ເບິ່ງຊອຣ໌ສສຳລັບ ສູດເຮລົງ

{{about|calculating the area of a triangle|calculating a square root|Heron's method}}
[[Image:Triangle with notations 2.svg|thumb|198px|ຮູບສາມແຈທີ່ມີຂ້າງ''a'', ''b'', and ''c''.]]
ໃນ[[ເລຂາຄະນິດ]] geometry, ສູດເຮລົງ '''Heron's (ຫຼື Hero's) formula''', ແມ່ນຖືກຕັ້ງຊື່ຕາມ [[ເຮລົງ ອະເລັກຊານເດຍ]](Heron of Alexandria), ໂດຍວ່າ [[ເນື້ອທີ່]] ''S'' ຂອງ [[ຮູບສາມແຈ]] ທີ່ປະກອບດ້ວຍຂ້າງ ''a'', ''b'', ແລະ ''c'' ແມ່ນ
:<math>S = \sqrt{p(p-a)(p-b)(p-c)}</math>

ເຊິ່ງ ''p'' ແມ່ນ ເຄິ່ງ[[ລວງຮອບ]] (semiperimeter) ຂອງຮູບສາມແຈ:

:<math>p=\frac{a+b+c}{2}.</math>

Heron's formula can also be written as:

:<math>S={\ \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\ \over 16}\,}</math>

:<math>S={\ \sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)\ \over 16}\,}</math>

:<math>S=\frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}.</math>

== ປະຫວັດ ==

ສູດນີ້ແມ່ນຜົນງານຂອງທ່ານ ເຮລົງ ອາເລັກຊານເດຍ, ແລະມີບົດພິສູດຢູ່ໃນປື້ມຂອງລາວຊື່ວ່າ ເມຕຣິກາ ''Metrica'', ຂຽນຂື້ນໃນ ຄ.ສ  60. ເຊິ່ງເຮັດໃຫ້ທ່ານ ອາຄີມີດີດ[[Archimedes]] ເປັນທີ່ຮູ້ຈັກ, ແລະນັບແຕ່ ປື້ມເມຕຮິກາ''Metrica'' ໄດ້ລວບລວມຄວາມຮູ້ກ່ຽວກັບຄະນິດສາດໃນຍຸກບູຮານ,ມັນເປັນໄປໄດ້ວ່ານີ້ເປັນປັດໃຈສຳຄັນໃນການຖືກຳເນີດໃນຜົນງານນີ້. .<ref>{{MathWorld |urlname=HeronsFormula |title=Heron's Formula}}</ref>

ສູດ ເຮລົງ ດັ້ງເດີມມີຮູບຮ່າງແມ່ນ:

:<math>S=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}</math>, where <math>a \ge b \ge c</math>

ຖືກຄົ້ນພົບໂດຍຄົນຈີນໃນປະເທດເກຣັກ.ແລະຖືກຕີພິມໃນປື້ມຊູຊູ ຈິວຈາງ ''Shushu Jiuzhang'' (“[[Mathematical Treatise in Nine Sections]]”), ຂຽນໂດຍ ທ່ານ ຊິນ ຈິວຊາວ [[Qin Jiushao]] ພິມໃນປີ  <small>ຄ.ສ.</small> 1247.

== ການພິສູດ ==

ການພິສູດລ່າສຸດແມ່ນນຳໃຊ້[[ພຶດຊະຄະນິດ]](algebra) ແລະ ບໍ່ຄ້າຍຄືກັບການພິສູດໂດຍທ່ານເຮລົງເອງ(ໃນໜັງສືເມຕຣິກາ)
ເຮົາຮູ້ວ່າ ''a'', ''b'', ''c'' ແມ່ນຂ້າງຂອງຮູບສາມແຈ ແລະ  ''A'', ''B'', ''C'' ແມ່ນມູມເຊິ່ງໜ້າຂອງຂ້າງດັ່ງກ່າວຕາມລຳດັບ. ເຮົາໄດ້:

