Another Triangle Problem

Let the unit equilateral triangle be $\triangle ABC$ and its incenter be the origin $O(0,0)$ ; and $A \left(-\frac 12, - \frac 1{2\sqrt 3}\right)$ , $B \left(0, \frac 1{\sqrt 3}\right)$ , and $C \left(\frac 12, - \frac 1{2\sqrt 3}\right)$ . Let the point on the incircle be $P(x,y)$ and $AP = a$ , $BP=b$ , and $CP=c$ . Then the incircle is given by $x^2 + y^2 = \frac 1{12}$ and by Pythagorean theorem :

$\begin{cases} a^2 = \left(x + \dfrac 12 \right)^2 + \left(y+\dfrac 1{2\sqrt 3}\right)^2 = x^2 + x + \dfrac 14 + y^2 + \dfrac y{\sqrt 3} + \dfrac 1{12} = \dfrac 5{12} + x + \dfrac y{\sqrt 3} \\ b^2 = \left(x - 0 \right)^2 + \left(y - \dfrac 1{\sqrt 3}\right)^2 = x^2 + y^2 - \dfrac {2y}{\sqrt 3} + \dfrac 13 = \dfrac 5{12} - \dfrac {2y}{\sqrt 3} \\ c^2 = \left(x - \dfrac 12 \right)^2 + \left(y+\dfrac 1{2\sqrt 3}\right)^2 = x^2 - x + \dfrac 14 + y^2 + \dfrac y{\sqrt 3} + \dfrac 1{12} = \dfrac 5{12} - x + \dfrac y{\sqrt 3} \end{cases}$

For geometric progression, we have $ac = b^2$ . Therefore,

$\begin{aligned} a^2 c^2 & = b^4 \\ \left(\frac 5{12} + x + \frac y{\sqrt 3}\right) \left(\frac 5{12} - x + \frac y{\sqrt 3}\right) & = \left(\frac 5{12} - \frac {2y}{\sqrt 3}\right)^2 \\ \left(\frac y{\sqrt 3} + \frac 5{12} \right)^2 - x^2 & = \frac {4y^2}3 - \frac {5y}{3\sqrt 3} + \frac {25}{144} \\ \frac {y^2}3 + \frac {5y}{6\sqrt 3} + \frac {25}{144} - \frac 1{12} + y^2 & = \frac {4y^2}3 - \frac {5y}{3\sqrt 3} + \frac {25}{144} \\ \frac {5y}{2\sqrt 3} & = \frac 1{12} \\ \implies y & = \frac {\sqrt 3}{30} \\ \implies b^2 & = \frac 5{12} - \frac 2{\sqrt 3} \times \frac {\sqrt 3}{30} = \frac 7{20} \\ \implies b & = \sqrt{\frac 7{20}} \end{aligned}$

Therefore $p+q = 7 + 20 = \boxed{27}$ .

Analytic approach:

Let $A=\left( -\frac { \sqrt { 3 } }{ 6 } ,-\frac { 1 }{ 2 } \right) ,B=\left( -\frac { \sqrt { 3 } }{ 6 } ,+\frac { 1 }{ 2 } \right) ,C=\left( +\frac { \sqrt { 3 } }{ 3 } ,0 \right)$ - three points forming a unit equilateral triangle. We need to find point $X$ on the incircle so that $a=AX, b=BX, c=CX$ were in a geometric progression. Let $X=\left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } ,\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } \right)$ , where $\alpha \in \left[ 0; 2 \pi \right]$ .

Now let's derive the distances:

$a=\left| \overrightarrow { AX } \right| =\left| \left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } +\frac { \sqrt { 3 } }{ 6 } ,\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } +\frac { 1 }{ 2 } \right) \right| =\sqrt { { \left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } +\frac { \sqrt { 3 } }{ 6 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 3 } }{ 6 } \sin { \alpha } +\frac { 1 }{ 2 } \right) }^{ 2 } } \\ a=\sqrt { \frac { 1 }{ 12 } \cos ^{ 2 }{ \alpha } +\frac { 1 }{ 6 } \cos { \alpha } +\frac { 1 }{ 12 } +\frac { 1 }{ 12 } \sin ^{ 2 }{ \alpha } +\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } +\frac { 1 }{ 4 } } =\sqrt { \frac { 5 }{ 12 } +\frac { 1 }{ 6 } \cos { \alpha } +\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } } \\ a=\sqrt { \frac { 5+2\cos { \alpha } +2\sqrt { 3 } \sin { \alpha } }{ 12 } }$

