The Midpoint Of The Points Of Tangency

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We start by computing the lengths of $\overline{BD}$ and $\overline{BE}$ . Let the incircle touch $\overline{AC}$ at $G$ . Since $AB, BC, CA$ are all tangents to the incircle, $\begin{array}{lcl} BD &=& BE \\ AE &=& AF \\ CD&=&CF \end{array}$ Note that $\begin{array}{lcl} BD+DC&=& BC \\ BE+EA &=& AB \\ AF+FC&=& AC \end{array}$ Summing them up, $\begin{array}{lrcl} & 2 (BD +CG + AE)&=& AB + BC + CA \\ \implies & BD+CG+AE &=& \dfrac{AB+BC+CA}{2} \end{array}$ Now, $\begin{array}{lcl} BD &=& BD+CG+AE - (BD+CG) \\ & = & \dfrac{AB+BC+CA}{2} - (BD+CD) \\ & = & \dfrac{AB+BC+CA}{2} - AC \\ &=& \dfrac{AB+BC-CA}{2} \end{array}$

Now, instead of computing the lengths $PQ, QR, RP$ , we shall find out the lengths of $AF, BF$ and $CF$ . Let $BC= a$ . Then, $AB = a \tan 60° = \sqrt{3}a$ and $BC = \dfrac{a}{\cos 60º} = 2a$ . Set the Cartesian coordinates $B= (0, 0), C= (a, 0), A= (0, \sqrt{3}a)$ .

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We have $BD = \dfrac{AB-BC+CA}{2} = \dfrac{\sqrt{3}a - a + 2a}{2} = \dfrac{\sqrt{3}+1}{2}$ . It follows that the coordinates for $D$ are $\left( \dfrac{\sqrt{3}+1}{2}a, 0 \right)$ . Similarly, the coordinates for $E$ are $\left(0, \dfrac{\sqrt{3}+1}{2}a \right)$ . Using Section rule, we find out that the coordinates of $F$ are $\left( \dfrac{\sqrt{3}+1}{4}a, \dfrac{\sqrt{3}+1}{4}a \right)$ . Computing the lengths $AF, BF$ and $CF$ is a straightforward job now. $\begin{array}{lcl} AF &=& \sqrt{\left(\dfrac{\sqrt{3}+1}{4}a \right) ^2 + \left( \dfrac{\sqrt{3}+1}{4}a - \sqrt{3}a \right)^2 } \\ &=& \dfrac{\sqrt{8-\sqrt{3}}}{2} a \\ BF &=& \sqrt{ \left(\dfrac{\sqrt{3}+1}{4}a \right) ^2 + \left( \dfrac{\sqrt{3}+1}{4}a \right) ^2 } \\ &=& \dfrac{\sqrt{3}+1}{2\sqrt{2}}a \\ CF &=& \sqrt{\left(\dfrac{\sqrt{3}+1}{4}a - a \right) ^2 + \left( \dfrac{\sqrt{3}+1}{4}a \right) ^2} \\ &=& \dfrac{\sqrt{4 - \sqrt{3}}}{2} a \end{array}$

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Since $\angle ARF= \angle FQA = 90°$ , quadrilateral $ARFQ$ is cyclic with diameter $AF$ . By sine rule on $\triangle ARQ$ , $\dfrac{RQ}{\sin A}= AF$ . Also notice that $AC$ is the diameter of the circumcircle of $\triangle ABC$ , since $\angle ABC = 90°$ . By Sine rule on $\triangle ABC$ , $\dfrac{BC}{\sin A} = AC = 2a$ . Eliminating $\sin A$ , we find out that $RQ = \dfrac{AF \times a}{2a} = \dfrac{AF}{2} = \dfrac{\sqrt{8-\sqrt{3}}}{4}a$ . Similarly, we find out that $PQ = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{4 - \sqrt{3}}}{2} a$ and $PR = \dfrac{\sqrt{3}+1}{2\sqrt{2}}a$ .

$k= \dfrac{PQ+QR+RP}{AB+BC+CA} = \boxed{\dfrac{ \dfrac{\sqrt{8-\sqrt{3}}}{4} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{4 - \sqrt{3}}}{2} + \dfrac{\sqrt{3}+1}{2\sqrt{2}}}{3+\sqrt{3}}} \approx 0.268$ Our desired answer is $\lfloor 100k \rfloor= \boxed{268}$ .