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 $5 \cdot \frac52 = \boxed{6.2}$.
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.
1 answer