$b=\left| \overrightarrow { BX } \right| =\left| \left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } +\frac { \sqrt { 3 } }{ 6 } ,\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } -\frac { 1 }{ 2 } \right) \right| =\sqrt { { \left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } +\frac { \sqrt { 3 } }{ 6 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 3 } }{ 6 } \sin { \alpha } -\frac { 1 }{ 2 } \right) }^{ 2 } } \\ b=\sqrt { \frac { 1 }{ 12 } \cos ^{ 2 }{ \alpha } +\frac { 1 }{ 6 } \cos { \alpha } +\frac { 1 }{ 12 } +\frac { 1 }{ 12 } \sin ^{ 2 }{ \alpha } -\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } +\frac { 1 }{ 4 } } =\sqrt { \frac { 5 }{ 12 } +\frac { 1 }{ 6 } \cos { \alpha } -\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } } \\ b=\sqrt { \frac { 5+2\cos { \alpha } -2\sqrt { 3 } \sin { \alpha } }{ 12 } }$

$c=\left| \overrightarrow { CX } \right| =\left| \left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } -\frac { \sqrt { 3 } }{ 3 } ,\frac { \sqrt { 3 } }{ 6 } \sin { \alpha } \right) \right| =\sqrt { { \left( \frac { \sqrt { 3 } }{ 6 } \cos { \alpha } -\frac { \sqrt { 3 } }{ 3 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 3 } }{ 6 } \sin { \alpha } \right) }^{ 2 } } \\ c=\sqrt { \frac { 1 }{ 12 } \cos ^{ 2 }{ \alpha } -\frac { 1 }{ 3 } \cos { \alpha } +\frac { 1 }{ 3 } +\frac { 1 }{ 12 } \sin ^{ 2 }{ \alpha } } =\sqrt { \frac { 5 }{ 12 } -\frac { 1 }{ 3 } \cos { \alpha } } \\ c=\sqrt { \frac { 5-4\cos { \alpha } }{ 12 } }$

Since $a, b, c$ are in a geometric progression, the following is true:

$\frac { a }{ b } =\frac { b }{ c } \\ a\cdot c= b\cdot b\\ { a }^{ 2 }\cdot { c }^{ 2 }={ b }^{ 2 } \cdot { b }^{ 2 }$

Now let's use the derived expressions:

$\frac { 5+2\cos { \alpha } +2\sqrt { 3 } \sin { \alpha } }{ 12 } \cdot \frac { 5-4\cos { \alpha } }{ 12 } =\frac { 5+2\cos { \alpha } -2\sqrt { 3 } \sin { \alpha } }{ 12 } \cdot \frac { 5+2\cos { \alpha } -2\sqrt { 3 } \sin { \alpha } }{ 12 } \\ 25+10\cos { \alpha } +10\sqrt { 3 } \sin { \alpha } -20\cos { \alpha } -8\cos ^{ 2 }{ \alpha } -8\sqrt { 3 } \sin { \alpha } \cos { \alpha } =25+4\cos ^{ 2 }{ \alpha } +12\sin ^{ 2 }{ \alpha } +20\cos { \alpha } -20\sqrt { 3 } \sin { \alpha } -8\sqrt { 3 } \sin { \alpha } \cos { \alpha } \\ 0=30\cos { \alpha } -30\sqrt { 3 } \sin { \alpha } +12\cos ^{ 2 }{ \alpha } +12\sin ^{ 2 }{ \alpha } \\ 2\cos { \alpha } -2\sqrt { 3 } \sin { \alpha } =-\frac { 4 }{ 5 }$

Finally now we can recall the expression of $b$ and find the value:

$b=\sqrt { \frac { 5+2\cos { \alpha } -2\sqrt { 3 } \sin { \alpha } }{ 12 } } \\ b=\sqrt { \frac { 5-\frac { 4 }{ 5 } }{ 12 } } =\sqrt { \frac { 25-4 }{ 60 } } =\sqrt { \frac { 21 }{ 60 } } =\sqrt { \frac { 7 }{ 20 } }$

Thus, $p=7,q=20$ and the answer is $\boxed { 27 }$