, but you might want to stop for a second and see if you can solve it from here. If you can't, one more hint is hidden in the gray box below.

DE is either outside $\triangle$D'E'F', or crosses it. Either way, can you say something similar about DF? About EF?

  • $\begingroup$ Yeah, proofing by contradiction is good choice.But is there any method of direct proof ? By transforming? $\endgroup$
    – yibotg
    Apr 20, 2012 at 7:14
  • $\begingroup$ It is unlikely for a transformation to help, since the key information AD=BE=CF is destroyed by most types of transformations, and those that preserve it leave the problem unchanged. $\endgroup$
    – Matt
    Apr 20, 2012 at 20:48
  • $\begingroup$ clearly explanation,Thank you. $\endgroup$
    – yibotg
    Apr 21, 2012 at 12:45

Suppose we fix $\ell = AD=BE=CF$, and let $x=\angle{BAD}$, $y=\angle{CBE}$, $z=\angle{ACF}$.

Then we can define $f : [0,\pi/6]\to[0,\pi/6]$ such that $y = f(x)$, $z = f(y)$, and $x = f(z)$. Note that $f$ is strictly decreasing: one good way to see this is by drawing the circles centered at $A,B,C$ with radius $\ell$. But then $f^3(x) = x$ forces $f(x) = x$ (and thus $x=y=z$); otherwise, if WLOG $x<f(x)$, then $$f(x)>f^2(x)\implies f^2(x)<f^3(x)=x\implies x=f^3(x)>f(x),$$ contradicting $x<f(x)$.

Note that this solution assumes that $x,y,z<\pi/6$, but the problem still holds as long as $x,y,z<\pi/3$. In particular, I think you can show that if one of $x,y,z$ is larger than $\pi/6$ (which requires $\ell>1/2$), the other two must be as well, so $f$ is still well-defined. Anyway, all of this is equivalent to the rotation method Matt mentioned in his answer, but I thought it would be good to include this more algebraic perspective.

fedja also has a long post about this kind of problem on AoPS.


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