Difference between revisions of "Geometric inequality"

Revision as of 19:49, 23 October 2019

A geometric inequality is an inequality involving various measures (angles, lengths, areas, etc.) in geometry.

The Triangle Inequality says that the sum of the lengths of any two sides of a nondegenerate triangle is greater than the length of the third side. This inequality is particularly useful and shows up frequently on Intermediate level geometry problems. It also provides the basis for the definition of a metric space in analysis.

Pythagorean Inequality

The Pythagorean Inequality is a generalization of the Pythagorean Theorem. The Theorem states that in a right triangle with sides of length $a \leq b \leq c$ we have $a^2 + b^2 = c^2$ . The Inequality extends this to obtuse and acute triangles. The inequality says:

For an acute triangle with sides of length $a \leq b \leq c$ , $a^2+b^2>c^2$ . For an obtuse triangle with sides $a \leq b \leq c$ , $a^2+b^2<c^2$ .

This inequality is a direct result of the Law of Cosines, although it is also possible to prove without using trigonometry.

Isoperimetric Inequality

The Isoperimetric Inequality states that if a figure in the plane has area $A$ and perimeter $P$ , then $\frac{4\pi A}{P^2} \le 1$ . This means that given a perimeter $P$ for a plane figure, the circle has the largest area. Conversely, of all plane figures with area $A$ , the circle has the least perimeter.

Trigonometric Inequalities

In $\triangle ABC$ , $\sin{A}+\sin{B}+\sin{C}\le \frac{3\sqrt{3}}{2}$ .

Proof: $\sin$ is a concave function from $0\le \theta \le \pi$ . Therefore we may use Jensen's inequality: $\frac{\sin{A}+\sin{B}+\sin{C}}{3}\le \sin{\left(\frac{A+B+C}{3}\right)}=\frac{\sqrt{3}}{2}$

Alternatively, we may use a method that can be called "perturbation". If we let all the angles $A,B,C$ be equal, we prove that if we make one angle greater and the other one smaller, we will decrease the total value of the expression. To prove this, all we need to show is if $0<A,B<180$ , then $\sin(A+B)+\sin(A-B)<2\sin A$ . This inequality reduces to $2\sin A \cos B<2\sin A$ , which is equivalent to $\cos B<1$ . Since this is always true for $0<B<180$ , this inequality is true. Therefore, the maximum value of this expression is when $A=B=C=60$ , which gives us the value $\sin{A}+\sin{B}+\sin {C}=\frac{3\sqrt{3}}{2}$ .

Similarly, in $\triangle ABC$ , $\cos{A}+\cos{B}+\cos{C}\le \frac{3}{2}$ .

Euler's inequality

Euler's inequality states that $R\ge2r$ with equality when $\triangle ABC$ is equailateral, where $R$ and $r$ denote the circumradius and inradius of triangle $ABC$ , respectively.

Proof: The distance $d$ from the circumcenter and incenter of a triangle can be expressed as $d^2=R(R-2r)$ , meaning $R-2r\ge 0$ or equivalently $R=2r$ with equality if and only if the incenter equals the circumcenter, namely the triangle is equilateral.

Ptolemy's inequality

Ptolemy's inequality states that for any quadrilateral $ABCD$ , $AB\cdot CD+BC\cdot DA\ge AC\cdot BD$ with equality when quadrilateral $ABCD$ is cyclic.

First Proof: Let P be the point such that $\triangle ABC\sim \triangle ADP$ . By SAS we also have that $\triangle ABD\sim \triangle ACP$ . By the triangle inequality, $PD+DC\ge PC$ . calculating the lengths, we obtain an equivalent statement: $BC\frac{DA}{AB}+CD\ge BD \frac{AC}{AB}$ . Multiplying by $AB$ we get the desired result with equality when P is on DC. This happens when $\angle ADP+\angle ADC=180^{\circ}$ . But $\angle ABC\cong \angle ADP$ so $\angle ABC+\angle ADC=180^{\circ}$ , or quadrilateral $ABCD$ is cyclic.

Second Proof (using inversion): Let the inversion $\psi(A,1)$ map B,C and D to B',C' and D' respectively. We then have $B'C'=\frac{BC}{AB\cdot AC}$ $C'D'=\frac{CD}{AC\cdot AD}$ $B'D'=\frac{BD}{AB\cdot AD}.$ By the triangle inequality, we have $B'C'+C'D'\ge B'D' \implies \frac{BC}{AB\cdot AC}+\frac{CD}{AC\cdot AD}\ge \frac{BD}{AB\cdot AD}.$ By multiplying $AB\cdot AC\cdot AD$ on both sides we get the desired result with equality when $B'C'D'$ is collinear, implying either ABCD is cyclic or collinear.

Erdos-Mordell inequality

The Erdős–Mordell inequality states that if $P$ lies in $ABC$ then $PA+PB+PC\ge 2(PD+PE+PF)$ where $D, E, F$ are the foot of the altitudes from $P$ to $AB, BC,$ and $AC.$

Proof: First, we prove a lemma.

