In geometry, Barrow's inequality is an inequality relating the distances between an arbitrary point within a triangle, the vertices of the triangle, and certain points on the sides of the triangle. It is named after David Francis Barrow.
Let P be an arbitrary point inside the triangle ABC. From P and ABC, define U, V, and W as the points where the angle bisectors of BPC, CPA, and APB intersect the sides BC, CA, AB, respectively. Then Barrow's inequality states that
PA+PB+PC\geq2(PU+PV+PW),
with equality holding only in the case of an equilateral triangle and P is the center of the triangle.
Barrow's inequality can be extended to convex polygons. For a convex polygon with vertices
A1,A2,\ldots,An
P
Q1,Q2,\ldots,Qn
\angleA1PA2,\ldots,\angleAn-1PAn,\angleAnPA1
A1A2,\ldots,An-1An,AnA1
n|PA | ||
\sum | \sec\left( | |
k|\geq |
\pi | |
n |
\right)
n|PQ | |
\sum | |
k| |
Here
\sec(x)
n=3
\sec\left(\tfrac{\pi}{3}\right)=2
Barrow's inequality strengthens the Erdős–Mordell inequality, which has identical form except with PU, PV, and PW replaced by the three distances of P from the triangle's sides. It is named after David Francis Barrow. Barrow's proof of this inequality was published in 1937, as his solution to a problem posed in the American Mathematical Monthly of proving the Erdős–Mordell inequality.[3] This result was named "Barrow's inequality" as early as 1961.
A simpler proof was later given by Louis J. Mordell.[4]