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'''Van der Waerden's theorem''' is a theorem in the branch of [[mathematics]] called [[Ramsey theory]].  Van der Waerden's theorem states that for any given positive [[integer]]s ''r'' and ''k'', there is some number ''N'' such that if the integers {1, 2, ..., ''N''} are [[Graph coloring|colored]], each with one of ''r'' different colors, then there are at least ''k'' integers in [[arithmetic progression]] all of the same color. The least such ''N'' is the [[Van der Waerden number]] ''W''(''r'',&nbsp;''k'').  It is named after the Dutch mathematician [[Bartel Leendert van der Waerden|B. L. van der Waerden]].<ref>{{cite journal |authorlink=Bartel Leendert van der Waerden |first=B. L. |last=van der Waerden |title={{lang|de|Beweis einer Baudetschen Vermutung}} |journal=Nieuw. Arch. Wisk. |volume=15 |year=1927 |issue= |pages=212–216 }}</ref>
:'' '''Incircle''' redirects here. For incircles of non-triangle polygons, see [[Tangential quadrilateral]] or [[Tangential polygon]].


For example, when ''r'' = 2, you have two [[:Category:Color|colors]], say [[red|<font color=red>red</font>]] and [[blue|<font color=blue>blue</font>]]. ''W''(2, 3) is bigger than 8, because you can color the integers from {1, ..., 8} like this:
[[Image:Incircle and Excircles.svg|right|thumb|300px|A triangle (black) with incircle (blue), [[incenter]] (I), excircles (orange), excenters (J<sub>A</sub>,J<sub>B</sub>,J<sub>C</sub>), internal [[angle bisector]]s (red) and external angle bisectors (green)]]
In [[geometry]], the '''incircle''' or '''inscribed circle''' of a [[triangle]] is the largest [[circle]] contained in the triangle; it touches (is [[tangent]] to) the three sides. The center of the incircle is called the triangle's '''incenter'''.


        '''1''' &nbsp;<u>2</u> &nbsp;<u>3</u> &nbsp;'''4''' &nbsp;'''5''' &nbsp;<u>6</u> &nbsp;<u>7</u> &nbsp;'''8'''
An '''excircle''' or '''escribed circle''' of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two.
        '''<font color=blue>B</font>''' &nbsp;<u><font color=red>R</font></u> &nbsp;<u><font color=red>R</font></u> &nbsp;'''<font color=blue>B</font>''' &nbsp;'''<font color=blue>B</font>''' &nbsp;<u><font color=red>R</font></u> &nbsp;<u><font color=red>R</font></u> &nbsp;'''<font color=blue>B</font>'''  
Every triangle has three distinct excircles, each tangent to one of the triangle's sides.


and no three integers of the same color form an [[arithmetic progression]]. But you can't add a ninth integer to the end without creating such a progression.  If you add a [[red|<font color=red>red 9</font>]], then the [[red|<font color=red>red 3</font>]], [[red|<font color=red>6</font>]], and [[red|<font color=red>9</font>]] are in arithmetic progression. Alternatively, if you add a [[blue|<font color=blue>blue 9</font>]], then the [[blue|<font color=blue>blue 1</font>]], [[blue|<font color=blue>5</font>]], and [[blue|<font color=blue>9</font>]] are in arithmetic progression.  In fact, there is no way of coloring 1 through 9 without creating such a progression.  Therefore, ''W''(2, 3) is 9.
The center of the incircle can be found as the intersection of the three internal [[angle bisector]]s.  
The center of an excircle is the intersection of the internal bisector of one angle and the external bisectors of the other two. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an [[orthocentric system]].


It is an open problem to determine the values  of ''W''(''r'', ''k'') for most values of ''r'' and ''k''. The proof of the theorem provides only an upper bound.  For the case of ''r'' = 2 and ''k'' = 3, for example, the argument given below shows that it is sufficient to color the integers {1, ..., 325} with two colors to guarantee there will be a single-colored arithmetic progression of length 3. But in fact, the bound of 325 is very loose; the minimum required number of integers is only 9.  Any coloring of the integers {1, ..., 9} will have three evenly spaced integers of one color.  
See also [[Tangent lines to circles]].


For ''r'' = 3 and ''k'' = 3, the bound given by the theorem is 7(2·3<sup>7</sup>&nbsp;+&nbsp;1)(2·3<sup>7·(2·3<sup>7</sup>&nbsp;+&nbsp;1)</sup>&nbsp;+&nbsp;1), or approximately 4.22·10<sup>14616</sup>. But actually, you don't need that many integers to guarantee a single-colored progression of length 3; you only need 27. (And it is possible to color {1, ..., 26} with three colors so that there is no single-colored arithmetic progression of length 3; for example, RRYYRRYBYBBRBRRYRYYBRBBYBY.)
== Relation to area of the triangle ==
The radii of the incircles and excircles are closely related to the [[area]] of the triangle.  Let ''A'' be the triangle's area and let ''a'', ''b'' and ''c'', be the lengths of its sides. By [[Heron's formula]], the area of the triangle is


Anyone who can reduce the general upper bound to any 'reasonable' function can win a large cash prize. [[Ronald Graham]] has offered a prize of [[US$]]1000 for showing ''W''(2,''k'')&lt;2<sup>''k''<sup>2</sup></sup>.<ref>{{cite journal |authorlink=Ronald Graham |first=Ron |last=Graham |title=Some of My Favorite Problems in Ramsey Theory |journal=INTEGERS (The Electronic Journal of Combinatorial Number Theory |url=http://www.integers-ejcnt.org/vol7-2.html |volume=7 |issue=2 |year=2007 |pages=#A2 }}</ref> The best-known upper bound is due to [[Timothy Gowers]],<ref>{{cite journal |authorlink=Timothy Gowers |first=Timothy |last=Gowers |title=A new proof of Szemerédi's theorem |journal=Geom. Funct. Anal. |volume=11 |issue=3 |pages=465–588 |year=2001 |url=http://www.dpmms.cam.ac.uk/~wtg10/papers.html |doi=10.1007/s00039-001-0332-9 }}</ref> who establishes
:<math>
A=\frac{1}{4}\sqrt{P(a-b+c)(b-c+a)(c-a+b)}=\sqrt{s(s-a)(s-b)(s-c)}
</math>


