Резюме. The present paper considers a proof of the Euler-Gergonne’s theorem and its application to two assertions.

Ключови думи: triangle; circumcentre; orthocentre; incentre; centroid; radius of the circumcircle; radius of the incircle of a triangle

The great Swiss mathematician Leonhard Euler (1707 – 1783) is well known as the discoverer of many interesting and important facts and theorems in Geometry. We will mention two of them in the sequel.

Theorem 1. The circumcentre \(O\), the centroid \(T\) and the orthocentre \(H\) of an arbitrary triangle \(\triangle A B C\) are colinear (the common line of the three points is known to be the Euler‘ s line of the triangle). The distance \(|H T|\) between the orthocentre \(H\) and the centroid \(T\) is two times greater than the distance \(|T O|\) between the centroid \(T\) and the circumcentre \(O\), i.e. \(|H T|=2|T O|\).

Theorem 2. The distance \(d\) between the circumcentre \(O\) and the incentre \(I\) of a traingle \(\triangle A B C\) is given by the formula:

\[ d^{2}=|O I|^{2}=R^{2}-2 R r \] where \(R\) and \(r\) are the radii of these circles, respectively.

The proofs of the above theorems could be found in textbooks and books like: (Arslanagić, 2005), (Lopandić, 1971), (Malcheski, Grozdev & Anevska, 2017) and (Palman, 1994). It is also known in the Geometry of traingle the theorem of Gergonne (Joseph Diaz Gergonne (1771 – 1859) is a French astronom and mathematician), which says:

Theorem 3. The lines which connect the vertices \(A, B\) and \(C\) of a triangle \(\triangle A B C\) with the points of tangency \(X, Y\) and \(Z\) of the incircle with the opposite sides of the triangle, respectively are conccurent (the commonn point \(G\) is known to be the Gergonne‘ s point).

The proof of this theorem could be found in (Lopandić, 1971) and (Palman, 1994).

We will mention another theorem in the present paper, known in the mathematical literature on the Geometry of triangle as Euler-Gergonne‘ s theorem. A proof of this theorem will be proposed and some interesting applications of it.

Theorem 4. Given is a triangle \(\triangle A B C\) and let the segments \(A X, B Y\) and \(C Z\) be conccurent, where \(X \in B C, Y \in A C\) and \(Z \in A B\). If \(u=\tfrac{A K}{K X}, v=\tfrac{B K}{K Y}\) and \(w=\tfrac{C K}{K Z}\), then:

(1) \[ \tfrac{1}{1+u}+\tfrac{1}{1+v}+\tfrac{1}{1+w}=1 \]

Proof:

Let \(F_{1}, F_{2}, F_{3}, F_{4}, F_{5}\) and \(F_{6}\) be the areas of the triangles \(\triangle A K Z, \triangle B K Z, \triangle B K X, \triangle C K X, \triangle C K Y\) and \(\triangle A K Y\) (Figure 1.). Using the formula for the area of a triangle, we get:

(2) \(u=\cfrac{A K}{K X}=\cfrac{F_{1}+F_{2}}{F_{3}}=\cfrac{F_{5}+F_{6}}{F_{4}},\)

(3) \(v=\cfrac{B K}{K Y}=\cfrac{F_{3}+F_{4}}{F_{5}}=\cfrac{F_{1}+F_{2}}{F_{6}},\)

(4) \(w=\cfrac{C K}{K Z}=\cfrac{F_{5}+F_{6}}{F_{1}}=\cfrac{F_{3}+F_{4}}{F_{2}} .\)

If \(F\) is the area of the triangle \(\triangle A B C\), then \(F=F_{1}+F_{2}+F_{3}+F_{4}+F_{5}+F_{6}\) and it follows that:

\[ \tfrac{1}{1+u}=\tfrac{F_{3}}{F_{1}+F_{2}+F_{3}}=\tfrac{F_{4}}{F_{4}+F_{5}+F_{6}}=\tfrac{F_{3}+F_{4}}{F_{1}+F_{2}+F_{3}+F_{4}+F_{5}+F_{6}}=\tfrac{F_{3}+F_{4}}{F} \]

\[ \begin{aligned} & \tfrac{1}{1+v}=\tfrac{F_{5}}{F_{3}+F_{4}+F_{5}}=\tfrac{F_{6}}{F_{6}+F_{1}+F_{2}}=\tfrac{F_{5}+F_{6}}{F_{3}+F_{4}+F_{5}+F_{6}+F_{1}+F_{2}}=\tfrac{F_{5}+F_{6}}{F} \\ & \tfrac{1}{1+w}=\tfrac{F_{1}}{F_{1}+F_{5}+F_{6}}=\tfrac{F_{2}}{F_{2}+F_{3}+F_{4}}=\tfrac{F_{1}+F_{2}}{F_{1}+F_{5}+F_{6}+F_{2}+F_{3}+F_{4}}=\tfrac{F_{1}+F_{2}}{F} \end{aligned} \]

Summing up the above three equations, we obtain (1).

