Резюме. A triangle with the property \(3 \alpha+2 \beta=180^{\circ}\) is considered in the paper. Five different ways are proposed to prove that \(a^{2}+b c=c^{2}\). The paper is of methodological character.

Ключови думи: triangle, problem solving, trigonometry, similarity.

To solve a mathematical problem in several ways is challenging and creative but also instructive and useful for a deep examination of the problem history and content to discover the initial idea of its author and to find the potential of the possible applications. This is exceptionally impotant for talented students who touch various mathematical domains by the different solutions, thus increasing their knowledge and capacity. Consider the following problem:

Using the standart notations for \(\triangle A B C\), prove that \(a^{2}+b c=c^{2}\) if \(3 \alpha+2 \beta=180^{0}\). We will give five solutions of this problem.

Solution 1. Accounting for the relations \(3 \alpha+2 \beta=180^{0}\) and \(\alpha+\beta+\gamma=180^{0}\), we have \(\gamma=180^{\circ}-(\alpha+\beta)=3 \alpha+2 \beta-(\alpha+\beta)=2 \alpha+\beta\). It follows that \(\gamma \gt \beta\) and hence \(c \gt b\). Take now a point \(D\) on the side \(A B\) such that \(\angle B C D=\angle A B C=\beta\) (Fig. 1). This implies that \(B D=A D\), i.e.

\(\triangle B C D\) is isosceles. If the lenghts of the legs are denoted by \(x\), then \(A D=c-x\).

By the law of sines and the law of cosines for \(\triangle A D C\) we get:

\[ \tfrac{x}{\sin \alpha}=\tfrac{c-x}{\sin 2 \alpha} \text { and } x^{2}=b^{2}+(c-x)^{2}-2 b(c-x) \cos \alpha . \]

Since \(\sin 2 \alpha=2 \sin \alpha \cos \alpha\), the first equality implies that \(\cos \alpha=\tfrac{c-x}{2 x}\). Now,

the second equality becomes\(x=\tfrac{c^{2}}{b+2 c}\). The law of sines for \(x^{2}=b^{2}+(c-x)^{2}-2 b(c-x) \tfrac{c-x}{2 x}\) \(\triangle B C D\) and the law of cosines forand consequently \(\triangle A B C\) give

\[ \tfrac{x}{\sin \beta}=\tfrac{a}{\sin \left(180^{0}-2 \beta\right)} \text { and } b^{2}=a^{2}+c^{2}-2 a c \cos \beta, \text { respectively. } \]

Since \(\sin \left(180^{\circ}-2 \beta\right)=\sin 2 \beta=2 \sin \beta \cos \beta\), it follows from the first equality that \(\cos \beta=\tfrac{a}{2 x}\). Now, the second equality becomes \(b^{2}=a^{2}+c^{2}-2 a c \tfrac{a}{2 x}\) and consequently \(x=\tfrac{a^{2} c}{a^{2}+c^{2}-b^{2}}\). We have obtained 2 expresions for \(x\), which give:

\[ \tfrac{c^{2}}{b+2 c}=\tfrac{a^{2} c}{a^{2}+c^{2}-b^{2}} \]

It is easy to check that the last is equivalent to \(a^{2}+b c=c^{2}\).

Solution 2. Take a point \(E\) on the side \(A B\) such that \(\angle A C E=\angle C A E=\alpha\) (Fig.2).

Now \(\triangle A E C\) is isosceles and if \(A E=C E=x\), then \(B E=c-x\). Applying the Mollweid’ s formula for \(\triangle B C E\), we get:

\[ \tfrac{(c-x)+x}{a}=\tfrac{\cos \tfrac{(\alpha+\beta)-\beta}{2}}{\sin \tfrac{2 \alpha}{2}}=\tfrac{\cos \tfrac{\alpha}{2}}{2 \sin \tfrac{\alpha}{2} \cos \tfrac{\alpha}{2}}=\tfrac{1}{2 \sin \tfrac{\alpha}{2}} . \]

Thus \(\tfrac{c}{a}=\tfrac{1}{2 \sin \tfrac{\alpha}{2}}\) and consequently \(\sin \tfrac{\alpha}{2}=\tfrac{a}{2 c}\). Further, apply the the law of cosines for \(\triangle A B C\) :

\[ \begin{aligned} a^{2} & =b^{2}+c^{2}-2 b c \cos \alpha=b^{2}+c^{2}-2 b c\left(1-2 \sin ^{2} \tfrac{\alpha}{2}\right)= \\ & =b^{2}+c^{2}-2 b c\left(1-\tfrac{a^{2}}{2 c^{2}}\right)=b^{2}+c^{2}-\tfrac{b}{c}\left(2 c^{2}-a^{2}\right) \end{aligned} \]

Now it is easy to check that the equality \(a^{2}=b^{2}+c^{2}-\tfrac{b}{c}\left(2 c^{2}-a^{2}\right)\) is equivalent to \(a^{2}+b c=c^{2}\), using that \(c \gt b\).

