Problem (Geometry):
Let $A B C$ be a triangle with $\mathrm{m} \angle B=\mathrm{m} \angle C=80^{\circ}$. Compute the number of points $P$ in the plane such that triangles $P A B, P B C$, and $P C A$ are all isosceles and non-degenerate. Note: the approximation $\cos 80^{\circ} \approx 0.17$ may be useful.

Solution:
Focus on $\triangle P B C$. Either $P B=P C$ or $P B=B C$ or $P C=B C$.

If $P B=P C$, then $P$ lies on the perpendicular bisector $l$ of side $\overline{B C}$. Considering now $\triangle P A B$, if $P A=P B$, then $P A=P C$, and $P$ must be the circumcenter of $\triangle A B C$; call this location $P_{1}$. If $P A=A B$, then $P A=A C$, and $P, B, C$ all lie on a circle with center $A$ and radius $A B$. There are two intersection points of that circle with $l$, one on each arc with endpoints $B$ and $C$; label the one on the major arc $P_{2}$ and on the minor $\operatorname{arc} P_{3}$. Finally, if $P B=A B$, then $P B=A C$ by the transitive property and $P C=A C$ by the perpendicular bisector theorem, so $P B A C$ is a rhombus; $P$ is the reflection of $A$ across $\overline{B C}$. Call this point $P_{4}$.

If $P B=B C$, then $P$ must lie on the circle centered at $B$ with radius $B C$. Considering $\triangle P A B$, if $P A=A B$, then $P$ lies on the circle centered at $A$ with radius $A B$. Now $\odot A$ and $\odot B$ intersect at two points, but one of them is $C$, so the other intersection must be the location of $P$, which is $P_{5}$. The condition $P B=A B$ is impossible, because it implies that $A B=B C$, which is false because in $\triangle A B C, \mathrm{~m} \angle C>\mathrm{m} \angle A=20^{\circ}$, so $A B>B C$. The third possibility for $\triangle P A B$ is that $P A=P B$, implying that the perpendicular bisector of $\overline{A B}$ intersects $\odot B$, which only occurs if $B C / A B \geq 1 / 2$ (although if $B C / A B=1 / 2$, the triangle is degenerate). But $B C / A B=2 \cos 80^{\circ}$, and the given approximation $\cos 80^{\circ} \approx 0.17$ implies that $B C / A B \approx 0.34$. Hence the perpendicular bisector of $\overline{A B}$ does not intersect $\odot B$. Thus the assumption $P B=B C$ yields only one additional location for $P, P_{5}$. Similarly, $P C=B C$ yields exactly one more location, $P_{6}$, for a total of $\mathbf{6}$ points. All six points, and their associated triangles, are pictured below.





Final answer: 6