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Math people:

I am looking for a reference or a nice proof of the following fact. I have proven it myself, but my proof is messy: let $\theta \in (0,1]$ and $\alpha \in (0, \frac{1}{2}\theta^2]$. Let $P$ and $S$ be two points on the unit circle $x^2 + y^2 = 1$ such that the arc length of ${\stackrel{\frown}{PS}}$ is $\theta$. Let $Q$ and $R$ be on ${\stackrel{\frown}{PS}}$ with $Q$ between $P$ and $R$, such that ${\stackrel{\frown}{PQ}}$ and ${\stackrel{\frown}{RS}}$ have equal arc length, and the arc length of ${\stackrel{\frown}{QR}}$ is $\alpha$. Let $T$ and $U$ be the points on the chord ${\overline {PS}}$ such that $\Delta PQT$ and $\Delta SRU$ are right triangles. See the picture below. arc, chord, points, and triangles

Obviously I would not include a hand-drawn picture in a paper I would submit to a journal.

I have proven that the the triangles have disjoint closures.

I proved it the following way: I assumed $P = (\cos(\frac{1}{2}\theta),\sin(\frac{1}{2}\theta))$, $S = (\cos(\frac{1}{2}\theta),-\sin(\frac{1}{2}\theta))$. Clearly it is good enough to take $\alpha = \frac{1}{2}\theta^2$. Then $Q = (\cos(\frac{1}{4}\theta^2),\sin(\frac{1}{4}\theta^2))$. I defined $V = (\cos(\frac{1}{2}\theta, 0)$. It suffices to show $\angle VQP$ is obtuse. I showed ${\vec{QP}} \cdot {\vec{QV}} < 0$. This required a long calculation that was not pretty. I am looking for a reference of this fact or a more elegant proof, perhaps a geometric proof.

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If you choose $V = (\cos(\frac{1}{2}\theta, 0)$ which means V is in the middle of the chord. Therefore, angle $PQV > \pi/4$. As a result cosine of that angle is <0. So ${\vec{QP}} \cdot {\vec{QV}} < 0$. Probably I did not understand. –  nikamed May 19 '13 at 19:41
    
Join the $Q$ and $R$, you will get a cyclic quadrilateral. Now you have a geometric proof. –  Inceptio May 19 '13 at 19:55
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The Taylor series of $\vec{QP}\cdot\vec{QV}$ begins with $-{\theta^3\over 8}$. This suggests that a proof will not be very simple. Note that squares of angles are not dealt with in elementary geometry. –  Christian Blatter May 19 '13 at 20:36

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No one has answered in a long time, so I'll answer myself. I'll start as I indicated in my question. It actually is not that bad. Using standard trig identities,

$${\vec{QP}}\cdot{\vec{QV}} = (\cos(\frac{1}{4}\theta^2)-\cos(\frac{1}{2}\theta))^2- (\sin(\frac{1}{2}\theta)-\sin(\frac{1}{4}\theta^2))\sin(\frac{1}{4}\theta^2)=\ldots= \frac{3}{2}+\frac{1}{2}\cos(\theta)-\frac{1}{2}\cos(\frac{\theta}{2}+\frac{\theta^2}{4})-\frac{3}{2}\cos(\frac{\theta}{2}-\frac{\theta^2}{4}). $$

Using the Maclaurin series of $\cos$ and properties of alternating series, $1-\frac{x^2}{2}<\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$ for all $x \in (0,1)$. Both $\frac{\theta}{2}+\frac{\theta^2}{4}$ and $\frac{\theta}{2}-\frac{\theta^2}{4}$ are between $0$ and $1$. Therefore

$${\vec{QP}}\cdot{\vec{QV}} < \frac{3}{2}+\frac{1}{2}(1-\frac{\theta^2}{2}+\frac{\theta^4}{24}) - \frac{1}{2}(1-(\frac{1}{2}(\frac{\theta}{2}+\frac{\theta^2}{4})^2)- \frac{3}{2}(1-(\frac{1}{2}(\frac{\theta}{2}-\frac{\theta^2}{4})^2) = -\frac{1}{8}\theta^3+\frac{1}{24}\theta^4<0.$$

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