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What triangles can be cut into three triangles with equal radii of the circumscribed circles around these triangles?

My work so far:

Case 1) let $ABC -$ an acute-angled triangle. Then radii of the triangles $ABH, BCH$ and $ACH$ are equal, where $H -$ orthocenter.

Case 2) Let $\angle C \ge 90^{\circ}$ and $AC>BC$.

$AD=BD, \angle AED= \angle C$. Sine rule radii of the triangles $ADE, BDE$ and $BDC$ are equal.

enter image description here

Case 3) Let $\angle C \ge 90^{\circ}$ and $AC=BC$.

I need help here.

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  • $\begingroup$ Try drawing these shapes on paper. You might be able to see the answer. $\endgroup$ – Jeffrey Young Apr 5 '16 at 8:07
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This is just a rough sketch without formal proof.

enter image description here

S is the axis of symmetry of $\triangle ABC$.

L (through B) and R (through C) are the perpendiculars to the base BC.

X and Y are the perpendicular bisectors of the legs.

The intersection of X and L forms the center of green circle passing A and B (and also P, see later). The blue circle is formed similarly. The two circles will intersect at P.

Z is the perpendicular bisector of PC. The intersection of Z and S forms the center of the black circle passing through B, C, and P.

These circles (including ABC) happen to be having the same circum-radius.

The picture is drawn with $\angle BAC$ acute. However, if we let A slide along S until it hits P, then $\angle BAC$ becomes obtuse. The same conclusion still true.


The following is an additional info from my finding:-

enter image description here

The triangles ($ABC$) in the three figures are identical. The only possible way that can meet the given requirement is to find points $B’$ and $C’$ on $BC$ such that $r_{ABB’} = R_{ABC}$. In order to have $r_{ABB’} = r_{ACC’}$, $B’$ and $C’$ must be symmetrically placed about the axis of symmetry through $A$.

The actual constructions of the perpendicular bisectors are clearly shown in Fig. 1.

In Fig. 3, if $B’$ is far away from $B$, we can clearly see that $R \lt r$.

If $B’$ is moved a bit closer to $B$ (as shown in Fig. 1), we still have $R \lt r$.

Fig. 2 shows that, even when $B’$ is very close to $B$, we still have $R \lt r$.

By symmetry, in order to have $R = r$, $B’$ must be coincide with $B$. In other words, if $\angle BAC$ is obtuse, there is no way that we can find two points on $BC$ such that the three triangles thus formed can have the same circum-radius.

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  • $\begingroup$ In your picture, $P$ is indeed the orthocenter of $\triangle ABC$. When $A$ moves, so does $P$. Thus your last statement doesn't apply. $\endgroup$ – Quang Hoang Apr 6 '16 at 3:32
  • $\begingroup$ @QuangHoang Agree that when A hits P, it has reached its critical point. At that instant, $\angle BAC = 90^0$. Going beyond that, $\angle BAC$ will be obtuse, but P has to be outside of $\triangle ABC$. Will leave my post on for the reference of the acute case. $\endgroup$ – Mick Apr 6 '16 at 5:21
  • $\begingroup$ @Roman83 Additional info has been added to my post. See above. $\endgroup$ – Mick Apr 10 '16 at 10:36

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