Consider the equilateral triangle $ABC$ with sides of length $8\sqrt{3}$ cm. A point in the interior of $ABC$ is said to be "special" if it is a distance of $3$ cm from one side of the triangle and a distance of $9$ cm from another side. Consider the convex polygon whose vertices consist of the special points. What is the area of this polygon? Express your answer as a decimal to the nearest tenth.
As shown in Solution 1, draw three lines parallel to the three sides, respectively 3 and 9 cm away. These lines split the triangle into several smaller triangles and quadrilaterals. Since these parallel lines are not the same length, the resulting polygon will not be a parallelogram.
[asy] size(7cm); import markers; pair A = (0,0), B = (8sqrt(3), 0), C = intersectionpoints(CR(A,8sqrt(3)), CR(B,8sqrt(3)))[1], D = (-0.2, 0.8), E = (1.8, 0.8), F = intersectionpoint(C--C + dir(D--C) * 8sqrt(3), D--E), P = intersectionpoints(CR(D, 3), CR(C, 9))[1], Q = intersectionpoints(CR(E, 3), CR(B, 9))[1], R = intersectionpoints(CR(F, 3), CR(A, 9))[1], A0 = intersectionpoint(A--A + 100 * abs(Q - A) * dir(D--E), C--F); fill(A--B--C--cycle^^(0.8A0+0.2P)--(0.8A0+0.2P+0.2Q-0.2P)--Q--R--F--(0.8A0+0.2P), lightgray); draw(A--B--C--A--D--E--cycle); draw(P--Q--R--F--cycle, linewidth(0.8) + bp); draw(F--C^^D--A^^E--B^^R--Q^^P--F, linewidth(0.5)); label("\small$A$", A, SW); label("\small$B$", B, SE); label("\small$C$", C, S); label("\small$D$", D, NW); label("\small$E$", E, NE); label("\small$F$", F, N); label("\small$P$", P, dir(D--C) + dir(C--B)); label("\small$Q$", Q, dir(A--B) + dir(B--C)); label("\small$R$", R, dir(A--C) + dir(C--D)); label("\small$O$", A0, NW); clip(CR(C, 1.8) -- CR(A, 6.2) -- cycle); [/asy]
Let $PQR$ be the desired polygon and $D, E, F$ be as shown above. Observe that $D$ lies on line segment $\overline{BC}$, $E$ lies on $CA$, and $F$ lies on $AB$.
As $\overline{CD}$ is longer than $\overline{FE}$ by 6 cm, we must have that $\overline{DF}$ is longer than $\overline{CE}$ by 6 cm. This means the line segments $\overline{DF}$ and $\overline{CE}$ intersect outside of $\triangle ABC$, which means the desired polygon is a quadrilateral.
By setting the coordinates of $C$ as $(0,0)$ and $A$ as $(8\sqrt{3}, 0)$, we can find the coordinates of $D$, $E$, and $F$ respectively as $\left(-\tfrac{3\sqrt3}5, \tfrac35\right),\left(\tfrac{11\sqrt3}5, \tfrac35\right),$ and $\left(\tfrac{\sqrt3}5, -\tfrac{16}5\right)$. Using Shoelace Theorem on $CEFD$, we compute that the polygon $PQRD$ has area $\frac52$.
By symmetry, the area of the entire desired palan (including the entire pentagon) is 5 times the area of $PQRD$; thus, the area is $5 \cdot \frac52 = \boxed{6.2}$.
To find the area of the convex polygon whose vertices consist of the special points, we first need to find the coordinates of these points.
Let's assume that the equilateral triangle $ABC$ is situated in the coordinate plane with $A$ at the origin. We can then find the coordinates of points $B$ and $C$ as $B(8\sqrt{3}, 0)$ and $C(4\sqrt{3}, 4\sqrt{3})$.
Now, let's find the coordinates of the special points.