By means of relation (1) we will prove the following two assertions:

Assertion 1. Given is an acute triangle \(\triangle A B C\). Let the points \(D, E\) and \(F\) be the feet of the altitudes from the vertices \(A, B\) and \(C\) to the sides \(B C, C A\) and \(A B\), respectively, and let \(H\) be the orthocentre of the triangle. Then:

(5) \[ \tfrac{A H}{A D}+\tfrac{B H}{B E}+\tfrac{C H}{C F}=2 \]

Proof: We will prove the more general assertion, which states, that:

(6) \(\tfrac{A K}{A X}+\tfrac{B K}{B Y}+\tfrac{C K}{C Z}=2\)

where the points \(X, Y, Z\) and \(K\) have the same meaning as in the proof of Theorem 4. If \(u=\tfrac{A K}{K X}, v=\tfrac{B K}{K Y}\) and \(w=\tfrac{C K}{K Z}\), then

(7) \[ \tfrac{A K}{A X}=\tfrac{A K}{A K+K X}=\tfrac{\tfrac{A K}{K X}}{1+\tfrac{A K}{K X}}=\tfrac{u}{1+u} \]

(8) \(\tfrac{B K}{B Y}=\tfrac{B K}{B K+K Y}=\tfrac{\tfrac{B K}{K Y}}{1+\tfrac{B K}{K Y}}=\tfrac{v}{1+v}\),

(9) \(\tfrac{C K}{C Z}=\tfrac{C K}{C K+K Z}=\tfrac{\tfrac{C K}{K Z}}{1+\tfrac{C K}{K Z}}=\tfrac{w}{1+w}\).

Further, using (1), we obtain:

\[ 3-1=\left(\tfrac{1+u}{1+u}+\tfrac{1+v}{1+v}+\tfrac{1+w}{1+w}\right)-\left(\tfrac{1}{1+u}+\tfrac{1}{1+v}+\tfrac{1}{1+w}\right) \Leftrightarrow 2=\tfrac{u}{1+u}+\tfrac{v}{1+v}+\tfrac{w}{1+w} . \] It follows from (7), (8) and (9), that:

\[ \tfrac{A K}{A X}+\tfrac{B K}{B Y}+\tfrac{C K}{C Z}=2 \]

and we are done. Take now \(K \equiv H, X \equiv D, Y \equiv E\) and \(Z \equiv F\). Thus, we get (5).

Assertion 2. Let \(P\) be an arbitrary point in the interior of a triangle \(\triangle A B C\) and let the lines \(A P, B P\) and \(C P\) intersect the sides \(B C, C A\) and \(A B\) in the points \(X, Y\) and \(Z\), respecvtively. Prove that:

(10) \[ \tfrac{P A}{P X} \cdot \tfrac{P B}{P Y}+\tfrac{P B}{P Y} \cdot \tfrac{P C}{P Z}+\tfrac{P C}{P Z} \cdot \tfrac{P A}{P X} \geq 12 . \]

Proof: Let \(u=\tfrac{P A}{P X}, v=\tfrac{P B}{P Y}\) and \(w=\tfrac{P C}{P Z}\). The inequality is equivalent to:

(11) \[ u v+v w+w u \geq 12 . \]

We have \[ \tfrac{1}{1+u}+\tfrac{1}{1+v}+\tfrac{1}{1+w}=1 \Leftrightarrow(1+v)(1+w)+(1+u)(1+w)+(1+u)(1+v)=(1+u)(1+v)(1+w), \] which is equivalent to

(12) \[ u+v+w+2=u v w . \]

Apply now the arithmetic-harmonic inequalty:

\[ \tfrac{(1+u)+(1+v)+(1+w)}{3} \geq \tfrac{3}{\tfrac{1}{1+u}+\tfrac{1}{1+v}+\tfrac{1}{1+w}} \Leftrightarrow(1+u+1+v+1+w)\left(\tfrac{1}{1+u}+\tfrac{1}{1+v}+\tfrac{1}{1+w}\right) \geq 9 \] By means of (1) we get \(3+u+v+w \geq 9\), i.e.

(13) \[ u+v+w \geq 6 . \]

By (12) and (13) we have:

(14) \[ u v w \geq 8 \]

Finally, we apply the arithmetic-geometric inequalty:

\[ \begin{gathered} \tfrac{u v+v w+w u}{3} \geq \sqrt[3]{u v \cdot v w \cdot w u}, \text { i.e. } \\ u v+v w+w u \geq 3 \sqrt[3]{u^{2} v^{2} w^{2}} \Leftrightarrow u v+v w+w u \geq 3 \sqrt[3]{64} \Leftrightarrow u v+v w+w u \geq 12, \end{gathered} \] and we are done.

The equality in (11) holds true iff \(u=v=w=2\), i.e. iff \(\tfrac{P A}{P X}=\tfrac{P B}{P Y}=\tfrac{P C}{P Z}=2\), i.e. when the point \(P\) is the centroid of the equilateral triangle \(\triangle A B C\).

REFERENCES

Arslanagić, Š. (2005). Matematika za nadarene. Sarajevo: Bosanska riječ. Gusić, I. (1995) Matematički riječnik. Zagreb: Element.

Lopandić, D. (1971). Zbirka zadataka iz osnova geometrije. Beograd: Univerzitet u Beogradu, Savez studenata Prirodnomatematičkog fakulteta.

Malcheski, R., S. Grozdev & K. Anevska (2017). Geometry of Complex Numbers. Skopje: Union of Mathematicians of Macedonia. (ISBN 9789989-646-94-2), 278 pages.

Palman, D. (1994). Trokut i kružnica. Zagreb: Element.