: <math>W(r,k) \leq 2^{2^{r^{2^{2^{k + 9}}}}},</math>
where <math>s= \tfrac{1}{2}(a+b+c)</math> is the semiperimeter and <math>P = 2s</math> is the perimeter.


by first establishing a similar result for [[Szemerédi's theorem]], which is a stronger version of Van der Waerden's theorem.  The previously best-known bound was due to [[Saharon Shelah]] and proceeded via first proving a result for the [[Hales&ndash;Jewett theorem]], which is another strengthening of Van der Waerden's theorem.
The radius of the incircle (also known as the '''inradius''', ''r'' ) is


The best-known lower bound for <math>W(2, k)</math> is that <math>W(2, k) > 2^k/k^\epsilon</math> for all positive <math>\epsilon</math>.<ref>{{cite book |title=Discrete Mathematics And Its Applications |editor=M. Sethumadhavan |last=Brown | first=Tom C. | pages=80 | chapter=A partition of the non-negative integers, with applications to Ramsey theory |authorlink= |coauthors= |year=2006 |publisher=Alpha Science Int'l Ltd. |location= |isbn=81-7319-731-8 }}</ref>
: <math>r = \frac{2A}{P} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.</math>


== Proof of Van der Waerden's theorem (in a special case) ==
Thus, the area ''A'' of a triangle may be found by multiplying the inradius by the semiperimeter:


The following proof is due to [[Ronald Graham|Ron Graham]] and B.L. Rothschild.<ref name="Graham1974">{{cite journal |authorlink=Ronald Graham |first=R. L. |last=Graham |first2=B. L. |last2=Rothschild |title=A short proof of van der Waerden's theorem on arithmetic progressions |journal=Proc. American Math. Soc. |volume=42 |issue=2 |year=1974 |pages=385–386 |doi=10.1090/S0002-9939-1974-0329917-8 }}</ref> [[A. Ya. Khinchin|Khinchin]]<ref>{{Cite document
:<math>\displaystyle A=rs.</math>
  | last1 = Khinchin  | first1 = A. Ya.
  | title = Three Pearls of Number Theory
  | publisher = Dover
  | location = Mineola, NY
  | date = 1998
  | isbn = 978-0-486-40026-6
  | postscript = .}}
</ref> gives a fairly simple proof of the theorem without estimating ''W''(''r'',&nbsp;''k'').


We will prove the special case mentioned above, that ''W''(2, 3) ≤ 325. Let ''c''(''n'') be a coloring of the integers {1, ..., 325}.  We will find three elements of {1, ..., 325} in arithmetic progression that are the same color.
The radii in the excircles are called the '''exradii'''. The excircle at side ''a'' has radius


Divide {1, ..., 325} into the 65 blocks {1, ..., 5}, {6, ..., 10}, ... {321, ..., 325}, thus each block is of the form {''b'' ·5 + 1, ..., ''b'' ·5 + 5} for some ''b'' in {0, ..., 64}. Since each integer is colored either red or blue, each block is colored in one of 32 different ways.  By the [[pigeonhole principle]], there are two blocks among the first 33 blocks that are colored identically. That is, there are  two integers ''b''<sub>1</sub> and ''b''<sub>2</sub>, both in {0,...,32}, such that
: <math>r_a = \frac{2A}{c-a+b} = \sqrt{\frac{s (s-b)(s-c)}{s-a}}.</math>


: ''c''(''b''<sub>1</sub>&middot;5 + ''k'') = ''c''(''b''<sub>2</sub>&middot;5 + ''k'')
Similarly the radii of the excircles at sides ''b'' and ''c'' are respectively


for all ''k'' in {1, ..., 5}.  Among the three integers ''b''<sub>1</sub>·5 + 1, ''b''<sub>1</sub>·5 + 2, ''b''<sub>1</sub>·5 + 3, there must be at least two that are the same color. (The [[pigeonhole principle]] again.) Call these ''b''<sub>1</sub>·5 + ''a''<sub>1</sub> and ''b''<sub>1</sub>·5 + ''a''<sub>2</sub>, where the ''a''<sub>''i''</sub> are in {1,2,3} and ''a''<sub>1</sub> &lt; ''a''<sub>2</sub>.  Suppose (without loss of generality) that these two integers are both red.  (If they are both blue, just exchange 'red' and 'blue' in what follows.)
: <math>r_b = \frac{2A}{a-b+c} = \sqrt{\frac{s (s-a)(s-c)}{s-b}}</math>


Let ''a''<sub>3</sub> = 2·''a''<sub>2</sub>&nbsp;&minus;&nbsp;''a''<sub>1</sub>. If ''b''<sub>1</sub>·5 + ''a''<sub>3</sub> is red, then we have found our arithmetic progression: ''b''<sub>1</sub>·5&nbsp;+&nbsp;''a''<sub>''i''</sub> are all red.
and


Otherwise, ''b''<sub>1</sub>·5 + ''a''<sub>3</sub> is blue. Since ''a''<sub>3</sub> ≤ 5,  ''b''<sub>1</sub>·5 + ''a''<sub>3</sub> is in the ''b''<sub>1</sub> block, and since the ''b''<sub>2</sub> block is colored identically, ''b''<sub>2</sub>·5 + ''a''<sub>3</sub> is also blue.
: <math>r_c = \frac{2A}{b-c+a} = \sqrt{\frac{s (s-a)(s-b)}{s-c}}.</math>


