Erdős–Mordell inequality

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.

Statement

Erdős–Mordell inequality

Let ${\displaystyle P}$ be an arbitrary point P inside a given triangle ${\displaystyle ABC}$, and let ${\displaystyle PL}$, ${\displaystyle PM}$, and ${\displaystyle PN}$ be the perpendiculars from ${\displaystyle P}$ 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 that

${\displaystyle PA+PB+PC\geq 2(PL+PM+PN)}$

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

${\displaystyle cr\geq ax+by.}$

This is equivalent to

${\displaystyle {\frac {c(r+z)}{2}}\geq {\frac {ax+by+cz}{2}}.}$

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:

${\displaystyle r\geq (a/c)y+(b/c)x,}$

${\displaystyle q\geq (a/b)z+(c/b)x,}$

${\displaystyle p\geq (b/a)z+(c/a)y.}$

Adding the three up, we get

${\displaystyle p+q+r\geq \left({\frac {b}{c}}+{\frac {c}{b}}\right)x+\left({\frac {a}{c}}+{\frac {c}{a}}\right)y+\left({\frac {a}{b}}+{\frac {b}{a}}\right)z.}$

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:

${\displaystyle PM+PN+PQ\geq 2(PD+PE+PF)}$

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

A generalization

Let ${\displaystyle A_{1}A_{2}...A_{n}}$ be a convex polygon, and ${\displaystyle P}$ be an interior point of ${\displaystyle A_{1}A_{2}...A_{n}}$. Let ${\displaystyle R_{i}}$ be the distance from ${\displaystyle P}$ to the vertex ${\displaystyle A_{i}}$ , ${\displaystyle r_{i}}$ the distance from ${\displaystyle P}$ to the side ${\displaystyle A_{i}A_{i+1}}$, ${\displaystyle w_{i}}$ the segment of the bisector of the angle ${\displaystyle A_{i}PA_{i+1}}$ from ${\displaystyle P}$ to its intersection with the side ${\displaystyle A_{i}A_{i+1}}$ then (Lenhard 1961):

${\displaystyle \sum _{i=1}^{n}R_{i}\geq \left(\sec {\frac {\pi }{n}}\right)\sum _{i=1}^{n}w_{i}\geq \left(\sec {\frac {\pi }{n}}\right)\sum _{i=1}^{n}r_{i}}$

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

• List of triangle inequalities

References

• Alsina, Claudi; Nelsen, Roger B. (2007), "A visual proof of the Erdős-Mordell inequality", Forum Geometricorum, 7: 99–102.
• Bankoff, Leon (1958), "An elementary proof of the Erdős-Mordell theorem", American Mathematical Monthly, 65 (7): 521, doi:10.2307/2308580, JSTOR 2308580.
• Dao, Thanh Oai; Nguyen, Tien Dung; Pham, Ngoc Mai (2016), "A strengthened version of the Erdős-Mordell inequality" (PDF), Forum Geometricorum, 16: 317–321, MR 3556993.
• Erdős, Paul (1935), "Problem 3740", American Mathematical Monthly, 42: 396, doi:10.2307/2301373, JSTOR 2301373.
• Kazarinoff, D. K. (1957), "A simple proof of the Erdős-Mordell inequality for triangles", Michigan Mathematical Journal, 4 (2): 97–98, doi:10.1307/mmj/1028988998. (See D. K. Kazarinoff's inequality for tetrahedra.)
• Lenhard, Hans-Christof (1961), "Verallgemeinerung und Verschärfung der Erdös-Mordellschen Ungleichung für Polygone", Archiv für Mathematische Logik und Grundlagenforschung, 12: 311–314, doi:10.1007/BF01650566, MR 0133060, S2CID 124681241.
• Marinescu, Dan Ștefan; Monea, Mihai (2017), "About a strengthened version of the Erdős-Mordell inequality" (PDF), Forum Geometricorum, 17: 197–202.
• Mordell, L. J.; Barrow, D. F. (1937), "Solution to 3740", American Mathematical Monthly, 44: 252–254, doi:10.2307/2300713, JSTOR 2300713.

• Pambuccian, Victor (2008), "The Erdős-Mordell inequality is equivalent to non-positive curvature", Journal of Geometry, 88 (1–2): 134–139, doi:10.1007/s00022-007-1961-4, S2CID 123082256.

External links

• Weisstein, Eric W. "Erdős-Mordell Theorem". MathWorld.
• Alexander Bogomolny, "Erdös-Mordell Inequality", from Cut-the-Knot.