Contoh Soal 1 (Segitiga dan Trigonometri)

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa}\\ &\cot 7\displaystyle \frac{1}{2}^{0}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\cot \alpha &=\displaystyle \frac{\cos \alpha }{\sin \alpha }=\displaystyle \frac{2\cos ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1+\cos 2\alpha }{\sin 2\alpha }\\ &=\displaystyle \frac{1+\cos 15^{\circ} }{\sin 15^{\circ} }\\ \cot 7\displaystyle \frac{1}{2}^{0}&=\displaystyle \frac{1+\cos (45-30) }{\sin (45-30) }\\\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}}{\displaystyle \frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}}\\ &=\displaystyle \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\\ &=\displaystyle \frac{\left ( 4+\sqrt{6}+\sqrt{2} \right )\left ( \sqrt{6}+\sqrt{2} \right )}{6-2}\\ &=\displaystyle \frac{4\sqrt{6}+4\sqrt{2}+6+2\sqrt{3}+2\sqrt{3}+2}{4}\\ &=\displaystyle \frac{8+4\sqrt{2}+4\sqrt{3}+4\sqrt{6}}{4}\\ &=2+\sqrt{2}+\sqrt{3}+\sqrt{6}\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan nilai eksak dari}\: \: \sin 18^{\circ}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \: 4(18^{\circ})=72^{\circ}=90^{\circ}-18^{\circ}\\ &\textrm{maka kita pilih}\: \: x=18^{\circ}.\: \textrm{Selanjutnya}\\ &\sin 4x=\sin (90^{\circ}-x)=\cos x\\ &\Leftrightarrow 2\sin 2x\cos 2x=\cos x\\ &\Leftrightarrow 2(2\sin x\cos x)(1-2\sin ^{2}x)=\cos x\\ &\Leftrightarrow 4\sin x(1-2\sin ^{2}x)=1\\ &\Leftrightarrow 4\sin x-8\sin ^{3}x-1=0\\ &\Leftrightarrow 8\sin ^{3}x-4\sin x+1=0\\ &\Leftrightarrow \left ( \sin x-1 \right )\left (4\sin ^{2}+2\sin x-1 \right )=0\\ &\Leftrightarrow 2\sin x=1\: \: \textrm{atau}\: \: 2\sin x=\displaystyle \frac{-1\pm \sqrt{5}}{2}\\ &\textrm{Nilai yang mungkin untuk}\: \: 2\sin x\: \: \textrm{dari}\\ &\textrm{ketiga nilai di atas adalah}\: \: \displaystyle \frac{-1+\sqrt{5}}{2}\\ &\textrm{Sehinga nilai dari}\\ &2\sin x=\displaystyle \frac{-1+\sqrt{5}}{2}\\ &\Leftrightarrow \sin x=\displaystyle \frac{-1+\sqrt{5}}{4}\\ &\textrm{karena}\: \: x=18^{\circ},\: \: \textrm{akan didapatkan}\\ &\textrm{nilai}\: \: \sin 18^{\circ}=\displaystyle \frac{-1+\sqrt{5}}{4}\\ &\textrm{Jadi, nilai eksak dari}\: \: \sin 18^{\circ}=\displaystyle \frac{\sqrt{5}-1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa}\\ & \tan 11\displaystyle \frac{1}{4}^{\circ}=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Langkah awal}\\ &\tan 22\frac{1}{2}^{\circ}=\displaystyle \frac{\sin 22\displaystyle \frac{1}{2}^{\circ}}{\cos 22\displaystyle \frac{1}{2}^{\circ}}=\displaystyle \frac{2\sin 22\displaystyle \frac{1}{2}^{\circ}\sin 22\displaystyle \frac{1}{2}^{\circ}}{2\sin 22\displaystyle \frac{1}{2}^{\circ}\cos 22\displaystyle \frac{1}{2}^{\circ}}\\ &\qquad =\displaystyle \frac{1-\cos 45^{\circ}}{\sin 45^{\circ}}=\displaystyle \frac{1-\displaystyle \frac{1}{2}\sqrt{2}}{\displaystyle \frac{1}{2}\sqrt{2}}=\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\\ &\qquad =\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{2\sqrt{2}-2}{2}\\ &\qquad =\sqrt{2}-1\\ &\textrm{Langkah berikutnya}\\ &\textrm{Misalkan}\: \: \tan 11\frac{1}{4}^{\circ}=x,\: \: \textrm{maka}\\ &\tan 22\displaystyle \frac{1}{2}^{\circ}=\displaystyle \frac{2\tan 