헤론의 공식

2016. 6. 27. 11:30

[예제] 수학의 정석 수1. 실력편. 1장 연습문제

가 삼각형의 세 변의 길이를 나타낼 때,

라 한다. 이 때 이 삼각형의 넓이를 구하여라.

헤론의 공식

1. 위키피디아에서

$A = \sqrt{s(s-a)(s-b)(s-c)},$

$s=\frac{a+b+c}{2}.$

A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}

A=\frac{1}{4}\sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}

[증명1]

$\cos \gamma = \frac{a^2+b^2-c^2}{2ab}$

$\sin \gamma = \sqrt{1-\cos^2 \gamma} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.$

$\begin{align} A & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\ & = \frac{1}{2} ab\sin \gamma \\ & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\ & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\ & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\ & = \sqrt{\frac{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)}{16}} \\ & = \sqrt{\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}\frac{(a + b + c)}{2}} \\ & = \sqrt{\frac{(a + b + c)}{2}\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}} \\ & = \sqrt{s(s-a)(s-b)(s-c)}. \end{align}$

[증명2]

$b^2=h^2+d^2$ $a^2=h^2+(c-d)^2$

$d=\frac{-a^2+b^2+c^2}{2c}$

$\begin{align} h^2 & = b^2-\left(\frac{-a^2+b^2+c^2}{2c}\right)^2\\ & = \frac{(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-c^2)}{4c^2}\\ & = \frac{((b+c)^2-a^2)(a^2-(b-c)^2)}{4c^2}\\ & = \frac{(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^2}\\ & = \frac{2(s-a)\cdot 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^2}\\ & = \frac{4s(s-a)(s-b)(s-c)}{c^2} \end{align}$

$\begin{align} A & = \frac{ch}{2}\\ & = \sqrt{\frac{c^2}{4}\cdot \frac{4s(s-a)(s-b)(s-c)}{c^2}}\\ & = \sqrt{s(s-a)(s-b)(s-c)} \end{align}$

[증명3]

https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Miscellaneous_--_the_triple_tangent_identity

the triple tangent identity

If $\psi + \theta + \phi = \pi =$ half circle (for example, $\psi$ , $\theta$ and $\phi$ are the angles of a triangle),

\tan(\psi) + \tan(\theta) + \tan(\phi) = \tan(\psi)\tan(\theta)\tan(\phi).

Proof:

\begin{align} \psi & = \pi - \theta - \phi \\ \tan(\psi) & = \tan(\pi - \theta - \phi) \\ & = - \tan(\theta + \phi) \\ & = \frac{- \tan\theta - \tan\phi}{1 - \tan\theta \tan\phi} \\ & = \frac{\tan\theta + \tan\phi}{\tan\theta \tan\phi - 1} \\ (\tan\theta \tan\phi - 1) \tan\psi & = \tan\theta + \tan\phi \\ \tan\psi \tan\theta \tan\phi - \tan\psi & = \tan\theta + \tan\phi \\ \tan\psi \tan\theta \tan\phi & = \tan\psi + \tan\theta + \tan\phi \\ \end{align}

the triple cotangent identity

If $\psi + \theta + \phi = \tfrac{\pi}{2} =$ quarter circle,

\cot(\psi) + \cot(\theta) + \cot(\phi) = \cot(\psi)\cot(\theta)\cot(\phi)

Proof:

Replace each of $\psi$ , $\theta$ , and $\phi$ with their complementary angles, so cotangents turn into tangents and vice versa.

Given

\psi + \theta + \phi = \tfrac{\pi}{2}\,

\therefore (\tfrac{\pi}{2}-\psi) + (\tfrac{\pi}{2}-\theta) + (\tfrac{\pi}{2}-\phi) = \tfrac{3\pi}{2} - (\psi+\theta+\phi) = \tfrac{3\pi}{2} - \tfrac{\pi}{2} = \pi

so the result follows from the triple tangent identity.

\begin{align} A &= 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{align}

A = rs

\begin{align} A &= 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{align}

A^2 = s(s-a)(s-b)(s-c)

[증명4]

https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Excircles

\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Combining this with the identity $\sin^2 A + \cos^2 A = 1$ , we have

\sin A = \frac{\sqrt{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2 c^2 + 2 a^2 c^2}}{2bc}

But $\Delta = \tfrac{1}{2}bc \sin A$ , and so

\begin{align} \Delta &= \frac{1}{4} \sqrt{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2 c^2 + 2 a^2 c^2} \\ &= \frac{1}{4} \sqrt{ (a+b+c) (-a+b+c) (a-b+c) (a+b-c) }\\ & = \sqrt{s(s-a)(s-b)(s-c)}, \end{align}

저작자표시 비영리 동일조건 (새창열림)

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