Trikotnik s stranicami a , b in c Heronova formula (tudi Heronova enačba ali Heronov obrazec ) je v ravninski geometriji formula za računanje ploščine trikotnika s podanimi stranicami , brez uporabe velikosti kotov. Poimenovana je po Heronu Aleksandrijskemu .[ 1]
Heronova formula pravi, da je ploščina trikotnika s stranicami
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
S
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
,
{\displaystyle S={\sqrt {s(s-a)(s-b)(s-c)}},}
kjer je
s
{\displaystyle s}
s
=
a
+
b
+
c
2
.
{\displaystyle s={\frac {a+b+c}{2}}.}
[ 2] Heronovo formulo lahko zapišemo tudi kot
S
=
1
4
(
a
+
b
+
c
)
(
−
a
+
b
+
c
)
(
a
−
b
+
c
)
(
a
+
b
−
c
)
{\displaystyle S={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}
S
=
1
4
2
(
a
2
b
2
+
a
2
c
2
+
b
2
c
2
)
−
(
a
4
+
b
4
+
c
4
)
{\displaystyle S={\frac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}}
S
=
1
4
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
{\displaystyle S={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}}
S
=
1
4
4
(
a
2
b
2
+
a
2
c
2
+
b
2
c
2
)
−
(
a
2
+
b
2
+
c
2
)
2
{\displaystyle S={\frac {1}{4}}{\sqrt {4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{2}+b^{2}+c^{2})^{2}}}}
S
=
1
4
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
.
{\displaystyle S={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}.}
Naj bo
△
A
B
C
{\displaystyle \bigtriangleup ABC}
a
=
4
{\displaystyle a=4}
b
=
13
{\displaystyle b=13}
c
=
15
{\displaystyle c=15}
s
{\displaystyle s}
s
=
1
2
(
a
+
b
+
c
)
=
1
2
(
4
+
13
+
15
)
=
16
{\displaystyle s={\frac {1}{2}}(a+b+c)={\frac {1}{2}}(4+13+15)=16}
ploščina pa je
S
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
16
⋅
(
16
−
4
)
⋅
(
16
−
13
)
⋅
(
16
−
15
)
=
16
⋅
12
⋅
3
⋅
1
=
576
=
24.
{\displaystyle S={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24.}
V tem zgledu so stranice in ploščina cela števila , zaradi česar dani trikotnik imenujemo Heronski trikotnik . Kljub temu Heronova formula deluje za poljuben trikotnik.
Formula je pripisana Heronu Aleksandrijskemu , njen dokaz najdemo v njegovi knjigi Metrica , napisani okoli leta 60. Domneva se, da je Arhimed formulo poznal že dve stoletji prej.[ 3] Metrica je namreč zbirka matematičnega znanja iz antičnega sveta in zato je možno, da je da je bila formula odkrita že prej.
Formulo ekvivalentno Heronovi
S
=
1
2
a
2
c
2
−
(
a
2
+
c
2
−
b
2
2
)
2
{\displaystyle S={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}}
so neodvisno od Grkov odkrili Kitajci. Objavljena je bila v Matimatičnem učbeniku v devetih poglavjih (Qin Jiushao , 1247).[ 4]
V originalnem dokazu je Heron uporabil tetivne štirikotnike . Drugi dokazi se oslanjajo na trigonometrijo, trikotniku včrtano in očrtano krožnico,[ 5] De Guajev izrek (za poseben primer ostrokotnih trikotnikov).[ 6]
Sledeči sodoben algebraični dokaz se precej razlikuje od Heronovega (Metrica ).[ 7]
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
γ
{\displaystyle \gamma }
koti . Z uporabo kosinusnega izreka dobimo
cos
γ
=
a
2
+
b
2
−
c
2
2
a
b
{\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}
S preoblikovanjem tega izraza dobimo
sin
γ
=
1
−
cos
2
γ
=
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
2
a
b
.
