$17 \cdot 155$ A point $P$ inside an equilateral triangle $ABC$ is at distances $PA=6, PB=8, PC=10$ from the three vertices. The integer closest to the area of $\triangle ABC$ is
(A) 159.
(B) 131.
(C) 95.
(D) 79.
(E) 50.
(18th American High School Mathematics Examination, 1967)
[Solution 1] As shown in the figure, we have already proved: If $P$ is a point inside the equilateral $\triangle ABC$, and $PA=a, PB=b, PC=c$, and the three segments cut by the lines through $P$ parallel to the three sides are $u, v, w$ respectively, then in the triangle $HIG$ with sides $a, b, c$, a point $Q$ inside the triangle forms angles of $120^{\circ}$ with the three vertices, and the lengths of these three segments are $u, v, w$. Returning to our proposition,
$$
\begin{aligned}
S_{\triangle ABC} & =2\left(S_{\triangle I}+S_{\triangle II}+S_{\triangle III}\right)+S_{\triangle I}+S_{\triangle V}+S_{\triangle (3)} \\
& =2 S_{\triangle IHG}+\frac{\sqrt{3}}{4}\left(u^{2}+v^{2}+w^{2}\right),
\end{aligned}
$$

In $\triangle QIH, \triangle QIG, \triangle QGH$, by the cosine rule we have
$$
\left\{\begin{array}{l}
u^{2}+v^{2}+uv=8^{2}, \\
u^{2}+w^{2}+uw=10^{2}, \\
v^{2}+w^{2}+vw=6^{2},
\end{array}\right.
$$
(1) + (2) + (3) gives $2\left(u^{2}+v^{2}+w^{2}\right)+uv+uw+vw=200$.
Also, $uv+vw+wu \leqslant u^{2}+v^{2}+w^{2}$ (by the rearrangement inequality), then
$$
2\left(u^{2}+v^{2}+u^{2}\right) \leqslant 200 \leqslant 3\left(u^{2}+v^{2}+w^{2}\right),
$$

i.e., $\frac{200}{3} \leqslant u^{2}+v^{2}+w^{2} \leqslant 100$.
Also, $2 S_{\triangle IHG}=6 \cdot 8=48$,
thus $48+\frac{\sqrt{3}}{4} \cdot \frac{200}{3} \leqslant S_{\triangle ABC} \leqslant 48+\frac{\sqrt{3}}{4} \cdot 100$.
i.e., $76.9 \leqslant S_{\triangle ABC} \leqslant 91.3$.
Therefore, the answer is $(D)$.
[Solution 2] Establish a coordinate system with the vertex $C$ of the equilateral $\triangle ABC$ as the origin, and the altitude on $AB$ as the x-axis.
If the side length of the equilateral $\triangle ABC$ is $a$, then the coordinates of points $A, B$ are
$$
\left(\frac{\sqrt{3}}{2} a, \frac{a}{2}\right),\left(\frac{\sqrt{3}}{2} a,-\frac{a}{2}\right) .
$$

From the problem, we have the equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=10^{2}, \\
\left(x-\frac{\sqrt{3}}{2} a\right)^{2}+\left(y+\frac{1}{2} a\right)^{2}=8^{2}, \\
\left(x-\frac{\sqrt{3}}{2} a\right)^{2}+\left(y-\frac{1}{2} a\right)^{2}=6^{2} .
\end{array}\right.
$$
(2)-(3) gives $2ay=28$ i.e., $ay=14$, substituting into (2) and using (1) we get
$10^{2}-\sqrt{3} ax+a^{2}+14=64$, then $ax=\frac{a^{2}+50}{\sqrt{3}}$.

Thus, from $(ax)^{2}+(ay)^{2}=a^{2}(x^{2}+y^{2})$ we get
$$
\left(\frac{a^{2}+50}{\sqrt{3}}\right)^{2}+14^{2}=10^{2} a^{2},
$$

i.e., $a^{4}-200 a^{2}+3088=0$.
Solving, we get $a^{2}=100 \pm 48 \sqrt{3}$ (discard $100-48 \sqrt{3}$, as $a^{2}>100$ ).
Then
$$
S_{\triangle ABC}=\frac{\sqrt{3}}{4} a^{2}=25 \sqrt{3}+36 \approx 79 \text {. }
$$

Therefore, the answer is $(D)$.
[Solution 3] Construct an equilateral $\triangle PAF$ outside the shape with $PA$ as a side, and connect $BF$.
Draw a perpendicular from $B$ to the extension of $AP$, with the foot of the perpendicular being $E$. Then
$$
\begin{array}{l}
\angle PAC=60^{\circ}-\angle PAB=\angle BAF . \\
\text { Also } BA=CA, AP=AF,
\end{array}
$$

Thus, $\triangle PAC \cong \triangle FAB$.
Therefore, $BF=CP=10$,

Then $\triangle BPF$ is a right triangle.
Also, $\angle BPE=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$, then $BE=4, EP=4 \sqrt{3}$.
Thus, $AB=\sqrt{BE^{2}+EA^{2}}=\sqrt{4^{2}+(6+4 \sqrt{3})^{2}}$
$$
=2 \sqrt{25+12 \sqrt{3}} \text {, }
$$

So
$$
\begin{aligned}
S_{\triangle ABC} & =\frac{\sqrt{3}}{4} AB^{2}=\sqrt{3}(25+12 \sqrt{3})=36+25 \sqrt{3} \\
& \approx 79.30 .
\end{aligned}
$$

Therefore, the answer is $(D)$.