In this paper we give a proof of Oppenheim’s theorem, and give some related results and problems.

Keywords: triangle, inequality, theorem, Oppenheim theorem, area, L.

There are many inequalities for triangles and for their elements (sides, height, bisector, radiuses of the circumscribed and inscribed circles, etc.). In many inequalities the elements of triangles appear symmetrically. Let us denote by the sides the radius of the circumscribed circle, the radius of the inscribed circle, semiperimeter, and area of the triangle respectively.

For instance,

But there are also inequalities which are not symmetric with regard to their elements. For instance, in [3], Parvardi showed that

(1)

These are often special cases of parametric families of inequalities with parameters not depending on the elements of triangle. For example:

If then prove that

(American Mathematical Monthly in1988)

In these inequalities we use Oppenheim theorem ([1], [2]) and it is very useful for many problems.

[Oppenheim’s theorem] For the triangle and real numbers , if then

(2)

The equality holds if and only if

Proof. (based on [2], page 1) Let, . It is known that,

so,

and we get

This inequality is equivalent to

(3)

We use Cauchy’s theorem for the left-hand side of the inequality (3):

(4)

we get,

Use cosine law and we get result:

By the Cauchy inequality in (4), the equality holds if and only if

So, as , the equality holds if and only if

Then, let

To prove the converse,

then equality holds in (2) if and only if

Note that in the theorem the parameters are not only positive numbers, they also can be negative numbers.

Now we study new and beautiful results of the theorem. As mentioned above, Pavardi [3] gave the inequality

Proof. Setting in (2), the result follows immediately.

Result.1 Prove that: , with equality only when

Proof. The usual proof of the Euler inequality uses hard geometric constructions and an explicit calculations of the distanse between the centers of the inscribed and circumscribed circles. We know give a quick proof using Oppenheim’s theorem and the well-known identity (see, Problems in plane and solid geometry, [9]) If we take then by Theorem 0 we have that

Squaring both sides and cancelling gives

Substituting and

(5)

the result follows.

Result 2 Prove that:

with equality only when

Proof. The usual proof of the Leibniz inequality uses hard geometric constructions and an explicit calculation of the distance between the center of the circumscribed circle and centroid of the triangle (see, [9]).Now we prove it algebraically using Theorem 0. Let , , and . We get

Substituting (5) we get

and again the equality holds only when

Result.3 If then prove that,

(Problem E3150, Am. Math. Monthly [5])

Proof. If instead of we consider

then we can easily check that they satisfy the conditions of the theorem.We get

By the Cauchy theorem

then,

References:

1. D.Sachelarie, Geometry of the triangle, Matrix-Rom-Bucuresti, 2000
2. Marcel Chirita, Mathematical Excalibur 17(5), 2013, p.1
3. Amir Hossein Parvardi, 150 Nice geometry problems, 2011
4. Dusan Djukic, Vladimir Jankovich, The IMO Compendium 2010
5. G. Tsintsifas, Problem E3150, Amer. Math. Monthly 93(5), 1986, p.400
6. Samer Seraj, Geometric Inequalities Marathon 2011, problem 4
7. V. E. Olhov, About applying trigonometric functions to demonstrate inequality, Mathematics in School (2), 1979, p.56
8. V.Nicula On an inequality in a triangle dinGM8,1984
9. Viktor Prasolov translated and edited by Dimitry Leites, Problems in plane and solid geometry, 2001