{\displaystyle \sin \gamma ={\sqrt {1-\cos ^{2}\gamma }}={\frac {\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}.}
Višina trikotnika na stranico
a
{\displaystyle a}
b
sin
γ
{\displaystyle b\,\sin \gamma }
S
=
1
2
(
stranica
)
(
višina na stranico
)
=
1
2
a
b
sin
γ
=
1
4
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
=
1
4
(
2
a
b
−
(
a
2
+
b
2
−
c
2
)
)
(
2
a
b
+
(
a
2
+
b
2
−
c
2
)
)
=
1
4
(
c
2
−
(
a
−
b
)
2
)
(
(
a
+
b
)
2
−
c
2
)
=
(
c
−
(
a
−
b
)
)
(
c
+
(
a
−
b
)
)
(
(
a
+
b
)
−
c
)
(
(
a
+
b
)
+
c
)
16
=
(
b
+
c
−
a
)
2
(
a
+
c
−
b
)
2
(
a
+
b
−
c
)
2
(
a
+
b
+
c
)
2
=
(
a
+
b
+
c
)
2
(
b
+
c
−
a
)
2
(
a
+
c
−
b
)
2
(
a
+
b
−
c
)
2
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle {\begin{aligned}S&={\frac {1}{2}}({\mbox{stranica}})({\mbox{višina na stranico}})\\&={\frac {1}{2}}ab\sin \gamma \\&={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\&={\frac {1}{4}}{\sqrt {(2ab-(a^{2}+b^{2}-c^{2}))(2ab+(a^{2}+b^{2}-c^{2}))}}\\&={\frac {1}{4}}{\sqrt {(c^{2}-(a-b)^{2})((a+b)^{2}-c^{2})}}\\&={\sqrt {\frac {(c-(a-b))(c+(a-b))((a+b)-c)((a+b)+c)}{16}}}\\&={\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 {s(s-a)(s-b)(s-c)}}.\end{aligned}}}
Na dveh mestih je bilo v dokazu je uporabljeno pravilo za razcep razlike kvadratov:
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
{\displaystyle a^{2}-b^{2}=(a+b)(a-b)}
Trikotnik z višino
h
{\displaystyle h}
c
{\displaystyle c}
d
+
(
c
−
d
)
.
{\displaystyle d+(c-d).}
Naslednji dokaz je zelo podoben dokazu, ki ga je objavil Raifaizen.[ 8] Pitagorovem izreku je
b
2
=
d
2
+
h
2
{\displaystyle b^{2}=d^{2}+h^{2}}
a
2
=
(
c
−
d
)
2
+
h
2
{\displaystyle a^{2}=(c-d)^{2}+h^{2}}
a
2
−
b
2
=
c
2
−
2
c
d
{\displaystyle a^{2}-b^{2}=c^{2}-2cd}
d
{\displaystyle d}
d
=
−
a
2
+
b
2
+
c
2
2
c
.
{\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}.}
Za višino trikotnika velja
h
2
=
b
2
−
d
2
{\displaystyle h^{2}=b^{2}-d^{2}}
d
{\displaystyle d}
h
2
=
b
2
−
(
−
a
2
+
b
2
+
c
2
2
c
)
2
=
(
2
b
c
−
a
2
+
b
2
+
c
2
)
(
2
b
c
+
a
2
−
b
2
−
c
2
)
4
c
2
=
(
(
b
+
c
)
2
−
a
2
)
(
a
2
−
(
b
−
c
)
2
)
4
c
2
=
(
b
+
c
−
a
)
(
b
+
c
+
a
)
(
a
+
b
−
c
)
(
a
−
b
+
c
)
4
c
2
=
2
(
s
−
a
)
⋅
2
s
⋅
2
(
s
−
c
)
⋅
2
(
s
−
b
)
4
c
2
=
4
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
c
2
.