Mordell's Lemma: $PA\sin A\ge PE\sin C+PF\sin B$

Proof of Lemma: Let $M$ and $N$ be the projections of $E$ and $F$ onto line $PD.$ $[asy] import geometry; import olympiad; size(400); point A = (2, 5), B = (0, 0), C = (10, 0), P = (3, 2); line c = line(A, B); line a = line(C, B); line b = line(A, C); point D = projection(a) * P; draw(P -- D); point e = projection(b) * P; draw(P -- e); point F = projection(c) * P; draw(P -- F); draw(A--B--C--A); draw(P--A); draw(P--B); draw(P--C); markrightangle(B, D, P); markrightangle(A, e, P); markrightangle(A, F, P); draw(circumcircle(A, P, e), dashed); line d = line(P, D); draw(d); point n = projection(d) * F; point M = projection(d) * e; draw(F--n); draw(e--M); label("A", A, N); label("B", B, SW); label("C", C, SE); label("D", D, SW); label("E", e - (0.1, 0), NE); label("F", F, W); label("P", P+(0.2, 0), S); label("N", n, E); label("M", M, W); [/asy]$ Note that $AFPE$ is cyclic with diameter $AP.$ By ELOS, $\dfrac{EF}{\sin A} = 2R=AP\implies EF = AP\sin A.$ Since $BDPF$ is cyclic, we have that $B$ and $\angle FPD$ are supplementary. Since $MPD$ is a line, $B = \angle FPM.$ This means that $\sin B = \sin FPN = \dfrac{FN}{FP}\implies FN = PF\sin B.$ Similarly, $EM = PE\sin C.$ So the problem is reduced to proving that $EF\ge FN+EM$ but this is obvious by the Pythagorean Theorem. $\blacksquare$

Now the rest of the problem is straightforward. We know that $\begin{align*} PA&\ge PE\dfrac{\sin C}{\sin A} + PF\dfrac{\sin B}{\sin A}\\ PB&\ge PF\dfrac{\sin A}{\sin B} + PD\dfrac{\sin C}{\sin B}\\ PC&\ge PD\dfrac{\sin B}{\sin C} + PE\dfrac{\sin A}{\sin C}.\\ \end{align*}$ Adding these cyclically implies $PA+PB+PC\ge PD\left(\frac{\sin B}{\sin C}+\frac{\sin C}{\sin B}\right)+PE\left(\frac{\sin C}{\sin A}+\frac{\sin A}{\sin C}\right)+PF\left(\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}\right).$ By AM-GM, $PA+PB+PC\ge 2(PD+PE+PF)$ with equality when ABC is equilateral and P is the center of it. $\blacksquare$

This article is a stub. Help us out by expanding it.

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 First Proof: Let P be the point such that <math>\triangle ABC\sim \triangle ADP</math>. By SAS we also have that <math>\triangle ABD\sim \triangle ACP</math>. By the triangle inequality, <math>PD+DC\ge PC</math>. calculating the lengths, we obtain an equivalent statement: <math>BC\frac{DA}{AB}+CD\ge BD \frac{AC}{AB}</math>. Multiplying by <math>AB</math> we get the desired result with equality when P is on DC. This happens when <math>\angle ADP+\angle ADC=180^{\circ}</math>. But <math>\angle ABC\cong \angle ADP</math> so <math>\angle ABC+\angle ADC=180^{\circ}</math>, or quadrilateral <math>ABCD</math> is cyclic.
-Second Proof (using inversion): Let the inversion <math>\psi(A,1)</math> map B,C and D to B',C' and D' respectively. We then have <cmath>B'C'=\frac{BC}{AB\cdot AC}\\ C'D'=\frac{CD}{AC\cdot AD}\\ B'D'=\frac{BD}{AB\cdot AD}\\ </cmath>. By the triangle inequality, we have <cmath>B'C'+C'D'\ge B'D' \implies \frac{BC}{AB\cdot AC}+\frac{CD}{AC\cdot AD}\ge \frac{BD}{AB\cdot AD}</cmath>. By multiplying <math>AB\cdot AC\cdot AD</math> on both sides we get the desired result with equality when <math>B'C'D'</math> is collinear, implying either ABCD is cyclic or collinear.
+Second Proof (using inversion): Let the inversion <math>\psi(A,1)</math> map B,C and D to B',C' and D' respectively. We then have <cmath>B'C'=\frac{BC}{AB\cdot AC}</cmath> <cmath>C'D'=\frac{CD}{AC\cdot AD}</cmath> <cmath>B'D'=\frac{BD}{AB\cdot AD}. </cmath> By the triangle inequality, we have <cmath>B'C'+C'D'\ge B'D' \implies \frac{BC}{AB\cdot AC}+\frac{CD}{AC\cdot AD}\ge \frac{BD}{AB\cdot AD}.</cmath> By multiplying <math>AB\cdot AC\cdot AD</math> on both sides we get the desired result with equality when <math>B'C'D'</math> is collinear, implying either ABCD is cyclic or collinear.
 ==Erdos-Mordell inequality==

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