Резюме. One more proof is proposed in the present paper concerning the the distance between the incentre \((I)\) and the orthocentre \((H)\) of the triangle.

Ключови думи: distance, incentre, orthocentre, sine law, cosine law, centroid of a triangle.

The following equality is considered in the paper:

(1) \[ |I H|^{2}=2 r^{2}-4 R^{2} \cos \alpha \cos \beta \cos \gamma, \]

where \(r\) and \(R\) are respectively the radii of the incircle and the circumcircle of the triangle \(\triangle A B C\) with angles \(\alpha, \beta\) and \(\gamma\). One proof of the equality, which is not simple at all, is propesed by the present author in his paper (Arslanagić, 2005, pp. 437-439). For a new proof the next two equalities will be used:

(2) \[ |I O|^{2}=R^{2}-2 R r, \]

where \(O\) is the circumcentre of the triangle and

(3) \[ |O H|^{2}=9 r^{2}-\left(a^{2}+b^{2}+c^{2}\right), \]

where \(a, b, c\) a are the lengths of its sides.

Two proofs of the equality (2) are proposed by the present author in (Arslanagić, 2005, pp. 432-434) and one my proof in (Arslanagić, 2009, pp. 79-80). Similarly, two proofs of the equality (3) are proposed in (Arslanagić, 2005, pp. 435-437).

One more equality will be used in the sequel:

(4) \[ |I T|^{2}=\tfrac{2}{3} r^{2}-\tfrac{4}{3} R r+\tfrac{1}{18}\left(a^{2}+b^{2}+c^{2}\right), \]

where \(T\) is the centroid of the triangle \(\triangle A B C\). Here is a proof of (4):

Proof: The idea is to apply Leibniz theorem for the triangle \(\triangle A B C\), see (Arslanagić, 2005, pp. 354-355) or (Grozdev, 2007), i.e.:

(5) \[ |M A|^{2}+|M B|^{2}+|M C|^{2}=3|M T|^{2}+|A T|^{2}+|B T|^{2}+|C T|^{2}, \]

where the point \(M\) is not necessary in the plane of the triangle \(\triangle A B C\). Since \(|A T|=\tfrac{2}{3} m_{a},|B T|=\tfrac{2}{3} m_{b},|C T|=\tfrac{2}{3} m_{c},\left(m_{a}, m_{b}, m_{c}\right.\) are the medians of the traingle \(\triangle A B C\) ), we have:

\[ |A T|^{2}+|B T|^{2}+|C T|^{2}=\tfrac{4}{9}\left(m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\right)=\tfrac{4}{9} \cdot \tfrac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)=\tfrac{1}{3}\left(a^{2}+b^{2}+c^{2}\right) . \]

By the substitutions \(|M A|=x,|M B|=y,|M C|=z\) we get:

\[ 9|M T|^{2}=3\left(x^{2}+y^{2}+z^{2}\right)-\left(a^{2}+b^{2}+c^{2}\right) . \]

If \(M\) is replaced by \(I\), then:

(6) \[ \begin{gathered} x^{2}=(s-a)^{2}+r^{2}, y^{2}=(s-b)^{2}+r^{2}, z^{2}=(s-c)^{2}+r^{2} ;\left(s=\tfrac{1}{2}(a+b+c)\right) \\ 9|I T|^{2}=3\left[(s-a)^{2}+r^{2}+(s-b)^{2}+r^{2}+(s-c)^{2}+r^{2}\right]-\left(a^{2}+b^{2}+c^{2}\right) \\ \Rightarrow 9|I T|^{2}=3\left[\left(3 s^{2}+2 s(a+b+c)+a^{2}+b^{2}+c^{2}+3 r^{2}\right)\right]-\left(a^{2}+b^{2}+c^{2}\right) \\ \Rightarrow 9|I T|^{2}=9 r^{2}-3 s^{2}+2\left(a^{2}+b^{2}+c^{2}\right) \end{gathered} \]

Since \(a^{2}+b^{2}+c^{2}=2 s^{2}-2 r^{2}-8 R r\), then.

(7) \[ \begin{gathered} 2 s^{2}=a^{2}+b^{2}+c^{2}+2 r^{2}+8 R r \\ \Rightarrow s^{2}=\tfrac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)+r^{2}+4 R r \end{gathered} \]

From (6) and (7) we get:

\[ \begin{aligned} 9|I T|^{2}= & 9 r^{2}-\tfrac{3}{2}\left(a^{2}+b^{2}+c^{2}\right)-3 r^{2}-12 R r+2\left(a^{2}+b^{2}+c^{2}\right) \\ & \Rightarrow|I T|^{2}=\tfrac{2}{3} r^{2}-\tfrac{4}{3} R r+\tfrac{1}{18}\left(a^{2}+b^{2}+c^{2}\right) \end{aligned} \] and this is the equality (4).

Further, we give the proof of the equality (1).