{\displaystyle {\begin{aligned}h^{2}&=b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)^{2}\\&={\frac {(2bc-a^{2}+b^{2}+c^{2})(2bc+a^{2}-b^{2}-c^{2})}{4c^{2}}}\\&={\frac {{\big (}(b+c)^{2}-a^{2}{\big )}{\big (}a^{2}-(b-c)^{2}{\big )}}{4c^{2}}}\\&={\frac {(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^{2}}}\\&={\frac {2(s-a)\cdot 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^{2}}}\\&={\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}.\end{aligned}}}
Ta rezultat zdaj uporabimo v formuli, ki ploščino podaja kot polovico produkta med stranico in pripadajočo višino:
S
=
c
h
2
=
c
2
4
⋅
4
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
c
2
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle {\begin{aligned}S&={\frac {ch}{2}}\\&={\sqrt {{\frac {c^{2}}{4}}\cdot {\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}}}
Geometrijski pomen s − a s − b s − c Iz prvega dela kotangensnega zakona[ 9]
S
=
r
(
(
s
−
a
)
+
(
s
−
b
)
+
(
s
−
c
)
)
=
r
2
(
s
−
a
r
+
s
−
b
r
+
s
−
c
r
)
=
r
2
(
cot
α
2
+
cot
β
2
+
cot
γ
2
)
{\displaystyle {\begin{aligned}S&=r{\big (}(s-a)+(s-b)+(s-c){\big )}=r^{2}\left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\&=r^{2}\left(\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}\right)\end{aligned}}}
ter
S
=
r
s
{\displaystyle S=rs}
π
2
{\displaystyle {\frac {\pi }{2}}}
cot
(
ψ
)
+
cot
(
θ
)
+
cot
(
ϕ
)
=
cot
(
ψ
)
cot
(
θ
)
cot
(
ϕ
)
{\displaystyle \cot(\psi )+\cot(\theta )+\cot(\phi )=\cot(\psi )\cot(\theta )\cot(\phi )}
S
=
r
2
(
cot
α
2
cot
β
2
cot
γ
2
)
=
r
2
(
s
−
a
r
⋅
s
−
b
r
⋅
s
−
c
r
)
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
r
.
{\displaystyle {\begin{aligned}S&=r^{2}\left(\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\right)=r^{2}\left({\frac {s-a}{r}}\cdot {\frac {s-b}{r}}\cdot {\frac {s-c}{r}}\right)\\&={\frac {(s-a)(s-b)(s-c)}{r}}.\end{aligned}}}
Ko združimo obe enačbi, dobimo
S
2
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle S^{2}=s(s-a)(s-b)(s-c).}
Heronova formula, kot je podana zgoraj, je numerično nestabilna za trikotnike z zelo majhnim kotom pri uporabi aritmetike s plavajočo vejico. Za stabilno alternativo[ 10] [ 11]
a
≥
b
≥
c
{\displaystyle a\geq b\geq c}
S
=
1
4
(
a
+
(
b
+
c
)
)
(
c
−
(
a
−
b
)
)
(
c
+
(
a
−
b
)
)
(
a
+
(
b
−
c
)
)
.
{\displaystyle S={\frac {1}{4}}{\sqrt {{\big (}a+(b+c){\big )}{\big (}c-(a-b){\big )}{\big (}c+(a-b){\big )}{\big (}a+(b-c){\big )}}}.}
Oklepaji v zgornji formuli preprečujejo numerično nestabilnost pri računanju.
Obstajajo tri druge formule za ploščino z enako strukturo kot Heronova formula, izražene z različnimi količinami. Težiščnice trikotnika s stranicami
a
{\displaystyle a}
b
{\displaystyle b}
c
{\displaystyle c}
t
a
{\displaystyle t_{a}}
t
b
{\displaystyle t_{b}}
t
c
{\displaystyle t_{c}}
t
a
+
t
b
+
t
c
{\displaystyle t_{a}+t_{b}+t_{c}}
σ
=
1
2
(
t
a
+
t
b
+
t
c
)
{\displaystyle \sigma ={\frac {1}{2}}(t_{a}+t_{b}+t_{c})}
[ 12]
S
=
4
3
σ
(
σ
−
t
a
)
(
σ
−
t
b
)
(
σ
−
t
c
)
.
{\displaystyle S={\frac {4}{3}}{\sqrt {\sigma (\sigma -t_{a})(\sigma -t_{b})(\sigma -t_{c})}}.}
Nato označimo višine na stranice kot
h
a
{\displaystyle h_{a}}
h
b
{\displaystyle h_{b}}
h
c
{\displaystyle h_{c}}
H
=
1
2
(
h
a
−
1
+
h
b
−
1
+
h
c
−
1
)
{\displaystyle H={\frac {1}{2}}(h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1})}
[ 13]
S
−
1
=
4
H
(
H
−
h
a
−
1
)
(
H
−
h
b
−
1
)
(
H
−
h
c
−
1
)
.
