![]() ![]() On the other hand, realizing that situation under the given constraints required some angles to be close to right. So as you can see, there was a solution which is very degenerate, with $\alpha_1$ really close to zero. $DPK$ is acute: $\angle PDK$CEP$ is a right triangle with $\angle CPE=\frac\pi2$.$CPD$ is a right triangle with $\angle DPC=\frac\pi2$.$ALE$ is acute: $\angle LAE=\alpha$, $\angle AELThen $ABC$ is partitioned into seven - almost acute - triangles as follows: Similarly, let $L$ be apoint that is both between $H$ and $F$ and between $H$ and $J$. Let $K$ be a point that is both between $G$ and $F$ and between $G$ and $I$. Then both $I$ and $F$ are between $A$ and $G$. Let $J$ be the intersection of $AB$ with the line trough $E$ perpendicular to $AC$. Let $I$ be the intersection of $AB$ with the line trough $D$ perpendicular to $BC$. Let $G,H$ be the projections of $D,E$ to $AB$. We may assume that we picked $P$ sufficiently close to $C$ such that $DE ![]() Let $F$ be the orthogonal projection of $C$ to $AB$ (which is between $A$ and $B$).įor any point $P$ between $C$ and $F$, let $\ell $ be the line through $P$ parallel to $AB$. Let $ABC$ be a triangle with angles $\alpha=\angle BAC<\frac\pi2,\beta=\angle CBA<\frac\pi2,\gamma=\angle ACB$. ![]()
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