Difference between revisions of "Geometric inequality"
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A '''geometric inequality''' is an [[inequality]] involving various measures ([[angle]]s, [[length]]s, [[area]]s, etc.) in [[geometry]]. | A '''geometric inequality''' is an [[inequality]] involving various measures ([[angle]]s, [[length]]s, [[area]]s, etc.) in [[geometry]]. | ||
+ | |||
+ | ===Triangle Inequality=== | ||
+ | The [[Triangle Inequality]] says that the sum of the lengths of any two sides of a non[[degenerate]] 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=== | ===Pythagorean Inequality=== | ||
Line 8: | Line 11: | ||
This inequality is a direct result of the [[Law of Cosines]], although it is also possible to prove without using [[trigonometry]]. | 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]] <math>A</math> and [[perimeter]] <math>P</math>, then <math>\frac{4\pi A}{P^2} \le 1</math>. This means that given a perimeter <math>P</math> for a plane figure, the [[circle]] has the largest area. Conversely, of all plane figures with area <math>A</math>, the circle has the least perimeter. | ||
+ | |||
+ | ===Trigonometric Inequalities=== | ||
+ | *In <math>\triangle ABC</math>, <math>\sin{A}+\sin{B}+\sin{C}\le \frac{3\sqrt{3}}{2}</math>. | ||
+ | Proof: <math>\sin</math> is a [[concave function]] from <math>0\le \theta \le \pi</math>. Therefore we may use [[Jensen's inequality]]: <math>\frac{\sin{A}+\sin{B}+\sin{C}}{3}\le \sin{\left(\frac{A+B+C}{3}\right)}=\frac{\sqrt{3}}{2}</math> | ||
− | === | + | Alternatively, we may use a method that can be called "perturbation". If we let all the angles <math>A,B,C</math> 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 <math>0<A,B<180</math>, then <math>\sin(A+B)+\sin(A-B)<2\sin A</math>. This inequality reduces to <math>2\sin A \cos B<2\sin A</math>, which is equivalent to <math>\cos B<1</math>. Since this is always true for <math>0<B<180</math>, this inequality is true. Therefore, the maximum value of this expression is when <math>A=B=C=60</math>, which gives us the value <math>\sin{A}+\sin{B}+\sin {C}=\frac{3\sqrt{3}}{2}</math>. |
− | + | ||
+ | Similarly, in <math>\triangle ABC</math>, <math>\cos{A}+\cos{B}+\cos{C}\le \frac{3}{2}</math>. | ||
+ | |||
+ | ==Euler's inequality== | ||
+ | |||
+ | [[Euler's inequality]] states that <math>R\ge2r</math> with equality when <math>\triangle ABC</math> is equailateral, where <math>R</math> and <math>r</math> denote the circumradius and inradius of triangle <math>ABC</math>, respectively. | ||
+ | |||
+ | Proof: The distance <math>d</math> from the circumcenter and incenter of a triangle can be expressed as <math>d^2=R(R-2r)</math>, meaning <math>R-2r\ge 0</math> or equivalently <math>R=2r</math> 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 <math>ABCD</math>, <math>AB\cdot CD+BC\cdot DA\ge AC\cdot BD</math> with equality when 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. | ||
+ | |||
+ | 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== | ||
+ | |||
+ | 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>AB, BC,</math> and <math>AC.</math> | ||
+ | |||
+ | |||
+ | <b>Proof: </b> First, we prove a lemma. | ||
+ | |||
+ | |||
+ | <b>Mordell's Lemma: </b> <math>PA\sin A\ge PE\sin C+PF\sin B</math> | ||
+ | |||
+ | |||
+ | <b>Proof of Lemma: </b> Let <math>M</math> and <math>N</math> be the projections of <math>E</math> and <math>F</math> onto line <math>PD.</math> | ||
+ | <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 <math>AFPE</math> is cyclic with diameter <math>AP.</math> By ELOS, <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 | ||
+ | <cmath>\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*}</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)</math> with equality when ABC is equilateral and P is the center of it. <math>\blacksquare</math> | ||
− | + | [[Category:Geometry]] | |
− | + | [[Category:Geometric Inequalities]] | |
+ | {{stub}} |
Latest revision as of 14:22, 22 October 2023
A geometric inequality is an inequality involving various measures (angles, lengths, areas, etc.) in geometry.
Contents
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
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 ELOS,
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.
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