5. One side of a certain triangle is twice as large as another, and the perimeter of this triangle is 60. The largest side of the triangle, when added to four times the smallest side, equals 71. Find the sides of this triangle.

(7 points)
Answer: 11, 22, 27.

## Solution:

Let $a, b$, and $c$ be the sides of the triangle, without loss of generality, we assume that $a \leq b \leq c$. Considering the conditions of the problem, we write the system of equations:

$$
\left\{\begin{array}{c}
a+b+c=60 \\
4 a+c=71
\end{array}\right.
$$

Since one of the sides of the triangle is twice as large as another, we consider three possible cases.

1) If $c = 2b$, then $a + b \leq 2b = c$. Therefore, the triangle inequality $a + b > c$, necessary for the existence of the triangle, is not satisfied.
2) If $c = 2a$, then from the second condition of the system we find $6a = 71$, $a = \frac{71}{6}$, then $c = \frac{71}{3}$.

Then we find the value of $b$ from the first equation $b = \frac{147}{6}$, from which it follows that $b > c$, which contradicts the condition that $a \leq b \leq c$.

3) If $b = 2a$, then the system can be written as:

$$
\left\{\begin{array}{l}
3a + c = 60 \\
4a + c = 71
\end{array}\right.
$$

From which $a = 11$, $c = 27$, $b = 22$. The obtained solution satisfies all the conditions of the problem.

Note: 1 point for a correct solution found by trial and error.

## MINISTRY OF SCIENCE AND HIGHER EDUCATION OF THE RF  COUNCIL OF RECTORS OF UNIVERSITIES OF TOMSK REGION  OPEN REGIONAL INTERUNIVERSITY OLYMPIAD 2018-2019  MATHEMATICS (8th GRADE)  INCLUSIVE STAGE  VARIANT 2  (ANSWERS)