Limit and Geometry 003

Let the intersection point of lines $AQ$ and $BP$ be $T$ . Near the limit, the arc of the semicircle joining $P$ and $Q$ can be approximated by a straight line, so that $r$ is approximately the inradius of the triangle $\Delta PTQ$ . Even better, this triangle is very nearly right-angled (which makes calculating $r$ easier).

Applying the sine rule in $\Delta ATB$ , we have $\frac{AB}{\sin 3\theta}=\frac{AT}{\sin 2\theta}=\frac{BT}{\sin \theta}$

so that $AT=\frac{2\sin 2\theta}{\sin 3\theta},\;\;\; BT=\frac{2\sin \theta}{\sin 3\theta}$

If we let the midpoint of $AB$ be $O$ , then the triangles $\Delta AOQ$ and $\Delta BOP$ are isosceles and we find $AQ=2\cos \theta,\;\;\;BP=2\cos 2\theta$

Putting these together, $QT=2\cos \theta - \frac{2\sin 2\theta}{\sin 3\theta},\;\;\;PT=2\cos 2\theta - \frac{2\sin \theta}{\sin 3\theta}$

Finally, it's also easy to show that $PQ=2\cos 3\theta$ .

These hold for any $\theta$ . However, we're interested in the limit as $\theta \to \frac{\pi}{6}^-$ ; so let $\theta=\frac{\pi}{6}-\varepsilon$ , where $\varepsilon$ is small and positive.

Now, in a right-angled triangle with legs $a,b$ and hypotenuse $c$ , the inradius is given by $\frac{a+b-c}{2}$ . Since $\Delta PTQ$ is very close to being a right-angled triangle, the quantity we want to find is $\begin{aligned} \lim_{\varepsilon \to 0} \frac{PT+QT-PQ}{2\varepsilon} &= \lim_{\varepsilon \to 0} \frac{2\cos 2\theta - \frac{2\sin \theta}{\sin 3\theta}+2\cos \theta - \frac{2\sin 2\theta}{\sin 3\theta}-2\cos 3\theta}{2\varepsilon} \\ &= \lim_{\varepsilon \to 0} \frac{\sin 3\theta \cdot \cos 2\theta - \sin \theta+\sin 3\theta \cdot\cos \theta - \sin 2\theta-\sin 3\theta \cdot\cos 3\theta}{\varepsilon\sin 3\theta} \end{aligned}$

Although $\varepsilon$ is small, $\theta$ is not, so we have to be a little careful with small angle substitutions. The easiest way to avoid problems is to expand the trig functions; for example, $3\theta=\frac{\pi}{2}-3\varepsilon$ so that $\sin 3\theta=\cos 3\varepsilon$ and so on.

Using the small-angle substitutions $\cos \varepsilon = 1-\frac12 \varepsilon^2 + \mathcal{O} \left(\varepsilon^4 \right)$ and $\sin \varepsilon = \varepsilon +\mathcal{O} \left(\varepsilon^3 \right)$ and dropping terms of $\mathcal{O} \left(\varepsilon^3 \right)$ and above, the limit becomes

$\begin{aligned} \lim_{\varepsilon \to 0} \frac{\cos 3\varepsilon \cdot \cos \left(\frac{\pi}{3}-2\varepsilon \right) - \sin \left(\frac{\pi}{6}-\varepsilon \right)+\cos 3\varepsilon \cdot\cos \left(\frac{\pi}{6}-\varepsilon \right) - \sin \left(\frac{\pi}{3}-2\varepsilon \right)-\sin 3\varepsilon \cdot\cos 3\varepsilon}{\varepsilon\cos 3\varepsilon} &= \lim_{\varepsilon \to 0} \frac{\frac32 \left(\sqrt3-1 \right)\varepsilon-\frac32 \left(\sqrt3+2\right)\varepsilon^2}{\varepsilon} \\ &=\frac32 \left(\sqrt3-1 \right) \end{aligned}$

so $p=\frac32$ , $q=-\frac32$ and the answer is $\boxed{18}$ .

Let's choose our coordinate system such that its origin is at the midpoint of $AB$ . At the point where the small circle is tangent to the big semi-circle, the normal passes through the center of the small circle and the center of the big semi-circle, which is the origin. Therefore, we can express the coordinates of the center of the small circle as follows:

$C = (1 - r) (\cos \phi, \sin \phi )$

To determine $r$ and $\phi$ , we use the fact that the distance between $C$ and the two line segments $AQ$ and $BP$ is the radius $r$ .