11\displaystyle \frac{1}{4}^{\circ}}{1-\tan ^{2}11\displaystyle \frac{1}{4}^{\circ}}\\ &\Leftrightarrow \sqrt{2}-1=\displaystyle \frac{2x}{1-x^{2}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{2}-1}=\displaystyle \frac{1-x^{2}}{2x}\\ &\Leftrightarrow x^{2}+2\left ( \sqrt{2}+1 \right )x-1=0\\ &\Leftrightarrow \: x_{_{1,2}}=\displaystyle \frac{-2\left ( \sqrt{2}+1 \right )\pm \sqrt{4\left ( \sqrt{2}+1 \right )^{2}+4}}{2}\\ &\Leftrightarrow \: x_{_{1,2}}= -\left ( \sqrt{2}+1 \right )\pm \sqrt{3+2\sqrt{2}+1}\\ &\Leftrightarrow \: x_{_{1,2}}=-\sqrt{2}-1\pm \sqrt{4+2\sqrt{2}}\\ &\textrm{Ambil yang nilai positif saja}\\ &\textrm{Sehingga}\\ & \tan 11\displaystyle \frac{1}{4}^{\circ}=-\sqrt{2}-1+\sqrt{4+2\sqrt{2}}\quad \blacksquare \\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\gamma }{2}\tan \displaystyle \frac{\alpha }{2}=1\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ & \textrm{atau}\: \: \displaystyle \frac{1}{2}(\alpha +\beta )=\displaystyle \frac{1}{2}\left ( 180^{\circ}-\gamma \right )=90^{\circ}-\displaystyle \frac{\gamma }{2}\\ &\textrm{Maka}\\ &\tan \left ( \displaystyle \frac{\alpha }{2}+\frac{\beta }{2} \right )=\tan \left ( 90^{\circ}-\displaystyle \frac{\gamma }{2} \right )=\cot \displaystyle \frac{\gamma }{2}\\ &\Leftrightarrow \tan \left ( \displaystyle \frac{\alpha }{2}+\frac{\beta }{2} \right )=\displaystyle \frac{1}{\tan \displaystyle \frac{\gamma }{2}}\\ &\Leftrightarrow \displaystyle \frac{\tan \displaystyle \frac{\alpha }{2}+\tan \displaystyle \frac{\beta }{2}}{1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}}=\displaystyle \frac{1}{\tan \displaystyle \frac{\gamma }{2}}\\ &\Leftrightarrow \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}=1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}\\ &\Leftrightarrow \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\gamma }{2}\tan \displaystyle \frac{\alpha }{2}=1\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\cos \alpha +\cos \beta +\cos \gamma =1+4\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ & \textrm{atau}\: \: \displaystyle \frac{1}{2}(\alpha +\beta )=\displaystyle \frac{1}{2}\left ( 180^{\circ}-\gamma \right )=90^{\circ}-\displaystyle \frac{\gamma }{2}\\ &\textrm{Maka}\\ &\cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad +\cos \left ( 180^{\circ}-\left ( \alpha +\beta \right ) \right )\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad -\cos (\alpha +\beta )\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad -2\cos ^{2}\left ( \displaystyle \frac{\alpha +\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left (\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )-\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right ) \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( -2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{-\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( 90^{\circ}-\displaystyle \frac{\gamma }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =4\sin \displaystyle \frac{\gamma }{2}\left ( \sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =1+4\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\qquad \blacksquare \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI

