David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$, which can each be inscribed in a circle with radius $1$. Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$, and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$, $\sin\varphi_B=\tfrac{3}{5}$, and $\sin\varphi_C=\tfrac{6}{7}$. All three quadrilaterals have the same area $K$, which can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Suppose our four sides lengths cut out arc lengths of $2a$, $2b$, $2c$, and $2d$, where $a+b+c+d=180^\circ$. Then, we only have to consider which arc is opposite $2a$. These are our three cases, so
\[\varphi_A=a+c\]
\[\varphi_B=a+b\]
\[\varphi_C=a+d\]
Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$, $\overarc{BC}=2b$, $\overarc{CD}=2c$, and $\overarc{DA}=2d$.
Then, by Law of Sines, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ and $BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)$. Therefore,
\[K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},\]
so our answer is $24+35=\boxed{059}$.
Note that the conditions of the problem are satisfied when the lengths of the four sticks are about $0.32, 0.91, 1.06, 1.82$.
By S.B.

Note
[CyclIntersect.png](https://artofproblemsolving.com/wiki/index.php/File:CyclIntersect.png)
The solution uses \[\varphi_A=a+c.\]
We can see that this follows because $\varphi_A = \frac12 (2a+2c)=a+c,$ where $a$ and $c$ are the central angles of opposite sides.