Geometry problem: Line intersecting  a semicircle Suppose we have a semicircle that rests on the negative x-axis and is tangent to the y-axis.A line intersects both axes and the semicircle. Suppose that the points of intersection create three segments of equal length. What is the slope of the line?
I have tried numerous tricks, none of which work sadly.
 A: In this kind of problem, it is inevitable that plain old analytic geometry will work.  A precise version of this assertion is an important theorem, due to Tarski.  If "elementary geometry" is suitably defined, then there is an algorithm that will determine, given any sentence of elementary geometry, whether that sentence is true in $\mathbb{R}^n$. 
So we might as well see what routine computation buys us. We can take the equation of the circle to be $(x+1)^2+y^2=1$, and the equation of the line to be (what else?) $y=mx+b$.  Let our semicircle be the upper half of the circle.
Substitute $mx+b$ for $y$ in the equation of the circle. We get
$$(1+m^2)x^2+2(1+mb)x +b^2=0. \qquad\qquad(\ast)$$
Let the root nearest the origin be $r_1$, and the next one $r_2$.  Note that the line meets the $x$-axis at $x=-b/m$.
From the geometry we can deduce that $-r_2=-2r_1$ and $b/m=-3r_1$, and therefore 
$$r_1=-\frac{b}{3m} \qquad\text{and} \qquad r_2=-\frac{2b}{3m}.$$
By looking at $(\ast)$ we conclude that
$$-\frac{b}{m}=-\frac{2(1+mb)}{1+m^2}  \qquad\text{and} \qquad \frac{2b^2}{9m^2}=\frac{b^2}{1+m^2}.$$
Thus the algebra gives  us the candidates $m=\pm\sqrt{\frac{2}{7}}$. (Of course, the first equation was not needed.) 
Sadly, we should not always believe what algebraic manipulation seems to tell us. I have checked out the details for the positive candidate for the slope, and everything is fine. Our line has equation $y=\sqrt{\frac{2}{7}}x+ \frac{2\sqrt{14}}{5}$. Pleasantly, the points $r_1$ and $r_2$ turn out to have rational coordinates. 
However, the negative candidate is not fine.  That can be checked by looking at the geometry. But it is also clear from the algebra, which has been symmetrical about the $x$-axis. The algebra was not told that we are dealing with a semicircle,  not a circle. So naturally it offered us a mirror symmetric list of configurations.   
We conclude that the slope is $\sqrt{\dfrac{2}{7}}$.
A: Let the line meet the axes at $A(-a,0)$ and $B(0,b)$, and the semi-circle at $A^\prime$ and $B^\prime$ (with $A^\prime$ the closer to $A$, and $B^\prime$ the closer to $B$). Let $O$ be the origin, and define $d := |AA^\prime|=|A^\prime B^\prime|=|B^\prime B| > 0$ as the common length of the segments.
The Power of a Point Theorem, applied to point $B$, tells us that 
$$\begin{eqnarray*}
|BB^\prime|\cdot|BA^\prime| &=& |BO|^2 \\
\implies d \cdot (2d) &=& b^2 \\
\implies 2 d^2 &=& b^2
\end{eqnarray*}$$
Also, Pythagoras tells us that
$$\begin{eqnarray*}
a^2 + b^2 = ( 3 d )^2 = 9 d^2
\end{eqnarray*}$$
Eliminating $b$, we have
$$a^2 = 7 d^2$$
so that the slope is
$$\frac{b}{a} = \frac{\sqrt{2}d}{\sqrt{7}d}=\sqrt{\frac{2}{7}}$$
NOTE. If we cared for the actual value of $d$, we could leverage the Power of Point $A$ (writing $P$ for the point $(-2,0)$):
$$\begin{eqnarray*}
|AP|\cdot|AO| &=& |AA^\prime|\cdot|AB^\prime|\\
\implies (2-a)\cdot a &=& 2 d^2 \\
\implies (2-d\sqrt{7})\cdot d\sqrt{7} &=& 2 d^2 \\
\implies \frac{2\sqrt{7}}{9} &=& d
\end{eqnarray*}$$