Lanjutan 3 Materi Ketaksamaan : Ketaksamaan Renata (Rearrangement)

1. Ketaksamaan Pengaturan Ulang / Renata (Rearrangement)

Diberikan  $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$ dan $c_{1}\leq c_{2}\leq c_{3}\leq \cdots \leq c_{n}$ bilangan real dan sembarang susunan $(b_{1},b_{2}, b_{3}, \cdots , b_{n})$ dari  $(a_{1},a_{2}, a_{3}, \cdots , a_{n})$, maka akan berlaku:

$\begin{aligned}&a_{1}c_{1}+ a_{2}c_{2}+a_{3}c_{3}+ \cdots + a_{n}c_{n}\\ &\geq b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}+...+b_{n}c_{n}\\ &\geq a_{n}c_{1}+a_{n-1}c_{2}+a_{n-2}c_{3}+...+a_{1}c_{n} \end{aligned}$.

Perhatikan bentuk $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$ dengan $(b_{1},b_{2}, b_{3}, \cdots , b_{n})$ adalah sembarang permutasi dari $a_{1}\leq a_{2}\leq a_{3}\leq \cdots \leq a_{n}$, maka akan memiliki akibat

$\begin{aligned}1.\: \: &a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}\geq a_{1}b_{1}+ a_{2}b_{2}+ \cdots + a_{n}b_{n}  \\ 2.\: \: &\displaystyle \frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}+\frac{b_{3}}{a_{3}}+\cdots +\frac{b_{n}}{a_{n}}\geq n \end{aligned}$.

Penjelasan akibat no.1

Misalkan $a_{1}, a_{2}, a_{3}, \cdots , a_{n}$ adalah bilangan real positif asumsikan monoton naik, dengan akibat no. 1 akan diperoleh

$\begin{aligned}a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{2}+ a_{2}a_{3}+ \cdots + a_{n}a_{1}\\ a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{3}+ a_{2}a_{4}+ \cdots + a_{n}a_{2}\\ a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{4}+ a_{2}a_{5}+ \cdots + a_{n}a_{3}\\ \vdots &\qquad\vdots\\ a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}&\geq a_{1}a_{n}+ a_{2}a_{1}+ \cdots + a_{n}a_{n-1}\\  \end{aligned}$.

Jika kedua ruas setelah dijumlahkan ditambahkan dengan $a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}$, maka akan didapatkan  bentuk:

