Obtuse Triangle Divided into Acute Triangles

== Theorem ==
Let $T$ be an obtuse triangle.
Let $T$ be dissected into $n$ acute triangles.
Then $n \ge 7$.
== Construction ==
Let $\triangle ABC$ be an obtuse triangle such that:
:$\angle ACB > 90^\circ$
:$\angle ACB - \angle CAB < 90^\circ$
:$\angle ACB - \angle CBA < 90^\circ$
Let $D$ be the incenter of $\triangle ABC$.
Let a circle be constructed whose center is $D$ and whose radius is $CD$.
The dissection into acute triangles is then performed as follows:
:
This gives a dissection into $7$ pieces.
If the conditions on $A$ and $B$ are not satisfied, then it is possible to draw a line from $C$ to $AB$ to dissect $\triangle ABC$ into an acute triangle and an obtuse triangle which does satisfy the conditions.
This gives a dissection into $8$ pieces.
== Proof ==
As $D$ is equidistant from $AC$, $CB$ and $BA$, it follows that $\angle CDH = \angle CDE = \angle FDG$.
As $CD = DE = DF = DG = DH$, it follows that each of $\triangle CDE$, $\triangle CDH$ and $\triangle FDG$ are isosceles.
From Triangle Side-Angle-Side Congruence, $\triangle CDE$, $\triangle CDH$ and $\triangle FDG$ are congruent.
Hence it follows that:
:$\angle ECD + \angle CED = \angle ACB$
and that both $\angle ECD$ and $\angle CED$ are acute.
As $\angle ACB$ is obtuse, it follows that $\angle ECD + \angle CED$ together are greater than a right angle.
From Sum of Angles of Triangle equals Two Right Angles, it follows that $\angle CDE$ is acute.
Thus $\triangle CDE$, $\triangle CDH$ and $\triangle FDG$ are all congruent acute triangles.
As $\angle ACB$ is obtuse, it follows that $\angle CAB$ and $\angle CBA$ are both acute.
As $AE = AF$ and $BG = BH$, it follows that $\triangle EAF$ and $\triangle GBH$ are both isosceles.
As $\angle AEF = \angle AFE$ and $\angle AEF + \angle AFE$ are less than $2$ right angles, each of $\angle AEF$ and $\angle AFE$ are acute.
Thus $\triangle EAF$ is an acute triangle.
By the same argument, $\triangle GBH$ is also an acute triangle.
Thus it has been established that for obtuse $\angle ACB$, five of the 
triangles into which $\triangle ABC$ has been dissected are acute.
The conditions under which the remaining triangles $\triangle DEF$ and $\triangle DGH$ are also acute triangle are still to be established.
By the same argument as for $\triangle CDE$ and $\triangle CDH$, we also have that $\triangle DEF$ and $\triangle DGH$ are isosceles.
By the same argument as for $\triangle EAF$ and $\triangle GBH$, it follows that $\angle DEF$ and $\angle DFE$ are both acute, as are $\angle DHG$ and $\angle DGH$.
We have that $\angle CED + \angle DEF + \angle AEF$ form $2$ right angles.
From Sum of Angles of Triangle equals Two Right Angles, and the fact that $\triangle AEF$ and $\triangle DEF$ are both isosceles:
:$2 \angle AEF = 180^\circ - \angle CAB$
:$2 \angle DEF = 180^\circ - \angle EDF$
and so:
:$2 \angle CED + \paren {180^\circ - \angle EDF} + \paren {180^\circ - \angle CAB} = 360^\circ$
which, when simplified, gives:
:$\angle EDF = 2 \angle CED - \angle CAB$
But:
:$2 \angle CED = \angle ACB$
as established above.
Hence $\triangle EDF$ is acute exactly when:
:$\angle ACB - \angle CAB < 90^\circ$
Similarly, $\triangle CDH$ is acute exactly when:
:$\angle ACB - \angle CBA < 90^\circ$
Thus it has been established that when:
:$\angle ACB > 90^\circ$
:$\angle ACB - \angle CAB < 90^\circ$
:$\angle ACB - \angle CBA < 90^\circ$
$\triangle ABC$ can be dissected into $7$ of acute triangles.
As seen above, when $\angle ACB - \angle CAB \ge 90^\circ$, or $\angle ACB - \angle CBA \ge 90^\circ$, the above construction does not work.
However, it will be demonstrated that there exists an $8$ piece dissection of such a triangle.
Let $\triangle ABC$ be an obtuse triangle such that either $\angle ACB - \angle CAB > 90^\circ$ or $\angle ACB - \angle CBA > 90^\circ$.
, suppose $\angle ACB - \angle CAB > 90^\circ$ and that $CAB \le CBA$.
Let $P$ be constructed on $AB$ such that $AP = AC$.
Then $\triangle ACP$ is an isosceles triangle with an acute apex.
Thus from Isosceles Triangle with Acute Apex is Acute Triangle, $\triangle ACP$ is an acute triangle.
:
We have that:
:$\angle CPB + \angle CPA = 180^\circ$
and:
:$2 \angle CPA + \angle CAB = 180^\circ$
Thus:
:$\angle CPB = 90^\circ + \dfrac {\angle CAB} 2$
Also:
Thus we have that $\triangle CPB$ fulfils the criteria to have a dissected into $7$ acute triangles.
Together with acute triangle $\triangle ACP$, it is seen that $\triangle ABC$ has a dissected into $8$ acute triangles.
It remains to be shown that at least $7$ pieces are always needed.
Let $\XX$ be a minimal dissection of $\triangle ABC$ into $n$ acute triangles.
Suppose one of those dissecting lines be drawn from any vertex of $\triangle ABC$ to the opposite side of $\triangle ABC$.
Then this divides $\triangle ABC$ into two triangles at least one of which is obtuse.
This obtuse triangle will itself require $n$ pieces.
Hence $\triangle ABC$ needs at least $n + 1$ pieces, contradicting its minimal nature.
Suppose one of its dissecting lines be drawn from $AC$ to $BC$.
This cuts off an obtuse triangle with $\angle ACB$ as a vertex.
This again needs $n$ pieces.
Hence $\triangle ABC$ needs at least $n + 1$ pieces, contradicting its minimal nature.
Let lines be drawn from $D$ on $AC$ and $E$ on $BC$ to a single point $P$ on $AB$.
Then either or both of $\triangle ADP$ or $\triangle BEP$ are themselves obtuse triangles.
This again needs $n$ pieces.
Hence $\triangle ABC$ needs at least $n + 1$ pieces, contradicting its minimal nature.
The remaining option is for two lines from $AC$ to $AB$ and $BC$ to $AB$ at two separate points such that the parts cut off are acute triangles, leaving a pentagonal shape in the middle.
From Dissection of Pentagon into 5 Acute Triangles, this needs at least $5$ acute triangles to dissect it.
Hence the result.
== Sources ==
* 
* : $7$
* 
* : $7$
Category:Recreational Mathematics
Category:Obtuse Triangles
Category:Acute Triangles
Category:Dissections