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Difference between revisions of "Geometric inequality"
m (→‎Ptolemy's inequality)
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Ptolemy's inequality states that for any quadrilateral <math>ABCD</math>, <math>AB\bullet CD+BC\bullet DA\ge AC\bullet BD</math> with equality when quadrilateral <math>ABCD</math> is cyclic.
 
Ptolemy's inequality states that for any quadrilateral <math>ABCD</math>, <math>AB\bullet CD+BC\bullet DA\ge AC\bullet BD</math> with equality when quadrilateral <math>ABCD</math> is cyclic.
  
−
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}. 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.
+
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.
  
 
==Erdos-Mordell inequality==
 
==Erdos-Mordell inequality==
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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, with equality being when the triangle is equilateral and P is the center.
 
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, with equality being when the triangle is equilateral and P is the center.
  
−
Proof: Let the perpendicular feet from P to BC, CA and AB be D, E and F, respectively. Because </math>AEFP<math> is cyclic with PA as the diameter, </math>PA=EF\sin A<math> or </math>EF=\frac{PA}{\sin A}<math>. Let the perpendicular feet from E and F to BC be M and N, respectively. Note that </math>DN=PF\sin B<math> and </math>DM=PE\sin C<math>. Because MN is the image of EF onto BC, </math>EF\ge MN<math>. This can be written as </math>\frac{PA}{\sin A}\ge PE\sin C+PF\sin B<math>, or </math>PA\ge PE \frac{\sin C}{\sin A}+PF\frac{\sin B}{\sin A}<math>. This is Mordell's Lemma.
+
Proof: Let the perpendicular feet from P to BC, CA and AB be D, E and F, respectively. Because <math>AEFP</math> is cyclic with PA as the diameter, <math>PA=EF\sin A</math> or <math>EF=\frac{PA}{\sin A}</math>. Let the perpendicular feet from E and F to BC be M and N, respectively. Note that <math>DN=PF\sin B</math> and <math>DM=PE\sin C</math>. Because MN is the image of EF onto BC, <math>EF\ge MN</math>. This can be written as <math>\frac{PA}{\sin A}\ge PE\sin C+PF\sin B</math>, or <math>PA\ge PE \frac{\sin C}{\sin A}+PF\frac{\sin B}{\sin A}</math>. This is Mordell's Lemma.
  
−
Because of symmetry, </math>PB\ge PF \frac{\sin A}{\sin B}+PD \frac{\sin C}{\sin B}<math> and </math>PC\ge PD \frac{\sin B}{\sin C}+PE \frac{\sin A}{\sin C}<math> Adding all of these together, we get
+
Because of symmetry, <math>PB\ge PF \frac{\sin A}{\sin B}+PD \frac{\sin C}{\sin B}</math> and <math>PC\ge PD \frac{\sin B}{\sin C}+PE \frac{\sin A}{\sin C}</math> Adding all of these together, we get
  
−
</math>PA+PB+PC\ge PD(\frac{\sin B}{\sin C}+\frac{\sin C}{\sin B})+PE(\frac{\sin C}{\sin A}+\frac{\sin A}{\sin C})+PF(\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A})<math> Because </math>a+\frac{1}{a}<math>, this is equivalent to
+
<math>PA+PB+PC\ge PD(\frac{\sin B}{\sin C}+\frac{\sin C}{\sin B})+PE(\frac{\sin C}{\sin A}+\frac{\sin A}{\sin C})+PF(\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A})</math> Because <math>a+\frac{1}{a}</math>, this is equivalent to
  
−
</math>PA+PB+PC\ge 2(PD+PE+PF)$
+
<math>PA+PB+PC\ge 2(PD+PE+PF)</math>
 
which restates the inequality.
 
which restates the inequality.
  
 
{{stub}}
 
{{stub}}

Revision as of 19:21, 27 August 2015

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

Triangle Inequality

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 $ABCD$, $AB\bullet CD+BC\bullet DA\ge AC\bullet BD$ with equality when quadrilateral $ABCD$ is cyclic.

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.

Erdos-Mordell inequality

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, with equality being when the triangle is equilateral and P is the center.

Proof: Let the perpendicular feet from P to BC, CA and AB be D, E and F, respectively. Because $AEFP$ is cyclic with PA as the diameter, $PA=EF\sin A$ or $EF=\frac{PA}{\sin A}$. Let the perpendicular feet from E and F to BC be M and N, respectively. Note that $DN=PF\sin B$ and $DM=PE\sin C$. Because MN is the image of EF onto BC, $EF\ge MN$. This can be written as $\frac{PA}{\sin A}\ge PE\sin C+PF\sin B$, or $PA\ge PE \frac{\sin C}{\sin A}+PF\frac{\sin B}{\sin A}$. This is Mordell's Lemma.

Because of symmetry, $PB\ge PF \frac{\sin A}{\sin B}+PD \frac{\sin C}{\sin B}$ and $PC\ge PD \frac{\sin B}{\sin C}+PE \frac{\sin A}{\sin C}$ Adding all of these together, we get

$PA+PB+PC\ge PD(\frac{\sin B}{\sin C}+\frac{\sin C}{\sin B})+PE(\frac{\sin C}{\sin A}+\frac{\sin A}{\sin C})+PF(\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A})$ Because $a+\frac{1}{a}$, this is equivalent to

$PA+PB+PC\ge 2(PD+PE+PF)$ which restates the inequality.

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