# Need help with Trigonometric Identity in solution to problem on Napoleon Triangles.

The problem is to show (in the diagram below) that $AO_1, BO_2, CO_3$ are concurrent.

The solution I am trying to follow applies the converse of Ceva's theorem, e.g. that $\frac{BX}{XC}\frac{CY}{YA}\frac{AZ}{ZB} = 1,$ so the three lines are concurrent. The first part of the solution is shown in the text on the diagram, e.g. that $\frac{BX}{XC} = \frac{(\Delta ABO_1)}{(\Delta CAO_1)},$ using the notation $(\Delta ABC)$ to refer to the area of $\Delta ABC$ (additionally, $a, b, c$ are the sides opposite to vertices $A, B, C).$

The part I do not follow is equating $\frac{c \cdot sin(\beta + 30^\circ)}{b \cdot sin(\gamma + 30^\circ)}$ to $\frac{BX}{XC} = \frac{(\Delta ABO_1)}{(\Delta CAO_1)}.$

In fact, I'm not even getting the numbers to add up, using the measurements as shown on the diagram! For example, on the lhs side of the diagram where the numbers $e_1$ and $w$ (both equal to 0.83) are shown, $e_1$ is $\frac{BX}{XC},$ and $w$ is $\frac{(\Delta ABO_1)}{(\Delta CAO_1)},$ so that part at least adds up as expected.

However $f_1 = 0.5$ is $\frac{c \cdot sin(\beta + 30^\circ)}{b \cdot sin(\gamma + 30^\circ)}$ and does not equate. Obviously I am miscalculating or misunderstanding something but have not been able to see what it is.

Help greatly appreciated as always.

For reference this is problem 3.3.2 (part ii) in Geometry Revisited by Coxeter and Greitzer.

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I don't know about your computations with lengths, but take the triangle $ABO_1$ and consider the altitude $O_1H$ with $H$ on $AB$. By standard trigonometry you have $$O_1H=BO_1\sin(\beta+30^\circ)$$ ad $A\widehat{B}O_1=\beta+30^\circ$. Same story in the other triangle $ACO_1$ with altitude $O_1K=CO_1\sin(\gamma+30^\circ$. Now, as the triangles you constructed outside are equilaterals (or should be, in Napoleon's theorem), you have that $CO_1=BO_1$, so $$\frac{(\triangle ABO_1)}{(\triangle ACO_1)}=\frac{O_1H\cdot AB}{O_1K\cdot AC}=\frac{c\sin(\beta+30^\circ)}{b\sin(\gamma+30^\circ)}\;.$$
I realize what happened now regarding the measurement inaccuracy: The diagram measurements were off because unbeknownst to me the point $O_1$ had shifted away from the centroid position of the equilateral triangle (undoubtedly I moved it accidentally, but not enough to notice visually). – Greg Hill Aug 20 '13 at 22:03