Given a triangle $ABC$, whose (one of the) longest side is $AC$, consider the two circles with centers in $A$ and $C$ passing by $B$.

(The part in italic is edited after clever observations pointed out buy some users: see below for details).

EDIT: You may be interested also in this other question Another conjecture about a circle intrinsically bound to any triangle.

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The two circles determine two points $D$ end $E$, where they intersect the side $AC$.

We draw two additional circles: one with center in $A$ and passing by $D$, and the other one with center in $C$ and passing by $E$.

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The new circles determines two points $F$ and $G$ where they intersect the sides $AB$ and $BC$, respectively.

My conjecture is that the points $BGEDF$ always determine a circle, whose center coincides with the incenter of the triangle.

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Is there an elementary proof for such conjecture?

Since I am not an expert in the field, this can be a very well known theorem. I apologize in that case. Thanks for your help.

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    $\begingroup$ Geogebra confirms that this should be true. $\endgroup$ – Arnaud Mortier Jul 4 '18 at 9:54
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    $\begingroup$ I think that you messed up a bit the inversion of the conjecture: the circle is not centered in $B$, it goes through it. But then if you replace centered by going through, it is clearly not true as such. You probably have to assume at least that $\overline{BF}=\overline{BG}$. $\endgroup$ – Arnaud Mortier Jul 4 '18 at 10:02
  • $\begingroup$ Sure, it is a mistake, sorry! I edit it now! Thanks! $\endgroup$ – user559615 Jul 4 '18 at 10:03
  • $\begingroup$ No worries. But again, as such, the inverted conjecture clearly doesn't hold. $\endgroup$ – Arnaud Mortier Jul 4 '18 at 10:04
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    $\begingroup$ I would say that the incenter of the triangle $ABC$ corresponds to the center of the circumference $BGDEF$ $\endgroup$ – user559615 Jul 4 '18 at 10:34

We have $AF=AD$ and $AB=AE$, so the triangles $AFD$ and $ABE$ are isosceles, so $FD\|EB$ and $BEDF$ is isosceles, thus inscriptible.

This shows $F$ is on the circle through $B,D,E$.

By analogy/symmetry, $G$ is also on it.

  • $\begingroup$ Wonderful! Brilliant and simple. Thanks, Dan! $\endgroup$ – user559615 Jul 4 '18 at 12:02
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    $\begingroup$ Your argument assumes that the point $E$ exists, but that is not guaranteed in general; take angle $C$ to be sufficiently obtuse, for example. $\endgroup$ – Allawonder Jul 4 '18 at 12:41
  • $\begingroup$ OP has found a way around this problem, so that your proof stands if we assume that $A$ and $C$ are acute (and this can always be done since every triangle has at least two acute angles). PS. A next interesting question is whether this circle is unique -- especially when triangle $ABC$ is not obtuse. $\endgroup$ – Allawonder Jul 4 '18 at 16:34

There can be no proof of the conjecture since it is false, for if the $\triangle ABC$ is obtuse, then one cannot guarantee that the circles will intersect the third side $AC$ at $D$ and $E.$

Unfortunately, one cannot bystep this by considering the line through $AC$ instead.

Edit: OP has found a way around this; he need only state as hypothesis that $A$ and $C$ be the acute angles of the triangle.

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    $\begingroup$ I see your point. However, in that case, one can still apply the same reasoning starting from another vertex. I should add that the whole process should applied to (one of) the vertex opposite to the longest sides of any triangle $ABC$? $\endgroup$ – user559615 Jul 4 '18 at 14:08
  • $\begingroup$ Or it should restricted to any triangle contained in the equilateral triangle with side $AC$. $\endgroup$ – user559615 Jul 4 '18 at 14:13
  • $\begingroup$ @andrea.prunotto Your first observation is true. Then you need only stipulate that the angles at $A$ and $C$ be the acute angles of the triangle $ABC.$ $\endgroup$ – Allawonder Jul 4 '18 at 16:24
  • $\begingroup$ @andrea.prunotto Another interesting question would then be about the uniqueness of this pentapunctual circle; obviously it is unique for obtuse triangles; it is not immediately clear however for right and acute triangles. $\endgroup$ – Allawonder Jul 4 '18 at 16:31

By construction $ADF$, $CEG$ and $BFG$ are isoceles so that the bisectors of $DF$, $EG$ and $FG$ are also bisectors of the angles of the triangle and meet at the incenter.

By symmetries, $IB=ID=IE=IF=IG$.

  • $\begingroup$ This proves that $D,E,F,G$ lie on a circle. Why does $B$ lie on this circle as well? $\endgroup$ – timon92 Jul 6 '18 at 8:25
  • $\begingroup$ @timon92: $BIF$ and $BIG$ are isoceles. $\endgroup$ – Yves Daoust Jul 6 '18 at 8:26
  • $\begingroup$ Cool! Now the first part of the solution is redundant. I would consider to write the chain of equalities in a different order so it is clear that it follows by symmetries: $IF = ID = IB = IE = IG$. $\endgroup$ – timon92 Jul 6 '18 at 8:32
  • $\begingroup$ @timon92: yep but need to show that $I$ exists (as a unique point) and is the incenter. $\endgroup$ – Yves Daoust Jul 6 '18 at 8:41
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    $\begingroup$ What I meant was that there's no need to say that $ADF$ and the other triangles are isosceles so the bisectors are [blah blah blah]. You can rephrase the solution as follows: Let $I$ be the incenter of $ABC$. Then by symmetries we get $IF=ID=IB=IE=IG$ so the points $F,D,B,E,G$ lie on a circle centered at $I$ with radius $IF$. $\endgroup$ – timon92 Jul 6 '18 at 10:20

I am writing this to add to both the conjecture and the proof given.

If the angle at $B$ is a greatest angle of the triangle $ABC,$ then the conjecture is true (provided we allow the points of the cyclic "pentagon" to coincide). Moreover, this pentapunctual circle is unique. This is clear when $ABC$ is scalene, for if one used $A$ or $C$ instead of $B,$ the largest angle, then necessarily the points $D$ and $E$ cannot exist since $AC$ is the longest side, thus it exceeds both $AB$ and $BC.$ If the triangle is isosceles (or even equilateral), then since $\hat B$ is equal to at least one other angle, the symmetry imposes that only one such circle exists.

I notice that you have added something about the identity of this "$5$-point" circle. Again, it is indeed the case that its centre coincides with the incentre $I$ of the $\triangle ABC,$ for since $DF$ and $EG$ are chords of this circle, their perpendicular bisectors must intersect at its centre $I'.$ But we also know that the triangles $ADF$ and $CEG$ are isosceles with $AD=AF$ and $CE=CG$ respectively. Therefore the perpendicular bisectors of the sides $DF$ and $EG$ must also be angle bisectors of $\hat A$ and $\hat C$ respectively. This shows that their intersection $I'$ is not different from $I.$

PS. This is not directly related, but let me point out an interesting relationship between the incircle and the circumcircle of any triangle $ABC$ which I had not learnt of before. Let the intersection of the incircle with $AB,BC,CA$ be $C',A',B'$ respectively. Then the lines $AA',BB',CC'$ intersect in a centre of the triangle $Q$ which I have called the quasicentroid. Of course, there are thousands of known triangle centres (cf. The Encyclopedia of Triangle Centres), but I have been unable to determine whether $Q$ is part of the categorised centres, and if so under what name, hence the tentative term "quasicentroid."

Edit: The point I called $Q,$ I subsequently found out, is more well known as the gergonne point of the triangle.


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