# 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.

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Are the three segments of equal length on the line or the semi-circle? –  user02138 Nov 26 '11 at 14:14
It would be helpful if you can post an image... –  pedja Nov 26 '11 at 14:18

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 $\sqrt{\dfrac{2}{7}}$.

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I never thought to use Vieta. (The solution I was working on was quite messier.) Neat! –  Guess who it is. Nov 26 '11 at 16:13
I prefer to break symmetry late, if at all. As a problem-solving strategy, it works more often than it should. –  André Nicolas Nov 26 '11 at 16:51

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 $(-2,0)$):

$$\begin{eqnarray*} |AP|\cdot|AO| &=& |AA^\prime|\cdot|AB^\prime|\\ \implies (2-a)\cdot a &=& 2 d^2 \\ \implies (2-d\sqrt{7})\cdot d\sqrt{7} &=& 2 d^2 \\ \implies \frac{2\sqrt{7}}{9} &=& d \end{eqnarray*}$$

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You made a typo in the last line before the Note... –  David Mitra Nov 26 '11 at 18:55
@David: Jumping Jiminy! A bit of fractional dyslexia. Fixed. –  Blue Nov 26 '11 at 22:27