Now let ''b''<sub>3</sub> = 2·''b''<sub>2</sub>&nbsp;&minus;&nbsp;''b''<sub>1</sub>. Then ''b''<sub>3</sub> ≤ 64. Consider the integer  ''b''<sub>3</sub>·5 + ''a''<sub>3</sub>, which must be ≤ 325. What color is it?
From these formulas one can see that the excircles are always larger than the incircle and that the largest excircle is the one tangent to the longest side and the smallest excircle is tangent to the shortest side. Further, combining these formulas with Heron's area formula yields:<ref>Baker, Marcus, "A collection of formulae for the area of a plane triangle," ''Annals of Mathematics'', part 1 in vol. 1(6), January 1885, 134-138. (See also part 2 in vol. 2(1), September 1885, 11-18.)</ref>


If it is red, then ''b''<sub>1</sub>·5 + ''a''<sub>1</sub>, ''b''<sub>2</sub>·5 + ''a''<sub>2</sub>, and ''b''<sub>3</sub>·5 + ''a''<sub>3</sub> form a red arithmetic progression. But if it is blue, then ''b''<sub>1</sub>·5 + ''a''<sub>3</sub>, ''b''<sub>2</sub>·5 + ''a''<sub>3</sub>, and ''b''<sub>3</sub>·5 + ''a''<sub>3</sub> form a blue arithmetic progression. Either way, we are done.
:<math>A=\sqrt{rr_ar_br_c}.</math>


A similar argument can be advanced to show that ''W''(3, 3) ≤ 7(2·3<sup>7</sup>+1)(2·3<sup>7·(2·3<sup>7</sup>+1)</sup>+1). One begins by dividing the integers into  2·3<sup>7·(2·3<sup>7</sup>&nbsp;+&nbsp;1)</sup>&nbsp;+&nbsp;1 groups of 7(2·3<sup>7</sup>&nbsp;+&nbsp;1) integers each; of the first 3<sup>7·(2·3<sup>7</sup>&nbsp;+&nbsp;1)</sup>&nbsp;+&nbsp;1 groups, two must be colored identically.
The ratio of the area of the incircle to the area of the triangle is less than or equal to <math>\frac{\pi}{3\sqrt{3}}</math>, with equality holding only for [[equilateral triangle]]s.<ref>Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", ''[[American Mathematical Monthly]]'' 115, October 2008, 679-689: Theorem 4.1.</ref>


Divide each of these two groups into 2·3<sup>7</sup>+1 subgroups of 7 integers each; of the first 3<sup>7</sup>&nbsp;+&nbsp;1 subgroups in each group, two of the subgroups must be colored identically. Within each of these identical subgroups, two of the first four integers must be the same color, say red; this implies either a red progression or an element of a different color, say blue, in the same subgroup.  
== Nine-point circle and Feuerbach point ==
The circle tangent to all three of the excircles as well as the incircle is known as the [[nine-point circle]]. The point where the nine-point circle touches the incircle is known as the Feuerbach point.


Since we have two identically-colored subgroups, there is a third subgroup, still in the same group that contains an element which, if either red or blue, would complete a red or blue progression, by a construction analogous to the one for ''W''(2, 3). Suppose that this element is yellow. Since there is a group that is colored identically, it must contain copies of the red, blue, and yellow elements we have identified; we can now find a pair of red elements, a pair of blue elements, and a pair of yellow elements that 'focus' on the same integer, so that whatever color it is, it must complete a progression.
== Gergonne triangle and point ==
[[Image:Intouch Triangle and Gergonne Point.svg|right|frame|200px|A triangle, Δ''ABC'', with incircle (blue), incenter (blue, ''I''), contact triangle (red, Δ''T''<sub>''a''</sub>''T''<sub>''b''</sub>''T''<sub>''c''</sub>) and Gergonne point (green, Ge)]]


The proof for ''W''(2, 3) depends essentially on proving that ''W''(32, 2) ≤ 33.  We divide the integers {1,...,325} into 65 'blocks', each of which can be colored in 32 different ways, and then show that two blocks of the first 33 must be the same color, and there is a block coloured the opposite way.  Similarly, the proof for ''W''(3, 3) depends on proving that
The '''Gergonne triangle'''(of ''ABC'') is defined by the 3 touchpoints of the incircle on the 3 sides.
Those vertices are denoted as ''T<sub>A</sub>'', etc.
The point that ''T<sub>A</sub>'' denotes, lies opposite to ''A''.


: <math>W(3^{7(2 \cdot 3^7+1)},2) \leq 3^{7(2 \cdot 3^7+1)}+1.</math>
This '''Gergonne triangle''' ''T<sub>A</sub>T<sub>B</sub>T<sub>C</sub>'' is also known as the '''contact triangle''' or '''intouch triangle''' of ''ABC''.


By a double [[mathematical induction|induction]] on the number of colors and the length of the progression, the theorem is proved in general.
The three lines ''AT<sub>A</sub>'', ''BT<sub>B</sub>'' and ''CT<sub>C</sub>'' intersect in a single point called '''Gergonne point''', denoted as ''Ge'' - [[Triangle center|''X(7)'']].