Solution 3. Take a point \(F\) on the line \(A C\) ( \(C\) is between \(A\) and \(F\) ) such that

\(\angle C B F=\alpha+\beta\). (Fig. 3)

Since \(\angle B C F=\alpha+\beta\), then \(\triangle B C F\) is isoceles. Also, we have that \(\angle B F C=\alpha\) and it follows that \(\triangle A B F\) is isoceles too, i.e. \(A B=B F=C F=c\). If \(B H\) is the height of \(\triangle A B F(H \in A F)\), then \(A H=\tfrac{1}{2} A F=\tfrac{b+c}{2}\) and consequently

\(C H=\tfrac{b+c}{2}-b=\tfrac{c-b}{2}\), using again that \(c \gt b\). Finally, apply the Pithagoras theorem to the right triangles \(\triangle B H F\) and \(\triangle B C H\). We have:

\[ B H^{2}=c^{2}-\left(\tfrac{b+c}{2}\right)^{2} \text { and } B H^{2}=a^{2}-\left(\tfrac{c-b}{2}\right)^{2} \]

Now, check that the equality \(c^{2}-\left(\tfrac{b+c}{2}\right)^{2}=a^{2}-\left(\tfrac{c-b}{2}\right)^{2}\) is equivalent to \(a^{2}+b c=c^{2}\).

Solution 4. Applying the Stewart’ s theorem to \(\triangle A B F\) (Fig. 3), we have:

\(A F \cdot\left(A C \cdot C F+B C^{2}\right)=A B^{2} \cdot C F+B F^{2} \cdot A C\), which gives, that

\((b+c)\left(b c+a^{2}\right)=c^{2} c+c^{2} b\), i.e. \((b+c)\left(b c+a^{2}\right)=(c+b) c^{2}\).

The last ie equivalent to \(a^{2}+b c=c^{2}\).

Solution 5. Let \(G \in A B\) be such point that \(\angle B C G=\angle B A C=\alpha\). (Fig. 4)

It follows that \(\angle A G C=\angle A C G=\alpha+\beta\) and \(\triangle A G C\) is isosceles. Hence, \(G B=c-b\). Since \(\angle G C B=\alpha\), the triangles \(A B C\) and \(C B G\) are similar. Thus, \(\tfrac{a}{c-b}=\tfrac{c}{a}\), which is equivalent to \(a^{2}+b c=c^{2}\).

Using trigonometry (the law of sines, the law of cosines and the Mollweid’s formula), solution 1 and solution 2 are rather complicated. On the contrary, solutions \(3-5\) seem to be shorter, avoiding trigonometry. They apply the Pythagoras theorem, the Stewart’ s theorem and similarity of triangles, which are deep mathematical facts. Maybe, this makes them more interesting and creative.

For exercise we recommend the following problems:

1. Given is \(\triangle A B C\). Prove that the relations \(\alpha=2 \beta\) and \(a^{2}=b(b+c)\) are equaivalent.

2. If the relation \(\alpha-\beta=90^{0}\) is valid for \(\triangle A B C\), prove that \(\left(a^{2}-b^{2}\right)^{2}=c^{2}\left(a^{2}+b^{2}\right)\).

(Such a triangle is called pseudoright triangle).

3. The angles \(\alpha, \beta\) and \(\gamma\) of \(\triangle A B C\) satisfy the equality \(2 \gamma=\alpha-\beta\). Prove that

\(c^{2}=a(a-b)\).

4. The proportion \(\alpha: \beta: \gamma=4: 2: 1\) is valid for \(\triangle A B C\). Prove that \(\tfrac{1}{a}+\tfrac{1}{b}=\tfrac{1}{c}\).

5. It is given for an acute \(\triangle A B C\) that \(\angle A C B=2 \angle A B C\). If \(D \in B C\) is such a point that \(2 \angle B A D=\angle A B C\), prove that \(\tfrac{1}{B D}=\tfrac{1}{A B}+\tfrac{1}{A C}\).

6. If the equality \(\alpha=3 \beta\) is valid for \(\triangle A B C\), prove that \(b c^{2}=(a-b)^{2}(a+b)\).

REFERENCES

1. Arslanagić, Š. (2004). Matematika za nadarene. Sarajevo: Bosanska riječ.

2. Blagojević, V. (2002). Teoreme i zadaci iz planimetrije. Sarajevo: Zavod za udžbenike i nastavna sredstva.

3 Marić, A. (1996). Planimetrija – Zbirka riješenih zadataka. Zagreb: Element.