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We consider a triangle $ABC$ whose angles are less then $120°$ and construct the equilateral triangles $ABC'$, $BCA'$ and $ACB'$, exterior to $ABC$. $I$ denotes the intersection of $(AA')$ and $(CC')$.

1) Show that $AA'=BB'=CC'$.
2) Show that $\widehat{BIC}=\widehat{BIA}=120°$.
3) Show that the lines $(AA')$, $(BB')$ and $(CC')$ intersect.

I have no idea how to attack this exercise. For point 3), it is clear that $I$ must be the point of intersection, but I do not know thow to prove any of these three steps.

Can anyone suggest how to proceed! Thank you in advance.

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    $\begingroup$ en.wikipedia.org/wiki/Fermat_point $\endgroup$ – Jack D'Aurizio Jan 18 '15 at 22:19
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    $\begingroup$ Start from proving triangles BAB` and CAC` are congruent. $\endgroup$ – Mick Jan 19 '15 at 4:40
  • $\begingroup$ I was able to show 1) and 2). But I do not see why this should imply 3) Any suggestions? $\endgroup$ – Mathoman Jan 19 '15 at 23:52
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Could you please write about you prove (1) and (2) without using the conclusion of (3)?

Here's my solution:

First, just connect $CC'$ but not $AA'$ $BB'$, draw the circumcircle of $ABC'$, and let the circle intersect $CC'$ at $I$. It's easy to see that angle $BIC'=60^o$. So $A',B,C,I$ are on the same circle and $B',A,I,C$ are on the same circle. Because angle $AIC=120^o$, angle $A'IC=60^o, AIA'$ are collinear. So do $BIB'$.

So we prove (2) and (3) are right and as Jack D'Aurizio pointed out, you can read Fermat point - Wikipedia to prove (1) is right.

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