== Proof ==
Interestingly, the Gergonne point of a triangle is the [[symmedian point]] of the Gergonne triangle. For a full set of properties of the Gergonne point see.<ref>{{Cite journal
  | last = Dekov
  | first = Deko
  | title = Computer-generated Mathematics : The Gergonne Point
  | journal = Journal of Computer-generated Euclidean Geometry
  | year = 2009
  | volume = 1
  | pages = 1&ndash;14. 
  | url = http://www.dekovsoft.com/j/2009/01/JCGEG200901.pdf}}</ref>


A [[Generalized arithmetic progression|''D-dimensional arithmetic progression'']] consists of
The touchpoints of the three excircles with segments ''BC,CA and AB'' are the vertices of the [[extouch triangle]].  The points of intersection of the interior angle bisectors of ''ABC'' with the segments ''BC,CA,AB'' are the vertices of the '''incentral triangle'''.
numbers of the form:
::<math> a + i_1 s_1 + i_2 s_2 ... + i_D s_D </math>
where a is the basepoint, the s's are the different step-sizes, and the i's range from 0 to L-1. A d-dimensional AP is ''homogenous'' for some coloring when it is all the same color.


A ''D-dimensional arithmetic progression with benefits'' is all numbers of the form above, but where you add on some of the "boundary" of the arithmetic progression, i.e. some of the indices i's can be equal to L. The sides you tack on are ones where the first k i's are equal to L, and the remaining i's are less than L.
== Nagel triangle and point ==
The '''Nagel triangle''' of ''ABC'' is denoted by the vertices ''X<sub>A</sub>'', ''X<sub>B</sub>'' and ''X<sub>C</sub>'' that are the three points where the excircles touch the reference triangle ''ABC'' and where ''X<sub>A</sub>'' is opposite of ''A'', etc. This triangle ''X<sub>A</sub>X<sub>B</sub>X<sub>C</sub>'' is also known as the '''extouch triangle''' of ''ABC''. The circumcircle of the extouch triangle ''X<sub>A</sub>X<sub>B</sub>X<sub>C</sub>'' is called the '''Mandart circle'''. The three lines ''AX<sub>A</sub>'', ''BX<sub>B</sub>'' and ''CX<sub>C</sub>'' are called the [[Splitter (geometry)|splitters]] of the triangle; they each bisect the perimeter of the triangle, and they intersect in a single point, the triangle's [[Nagel point]] ''Na'' - [[Triangle center|''X(8)'']].


The boundaries of a D-dimensional AP with benefits are these additional arithmetic progressions of dimension d-1,d-2,d-3,d-4, down to 0. The 0 dimensional arithmetic progression is the single point at index value (L,L,L,L...,L). A D-dimensional AP with benefits is ''homogenous'' when each of the boundaries are individually homogenous, but different boundaries do not have to necessarily have the same color.
[[Trilinear coordinates]] for the vertices of the intouch triangle are given by
*<math> A-\text{vertex}= 0 : \sec^2 \left(\frac{B}{2}\right) :\sec^2\left(\frac{C}{2}\right)</math>
*<math> B-\text{vertex}= \sec^2 \left(\frac{A}{2}\right):0:\sec^2\left(\frac{C}{2}\right)</math>
*<math> C-\text{vertex}= \sec^2 \left(\frac{A}{2}\right) :\sec^2\left(\frac{B}{2}\right):0</math>


Next define the quantity MinN(L, D, N) to be the least integer so
Trilinear coordinates for the vertices of the extouch triangle are given by
that any assignment of N colors to an interval of length MinN or more
* <math> A-\text{vertex} = 0 : \csc^2\left(\frac{B}{2}\right) : \csc^2\left(\frac{C}{2}\right)</math>
necessarily contains a homogenous D-dimensional arithmetical progression with benefits.
* <math> B-\text{vertex} = \csc^2\left(\frac{A}{2}\right) : 0 : \csc^2\left(\frac{C}{2}\right)</math>
* <math> C-\text{vertex} = \csc^2\left(\frac{A}{2}\right) : \csc^2\left(\frac{B}{2}\right) : 0</math>


The goal is to bound the size of MinN. Note that MinN(L,1,N) is an upper bound for Van-Der-Waerden's
Trilinear coordinates for the vertices of the incentral triangle are given by
number. There are two inductions steps, as follows:
* <math>\ A-\text{vertex} = 0 : 1 : 1</math>
* <math>\ B-\text{vertex} = 1 : 0 : 1</math>
* <math>\ C-\text{vertex} = 1 : 1 : 0</math>


1. Assume MinN is known for a given lengths L for all dimensions of arithmetic progressions with benefits up to D. This formula gives a bound on MinN when you increase the dimension to D+1:
Trilinear coordinates for the vertices of the excentral triangle are given by
* <math>\ A-\text{vertex}= -1 : 1 : 1 </math>
* <math>\ B-\text{vertex}= 1 : -1 : 1 </math>
* <math>\ C-\text{vertex}= 1 : -1 : -1 </math>


let <math> M = {\mathrm MinN}(L,D,n)</math>
Trilinear coordinates for the Gergonne point are given by
: <math>\sec^2\left(\frac{A}{2}\right) : \sec^2 \left(\frac{B}{2}\right) : \sec^2\left(\frac{C}{2}\right)</math>,


::<math> {\mathrm MinN}(L, D+1 , n) \le  M*{\mathrm MinN}(L,1,n^M)</math>
or, equivalently, by the [[Law of Sines]],
: <math>\frac{bc}{b+ c - a} : \frac{ca}{c + a-b} : \frac{ab}{a+b-c}</math>.


Proof:
Trilinear coordinates for the Nagel point are given by
First, if you have an n-coloring of the interval 1...I, you can define a ''block coloring'' of k-size
: <math>\csc^2\left(\frac{A}{2}\right) : \csc^2 \left(\frac{B}{2}\right) : \csc^2\left(\frac{C}{2}\right)</math>,
blocks. Just consider each sequence of k colors in each k block to define a unique color. Call this ''k-blocking'' an n-coloring. k-blocking an n coloring of length l produces an n^k coloring of length l/k.


