10.6. In an equilateral triangle $ABC$, a point $P$ is chosen such that $AP=10, BP=8, CP=6$. Find the area of this triangle.

#
# Solution:

Method 1. Rotate triangle $A P B$ by $60^{\circ}$ around point $A$, triangle $B P C$ by $60^{\circ}$ around point $B$, and triangle $C P A$ by $60^{\circ}$ around point $C$ (all rotations counterclockwise) - see the left figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_d79b1c9dd7e6003b3f28g-08.jpg?height=414&width=876&top_left_y=1148&top_left_x=592)

For the solution to problem 10.6, method 1

We obtained a hexagon $A F B D C E$, the area of which is twice the area of triangle $A B C$. Let's cut this hexagon differently: into triangles $A P F, F P B, B P D, D P C, C P E$, and $E P A$ - see the right figure. Triangles $C P D, B P F$, and $A P E$ are isosceles with an angle of $60^{\circ}$, so they are equilateral. Then triangles $C P E, B P D$, and $A P F$ are equal by three sides. Each of them has sides of 6, 8, and 10. Thus, these triangles are right-angled, and the area of each is 24. Finally, we have

$$
S_{A B C}=\frac{3 \cdot 24+9 \sqrt{3}+16 \sqrt{3}+25 \sqrt{3}}{2}=36+25 \sqrt{3}
$$

Method 2. Let the angle $P A B$ be $\alpha$. Then $\angle P A C=60^{\circ}-\alpha$. Writing the cosine theorems for triangles $C A P$ and $P A B$, we get the system of equations:

$$
\left\{\begin{array}{l}
36=100+a^{2}-20 a \cos \left(60^{\circ}-\alpha\right) \\
64=100+a^{2}-20 a \cos \alpha
\end{array}\right.
$$

From the second equation of the system, we find that

$$
\cos \alpha=\frac{a^{2}+36}{20 a}
$$

Then the first equation of the system transforms as follows:

![](https://cdn.mathpix.com/cropped/2024_05_06_d79b1c9dd7e6003b3f28g-09.jpg?height=626&width=1590&top_left_y=224&top_left_x=130)

![](https://cdn.mathpix.com/cropped/2024_05_06_d79b1c9dd7e6003b3f28g-09.jpg?height=488&width=459&top_left_y=236&top_left_x=136)

For the solution to problem 10.6, method 2

$$
\begin{gathered}
36=100+a^{2}-20 a \cos \left(60^{\circ}-\alpha\right) \\
0=64+a^{2}-20 a\left(\frac{1}{2} \cos \alpha+\frac{\sqrt{3}}{2} \sin \alpha\right) \\
0=64+a^{2}-\frac{1}{2}\left(a^{2}+36\right)-10 a \sqrt{3} \sin \alpha \\
20 a \sqrt{3} \sin \alpha=92+a^{2} \\
\sin \alpha=\frac{92+a^{2}}{20 a \sqrt{3}}
\end{gathered}
$$

According to the fundamental trigonometric identity, we have:

$$
\begin{gathered}
\frac{\left(a^{2}+36\right)^{2}}{400 a^{2}}+\frac{\left(92+a^{2}\right)^{2}}{1200 a^{2}}=1 \\
3\left(a^{4}+72 a^{2}+1296\right)+\left(a^{4}+184 a^{2}+8464\right)=1200 a^{2} \\
4 a^{4}-800 a^{2}+12352=0 \\
a^{4}-200 a^{2}+3088=0 \\
a^{4}-200 a^{2}+10000=10000-3088=6912 \\
\left(a^{2}-100\right)^{2}=2^{8} \cdot 3^{3} \\
a^{2}=100 \pm 48 \sqrt{3}
\end{gathered}
$$

Since $\angle A P B > \angle A C B$, in triangle $A P B$ the side $A B$ is the largest and $a > 5$. Then $a^{2} > 25$ and the value $a^{2}=100-48 \sqrt{3}$ is not valid. Therefore, $a^{2}=100+48 \sqrt{3}$. Then $S_{A B C}=\frac{a^{2} \sqrt{3}}{4}=25 \sqrt{3}+36$.

Method 3. Apply the coordinate method, denoting the side of the equilateral triangle by $a$, assuming that points $A, B$, and $C$ have the following coordinates:

$$
A\left(-\frac{a}{2} ; 0\right), \quad B\left(\frac{a}{2} ; 0\right), \quad C\left(0 ; \frac{a \sqrt{3}}{2}\right)
$$

Let point $P$ have coordinates $P(x ; y)$. By the condition, we have a system of three equations (from which we will then eliminate the unknowns $x, y$):

$$
\left\{\begin{aligned}
(x-0)^{2}+\left(y-\frac{a \sqrt{3}}{2}\right)^{2} & =36 \\
\left(x+\frac{a}{2}\right)^{2}+(y-0)^{2} & =100 \\
\left(x-\frac{a}{2}\right)^{2}+(y-0)^{2} & =64
\end{aligned}\right.
$$

Subtracting the third equation from the second, and then the first equation from the second, we get:

$$
a x=18, a y \sqrt{3}=46+\frac{a^{2}}{2}
$$

Substituting the values of $x, y$ from these equations into the first equation, we derive a quadratic equation in terms of the number $a^{2}$:

$$
a^{4}-200 a^{2}+3088=0
$$

Since by the condition of the problem $a > 10$ (point $P$ is inside triangle $A B C$), then

$$
a^{2}=100+\sqrt{6912}=100+48 \sqrt{3}
$$

Therefore, the area of triangle $A B C$ is

$$
S=\frac{a^{2} \sqrt{3}}{4}=36+25 \sqrt{3}
$$

Answer: $36+25 \sqrt{3}$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| There is an arithmetic error in the correct solution, possibly leading to an incorrect answer | 6 points |
| The side of triangle $A B C$ is correctly found | 5 points |
| The cosine of one of the angles $A B P, B P C, C P A$ is correctly found | 2 points |
| Incorrect solution or its absence | 0 points |