First, we'll find the point that is 3 cm away from side $AB$. Since $AB$ is the x-axis, the y-coordinate of this point is 3. Let's call this point $P$. The x-coordinate of $P$ can be found using the distance formula:
$AP = 3$
$x_P^2 + y_P^2 = 3^2$
Substituting the coordinates of $A$, we have:
$x_P^2 + 3^2 = 3^2$
$x_P^2 + 9 = 9$
$x_P^2 = 0$
This means that the point $P$ is at $(0,3)$.
Next, let's find the point that is 9 cm away from side $AC$. Let's call this point $Q$. The line $AC$ has equation $y = x$ because it forms a 45-degree angle with the x-axis. Using the distance formula:
$AQ = 9$
$x_Q^2 + y_Q^2 = 9^2$
Substituting the coordinates of $A$, we have:
$x_Q^2 + x_Q^2 = 9^2$
$2x_Q^2 = 81$
$x_Q^2 = 40.5$
Since $x_Q > 0$ (as $Q$ is in the interior of triangle $ABC$), we have:
$x_Q = \sqrt{40.5} \approx 6.4$
This means that the point $Q$ is approximately $(6.4, 6.4)$.
Finally, let's find the point that is 9 cm away from side $BC$. First, let's find the equation of line $BC$. The slope of line $BC$ is $\frac{4\sqrt{3} - 0}{4\sqrt{3} - 8\sqrt{3}} = \frac{4\sqrt{3}}{-4\sqrt{3}} = -1$. Since the midpoint of $BC$, which we'll call $M$, is $(6\sqrt{3}, 2\sqrt{3})$, the equation of line $BC$ can be written as:
$y - 2\sqrt{3} = -1(x - 6\sqrt{3})$
$y = -x + 8\sqrt{3}$
Now, let's find the point $R$ that is 9 cm away from line $BC$. The distance from point $(x,y)$ to line $y = mx + b$ is given by:
$\frac{|y - mx - b|}{\sqrt{m^2 + 1^2}}$
Using this formula, we have:
$\frac{|y - (-1)x - 8\sqrt{3}|}{\sqrt{(-1)^2 + 1^2}} = 9$
$|y + x - 8\sqrt{3}| = \sqrt{2} \cdot 9$
$|y + x - 8\sqrt{3}| = 9\sqrt{2}$
To find the two possible values for $y$, we have:
$y + x - 8\sqrt{3} = 9\sqrt{2}$
$y + x = 9\sqrt{2} + 8\sqrt{3} = 9\sqrt{2} + 8\sqrt{2}\sqrt{3} = \sqrt{2} \cdot (9 + 8\sqrt{3})$
$y + x - 8\sqrt{3} = -9\sqrt{2}$
$y + x = -9\sqrt{2} + 8\sqrt{3} = -\sqrt{2} \cdot (9 - 8\sqrt{3})$
Simplifying both equations:
$y = \sqrt{2} \cdot (9 + 8\sqrt{3}) - x$
$y = -\sqrt{2} \cdot (9 - 8\sqrt{3}) - x$
These two equations represent two lines. The vertices of the polygon are the intersection points of these lines. Solving these equations simultaneously, we have:
$\sqrt{2} \cdot (9 + 8\sqrt{3}) - x = -\sqrt{2} \cdot (9 - 8\sqrt{3}) - x$
$\sqrt{2} \cdot (9 + 8\sqrt{3}) = -\sqrt{2} \cdot (9 - 8\sqrt{3})$
$(9 + 8\sqrt{3}) = -(9 - 8\sqrt{3})$
Simplifying, we get:
$8\sqrt{3} = -8\sqrt{3}$
This is not possible, so there are no intersection points between these two lines. Therefore, the convex polygon formed by the special points is actually just a triangle $APQ$.
The area of an equilateral triangle can be found using the formula:
$A = \frac{\sqrt{3}}{4} \cdot \text{side length}^2$
Plugging in the given side length of $8\sqrt{3}$, we have:
$A = \frac{\sqrt{3}}{4} \cdot (8\sqrt{3})^2$
$A = \frac{\sqrt{3}}{4} \cdot 288$
$A \approx 49.6$
Therefore, the area of the convex polygon formed by the special points is approximately $49.6$ square units to the nearest tenth.