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diagram

Given a circle, a point $H$ outside the circle, segments $\overline{HE}$ and $\overline{HT}$ tangent to the circle at $E$ and $T$, respectively, and points $I$ and $G$ on the circle such that $I$, $G$, and $H$ are collinear (all as shown above), knowing the measures of $\angle EHG$ and $\angle GHT$ (call them $\alpha$ and $\beta$, respectively, for convenience) determines the measures of each of the four arcs on the circle.

Is it possible to compute the measures of the minor arcs $EI$, $IT$, $TG$, and $GE$ in terms of $\alpha$ and $\beta$ without using trigonometry?

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I have a dumb question: "without using trigonometry" - why? –  J. M. Dec 8 '10 at 3:24
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@J.M.: Because I was constructing a problem aimed at the high school geometry level. I wrote it, thought I had enough information, failed to solve the problem using the methods I had expected to work, then verified that the problem was actually determined. –  Isaac Dec 8 '10 at 3:42
    
See, I thought I had something, but then you said trig is verboten, so I got curious about this restriction. If so, I got nothin'... –  J. M. Dec 8 '10 at 3:46
    
Another thing: I parse from your comment that the system there is that trig comes much later than "high school geometry". Am I correct? –  J. M. Dec 8 '10 at 3:47
    
@J.M.: It depends on how one defines "trig"--technically, trig as applied only to right triangles, would be allowed, but still isn't what I wanted in the problem. I'd describe as typical for the U.S. to cover definitions of sine, cosine, and tangent relative only to acute angles in right triangles as standard in Geometry (typically in grade 9 or 10); the definition of secant, cosecant, and cotangent, and the extension to circular functions (functions of rotations rather than of acute angles) is usually in Advanced Algebra or Precalculus (the year following Geometry or the year after that). –  Isaac Dec 8 '10 at 3:50
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3 Answers

I'm within one, so I hope it prompts your thinking. Basic geometry was a lot of years ago. The theorems I remember are $\stackrel{\frown}{EGT} = 2\angle EIT$, $\beta=(\stackrel{\frown}{IT} - \stackrel{\frown}{TG})/2$, and $ET=\pi-\alpha-\beta$, which you can see by drawing radii from E and T and the line from H to the center of the circle. Then if $\angle EIG=\gamma$ we have $EG=2\gamma, GT=\pi-\alpha-\beta-2\gamma, TI=\pi-\alpha+\beta-2\gamma, IE=2\gamma+2\alpha$

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@J.M.: I guess I should thank you for frowning on my answer :) I couldn't find that in my LaTeX cheat sheet. stackrel is what puts the frown over everything in the brackets? –  Ross Millikan Dec 8 '10 at 5:45
    
I personally consider stackrel{}{} a bit of a cheat, but as I recall $\LaTeX$ never bothered to supply an arc symbol. No prob with the fix; I'd upvote this if I hadn't been vote-crazy again and used up my votes in the span of an hour... –  J. M. Dec 8 '10 at 5:54
    
I think this is about as far as I was able to get. It really seems like there ought to be a geometric solution, though. –  Isaac Dec 8 '10 at 18:33
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With trig, I get that the arcs have measure $\theta \pm \alpha$ and $\pi - \theta \pm \beta$, where $$\cos\theta = \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} = \frac{\cos\beta-\cos\alpha}{1-\cos\left(\alpha+\beta\right)}$$

By the "far arc / near arc" principle, $\stackrel{\frown}{EI}-\stackrel{\frown}{EG} = 2\alpha$, so we can write $\stackrel{\frown}{EI} = \theta+\alpha$ and $\stackrel{\frown}{EG}=\theta-\alpha$.

Let's go analytical ...

Let the circle to have radius $1$, with its center, $O$, at the origin; and let $H$ lie on the positive $x$-axis. In quadrilateral $OEHT$, angles at $E$ and $T$ are right angles, and the angle at $H$ has measure $\alpha+\beta$, so that the angle at $O$ has measure $2 \gamma := \pi-(\alpha+\beta)$. Thus, $\angle HOE = \gamma$ and $|OH| = \sec\gamma$. Moving around the circle by arcs of $\theta \pm \alpha$ from $E$ gets us to $I$ and $G$, and we determine coordinates for these points, writing "$\rm{cis}\cdot$" for "$(\cos\cdot,\sin\cdot)$":

$$I = \rm{cis}\left(\gamma + ( \theta + \alpha )\right) = \rm{cis}\left(\gamma+\alpha+\theta\right)$$ $$G = \rm{cis}\left(\gamma-(\theta-\alpha)\right) = \rm{cis}\left(\gamma+\alpha-\theta\right)$$

