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ABC is an equilateral triangle,and AD = BE = CF,Prove DEF is an equilateral triangle.

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Has anyone considered either of these approaches? $$1. \text{Drawing "auxiliary" lines}$$ $$2. \text{Assuming that DEF is not equilateral and reaching a contradiction}$$ –  The Chaz 2.0 Apr 18 '12 at 16:24
    
approach 2 is OK , But how to draw the auxiliary lines?I have no ideas. –  tan9p Apr 21 '12 at 12:48
    
I have racked my brain for a solution that doesn't involve circles, with no success. Good luck to you! –  The Chaz 2.0 Apr 21 '12 at 14:02
    
OK,thanks a lot. –  tan9p Apr 23 '12 at 10:38
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3 Answers

up vote 3 down vote accepted

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?

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Yeah, proofing by contradiction is good choice.But is there any method of direct proof ? By transforming? –  tan9p Apr 20 '12 at 7:14
    
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. –  Matt Apr 20 '12 at 20:48
    
clearly explanation,Thank you. –  tan9p Apr 21 '12 at 12:45
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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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Here is an attempt I have made. It is far from rigorous and may contain fatal flaws. Nevertheless,

Let $ABC$ be an equilateral triangle, with centre $O$. Draw lines $AO, BO$ and $CO$ and extend them all by a similar amount to points $A', B', C'$. Draw lines $AC', BA'$ and $CB'$, these have the same length since $AOC', BOA'$ and $COB'$ are congruent triangles.

Extend these lines to meet each other, and call their points of intersection $E,F$ and $G$. I.e extend $AC'$ to $AE, BA'$ to $BF$ and $CB'$ to $CG$. It is clear that $EC'=FA'=GB'$ because if we were to extend these lines to the edge of the triangle, this distance extended would be the same in each case, and since $AC'=BA'=CB'$ it follows that $EC'=FA'=GB'$.

Finally, since $A'AE, B'BF$ and $C'CG$ are similar triangles, clearly $C'G=A'E=B'F$ and hence $EF=FG=GE$.

Alternatively, the following image is invariant under rotation by 60 degrees:

enter image description here

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You're not allowed to pick the points $EFG$ that you want, though you've shown there is some family of such triangles that has the desired property. –  Thomas Belulovich Apr 18 '12 at 6:27
    
I agree with Thomas. You have to show that whenever AD = BE = CF, then DEF is equilateral. –  TonyK Apr 18 '12 at 21:59
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