Dividing an obtuse triangle into acute triangles Can an obtuse triangle be subdivided into only acute triangles (right triangles are not allowed)? 
Any number of subdivisions can be made as long as all of the angles in all resulting triangles are less than 90 degrees.
Example of an incorrect answer: Splitting a triangle with angles {120, 30, 30} by bisecting the obtuse angle results in two right triangles with angles {90, 60, 30}. This is incorrect because there are two angles which are not less than 90 degrees.
 A: Experimentally it appears that seems that something close to Hagen von Eitzen's comment of the construction using a regular pentagon

works for all obtuse triangles.  Here is an example, with two isosceles triangles cutting off the sharp angles, and then choosing a suitable point inside the remaining pentagon to give $7$ acute angled triangles in all.  Not all pairs of isosceles triangles allow there to be a suitable interior point, but I could not find an obtuse angled triangle without some solution.

I would be surprised if there was a solution with fewer acute angled triangles.
A: Let $ABC$ be a triangle with angles $\alpha=\angle BAC<\frac\pi2,\beta=\angle CBA<\frac\pi2,\gamma=\angle ACB$.
Let $F$ be the orthogonal projection of $C$ to $AB$ (which is between $A$ and $B$).
For any point $P$ between $C$ and $F$, let $\ell $ be the line through $P$ parallel to $AB$.
Let $E$ be the intersection of $\ell$ and $AC$, $D$ the intersection of $\ell$ and $BC$.
We may assume that we picked $P$ sufficiently close to $C$ such that $DE<PF$ holds (indeed, $DP, PE\to0$ and $FP\to FC$ as $P\to C$, so a continuity argument applies).
Let $G,H$ be the projections of $D,E$ to $AB$.
Let $I$ be the intersection of $AB$ with the line trough $D$ perpendicular to $BC$.
Let $J$ be the intersection of $AB$ with the line trough $E$ perpendicular to $AC$. Then both $I$ and $F$ are between $A$ and $G$. Let $K$ be a point that is both between $G$ and $F$ and between $G$ and $I$. Similarly, let $L$ be apoint that is both between $H$ and $F$ and between $H$ and $J$.

Then $ABC$ is partitioned into seven - almost acute - triangles as follows:


*

*$ALE$ is acute: $\angle LAE=\alpha$, $\angle AEL<\angle AEJ=\frac\pi2$, $\angle ELA<\angle EHA=\frac\pi2$.

*$BDK$ is acute by the same reasoning

*$CPD$ is a right triangle with $\angle DPC=\frac\pi2$

*$CEP$ is a right triangle with $\angle CPE=\frac\pi2$

*$DPK$ is acute: $\angle PDK<\angle PDG=\frac\pi2$, $\angle KPD<\angle FPD=\frac\pi2$, $\angle DKP$ is also acute because it is opposed to the shortest side: $DP<DE<PF<\min\{KP,KD\}$

*$ELP$ is acute by the same reasoning

*$KPL$ is acute: $\angle PKL<\angle PFL=\frac\pi2$, $\angle KLP<\angle KFP=\frac\pi2$, $\angle LPK$ is also acute because it is oppsed to the shortest side: $KL<GH=DE<PF<\min\{PK,PL\}$


The Thales circles over $CD$ and $CE$ intersect in $C$ and $P$. Hence for any $Q$ between $P$ and $F$, the angles $\angle DQC$ and $\angle CQE$ are acute.
As long as $Q$ is sufficiently close to $P$, the triangles $CQD$, $CEQ$, $DQK$, $ELQ$, $KQL$ are acute, thus giving us a partition of $ABC$ into seven acute triangles.