Since $H$, $G$, and $I$ are collinear, we must have equal slopes for segments $HI$ and $HG$:

$$\frac{ \sin(\gamma+\alpha+\theta) }{ \cos(\gamma+\alpha+\theta)-\sec\gamma } = \frac{ \sin(\gamma+\alpha-\theta) }{ \cos(\gamma+\alpha-\theta)-\sec\gamma }$$

Therefore,

$$\sin(\gamma+\alpha+\theta)\left(\cos(\gamma+\alpha-\theta)-\sec\gamma\right) = \sin(\gamma+\alpha-\theta) \left( \cos(\gamma+\alpha+\theta)-\sec\gamma\right) $$

$$\begin{eqnarray*} &&\sin(\gamma+\alpha+\theta)\cos(\gamma+\alpha-\theta)- \sin(\gamma+\alpha-\theta) \cos(\gamma+\alpha+\theta) \\ &=& \sec{\gamma} \; \left( \sin(\gamma+\alpha+\theta) -\sin(\gamma+\alpha-\theta) \right) \\ \sin 2\theta &=& 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \\ 2 \sin\theta \cos\theta &=& 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \end{eqnarray*}$$ If $\theta\ne 0$ and $\theta \ne \pi$ [*], then we can cancel $\sin\theta$ and finish-up: $$\begin{eqnarray*} \cos\theta &=& \frac{\cos\left(\gamma+\alpha\right)}{\cos\gamma} = \frac{\cos\frac{\pi+\alpha-\beta}{2}}{\cos\frac{\pi-(\alpha+\beta)}{2}}= \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} \\ \end{eqnarray*}$$

There's certainly a way to arrive at this result without using coordinates and slopes (I got lazy), but if there's a way to get there "without using trigonometry", I'm not seeing it.

Edit. I'll just point out that, if $F$ is the foot of the perpendicular from the circle's center to segment $HI$, then

$$\frac{|OF|}{|OE|}=\frac{|OH| \sin\angle FHO}{|OH|\sin\angle EHO}=\frac{ \sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}=\cos\left(\pi-\theta\right)$$

Also, there are points $P$ and $Q$ on the circle, with $P$ on $\stackrel{\frown}{IE}$ and $G$ on $\stackrel{\frown}{QE}$, such that $\stackrel{\frown}{IP} = \stackrel{\frown}{GQ} = \alpha$, whence $\stackrel{\frown}{PE}=\stackrel{\frown}{QE}=\theta$. So, all the components of the final relation appear in the figure somewhere. I haven't looked hard enough to determine if it's possible to "see" the relation geometrically. End edit.

[*] If $\theta = 0$, then $\alpha$ itself must be $0$ (so that $\theta-\alpha$ is a non-negative arc measure), with $I$ and $G$ coinciding with $E$. Likewise, $\theta = \pi$ implies $\beta = 0$, with $I$ and $G$ coinciding with $T$. Analysis of these cases is straightforward, with the same formula resulting as in the general case.

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I never actually carried through with my ideas for a trigonometric solution. I'd hoped to avoid doing it analytically if I'd gone that route, but I don't know that it would have been possible. It wouldn't surprise me if it can't be done in a non-trigonometric way. (For the situation I was constructing when this came up, I ended up giving an additional piece of information and making sure that I didn't overdetermine the problem.) –  Isaac Dec 8 '10 at 23:09
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There is no solution without inverse trigonometric functions, because any elementary geometric argument would, for a given value of the angle between the tangent lines, calculate all 4 arc lengths as linear functions of $\alpha$. This would imply that, as the (secant) line through $H$ is rotated at uniform angular speed, it sweeps out arclength on the circle at a uniform rate. But this is impossible because the distance from $H$ to the intersection point, $|HG|$, is not constant. It is not hard to make this argument rigorous using calculus.

If the arcs, in the order listed in the question, are denoted $A, B, b, a$, then elementary geometry does express $(A-a)$, $(B-b)$, and $(A+B+b+a)$ as linear functions of $\alpha, \beta$ and $1$, as in the other solutions. I think this set of 3 equations in 4 unknowns is what the OP meant by "not enough equations" to solve everything purely geometrically.

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