## Task 4 - 240714

(a) It is assumed about the measurements in centimeters of the side lengths of a triangle:

(1) These measurements are three consecutive natural numbers.

(2) The perimeter of the triangle is 25 cm longer than the shortest side of the triangle. Determine the three side lengths from these assumptions!

(b) Solve the problem if assumption (2) is replaced by the following assumption (2')!

(2') Let $\mathrm{n}$ be a given natural number. The perimeter of the triangle is $\mathrm{n}$ centimeters longer than the shortest side of the triangle. The three side lengths to be found are to be expressed in terms of $\mathrm{n}$.

(c) Investigate which natural numbers $\mathrm{n}$ must be specified in (2') so that a solvable problem results in (b)!
(a) Since the perimeter is the sum of the lengths of the shortest side of the triangle and the other two sides, the sum of the lengths of the other two sides is 25 cm according to (2). Therefore, 25 is the sum of two consecutive natural numbers according to (1). This is only possible if the numbers are 12 and 13. Consequently (and again because of (1)), the sought side lengths are 11 cm, 12 cm, 13 cm.

(b) As in (a), it follows that \( n \) is the sum of two consecutive natural numbers.

If \( m \) is the number in the middle between these two consecutive natural numbers, they are \( m - \frac{1}{2} \) and \( m + \frac{1}{2} \); their sum is thus \( 2m \).

Since this sum is \( n \), \( m = \frac{n}{2} \) must hold. The two consecutive natural numbers are therefore \( \frac{n}{2} - \frac{1}{2} \) and \( \frac{n}{2} + \frac{1}{2} \), and the measures of the three sought side lengths can only be \( \frac{n}{2} - \frac{3}{2} \), \( \frac{n}{2} - \frac{1}{2} \), \( \frac{n}{2} + \frac{1}{2} \).

(c) The problem (b) is solvable if and only if the last-found numbers are natural numbers greater than 0 and also satisfy the triangle inequality condition that the largest of the three measures is less than the sum of the other two measures.

(I) The three numbers are natural numbers exactly when the (pre-given natural) number \( n \) is odd and \( n \geq 3 \).

(II) They are greater than 0 exactly when \( n > 3 \).

(III) For \( n = 5 \), the three measures are 1, 2, 3; they do not satisfy the triangle inequality.

For \( n = 7 \), they are 2, 3, 4 and thus satisfy the triangle inequality.

If \( n \) is increased further, the three measures always increase by a uniform amount. Therefore, the sum of the two shortest lengths increases by twice the amount as the longest length; thus, the triangle inequality remains valid.

With (I), (II), (III), it is proven: In (b), a solvable problem arises exactly when the natural number \( n \) is given as an odd number \( n \geq 7 \).

Solutions of the 1st Round 1984 taken from [5]

### 4.26.2 2nd Round 1984, Class 7