Now the unit normal vector to $AQ$ that is pointing to point $C$ , is $(- \sin \theta , \cos \theta )$ , hence,

$r = - \sin \theta ( (1 - r) \cos \phi + 1 ) + \cos \theta (1 - r) \sin \phi \hspace{12pt} (1)$

Similarly, the unit normal vector to $BP$ that is pointing to point $C$ , is $(\sin 2 \theta, \cos 2 \theta)$ , hence,

$r = \sin 2 \theta ( (1 - r) \cos \phi - 1 ) + \cos 2 \theta (1 - r) \sin \phi \hspace{12pt} (2)$

Equations $(1), (2)$ are linear equations in $\cos \phi$ and $\sin \phi$ . If we define the vector $u = ( \cos \phi, \sin \phi )$ , then equations $(1), (2)$ can be written compactly as $A u = b$ , where

$A = (1 - r) \begin{bmatrix} - \sin \theta && \cos \theta \\ \sin 2 \theta && \cos 2 \theta \end{bmatrix}$

And,

$b = \begin{bmatrix} r + \sin \theta \\ r + \sin 2 \theta \end{bmatrix}$

The vector $u = A^{-1} b$ , and since $\cos^2 \phi + \sin^2 \phi = 1$ , then $b^T A^{-T} A^{-1} b = 1$

Matrix $A$ is a $2 \times 2$ matrix that is easy to invert.

$A^{-1} = \dfrac{-1}{(1 -r) \sin 3 \theta } \begin{bmatrix} \cos 2 \theta && -\cos \theta \\ -\sin 2 \theta && - \sin \theta \end{bmatrix}$

Hence,

$A^{-T} A^{-1} = \dfrac{1}{(1 - r)^2 \sin^2 3 \theta } \begin{bmatrix} 1 && - \cos 3 \theta \\ -\cos 3 \theta && 1 \end{bmatrix}$

and therefore,

$b^T A^{-T} A^{-1} b =\dfrac{1}{(1 - r)^2 \sin^2 3 \theta } ( (r + \sin \theta)^2 + (r + \sin 2 \theta)^2 - 2 \cos 3 \theta (r + \sin \theta)( r + \sin 2 \theta) ) = 1$

from which,

$(r + \sin \theta)^2 + (r + \sin 2 \theta)^2 - 2 \cos 3 \theta (r + \sin \theta)( r + \sin 2 \theta) = (1 - r)^2 \sin^2 3 \theta \hspace{12pt} (3)$

Now we have $r$ has a function of $\theta$ defined implicity by the above equation, which is just a quadratic in $r$ .

Taking the limit as $\theta \to \frac{\pi}{6}$ , we can see that the limiting $r$ is given by,

$( r + \frac{1}{2})^2 + (r + \frac{\sqrt{3}}{2} )^2 = (1 - r)^2$

whose only solution is $r = 0$ . At this point I used L'Hopital's rule to find the required limit, because we have a $\dfrac{0}{0}$ case, and the required limit will be $\displaystyle - \lim_{\theta \to {\frac{\pi}{6}}^- } \dfrac{dr}{d\theta}$ .

Differentiating equation $(3)$ implicitly,

$2 (r + \sin \theta)( r' + \cos \theta ) + 2 (r + \sin 2 \theta) (r' + 2 \cos 2 \theta) + 6 \sin 3 \theta (r + \sin \theta)(r + \sin 2 \theta) - 2 \cos 3 \theta (r' + \cos \theta)(r + \sin 2 \theta) - 2 \cos 3 \theta (r + \sin \theta) (r' + 2 \cos 2 \theta) = 2 (1 - r) (-r') \sin^2 3 \theta + 6 (1 - r)^2 \sin 3 \theta \cos 3 \theta$

setting $\theta = \dfrac{\pi}{6} , r = 0$ , we get,

$(r' + \dfrac{\sqrt{3}}{2} ) + 2 ( \dfrac{\sqrt{3}}{2} ) (r' + 1 ) + 6 ( \dfrac{\sqrt{3}}{4} ) = -2 r'$

so that,

$r' (3 + \sqrt{3} ) = -3 \sqrt{3}$

$r' = \dfrac{ - 3 \sqrt{3} }{(3 + \sqrt{3} ) } = -\dfrac{(9 \sqrt{3} - 9 )}{6} = -\frac{3}{2} \sqrt{3} + \frac{3}{2}$

Therefore, the required limit is $- r' = \frac{3}{2} \sqrt{3} - \frac{3}{2}$ , so that $p = \frac{3}{2}$ and $q = -\frac{3}{2}$ , and thus $4(p^2 + q^2 ) = \boxed{18}$