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Erdős–Mordell inequality

From Wikipedia, the free encyclopedia

In Euclidean geometry, the Erdős–Mordell inequality states that for any triangle ABC and point P inside ABC, the sum of the distances from P to the sides is less than or equal to half of the sum of the distances from P to the vertices. It is named after Paul Erdős and Louis Mordell. Erdős (1935) posed the problem of proving the inequality; a proof was provided two years later by Mordell and D. F. Barrow (1937). This solution was however not very elementary. Subsequent simpler proofs were then found by Kazarinoff (1957), Bankoff (1958), and Alsina & Nelsen (2007).

Barrow's inequality is a strengthened version of the Erdős–Mordell inequality in which the distances from P to the sides are replaced by the distances from P to the points where the angle bisectors of ∠APB, ∠BPC, and ∠CPA cross the sides. Although the replaced distances are longer, their sum is still less than or equal to half the sum of the distances to the vertices.

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  • The Erdos-Mordell Inequality
  • Intro to Mordell Theorem
  • Description of the Triangle Inequality

Transcription

Statement

Erdős–Mordell inequality

Let be an arbitrary point P inside a given triangle , and let , , and be the perpendiculars from to the sides of the triangles. (If the triangle is obtuse, one of these perpendiculars may cross through a different side of the triangle and end on the line supporting one of the sides.) Then the inequality states that

Proof

Let the sides of ABC be a opposite A, b opposite B, and c opposite C; also let PA = p, PB = q, PC = r, dist(P;BC) = x, dist(P;CA) = y, dist(P;AB) = z. First, we prove that

This is equivalent to

The right side is the area of triangle ABC, but on the left side, r + z is at least the height of the triangle; consequently, the left side cannot be smaller than the right side. Now reflect P on the angle bisector at C. We find that cray + bx for P's reflection. Similarly, bqaz + cx and apbz + cy. We solve these inequalities for r, q, and p:

Adding the three up, we get

Since the sum of a positive number and its reciprocal is at least 2 by AM–GM inequality, we are finished. Equality holds only for the equilateral triangle, where P is its centroid.

Another strengthened version

Let ABC be a triangle inscribed into a circle (O) and P be a point inside of ABC. Let D, E, F be the orthogonal projections of P onto BC, CA, AB. M, N, Q be the orthogonal projections of P onto tangents to (O) at A, B, C respectively, then:

Equality hold if and only if triangle ABC is equilateral (Dao, Nguyen & Pham 2016; Marinescu & Monea 2017)

A generalization

Let be a convex polygon, and be an interior point of . Let be the distance from to the vertex , the distance from to the side , the segment of the bisector of the angle from to its intersection with the side then (Lenhard 1961):

In absolute geometry

In absolute geometry the Erdős–Mordell inequality is equivalent, as proved in Pambuccian (2008), to the statement that the sum of the angles of a triangle is less than or equal to two right angles.

See also

References

External links

This page was last edited on 2 March 2024, at 17:26
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