Difference between revisions of "Geometric inequality"
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==Erdos-Mordell inequality== | ==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. | + | 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 foot 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 | ||
+ | |||
+ | <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> | ||
+ | which restates the inequality. | ||
{{stub}} | {{stub}} |
Revision as of 00:56, 6 June 2015
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 , where
and
denote the circumradius and inradius of triangle
, respectively.
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 foot from P to BC, CA and AB be D, E and F, respectively. Because is cyclic with PA as the diameter,
or
. Let the perpendicular feet from E and F to BC be M and N, respectively. Note that
and
. Because MN is the image of EF onto BC,
. This can be written as
, or
. This is Mordell's Lemma.
Because of symmetry, and
Adding all of these together, we get
Because
, this is equivalent to
which restates the inequality.
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