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

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

I need help here.

• Try drawing these shapes on paper. You might be able to see the answer. Apr 5, 2016 at 8:07

This is just a rough sketch without formal proof. 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:- 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.

• In your picture, $P$ is indeed the orthocenter of $\triangle ABC$. When $A$ moves, so does $P$. Thus your last statement doesn't apply. Apr 6, 2016 at 3:32
• @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.
– Mick
Apr 6, 2016 at 5:21
• @Roman83 Additional info has been added to my post. See above.
– Mick
Apr 10, 2016 at 10:36