Erdős–Mordell inequality explained

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. posed the problem of proving the inequality; a proof was provided two years later by . This solution was however not very elementary. Subsequent simpler proofs were then found by,, and .

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.

Statement

Let

P be an arbitrary point P inside a given triangle ABC , and let PL , PM , and PN be the perpendiculars from P 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 thatPA + PB + PC\geq 2 (PL + PM + PN)

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

cr\geq ax + by.

This is equivalent to

(c(r + z))/(2 \geq) (ax + by + cz)/(2 .)

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 cr ≥ ay + bx for P's reflection. Similarly, bq ≥ az + cx and ap ≥ bz + cy. We solve these inequalities for r, q, and p:

r\geq (a/c)y + (b/c)x, q\geq (a/b)z + (c/b)x, p\geq (b/a)z + (c/a)y.

Adding the three up, we get

p + q + r \geq \left( (b)/(c) + (c)/(b) \right) x + \left( (a)/(c) + (c)/(a) \right) y + \left( (a)/(b) + (b)/(a) \right) z.

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:

PM + PN + PQ \ge 2 (PD + PE + PF) Equality hold if and only if triangle ABC is equilateral (;)

A generalization

Let

A _{1A} _{2...A} _{n} be a convex polygon, and P be an interior point of A _{1A} _{2...A} _{n} . Let R _{i} be the distance from P to the vertex A _{i} , r _{i} the distance from P to the side A _{iA} _{i+1} , w _{i} the segment of the bisector of the angle A _{iPA} _{i+1} from P to its intersection with the side A _{iA} _{i+1} then :\sum ^{n} _{i =1} R _{i} \ge \left(\sec{ (\pi)/(n) }\right)\sum_^ w_i \ge \left(\sec\right)\sum_^ r_i

In absolute geometry

In absolute geometry the Erdős–Mordell inequality is equivalent, as proved in, to the statementthat the sum of the angles of a triangle is less than or equal to two right angles.

See also

• List of triangle inequalities

References

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External links

• • Alexander Bogomolny, "Erdös-Mordell Inequality", from Cut-the-Knot.