Hexagon Interior Point

Extend sides $AB$ , $CD$ , and $EF$ to make equilateral $\triangle MNP$ , as shown below:

$\triangle MNP$ is made up of $9$ equilateral triangles, $6$ of which make up hexagon $ABCDEF$ . Therefore, $A_{\triangle MNP} = \frac{3}{2}A_{ABCDEF}$ .

$\triangle MNP$ is also made up of triangles $\triangle MQN$ , $\triangle MQP$ , and $\triangle NQP$ , so $A_{\triangle MNP} = A_{\triangle MQN} + A_{\triangle MQP} + A_{\triangle NQP}$ .

$\triangle AQB$ has the same height but a base that is $\frac{1}{3}$ that of $\triangle NQP$ , so $A_{\triangle AQB} = \frac{1}{3}A_{\triangle NQP}$ . Likewise, $A_{\triangle CQD} = \frac{1}{3}A_{\triangle MQP}$ and $A_{\triangle EQF} = \frac{1}{3}A_{\triangle MQN}$ .

Therefore, $A_{\triangle AQB} + A_{\triangle CQD} + A_{\triangle EQF}$ $=$ $\frac{1}{3}(A_{\triangle NQP} + A_{\triangle MQP} + A_{\triangle MQN})$ $=$ $\frac{1}{3}(A_{\triangle MNP})$ $=$ $\frac{1}{3} (\frac{3}{2}A_{ABCDEF})$ $=$ $\frac{1}{2}A_{ABCDEF}$ .

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In the problem, $A_{\triangle AQB} = 10$ , $A_{\triangle CQD} = 17$ , $A_{\triangle EQF} = 5$ , and $A_{ABCDEF} = \frac{3\sqrt{3}}{2}R^2$ , so $10 + 17 + 5 = \frac{1}{2} \cdot \frac{3\sqrt{3}}{2}R^2$ which solves to $R = \frac{8\sqrt[4]{12}}{3}$ .

$\triangle EQF$ has an area of $A_{\triangle EQF} = 5$ , a base of $b_{\triangle EQF} = R$ , and a height of $h_{\triangle EQF} = Q_y + \frac{\sqrt{3}}{2}R$ , and using $A = \frac{1}{2}bh$ and $R = \frac{8\sqrt[4]{12}}{3}$ this solves to $Q_y = -\frac{17\sqrt[4]{108}}{24}$ .

$\triangle AQB$ has an area of $A_{\triangle AQB} = 10$ , a base of $b_{\triangle AQB} = R$ , and using $A = \frac{1}{2}bh$ it has a height of $h_{\triangle AQB} = \frac{20}{R}$ . As a side of an equilateral triangle with a length of $3R$ on centered at the origin, $NP$ has a slope of $m = \tan 120° = -\sqrt{3}$ and a $y$ -intercept of $\frac{2}{3} \cdot \frac{\sqrt{3}}{2} 3R = \sqrt{3}R$ , for an equation of $y = -\sqrt{3}x + \sqrt{3}R$ . A line parallel to it with $Q$ on it will have the same slope as $NP$ but be $\frac{20}{R}$ away on its perpendicular, which is $\frac{40}{R}$ less than the $NP$ 's $y$ -intercept of $\sqrt{3}R$ . Therefore, $Q$ is on the line $y = -\sqrt{3}x + \sqrt{3}R - \frac{40}{R}$ . Since $Q_y = -\frac{17\sqrt[4]{108}}{24}$ and $R = \frac{8\sqrt[4]{12}}{3}$ , $Q_x$ solves to $Q_x = \frac{7\sqrt[4]{12}}{8}$ .

Therefore, $\lfloor 1000(R + Q_x + Q_y) \rfloor = \lfloor 1000(\frac{8\sqrt[4]{12}}{3} + \frac{7\sqrt[4]{12}}{8} + -\frac{17\sqrt[4]{108}}{24}) \rfloor = \boxed{4308}$ .

Frankly, I suspect, the two sums of areas of alternate internal triangles of a regular polygon of even number of sides are equal. The proof for hexagon is given below. Therefore $10+17+5 = 32$ is half the area of the regular hexagon. Then the area of the regular hexagon is $64$ and the area of one of the six central equilateral triangle is $\dfrac {64}6$ and $\dfrac {\sqrt 3} 4R^2 = \dfrac {64}6$ $\implies R = \dfrac {8\sqrt{2\sqrt 3}}3 \approx 4.963225915$ .

Consider $\triangle QEF$ , let its height be $h$ , then $[QEF] = \dfrac 12 h \overline{EF} = 5$ . Since $\overline{EF} = R$ , $\implies h = \dfrac {10}R$ . Therefore, $Q_y$ $= - \dfrac {\sqrt 3}2R + h$ $= - \dfrac {\sqrt 3}2R + \dfrac {10}R$ $\approx - 2.283461105$ .

Consider $\triangle QAB$ , we can find its area using the coordinates of its vertices as follows:

$\begin{aligned} \frac {(Q_x-A_x)(Q_y+A_y)}2 + \frac {(A_x-B_x)(A_y+B_y)}2 + \frac {(B_x-Q_x)(B_y+Q_y)}2 & = [QAB] = 10 \\ \left(O_x - R\right)\left(O_y + 0 \right) + \left(R - \frac R2 \right)\left(0 + \frac {\sqrt 3R}2 \right) + \left(\frac R2 - Q_x \right)\left(\frac {\sqrt 3R}2 + Q_y \right) & = 20 \\ \frac {\sqrt 3}2R^2 - \frac {\sqrt 3}2RQ_x - \frac 12 RQ_y & = 20 \\ R - Q_x - \frac {Q_y}{\sqrt 3} & = \frac {40}{\sqrt 3R} \\ \implies Q_x & = \frac 32 R - \frac {50}{\sqrt 3R} \\ & \approx 1.628558503 \end{aligned}$

Therefore, $\left \lfloor 1000 (R+Q_x+Q_y)\right \rfloor = \boxed{4308}$ .

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Proof: Using coordinates of vertices to find the area of triangle:

$\begin{cases} [QAB] = \frac {\sqrt 3}4R^2 - \frac 14 RQ_y - \frac {\sqrt 3}4RQ_x \\ [QCD] = \frac {\sqrt 3}4R^2 - \frac 14 RQ_y + \frac {\sqrt 3}4RQ_x \\ [QEF] = \frac {\sqrt 3}4R^2 + \frac 12 RQ_y \end{cases}$

$\implies [QAB] + [QCD] + [QEF] = \frac {3\sqrt 3}4 R^2 = \frac 12 [ABCDEF]$ .