{\displaystyle S^{-1}=4{\sqrt {H(H-h_{a}^{-1})(H-h_{b}^{-1})(H-h_{c}^{-1})}}.}
Na koncu polovično vsoto sinusov kotov označimo kot
K
=
1
2
(
sin
α
+
sin
β
+
sin
γ
)
{\displaystyle K={\frac {1}{2}}(\sin \alpha +\sin \beta +\sin \gamma )}
[ 14]
S
=
R
2
K
(
K
−
sin
α
)
(
K
−
sin
β
)
(
K
−
sin
γ
)
{\displaystyle S=R^{2}{\sqrt {K(K-\sin \alpha )(K-\sin \beta )(K-\sin \gamma )}}}
kjer je
R
{\displaystyle R}
R
=
a
sin
α
=
b
sin
β
=
c
sin
γ
{\displaystyle R={\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}}
Heronova formula je poseben primer Brahmaguptine formule , za računanje ploščin tetivnih štirikotnikov . Heronova formula in Brahmaguptina formula pa sta posebna primera Bretschneiderjeve formule za ploščino poljubnega štirikotnika . Heronovo formulo lahko dobimo iz Brahmaguptine formule ali Bretschneiderjeve formule tako, da eno od stranic štirikotnika nastavimo na nič.
Heronova formula je tudi poseben primer formule za ploščino trapeza , izračunano zgolj iz stranic. Heronovo formulo dobimo tako, da krajšo osnovnico nastavimo na nič.
Izražanje Heronove formule s Cayley-Mengerjevo determinanto kot kvadrati razdalj med tremi presečišči,[ 15]
S
=
1
4
−
|
0
a
2
b
2
1
a
2
0
c
2
1
b
2
c
2
0
1
1
1
1
0
|
{\displaystyle S={\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}}}}}
ponazarja njeno podobnost s Tartagliovo enačbo za prostornino tetraedra.
David P. Robbins je odkril še eno posplošitev Heronove formule na včrtane petkotnike in šestkotnike.[ 16]
Če so
U
{\displaystyle U}
V
{\displaystyle V}
W
{\displaystyle W}
u
{\displaystyle u}
v
{\displaystyle v}
w
{\displaystyle w}
u
{\displaystyle u}
U
{\displaystyle U}
[ 17]
prostornina
=
(
−
a
+
b
+
c
+
d
)
(
a
−
b
+
c
+
d
)
(
a
+
b
−
c
+
d
)
(
a
+
b
+
c
−
d
)
192
u
v
w
{\displaystyle {\text{prostornina}}={\frac {\sqrt {\,(-a+b+c+d)\,(a-b+c+d)\,(a+b-c+d)\,(a+b+c-d)}}{192\,u\,v\,w}}}
kjer so
a
=
x
Y
Z
b
=
y
Z
X
c
=
z
X
Y
d
=
x
y
z
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
)
.
{\displaystyle {\begin{aligned}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{aligned}}}
↑ »fórmula de Herón« . recursostic.educacion.es . Pridobljeno 1. julija 2021 .↑ Kendig, Keith (2000). »Is a 2000-Year-Old Formula Still Keeping Some Secrets?« . Amer. Math. Monthly . 107 : 402–415. doi :10.2307/2695295 . ↑ Heath, Thomas L. (1921). A History of Greek Mathematics . Zv. II. Oxford University Press. str. 321–323. ↑ 秦, 九韶 (1773). »卷三上, 三斜求积« . 數學九章 (四庫全書本) . ↑ »Personal email communication between mathematicians John Conway and Peter Doyle« . 15. december 1997. Pridobljeno 25. septembra 2020 .↑ Lévy-Leblond, Jean-Marc (14. september 2020). »A Symmetric 3D Proof of Heron's Formula« . The Mathematical Intelligencer (v angleščini). doi :10.1007/s00283-020-09996-8 . ISSN 0343-6993 . ↑ Niven, Ivan (1981). Maxima and Minima Without Calculus 7–8 . ↑ Raifaizen, Claude H. (1971). »A Simpler Proof of Heron's Formula« . Mathematics Magazine . 44 : 27–28. ↑ The second part of the Law of cotangents proof depends on Heron's formula itself, but this article depends only on the first part.
↑ Sterbenz, Pat H. (1. maj 1974). Floating-Point Computation Prentice Hall . ISBN 0-13-322495-3 ↑ William M. Kahan (24. marec 2000). »Miscalculating Area and Angles of a Needle-like Triangle« (PDF) . ↑ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette" 87, July 2003, 324–326.
↑ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
↑ Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.
↑ Weisstein, Eric W. »Cayley-Menger Determinant« . mathworld.wolfram.com (v angleščini). Pridobljeno 1. julija 2021 . ↑ D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.
↑ W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?",, pp. 16–17.