So given a n-coloring of an interval I of size M*MinN(L,1,n^M)) you can M-block it into an n^M coloring
or, equivalently, by the [[Law of Sines]],
of length MinN(L,1,n^M). But that means, by the definition of MinN, that you can find a 1-dimensional arithmetic sequence (with benefits) of length L in the block coloring, which is a sequence of blocks equally spaced, which are all the same block-color, i.e. you have a bunch of blocks of length M in the original sequence, which are equally spaced, which have exactly the same sequence of colors inside.
: <math>\frac{b+ c - a}{a} : \frac{c + a-b}{b} : \frac{a+b-c}{c}</math>.


Now, by the definition of M, you can find a d-dimensional arithmetic sequence with benefits in any one of these blocks, and since all of the blocks have the same sequence of colors, the same d-dimensional AP with benefits appears in all of the blocks, just by translating it from block to block. This is the definition of a d+1 dimensional arithmetic progression, so you have a homogenous d+1 dimensional AP. The new stride parameter s_{D+1} is defined to be the distance between the blocks.
It is the isotomic conjugate of the Gergonne point.


But you need benefits. The boundaries you get now are all old boundaries, plus their translations into identically colored blocks, because i_{D+1} is always less than L. The only boundary which is not like this is the 0 dimensional point when <math>i_1=i_2=...=i_{D+1}=L</math>. This is a single point, and is automatically homogenous.
==Coordinates of the incenter==
The [[Cartesian coordinates]] of the incenter are a weighted average of the coordinates of the three vertices using the side lengths of the triangle as weights. (The weights are positive so the incenter lies inside the triangle as stated above.)  If the three vertices are located at <math>(x_a,y_a)</math>, <math>(x_b,y_b)</math>, and <math>(x_c,y_c)</math>, and the sides opposite these vertices have corresponding lengths <math>a</math>, <math>b</math>, and <math>c</math>, then the incenter is at
:<math>\bigg(\frac{a x_a+b x_b+c x_c}{P},\frac{a y_a+b y_b+c y_c}{P}\bigg) = \frac{a(x_a,y_a)+b(x_b,y_b)+c(x_c,y_c)}{P}</math>
where <math>\ P = a + b + c.</math>


2. Assume MinN is known for one value of L and all possible dimensions D. Then you can bound MinN for length L+1.
[[Trilinear coordinates]] for the incenter are given by
:<math>\ 1 : 1 : 1.</math>
[[Barycentric coordinates (mathematics)|Barycentric coordinates]] for the incenter are given by
:<math>\ a : b : c</math>


::<math>{\mathrm MinN}(L+1,D,n) \le 2{\mathrm MinN}(L,n,n)</math>
or equivalently


proof:
:<math>\sin(A):\sin(B):\sin(C).</math>
Given an n-coloring of an interval of size MinN(L,n,n), by definition, you can find an arithmetic sequence with benefits of dimension n of length L. But now, the number of "benefit" boundaries is equal to the number of colors, so one of the homogenous boundaries, say of dimension k, has to have the same color as another one of the homogenous benefit boundaries, say the one of dimension p<k. This allows a length L+1 arithmetic sequence (of dimension 1) to be constructed, by going along a line inside the k-dimensional boundary which ends right on the p-dimensional boundary, and including the terminal point in the p-dimensional boundary. In formulas:


if
==Equations for four circles==
::<math> a+ L s_1 +L s_2... + L s_{D-k}</math> has the same color as
Let x : y : z be a variable point in [[trilinear coordinates]], and let u = cos<sup>''2''</sup>''(A/2)'', v = cos<sup>''2''</sup>''(B/2)'', w = cos<sup>''2''</sup>''(C/2)''.  The four circles described above are given by these equations:
::<math> a + L s_1 +L s_2 ... +L s_{D-p}</math>
:* Incircle:
then
::<math>\ u^2x^2+v^2y^2+w^2z^2-2vwyz-2wuzx-2uvxy=0</math>
::<math> a + L*(s_1 ... +s_{D-k}) + u *(s_{D-k+1} ... +s_p) </math> have the same color
:* ''A-''excircle:
::<math> u = 0,1,2,...,L-1,L </math> i.e. u makes a sequence of length L+1.
::<math>\ u^2x^2+v^2y^2+w^2z^2-2vwyz+2wuzx+2uvxy=0</math>
:* ''B-''excircle:
::<math>\ u^2x^2+v^2y^2+w^2z^2+2vwyz-2wuzx+2uvxy=0</math>
:* ''C-''excircle:
::<math>\ u^2x^2+v^2y^2+w^2z^2+2vwyz+2wuzx-2uvxy=0</math>


This constructs a sequence of dimension 1, and the "benefits" are automatic, just add on another point of whatever color. To include this boundary point, one has to make the interval longer by the maximum possible value of the stride, which is certainly less than the interval size. So doubling the interval size will definitely work, and this is the reason for the factor of two. This completes the induction on L.
==Euler's theorem==
[[Euler's theorem in geometry|Euler's theorem]] states that in a triangle:
:<math>(R-r_{in})^2=d^2+r_{in}^2,</math>
where ''R'' and ''r''<sub>''in''</sub> are the circumradius and inradius respectively, and ''d'' is the distance between the circumcenter and the incenter.