By the cosine law for the triangle \(\triangle T I O\) :

(8) \[ |I T|^{2}=|I O|^{2}+|O T|^{2}-2|I O| \cdot|O T| \cos \varphi \]

where \(\varphi=\measuredangle I O T\).

It is well known that the points \(O, T\) and \(H\) are collinear, lying on the Euler’s line. We will use the corresponding Euler’s theorem:

(9) \[ |O T|=\tfrac{1}{3}|O H| \]

From (8) and (9) it follows, that

\[ \cos \varphi=\tfrac{|I O|^{2}+|O T|^{2}-|I T|^{2}}{2|I O| \cdot|O T|} \]

(10) \[ \begin{aligned} & \Rightarrow \cos \varphi=\tfrac{|I O|^{2}+\tfrac{1}{9}|O H|^{2}-|I T|^{2}}{2|I O| \cdot \tfrac{1}{3}|O H|} \\ & \Rightarrow \cos \varphi=\tfrac{9|I O|^{2}+|O H|^{2}-9|I T|^{2}}{6|I O| \cdot|O H|} \end{aligned} \]

Apply now the cosine law for the triangle \(\triangle I H O\) : \[ |I H|^{2}=|I O|^{2}+|O H|^{2}-2|I O| \cdot|O H| \cos \varphi, \]

and from here, because of (10), we have:

(11) \[ \begin{gathered} |I H|^{2}=|I O|^{2}+|O H|^{2}-2|I O| \cdot|O H| \cdot \tfrac{9|I O|^{2}+|O H|^{2}-9|I T|^{2}}{6|I O| \cdot|O H|} \\ \Rightarrow|I H|^{2}=|I O|^{2}+|O H|^{2}-3|I O|^{2}-\tfrac{1}{3}|O H|^{2}+3|I T|^{2} \\ \Rightarrow|I H|^{2}=-2|I O|^{2}+\tfrac{2}{3}|O H|^{2}+3|I T|^{2} \end{gathered} \]

Finally, from (11) on the base of (2), (3) and (4), we obtain:

\[ \begin{aligned} |I H|^{2}=-2 R^{2} & +4 R r+6 R^{2}-\tfrac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)+2 r^{2}-4 R r+\tfrac{1}{6}\left(a^{2}+b^{2}+c^{2}\right) \\ & \Rightarrow|I H|^{2}=4 R^{2}+2 r^{2}-\tfrac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) \end{aligned} \] next by the sine law:

\[ \Rightarrow|I H|^{2}=4 R^{2}+2 r-\tfrac{1}{2} \cdot 4 R^{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right), \]

and by the well known identity \[ \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2+2 \cos \alpha \cos \beta \cos \gamma \] we have:

\[ \begin{gathered} |I H|^{2}=4 R^{2}+2 r^{2}-2 R^{2}(2+2 \cos \alpha \cos \beta \cos \gamma), \text { i.e. } \\ |I H|^{2}=2 r^{2}-4 R^{2} \cos \alpha \cos \beta \cos \gamma \end{gathered} \]

Corollary. Prove the inequality 2.23 from (Bottema & al., 1969, p. 25), i.e.

(12) \[ \cos \alpha \cos \beta \cos \gamma \leq \tfrac{1}{8}, \]

where the equality holds true if and only if \(\alpha=\beta=\gamma\), i.e. if the triangle is equilateral.

Proof: Since \(|I H|^{2} \geq 0\), it follows from (1), that:

(13) \[ \begin{gathered} 2 r^{2}-4 R^{2} \cos \alpha \cos \beta \cos \gamma \geq 0 \\ \Rightarrow \cos \alpha \cos \beta \cos \gamma \leq \tfrac{r^{2}}{2 R^{2}} \end{gathered} \]

It is well known the inequality \(R \geq 2 r\) (it follows from (2): \(|I O|^{2} \geq 0 \Rightarrow R^{2}-2 R r \geq 0 \Rightarrow R \geq 2 r\) ). Thus, \(\tfrac{r}{R} \leq \tfrac{1}{2} \Rightarrow \tfrac{r^{2}}{R^{2}} \leq \tfrac{1}{4} \Rightarrow \tfrac{r^{2}}{2 R^{2}} \leq \tfrac{1}{8}\) and we conclude that the inequality (13) is stronger than the inequality (12). The equality in (13) holds true for the equilateral triangle too.

REFERENCES

Arslanagić, Š. (2005). Matematika za nadarene, Sarajevo: Bosanska riječ.

Grozdev, S. (2007). For High Achievements in Mathematics. The Bulgarian Experience (Theory and Practice) . Sofia: ADE. (ISBN 978-954-92139-1-1), 295 pages.

Arslanagić, Š. (2009). Matematička čitanka 1, Sarajevo: Grafičar promet d.o.o..

Bottema, O., R. Z. Đorđević, R. R. Janić, D. S. Mitrinović & P. M. Vasić (1969). Geometric Inequalities, Gronningen (The Netherlands): Wolters-Noordhoff Publishing.