A “primer” on geometric inequalities for pre-RMO and RMO

The comparison of lengths is more basic than comparison of other geometric quantities (such as angles, areas and volumes). A geometric inequality that involves only the lengths is called a distance inequality.

Some simple axioms and theorems on inequalities in Euclidean geometry are usually the starting point to solve the problem of distance inequality, in which most frequently used tools are:

Proposition I:

The shortest line connecting point A with point B is the segment AB.

The direct corollary of proposition 1 is as follows:

Proposition 2:

(Triangle inequality)

For arbitrary three points A, B and C (lying in the same plane), we have AB \leq AC +CB, the equality holds iff the three points are collinear.

Remark:

In most literature, any symbol of a geometric object also denotes its quantity according to the context.

Proposition 3:

In a triangle, the longer side has the larger opposite angle. And, conversely, the longer angle has the longer opposite side.

Proposition 4:

The median of a triangle on a side is shorter than the half-sum of the other two sides.

Proposition 5:

If a convex polygon is within another one, then the outside convex polygon’s perimeter is larger.

Proposition 6:

Any segment in a convex polygon is either less than the longest side or the longest diagonal of the convex polygon.

Here, is a classic example:

Example 1:

Let a, b, c be the sides of a \triangle ABC. Prove that \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}<2.

Solution 1:

By the triangle inequality, a<b+c yields \frac{a}{b+c} = \frac{2a}{2(b+c)} < \frac{2a}{a+b+c}

Similarly, \frac{b}{c+a} < \frac{2b}{a+b+c} and \frac{c}{b+a} < \frac{2c}{a+b+c}

Adding up the above three inequalities, we get the required inequality.

Example 2:

Let AB be the longest side of \triangle ABC, and P a point in the triangle, prove that PA+PB>PC.

Solution/Proof 2:

Let D be the intersection point of CP and AB. (Note P is in the interior of the triangle ABC). Then, \angle ADC or \angle BDC is not acute. Without loss of generality, we assume that \angle ADC is not acute. Applying proposition 3 to \triangle ADC, we obtain AC \geq CD. Therefore, AB \geq AC \geq CD > PC….call this asĀ  relationship “a”.

Furthermore, applying triangle inequality to \triangle PAB, we have PA+PB>AB…call this as relationship “b”.

Combining “a” and “b”, we obtain the required inequality immediately. QED.

Remarks: (1) If AB is not the longest, then the conclusion may not be true. (2) If point P on the plane of regular triangle ABC, P is not on the circumcircle of the triangle, then the sum of any two of PA, PB, and PC is longer than the remaining one. That is, PA, PB and PC consist of a triangle’s three sides.

Quiz:

Prove that a closed polygonal line with perimeter 1 can be put inside a circle with radius 0.25.

Reference:

Geometric Inequalities, Vol 12, Gangsong Leng, translated by Yongming Liu.

Amazon India link:

Cheers,

Nalin Pithwa.

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