Base case: MinN(1,d,n)=1, i.e. if you want a length 1 homogenous d-dimensional arithmetic sequence, with or without benefits, you have nothing to do. So this forms the base of the induction. The VanDerWaerden theorem itself is the assertion that MinN(L,1,N) is finite, and it follows from the base case and the induction steps.<ref name="Graham1974" />
For excircles the equation is similar:
:<math>(R+r_{ex})^2=d^2+r_{ex}^2,</math>
where ''r''<sub>''ex''</sub> is the radius one of the excircles, and ''d'' is the distance between the circumcenter and this excircle's center.
<ref name=Nelson>Nelson, Roger, "Euler's triangle inequality via proof without words," ''Mathematics Magazine'' 81(1), February 2008, 58-61.</ref>
<ref>Johnson, R. A. ''Modern Geometry'', Houghton Mifflin, Boston, 1929: p. 187.</ref>
<ref>[http://forumgeom.fau.edu/FG2001volume1/FG200120.pdf Emelyanov, Lev, and Emelyanova, Tatiana. "Euler’s formula and Poncelet’s porism", ''Forum Geometricorum'' 1, 2001: pp. 137–140.]</ref>


==See also==
==Other incircle properties==
* [[Van der Waerden number]]s for all known values for ''W''(''n'',''r'') and the best-known bounds for unknown values
Suppose the tangency points of the incircle divide the sides into lengths of ''x'' and ''y'', ''y'' and ''z'', and ''z'' and ''x''.  Then the incircle has the radius<ref>Chu, Thomas, ''The Pentagon'', Spring 2005, p. 45, problem 584.</ref>
 
:<math> r = \sqrt{\frac{xyz}{x+y+z}}</math>
 
and the area of the triangle is
 
:<math>K=\sqrt{xyz(x+y+z)}.</math>
 
If the altitudes from sides of lengths ''a'', ''b'', and ''c'' are ''h<sub>a</sub>'', ''h<sub>b</sub>'', and ''h<sub>c</sub>'' then the inradius ''r'' is one-third of the harmonic mean of these altitudes, i.e.
 
:<math> r = \frac{1}{h_a^{-1}+h_b^{-1}+h_c^{-1}}.</math>
 
The product of the incircle radius ''r'' and the circumcircle radius ''R'' of a triangle with sides ''a'', ''b'', and ''c'' is<ref>Johnson, Roger A., ''Advanced Euclidean Geometry'', Dover, 2007 (orig. 1929), p. 189, #298(d).</ref>
 
:<math>rR=\frac{abc}{2(a+b+c)}.</math>
 
Some relations among the sides, incircle radius, and circumcircle radius are:<ref name=Bell/>
:<math>ab+bc+ca=s^2+(4R+r)r,</math>
:<math>a^2+b^2+c^2=2s^2-2(4R+r)r.</math>
 
Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.<ref>Kodokostas, Dimitrios, "Triangle Equalizers," ''Mathematics Magazine'' 83, April 2010, pp. 141-146.</ref>
 
The distance from the incenter to the [[centroid]] is less than one third the length of the longest median of the triangle.<ref name=Franzsen>[http://forumgeom.fau.edu/FG2011volume11/FG201126.pdf Franzsen, William N.. "The distance from the incenter to the Euler line", ''Forum Geometricorum'' 11 (2011): 231–236.]</ref>
 
Denoting the distance from the incenter to the Euler line as ''d'', the length of the longest median as ''v'', the length of the longest side as ''u'', and the semiperimeter as ''s'', the following inequalities hold:<ref name=Franzsen/>
 
:<math>\frac{d}{s} < \frac{d}{u} < \frac{d}{v} < \frac{1}{3}.</math>
 
Denoting the center of the incircle of triangle ''ABC'' as ''I'', we have<ref>Allaire, Patricia R.; Zhou, Junmin; and Yao, Haishen, "Proving a nineteenth century ellipse identity", ''[[Mathematical Gazette]]'' 96, March 2012, 161-165.</ref>
 
:<math>\frac{\overline{IA} \cdot \overline{IA}}{\overline{CA} \cdot \overline{AB}} + \frac{\overline{IB} \cdot \overline{IB}}{\overline{AB} \cdot \overline{BC}} + \frac{\overline{IC} \cdot \overline{IC}}{\overline{BC} \cdot \overline{CA}} = 1.</math>
 
==Other excircle properties==
 
The circular [[convex hull|hull]] of the excircles is internally tangent to each of the excircles, and thus is an [[Problem of Apollonius|Apollonius circle]].<ref>[http://forumgeom.fau.edu/FG2002volume2/FG200222.pdf Grinberg, Darij, and Yiu, Paul, "The Apollonius Circle as a Tucker Circle", ''Forum Geometricorum'' 2, 2002: pp. 175-182.]</ref> The radius of this Apollonius circle is <math>\frac{r^2+s^2}{4r}</math> where ''r'' is the incircle radius and ''s'' is the semiperimeter  of the triangle.<ref>[http://forumgeom.fau.edu/FG2003volume3/FG200320.pdf Stevanovi´c, Milorad R., "The Apollonius circle and related triangle centers", ''Forum Geometricorum'' 3, 2003, 187-195.]</ref>
 
The following relations hold among the inradius ''r'', the circumradius ''R'', the semiperimeter ''s'', and the excircle radii ''r''<sub>'a''</sub>, ''r''<sub>''b''</sub>, ''r''<sub>''c''</sub>:<ref name=Bell>[http://forumgeom.fau.edu/FG2006volume6/FG200639.pdf Bell, Amy, "Hansen’s right triangle theorem, its converse and a generalization", ''Forum Geometricorum'' 6, 2006, 335–342.]</ref>
 
:<math>r_a+r_b+r_c=4R+r,</math>
:<math>r_a r_b+r_br_c+r_cr_a = s^2,</math>
:<math>r_a^2 + r_b^2 + r_c^2 = (4R+r)^2 -2s^2,</math>
 
The circle through the centers of the three excircles has radius 2''R''.<ref name=Bell/>
 
If ''H'' is the [[orthocenter]] of triangle ''ABC'', then<ref name=Bell/>
 
:<math>r_a+r_b+r_c+r=AH+BH+CH+2R,</math>
:<math>r_a^2+r_b^2+r_c^2+r^2=AH^2+BH^2+CH^2+(2R)^2.</math>
 
==Incircle in a quadrilateral==
Some (but not all) [[quadrilateral]]s have an incircle. These are called [[tangential quadrilateral]]s. Among their many properties perhaps the most important is that their opposite sides have equal sums. This is called the [[Pitot theorem]].
 