$\begin{aligned} &n\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}  \right )\geq \left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}\\ &\Leftrightarrow \displaystyle \frac{n\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}  \right )}{n^{2}}\geq \displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}}{n^{2}}\\ &\Leftrightarrow \displaystyle \frac{\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}  \right )}{n}\geq \displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}}{n^{2}}\\ &\Leftrightarrow \sqrt{\displaystyle \frac{\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}  \right )}{n}}\geq \sqrt{\displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )^{2}}{n^{2}}}\\ &\Leftrightarrow \color{blue}\sqrt{\displaystyle \frac{\left (a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}  \right )}{n}}\color{black}\geq \color{blue}\displaystyle \frac{\left ( a_{1}+a_{2}+\cdots +a_{n} \right )}{n}\\ &\textrm{Dan bentuk terakhir di atas adalah bentuk}\: \color{red}\textrm{QM-AM} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui} \: \: a,b\: \: \textrm{bilangan real positif}\\&\textrm{Tunjukkan bahwa}\: \: a^{2}+b^{2}\geq 2ab\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{red}\textbf{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\\ &(a-b)^{2}\geq 0\Leftrightarrow a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare  \end{aligned} \\  &\begin{aligned}&\color{red}\textbf{Alternatif 2}\\ &\textrm{Dengan AM-GM diperoleh}\\  &\displaystyle \frac{a^{2}+b^{2}}{2}\geq \sqrt{(ab)^{2}}\\ &\Leftrightarrow \frac{a^{2}+b^{2}}{2}\geq ab\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare  \end{aligned} \\ &\color{red}\textbf{Alternatif 3}\\ &\textrm{Asumsikan bahwa}\: \: a\geq b,\: \: \textrm{dengan}\\ & \textbf{ketaksamaan Renata}\: \textrm{diperoleh}\\  &a.a+b.b\geq a.b+b.a\\ &\Leftrightarrow a^{2}+b^{2}\geq ab+ab\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui} \: \: a,b\: \: \textrm{bilangan real positif}\\&\textrm{Tunjukkan bahwa}\: \: \displaystyle \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b\\\\ &\textbf{Bukti}\\ &\textrm{Asumsikan bahwa}\: \: a\geq b,\: \textrm{maka}\: \: a^{2}\geq b^{2}\\&\textrm{dan}\: \: \displaystyle \frac{1}{b}\geq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa baik}\: \left ( a^{2},b^{2} \right )\: \textrm{dan}\: \left ( \displaystyle \frac{1}{b}, \frac{1}{a} \right)\\ &\textrm{adalah kumpulan dua barisan yang monoton}\\ &\textrm{sama yaitu sama-sama naik. Sehingga}\\ &\textrm{dengan}\: \: \textbf{ketaksamaan Renata}\: \textrm{diperoleh}\\ &a^{2}.\displaystyle \frac{1}{b}+b^{2}.\displaystyle \frac{1}{a}\geq a^{2}.\displaystyle \frac{1}{a}+b^{2}.\displaystyle \frac{1}{b}\\ &\Leftrightarrow \displaystyle \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\  &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan}\: \: \color{red}\textbf{Cauchy-Schwarz}\\ &(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+a^{2})\geq (ab+bc+ca)^{2}\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare\\ &\color{blue}\textrm{Alternatif 3}\\  &\textrm{Untuk barisan}\: \: (a,b,c),\: \textrm{asumsikan}\: a\geq b\geq c\\ &\textrm{maka dengan}\: \: \color{red}\textbf{Ketaksamaan Renata}\: \: \color{black}\textrm{diperoleh}\\ &a.a+b.b+c.c\geq ab+bc+ca\\ &a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan bahwa}\\ &\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan AM-GM  diperoleh}\\ &\bullet \: \: \displaystyle \frac{a}{c}+\frac{c}{a}\geq 2\Leftrightarrow \displaystyle \frac{c}{a}\geq 2-\displaystyle \frac{a}{c}\\  &\bullet \: \: \displaystyle \frac{b}{c}+\frac{c}{b}\geq 2\Leftrightarrow \displaystyle \frac{b}{c}\geq 2-\displaystyle \frac{c}{b}\\ &\bullet \: \: \displaystyle \frac{a}{b}+\frac{b}{a}\geq 2\Leftrightarrow \displaystyle \frac{a}{b}\geq 2-\displaystyle \frac{b}{a}\\ &\textrm{Sehingga}\\ &\begin{aligned}\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq a\left ( 2-\displaystyle \frac{c}{b} \right )+b\left ( 2-\displaystyle \frac{a}{c} \right )+c\left ( 2-\displaystyle \frac{b}{a} \right )\\ &=2a-\displaystyle \frac{ac}{b}+2b-\displaystyle \frac{ab}{c}+2c-\displaystyle \frac{bc}{a}\\ \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq 2(a+b+c)-\left ( \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \right )\\ 2&\left ( \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \right )\geq 2(a+b+c)\\ \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}&\geq a+b+c\qquad \blacksquare  \end{aligned}\\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan Renata}\\ &\begin{aligned}&\textrm{Asumsikan}\: \: a\geq b\geq c,\: \textrm{maka}\: \: ab\geq ca\geq bc\\ &\textrm{dan}\: \: \displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}.\\ &\textrm{Perhatikan bahwa}\\ & (ab\geq ca\geq bc)\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a} \right )\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &ab.\displaystyle \frac{1}{c}+ac.\displaystyle \frac{1}{b}+bc.\displaystyle \frac{1}{a}\geq ab.\displaystyle \frac{1}{b}+ac.\displaystyle \frac{1}{a}+bc.\displaystyle \frac{1}{c}\\ &\Leftrightarrow \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+c+b\\ &\Leftrightarrow \displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan kebenaran}\\ &\textbf{ketaksamaan Nesbitt}\: \textrm{berikut}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan AM-GM  diperoleh}\\ &\displaystyle \frac{(a+b)+(b+c)+(c+a)}{3}\geq \displaystyle \frac{3}{\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}\\ &\Leftrightarrow ((a+b)+(b+c)+(c+a))\left ( \displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )\geq 9\\ &\Leftrightarrow 2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )+6\geq 9\\ &\Leftrightarrow 2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq 3\\ &\Leftrightarrow \left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare \\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan Renata}\\ &\begin{aligned}&\textrm{Asumsikan}\: \: a\leq  b\leq  c,\: \textrm{maka}\: \: a+b\leq a+c\leq b+c\\ &\textrm{dan}\: \: \displaystyle \frac{1}{b+c}\leq  \frac{1}{a+c}\leq  \frac{1}{a+b}.\\ &\textrm{Perhatikan bahwa}\\ & (a\leq  b\leq  c)\: \: \textrm{dan}\: \: \displaystyle \frac{1}{b+c}\leq  \frac{1}{a+c}\leq  \frac{1}{a+b}\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &a.\displaystyle \frac{1}{b+c}+b.\displaystyle \frac{1}{a+c}+c.\displaystyle \frac{1}{a+b}\geq b.\displaystyle \frac{1}{b+c}+c.\displaystyle \frac{1}{a+c}+a.\displaystyle \frac{1}{a+b}\\ &\textrm{dan}\\ &a.\displaystyle \frac{1}{b+c}+b.\displaystyle \frac{1}{a+c}+c.\displaystyle \frac{1}{a+b}\geq c.\displaystyle \frac{1}{b+c}+a.\displaystyle \frac{1}{a+c}+b.\displaystyle \frac{1}{a+b}\\ &\textrm{Jika dijumlahkan keduanya, maka}\\ &2\left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq 3\\ &\Leftrightarrow \left ( \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&(\textbf{OSN 2015})\\ &\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{Buktikan bahwa}\\ &\sqrt{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\displaystyle\frac{c}{a+b}+ \frac{a}{b+c}}\geq 3\\\\ &\textbf{Bukti}:\\  &\textrm{Perhatikan bukti soal no. 4 di atas}\\ &\textrm{Dengan}\: \: \textbf{keksamaan Renata}\: \: \textrm{dapat diperoleh}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{b}{b+c}+\frac{a}{a+c}\\ &\textrm{Misalkan}\\ &x=b+c,\: y=c+a,\: y=a+b,\: \textrm{maka}\\ &\displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{b}{b+c}+\frac{a}{a+c}\\ &\Leftrightarrow \displaystyle \frac{a}{b+c}+\frac{b}{a+c}\geq \frac{y+z-x}{2x}+\frac{x+z-y}{2y}\\ &\begin{aligned}  &\Leftrightarrow \sqrt{ \displaystyle \frac{a}{b+c}+\frac{b}{a+c}}\geq \sqrt{\frac{y+z-x}{2x}+\frac{x+z-y}{2y}}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \:  = \sqrt{\displaystyle \frac{1}{2}}\sqrt{\displaystyle \frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \: \:  \textrm{dengan AM-GM}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \:  \geq \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}+2\sqrt{\frac{y}{x}.