Problem 8.4. The distances from a point $P$, lying inside an equilateral triangle, to its vertices are 3, 4, and 5. Find the area of the triangle.
Answer: $9+\frac{25 \sqrt{3}}{4}$.

Solution. Let the vertices of the triangle be denoted by $A, B$, and $C$ such that $PA = 3$, $PB = 4$, and $PC = 5$. Construct point $C_1$ such that $ACC_1$ is an equilateral triangle and mark a point $P_1$ inside it such that $P_1A = 3$, $P_1C = 4$, and $P_1C_1 = 5$ (Fig. 1). Triangles $AP_1C$ and $APB$ are congruent by the three sides, so $\angle P_1AC = \angle PAB$. Therefore, $\angle PAP_1 = 60^\circ$, meaning triangle $APP_1$ is equilateral. (The same conclusion can be reached by rotating around point $A$ by $60^\circ$.) Consequently, we have $PP_1 = 3$, $PC = 5$, and $P_1C = 4$. From this, by the Pythagorean theorem, triangle $PP_1C$ is a right triangle with legs 3 and 4 and hypotenuse 5, and its area is 6. The area of the equilateral triangle $PP_1A$ with side 3 is $9\sqrt{3}/4$. Therefore, considering the congruence of triangles $APB$ and $AP_1C$, we get

$$
S_{APB} + S_{CPA} = S_{AP_1C} + S_{CPA} = S_{CPP_1} + S_{APP_1} = 6 + \frac{9\sqrt{3}}{4}
$$

## Kurchatov School Olympiad in Mathematics - 2019

Final stage. ![](https://cdn.mathpix.com/cropped/2024_05_06_94f52a7a428b58ebce23g-3.jpg?height=520&width=826&top_left_y=217&top_left_x=615)

Fig. 1: to the solution of problem 8.4

By similar reasoning, we arrive at the equalities

$$
S_{CPA} + S_{BPC} = 6 + \frac{25\sqrt{3}}{4}, \quad S_{BPC} + S_{APB} = 6 + \frac{16\sqrt{3}}{4}
$$

Adding the three obtained relations and dividing by two, we get

$$
S_{ABC} = 9 + \frac{25\sqrt{3}}{4}
$$

## Criteria

The largest suitable criterion is used:

## 7 6. Any correct solution to the problem.

4 6. Any of the sums of areas $S_{APB} + S_{CPA}$, $S_{CPA} + S_{BPC}$, or $S_{BPC} + S_{APB}$ is correctly found.

3 6. The correct algorithm for finding any of the above sums of areas is provided, but an arithmetic error is made.

## 26 . The correct answer is provided.