== See also ==
*[[Altitude (triangle)]]
*[[Circumscribed circle]]
*[[Ex-tangential quadrilateral]]
*[[Harcourt's theorem]]
*[[Inscribed sphere]]
*[[Power of a point]]
*[[Steiner inellipse]]
*[[Tangential quadrilateral]]
*[[Triangle center]]


==References==
==References==
{{reflist}}
{{reflist}}
*Clark Kimberling, "Triangle Centers and Central Triangles," ''Congressus Numerantium'' 129 (1998) i-xxv and 1-295.
*Sándor Kiss, "The Orthic-of-Intouch and Intouch-of-Orthic Triangles," ''Forum Geometricorum'' 6 (2006) 171-177.


==External links==
==External links==
* [http://www.math.uga.edu/~lyall/REU/ramsey.pdf Proof of Van der Waerden's theorem]
* [http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Derivation of formula for radius of incircle of a triangle]
* {{MathWorld |title=Incircle |urlname=Incircle}}
 
===Interactive===
*[http://www.mathopenref.com/triangleincenter.html Triangle incenter]&nbsp; &nbsp;[http://www.mathopenref.com/triangleincircle.html Triangle incircle]&nbsp;&nbsp;[http://www.mathopenref.com/polygonincircle.html Incircle of a regular polygon]&nbsp; &nbsp;With interactive animations
*[http://www.mathopenref.com/constincircle.html Constructing a triangle's incenter / incircle with compass and straightedge] An interactive animated demonstration
* [http://www.cut-the-knot.org/Curriculum/Geometry/AdjacentIncircles.shtml Equal Incircles Theorem] at [[cut-the-knot]]
* [http://www.cut-the-knot.org/Curriculum/Geometry/FourIncircles.shtml Five Incircles Theorem] at [[cut-the-knot]]
* [http://www.cut-the-knot.org/Curriculum/Geometry/IncirclesInQuadri.shtml Pairs of Incircles in a Quadrilateral] at [[cut-the-knot]]
*[http://www.uff.br/trianglecenters/X0001.html An interactive Java applet for the incenter]


[[Category:Ramsey theory]]
{{DEFAULTSORT:Incircle And Excircles Of A Triangle}}
[[Category:Theorems in discrete mathematics]]
[[Category:Circles]]
[[Category:Articles containing proofs]]
[[Category:Triangle geometry]]


[[de:Satz von Van der Waerden]]
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[[ja:ファン・デル・ヴェルデンの定理]]
[[uk:Теорема ван дер Вардена]]
[[zh:范德瓦尔登定理]]

Revision as of 08:05, 10 August 2014

Incircle redirects here. For incircles of non-triangle polygons, see Tangential quadrilateral or Tangential polygon.
A triangle (black) with incircle (blue), incenter (I), excircles (orange), excenters (JA,JB,JC), internal angle bisectors (red) and external angle bisectors (green)

In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is called the triangle's incenter.

An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.

The center of the incircle can be found as the intersection of the three internal angle bisectors. The center of an excircle is the intersection of the internal bisector of one angle and the external bisectors of the other two. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.

See also Tangent lines to circles.

Relation to area of the triangle

The radii of the incircles and excircles are closely related to the area of the triangle. Let A be the triangle's area and let a, b and c, be the lengths of its sides. By Heron's formula, the area of the triangle is

where is the semiperimeter and is the perimeter.

The radius of the incircle (also known as the inradius, r ) is

Thus, the area A of a triangle may be found by multiplying the inradius by the semiperimeter:

The radii in the excircles are called the exradii. The excircle at side a has radius

Similarly the radii of the excircles at sides b and c are respectively

and

From these formulas one can see that the excircles are always larger than the incircle and that the largest excircle is the one tangent to the longest side and the smallest excircle is tangent to the shortest side. Further, combining these formulas with Heron's area formula yields:[1]

The ratio of the area of the incircle to the area of the triangle is less than or equal to , with equality holding only for equilateral triangles.[2]

Nine-point circle and Feuerbach point

The circle tangent to all three of the excircles as well as the incircle is known as the nine-point circle. The point where the nine-point circle touches the incircle is known as the Feuerbach point.

Gergonne triangle and point

A triangle, ΔABC, with incircle (blue), incenter (blue, I), contact triangle (red, ΔTaTbTc) and Gergonne point (green, Ge)

The Gergonne triangle(of ABC) is defined by the 3 touchpoints of the incircle on the 3 sides. Those vertices are denoted as TA, etc. The point that TA denotes, lies opposite to A.

This Gergonne triangle TATBTC is also known as the contact triangle or intouch triangle of ABC.

The three lines ATA, BTB and CTC intersect in a single point called Gergonne point, denoted as Ge - X(7).

Interestingly, the Gergonne point of a triangle is the symmedian point of the Gergonne triangle. For a full set of properties of the Gergonne point see.[3]

The touchpoints of the three excircles with segments BC,CA and AB are the vertices of the extouch triangle. The points of intersection of the interior angle bisectors of ABC with the segments BC,CA,AB are the vertices of the incentral triangle.

