8.5. How many different (i.e., not equal to each other) acute triangles with integer side lengths and a perimeter of 24 exist? List the lengths of the three sides of all these triangles and prove that there are no others.

ANSWER: 6.

CRITERIA

b - correct criterion for acuteness and either a correct list of acute triangles but without full justification (it is stated without justification that only these sets of lengths satisfy the criterion, or the enumeration is incomplete), or a complete enumeration with an arithmetic error leading to the erroneous inclusion or exclusion of a triangle.

$d$ - correct criterion for acuteness and either an incorrect list without detailed proof or with significant errors in the proof, or no list.

*     - a complete list of all triangles, not just acute ones.
Solution. Note that a triangle with sides $a \geqslant b \geqslant c$ is obtuse if and only if $a^{2}>b^{2}+c^{2}$. Indeed, if $A B C$ is a triangle with sides $a \geqslant b \geqslant c$ opposite vertices $A, B$, and $C$ respectively, and $A B^{\prime} C$ is a triangle with a right angle $\angle A$ and $A B^{\prime}=A B$, then $B$ and $C$ lie on opposite sides of the line $A B^{\prime}$, otherwise in the triangle $C B B^{\prime}$ it would be $\angle C B B^{\prime}>\angle A B B^{\prime}=\angle A B^{\prime} B>\angle B B^{\prime} C$, and $B C=ab^{2}+c^{2}$ follows that $\angle C A B>\angle C A B^{\prime}=90^{\circ}$, the converse is proved similarly.

Let's count the triples of natural numbers satisfying the conditions $a>b \geqslant c, a^{2}a^{2}$.

For $a=8$ there is one pair of numbers $(b, c)$ such that $8 \geqslant b \geqslant c$ and $b+c=16$: this is $(8,8)$, and the corresponding triangle is equilateral.

For $a=9$ there are two pairs of numbers $(b, c)$ such that $9 \geqslant b \geqslant c$ and $b+c=15$: these are $(8,7)$ and $(9,6)$, both satisfy the inequality $b^{2}+c^{2}>a^{2}$.

For $a=10$ there are four pairs of numbers $(b, c)$ such that $10 \geqslant b \geqslant c$ and $b+c=14$: these are $(7,7),(8,6),(9,5)$ and $(10,4)$. The last two satisfy the inequality $b^{2}+c^{2}>a^{2}$, while the first two do not.

For $a=11$ there are five pairs of numbers $(b, c)$ such that $11 \geqslant b \geqslant c$ and $b+c=13$: these are $(7,6),(8,5),(9,4),(10,3)$ and $(11,2)$. Only the last one satisfies the inequality $b^{2}+c^{2}>a^{2}$.