Difference between revisions of "Geometric inequality"
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==Ptolemy's inequality== | ==Ptolemy's inequality== | ||
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Ptolemy's inequality states that for any quadrilateral <math>ABCD</math>, <math>AB\cdot CD+BC\cdot DA\ge AC\cdot BD</math> with equality when quadrilateral <math>ABCD</math> is cyclic. | Ptolemy's inequality states that for any quadrilateral <math>ABCD</math>, <math>AB\cdot CD+BC\cdot DA\ge AC\cdot 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}</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. | + | 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. |
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+ | 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== | ==Erdos-Mordell inequality== | ||
− | The Erdős–Mordell inequality states that if <math>P</math> lies in <math>ABC</math> then <math>PA+PB+PC\ge 2(PD+PE+PF)</math> where <math>D, E, F</math> are the foot of the altitudes from <math>P</math> to <math> | + | The Erdős–Mordell inequality states that if <math>P</math> lies in <math>ABC</math> then <math>PA+PB+PC\ge 2(PD+PE+PF)</math> where <math>D, E, F</math> are the foot of the altitudes from <math>P</math> to <math>BC, AC,</math> and <math>AB</math>, respectively. |
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label("M", M, W); | label("M", M, W); | ||
</asy> | </asy> | ||
− | Note that <math>AFPE</math> is cyclic with diameter <math>AP.</math> By | + | Note that <math>AFPE</math> is cyclic with diameter <math>AP.</math> By the [[Law of Sines]], <math>\dfrac{EF}{\sin A} = 2R=AP\implies EF = AP\sin A.</math> Since <math>BDPF</math> is cyclic, we have that <math>B</math> and <math>\angle FPD</math> are supplementary. Since <math>MPD</math> is a line, <math>B = \angle FPM.</math> This means that <math>\sin B = \sin FPN = \dfrac{FN}{FP}\implies FN = PF\sin B.</math> Similarly, <math>EM = PE\sin C.</math> So the problem is reduced to proving that <math>EF\ge FN+EM</math> but this is obvious by the Pythagorean Theorem. <math>\blacksquare</math> |
Now the rest of the problem is straightforward. We know that | Now the rest of the problem is straightforward. We know that | ||
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PC&\ge PD\dfrac{\sin B}{\sin C} + PE\dfrac{\sin A}{\sin C}.\ | PC&\ge PD\dfrac{\sin B}{\sin C} + PE\dfrac{\sin A}{\sin C}.\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Adding these cyclically implies <cmath>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).</cmath> By AM-GM, <math>PA+PB+PC\ge 2(PD+PE+PF) with equality when ABC is equilateral and P is the center of it. | + | Adding these cyclically implies <cmath>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).</cmath> By AM-GM, <math>PA+PB+PC\ge 2(PD+PE+PF)</math> with equality when ABC is equilateral and P is the center of it. <math>\blacksquare</math> |
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+ | [[Category:Geometry]] | ||
+ | [[Category:Geometric Inequalities]] | ||
{{stub}} | {{stub}} |
Latest revision as of 00:11, 28 October 2024
A geometric inequality is an inequality involving various measures (angles, lengths, areas, etc.) in geometry.
Contents
[hide]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 we have . The Inequality extends this to obtuse and acute triangles. The inequality says:
For an acute triangle with sides of length , . For an obtuse triangle with sides , .
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 and perimeter , then . This means that given a perimeter for a plane figure, the circle has the largest area. Conversely, of all plane figures with area , the circle has the least perimeter.
Trigonometric Inequalities
- In , .
Proof: is a concave function from . Therefore we may use Jensen's inequality:
Alternatively, we may use a method that can be called "perturbation". If we let all the angles 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 , then . This inequality reduces to , which is equivalent to . Since this is always true for , this inequality is true. Therefore, the maximum value of this expression is when , which gives us the value .
Similarly, in , .
Euler's inequality
Euler's inequality states that with equality when is equailateral, where and denote the circumradius and inradius of triangle , respectively.
Proof: The distance from the circumcenter and incenter of a triangle can be expressed as , meaning or equivalently 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 , with equality when quadrilateral is cyclic.
First Proof: Let P be the point such that . By SAS we also have that . By the triangle inequality, . calculating the lengths, we obtain an equivalent statement: . Multiplying by we get the desired result with equality when P is on DC. This happens when . But so , or quadrilateral is cyclic.
Second Proof (using inversion): Let the inversion map B,C and D to B',C' and D' respectively. We then have By the triangle inequality, we have By multiplying on both sides we get the desired result with equality when is collinear, implying either ABCD is cyclic or collinear.
Erdos-Mordell inequality
The Erdős–Mordell inequality states that if lies in then where are the foot of the altitudes from to and , respectively.
Proof: First, we prove a lemma.
Mordell's Lemma:
Proof of Lemma: Let and be the projections of and onto line
Note that is cyclic with diameter By the Law of Sines, Since is cyclic, we have that and are supplementary. Since is a line, This means that Similarly, So the problem is reduced to proving that but this is obvious by the Pythagorean Theorem.
Now the rest of the problem is straightforward. We know that Adding these cyclically implies By AM-GM, with equality when ABC is equilateral and P is the center of it. This article is a stub. Help us out by expanding it.