Nagel triangle and point

The Nagel triangle of ABC is denoted by the vertices XA, XB and XC that are the three points where the excircles touch the reference triangle ABC and where XA is opposite of A, etc. This triangle XAXBXC is also known as the extouch triangle of ABC. The circumcircle of the extouch triangle XAXBXC is called the Mandart circle. The three lines AXA, BXB and CXC are called the splitters of the triangle; they each bisect the perimeter of the triangle, and they intersect in a single point, the triangle's Nagel point Na - X(8).

Trilinear coordinates for the vertices of the intouch triangle are given by

Trilinear coordinates for the vertices of the extouch triangle are given by

Trilinear coordinates for the vertices of the incentral triangle are given by

Trilinear coordinates for the vertices of the excentral triangle are given by

Trilinear coordinates for the Gergonne point are given by

,

or, equivalently, by the Law of Sines,

.

Trilinear coordinates for the Nagel point are given by

,

or, equivalently, by the Law of Sines,

.

It is the isotomic conjugate of the Gergonne point.

Coordinates of the incenter

The Cartesian coordinates of the incenter are a weighted average of the coordinates of the three vertices using the side lengths of the triangle as weights. (The weights are positive so the incenter lies inside the triangle as stated above.) If the three vertices are located at , , and , and the sides opposite these vertices have corresponding lengths , , and , then the incenter is at

where

Trilinear coordinates for the incenter are given by

Barycentric coordinates for the incenter are given by

or equivalently

Equations for four circles

Let x : y : z be a variable point in trilinear coordinates, and let u = cos2(A/2), v = cos2(B/2), w = cos2(C/2). The four circles described above are given by these equations:

  • Incircle:
  • A-excircle:
  • B-excircle:
  • C-excircle:

Euler's theorem

Euler's theorem states that in a triangle:

where R and rin are the circumradius and inradius respectively, and d is the distance between the circumcenter and the incenter.

For excircles the equation is similar:

where rex is the radius one of the excircles, and d is the distance between the circumcenter and this excircle's center. [4] [5] [6]

Other incircle properties

Suppose the tangency points of the incircle divide the sides into lengths of x and y, y and z, and z and x. Then the incircle has the radius[7]

and the area of the triangle is

If the altitudes from sides of lengths a, b, and c are ha, hb, and hc then the inradius r is one-third of the harmonic mean of these altitudes, i.e.

The product of the incircle radius r and the circumcircle radius R of a triangle with sides a, b, and c is[8]

Some relations among the sides, incircle radius, and circumcircle radius are:[9]

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.[10]

The distance from the incenter to the centroid is less than one third the length of the longest median of the triangle.[11]

Denoting the distance from the incenter to the Euler line as d, the length of the longest median as v, the length of the longest side as u, and the semiperimeter as s, the following inequalities hold:[11]

Denoting the center of the incircle of triangle ABC as I, we have[12]

Other excircle properties

The circular hull of the excircles is internally tangent to each of the excircles, and thus is an Apollonius circle.[13] The radius of this Apollonius circle is where r is the incircle radius and s is the semiperimeter of the triangle.[14]

The following relations hold among the inradius r, the circumradius R, the semiperimeter s, and the excircle radii r'a, rb, rc:[9]

The circle through the centers of the three excircles has radius 2R.[9]

If H is the orthocenter of triangle ABC, then[9]

Incircle in a quadrilateral

Some (but not all) quadrilaterals have an incircle. These are called tangential quadrilaterals. Among their many properties perhaps the most important is that their opposite sides have equal sums. This is called the Pitot theorem.

See also

References

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

  • Clark Kimberling, "Triangle Centers and Central Triangles," Congressus Numerantium 129 (1998) i-xxv and 1-295.
  • Sándor Kiss, "The Orthic-of-Intouch and Intouch-of-Orthic Triangles," Forum Geometricorum 6 (2006) 171-177.

External links

  • Derivation of formula for radius of incircle of a triangle


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  1. Baker, Marcus, "A collection of formulae for the area of a plane triangle," Annals of Mathematics, part 1 in vol. 1(6), January 1885, 134-138. (See also part 2 in vol. 2(1), September 1885, 11-18.)
  2. Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", American Mathematical Monthly 115, October 2008, 679-689: Theorem 4.1.
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  4. Nelson, Roger, "Euler's triangle inequality via proof without words," Mathematics Magazine 81(1), February 2008, 58-61.
  5. Johnson, R. A. Modern Geometry, Houghton Mifflin, Boston, 1929: p. 187.
  6. Emelyanov, Lev, and Emelyanova, Tatiana. "Euler’s formula and Poncelet’s porism", Forum Geometricorum 1, 2001: pp. 137–140.
  7. Chu, Thomas, The Pentagon, Spring 2005, p. 45, problem 584.
  8. Johnson, Roger A., Advanced Euclidean Geometry, Dover, 2007 (orig. 1929), p. 189, #298(d).
  9. 9.0 9.1 9.2 9.3 Bell, Amy, "Hansen’s right triangle theorem, its converse and a generalization", Forum Geometricorum 6, 2006, 335–342.
  10. Kodokostas, Dimitrios, "Triangle Equalizers," Mathematics Magazine 83, April 2010, pp. 141-146.
  11. 11.0 11.1 Franzsen, William N.. "The distance from the incenter to the Euler line", Forum Geometricorum 11 (2011): 231–236.
  12. Allaire, Patricia R.; Zhou, Junmin; and Yao, Haishen, "Proving a nineteenth century ellipse identity", Mathematical Gazette 96, March 2012, 161-165.
  13. Grinberg, Darij, and Yiu, Paul, "The Apollonius Circle as a Tucker Circle", Forum Geometricorum 2, 2002: pp. 175-182.
  14. Stevanovi´c, Milorad R., "The Apollonius circle and related triangle centers", Forum Geometricorum 3, 2003, 187-195.