:<math>\cos \widehat C = \frac{a^2+b^2-c^2}{2ab}</math> ອີງຕາມ[[ກົດເກນຂອງໂກຊິນ]](law of cosines),
ຈາກນັ້ນ  ເຮົາສາມາດຖອນເອົາຄ່າຂອງ ຊິນ:

:<math>\sin \widehat C = \sqrt{1-\cos^2 \widehat C} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.</math>

 [[ລວງສູງ]] ຂອງຮູບສາມແຈ ແມ່ນເທົ່າກັບຜົນຄູນລະຫວ່າງ ຂ້າງພື້ນ ''a'' ກັບຂ້າງ ''b'' ຄູນໃຫ້ sin(''C''),(ອີງຕາມການພົວພັນມູມໃນຮູບສາມແຈ). ເຮົາໄດ້:

: <math>
\begin{align}
S & = \frac{1}{2}ah \\
& = \frac{1}{2} ab\sin \widehat C \\
& = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\
& = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\
& = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\
& = \frac{1}{4}\sqrt{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)} \\
& = \sqrt{\frac{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)}{16}} \\
& = \sqrt{\frac{(c -(a -b))}{2}\frac{(c +(a -b))}{2}\frac{((a +b) -c)}{2}\frac{((a +b) +c)}{2}} \\
& = \sqrt{\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}\frac{(a + b + c)}{2}} \\
& = \sqrt{\frac{(a + b + c)}{2}\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}} \\
& = \sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}.
\end{align}
</math>

==Proof using the Pythagorean theorem==
[[Image:Triangle with notations 3.svg|thumb|270px|Triangle with altitude ''h'' cutting base ''c'' into ''d''&nbsp;+&nbsp;(''c''&nbsp;&minus;&nbsp;''d'').]]Heron's original proof made use of [[cyclic quadrilateral]]s, while other arguments appeal to [[trigonometry]] as above, or to the [[incenter]] and one [[excircle]] of the triangle [http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt].  The following argument reduces Heron's formula directly to the [[Pythagorean theorem]] using only elementary means.

In the form 4''T''<sup>&nbsp;2</sup> = 4''s''(''s''&nbsp;&minus;&nbsp;''a'')(''s''&nbsp;&minus;&nbsp;''b'')(''s''&nbsp;&minus;&nbsp;''c''), Heron's formula reduces on the left to (''ch'')<sup>2</sup>, or

:<math>(cb)^2-(cd)^2</math> which is the same as <math>c^2(b^2-d^2)</math>

using ''b''<sup>&nbsp;2</sup>&nbsp;&minus;&nbsp;''d''<sup>&nbsp;2</sup> = ''h''<sup>&nbsp;2</sup> by the [[Pythagorean theorem]], and on the right to

:<math>\displaystyle (s(s-a)+(s-b)(s-c))^2 - (s(s-a)-(s-b)(s-c))^2</math>

via the identity (''p''&nbsp;+&nbsp;''q'')<sup>&nbsp;2</sup>&nbsp;&minus;&nbsp;(''p''&nbsp;&minus;&nbsp;''q'')<sup>&nbsp;2</sup>&nbsp;=&nbsp;4''pq''.  It therefore suffices to show

: <math> cb=s(s-a)+(s-b)(s-c), \, </math>

and

: <math> cd = s(s-a)-(s-b)(s-c). \, </math>

Then expanding the former you get the following: 

: <math> 2s^2 - s(a + b + c) + cb. \, </math>
 
and that reduces to <math>cb</math> by substituting ''2s'' = (''a''&nbsp;+&nbsp;''b''&nbsp;+&nbsp;''c'') and simplifying.   Submitting for ''s'' the latter  ''s''(''s''&nbsp;&minus;&nbsp;''a'')&nbsp;&minus;&nbsp;(''s''&nbsp;&minus;&nbsp;''b'')(''s''&nbsp;&minus;&nbsp;''c'') reduces only as far as (''b''<sup>&nbsp;2</sup>&nbsp;+&nbsp;''c''<sup>&nbsp;2</sup>&nbsp;&minus;&nbsp;''a''<sup>&nbsp;2</sup>)/2.  But if we replace ''b''<sup>&nbsp;2</sup> by ''d''<sup>&nbsp;2</sup>&nbsp;+&nbsp;''h''<sup>&nbsp;2</sup> and ''a''<sup>&nbsp;2</sup> by (''c''&nbsp;&minus;&nbsp;''d'')<sup>&nbsp;2</sup>&nbsp;+&nbsp;''h''<sup>&nbsp;2</sup>, both by Pythagoras, simplification then produces ''cd'' as required.

