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.

• If there is some obtuse angle, you can just consider the height from that vertex and split the triangle accordingly. – Jack D'Aurizio Dec 4 '14 at 22:05
• Construct a regular pentagon, connect each vertex to its center, and prolong three of its edges (all but two non-adjacent edges). This gives you a (108,36,36) triangle partitioned into seven acute triangles. You can solve a wide range of cases by distorting this figure. Hopefuly, one can manage to solve every right triangle this way - for any obtuse triangle can be partitioned into two right triangles. – Hagen von Eitzen Dec 4 '14 at 22:18
• No triangle of any kind can be subdivided into two other triangles with all angles $<90^\circ$, since we'd have $x+y=180^\circ$, while, at the same time, and $x<90^\circ$ and $y<90^\circ=>$ contradiction. – Lucian Dec 4 '14 at 22:47
• You're not limited to two triangles, you can split it into any number of triangles you need to solve the problem. – Joe C Dec 4 '14 at 22:56
• I didn't say I was, I was merely pointing out an observation. Furthermore, no triangle can be split only into acute triangles merely by drawing any number of lines originating from the same vertex. – Lucian Dec 4 '14 at 23:09

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.

• There isn't a solution with fewer triangles. – Edward Jiang Dec 5 '14 at 1:20
• @EdwardJiang How to prove your claim that seven is minimum? – user1551 Dec 7 '14 at 19:16

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. Construct a regular pentagon, connect each vertex to its center, and prolong three of its edges (all but two non-adjacent edges). This gives you a $(108°,36°,36°)$ triangle partitioned into seven acute triangles. You can solve a wide range of cases by distorting this figure. Hopefuly, one can manage to solve every right triangle this way - for any obtuse triangle can be partitioned into two right triangles.

So can we solve every right triangle that way? To answer this question, I'll label all my angles: Now one can formulate a linear program in these angles $\alpha_1$ through $\alpha_{21}$ and one extra angle $\varepsilon$. The rules of the linear program are as follows:

• The sum of angles for every one of the seven small triangles must be $180°$.
• At every vertex which is incident with more than one triangle, the incident angles must add up to $180°$ except for $\alpha_6+\alpha_7=90°$ and $\alpha_{17}+\alpha_{18}+\alpha_{19}+\alpha_{20}+\alpha_{21}=360°$.
• All angles must be strictly greater than $0°$, which I express as $\alpha_i\ge\varepsilon$.
• All angles must be strictly less than $90°$, which I express as $\alpha_i\le90°-\varepsilon$.
• $\varepsilon$ must be strictly greater than zero, which we'll have to ensure by including it into our objective function.

Now one can look for solutions of this linear program. I did so trying to minimize $10\alpha_1-\varepsilon$. Or, phrased differently, I tried to make the right triangle far from isosceles while at the same time trying to avoid both zero angles and right angles.

So as you can see, there was a solution which is very degenerate, with $\alpha_1$ really close to zero. On the other hand, realizing that situation under the given constraints required some angles to be close to right. Of course, the above numerical evidence is no proof, but I'd be really surprised if something which works for such small angles doesn't work for arbitrary small angles, as long as you make $\varepsilon$ sufficient small as well.
Come to think of it, the small deviations from $0°$ resp. $90°$ in the above numbers are likely an artifact of how the solver works. A different solver gave me results where the angles were exactly $0°$ resp. $90°$. And it makes sense to assume that you can achieve $0°$ if you allow $90°$, and that doing so is better in terms of the objective function than the above. To actually forbid those solutions for all solvers would likely require a fixed non-zero lower bound on $\varepsilon$.