== Numerical Stability ==
Heron's formula as given above is [[Numerical_stability|numerically unstable]] for triangles with a very small angle. A stable alternative involves arranging the lengths of the sides so that <math>a \ge b \ge c</math> and computing
:<math>T = \frac{1}{4}\sqrt{(a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c))}.</math>
The brackets in the above formula are required in order to prevent numerical instability in the evaluation.

== Generalizations ==
Heron's formula is a special case of [[Brahmagupta's formula]] for the area of a [[cyclic quadrilateral]]. Heron's formula and Brahmagupta's formula are both special cases of [[Bretschneider's formula]] for the area of a [[quadrilateral]]. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero.

Heron's formula is also a special case of the formula of the area of the [[trapezoid]] based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero.

Expressing Heron's formula with a [[Cayley–Menger determinant]] in terms of the squares of the [[distance]]s between the three given vertices,
:<math> T =  \frac{1}{4} \sqrt{- \begin{vmatrix} 
  0 & a^2 & b^2 & 1 \\
a^2 & 0   & c^2 & 1 \\
b^2 & c^2 & 0   & 1 \\
  1 &   1 &   1 & 0
\end{vmatrix} } </math>
illustrates its similarity to [[Tartaglia's formula]] for the [[volume]] of a [[Simplex|three-simplex]].

Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by [[David P. Robbins]].<ref>D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.</ref>

=== Heron-type formula for the volume of a tetrahedron ===
If ''U'', ''V'', ''W'', ''u'', ''v'', ''w'' are lengths of edges of the tetrahedron (first three form a triangle; ''u'' opposite to ''U'' and so on), then<ref>W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?", [http://www.cs.berkeley.edu/~wkahan/VtetLang.pdf], pp. 16-17.</ref>
:<math>
\text{volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}</math>

where

: <math>
\begin{align}
a & = \sqrt {xYZ} \\
b & = \sqrt {yZX} \\
c & = \sqrt {zXY} \\
d & = \sqrt {xyz} \\
X & = (w - U + v)\,(U + v + w) \\
x & = (U - v + w)\,(v - w + U) \\
Y & = (u - V + w)\,(V + w + u) \\
y & = (V - w + u)\,(w - u + V) \\
Z & = (v - W + u)\,(W + u + v) \\
z & = (W - u + v)\,(u - v + W).
\end{align}
</math>

== ເບິ່ງຍັງ ==
* [[Brahmagupta's formula]]
* [[Heronian triangle]]
* [[Synthetic geometry]]

== ອ້າງອີງ ==
<references/>

== ບັນນານຸກົມ ==
* {{cite book
| author=Heath, Thomas L.
| title=A History of Greek Mathematics (Vol II)
| publisher=Oxford University Press
| year=1921
| pages=321–323}}

== ແຫຼ່ງຂໍ້ມູນອື່ນ ==
*[http://www.cut-the-knot.org/pythagoras/herons.shtml A Proof of the Pythagorean Theorem From Heron's Formula] at [[cut-the-knot]]
*[http://www.mathopenref.com/heronsformula.html Interactive applet and area calculator using Heron's Formula]
*[http://www.math.dartmouth.edu/~doyle/docs/heron/heron.txt J.H. Conway discussion on Heron's Formula]
*{{MathPages|id=home/kmath196/kmath196|title=Heron's Formula and Brahmagupta's Generalization}} 
*[http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html A Geometric Proof of Heron's Formula]

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[[Category:Triangle geometry]]
[[Category:Articles containing proofs]]
[[Category:Area]]
[[Category:Theorems in plane geometry]]