\frac{x}{y}}-2}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \:  = \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}+2-2}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \:  \geq \displaystyle \frac{1}{\sqrt{2}}\sqrt{\displaystyle \frac{z}{x}+\frac{z}{y}}\\ &\Leftrightarrow \quad\quad\quad\quad \quad\quad\quad\: \:  \geq \displaystyle \frac{\sqrt{2}}{\sqrt{2}}\sqrt{\displaystyle \frac{z^{2}}{xy}}=\displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}\\ \end{aligned}\\ &\begin{aligned} &\bullet \:  \sqrt{ \displaystyle \frac{a}{b+c}+\frac{b}{a+c}}  \geq \displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}\\ &\bullet \:  \sqrt{ \displaystyle \frac{b}{c+a}+\frac{c}{a+b}}\geq \sqrt{\displaystyle \frac{x^{2}}{yz}}\\ &\bullet \:  \sqrt{ \displaystyle \frac{c}{a+b}+\frac{a}{b+c}}\geq \sqrt{\displaystyle \frac{y^{2}}{xz}} \end{aligned}\\ &\begin{aligned} &\textrm{Selanjutnya}\\ &\sqrt{\displaystyle \frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\displaystyle \frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\displaystyle\frac{c}{a+b}+ \frac{a}{b+c}}\\ &\geq \displaystyle \sqrt{\displaystyle \frac{z^{2}}{xy}}+\sqrt{\displaystyle \frac{x^{2}}{yz}}+\sqrt{\displaystyle \frac{y^{2}}{xz}}\\ &\textrm{Dengan AM-GM lagi}\\ &\geq 3\sqrt[3]{\sqrt{\displaystyle \frac{z^{2}}{xy}}\times \sqrt{\displaystyle \frac{x^{2}}{yz}}\times \sqrt{\displaystyle \frac{y^{2}}{xz}}}\\ &\geq 3\sqrt[3]{\sqrt{\displaystyle \frac{(xyz)^{2}}{(xyz)^{2}}}}\\ &\geq 3\qquad \blacksquare  \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diberikan}\: \: a,b,c>0,\: \: \textrm{tunjukkan bahwa}\\  &\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\textrm{Dengan mengaplikasikan AM-GM-HM}\\ &\textrm{pada}\: \: \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\: \: \textrm{kita dapat menemukan}\\ &\color{blue}\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\color{black}\geq \displaystyle \frac{3}{(abc)^{.^{\frac{1}{3}}}}\geq \color{blue}\displaystyle \frac{9}{a+b+c}\\ &\textrm{Jika kedua ruas dikalikan dengan}\: \: \color{red}a+b+c\color{black},\\ &\textrm{maka}\\ &\color{blue}3+\displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\color{black}\geq \color{blue}\displaystyle \frac{9(a+b+c)}{a+b+c}\\ &\Leftrightarrow \displaystyle \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 6\qquad \blacksquare \\ &\begin{aligned}&\color{red}\textrm{Alternatif 2}\\ &\textrm{Asumsikan}\: \: a\leq  b\leq  c,\: \textrm{maka}\: \: a+b\leq a+c\leq b+c\\ &\textrm{dan}\: \: \displaystyle \frac{1}{c}\leq  \frac{1}{b}\leq  \frac{1}{a}.\\ &\textrm{Perhatikan bahwa}\\ & (a+b\leq  a+c\leq  b+c)\: \: \textrm{dan}\: \: \displaystyle \frac{1}{c}\leq  \frac{1}{b}\leq  \frac{1}{a}\\ &\textrm{memiliki kemonotonan yang sama}\\ &\textrm{maka dengan}\: \: \textbf{ketaksamaan Renata}\\ &\textrm{dapat diperoleh bentuk}\\ &(b+c).\displaystyle \frac{1}{a}+(c+a).\displaystyle \frac{1}{b}+(a+b).\displaystyle \frac{1}{c}\geq (b+c).\displaystyle \frac{1}{b}+(c+a).\displaystyle \frac{1}{c}+(a+b).\displaystyle \frac{1}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 1+\displaystyle \frac{c}{b}+1+\frac{a}{c}+1+\frac{b}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+3\left ( \displaystyle \frac{abc}{abc} \right )^{.^{\frac{1}{3}}}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 3+3\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \doteq 6\qquad \blacksquare   \end{aligned}   \end{array}$.

DAFTAR PUSTAKA

1. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

WEBSITE

1. https://holdenlee.github.io/high_school/omc/23-rearrange.pdf diakses 18 Januari 2022.
2. https://www.gotohaggstrom.com/Advanced%20inequality%20manipulations.pdf  diakses 20 Januari 2022

Lanjutan 2 Materi Segitiga dan Ketaksamaan Cauchy-Schwarz-Engel

3. 2 Ketaksamaan Cauchy-Schwars

Jika diberikan sebarang bilangan real  $x_{1},x_{2},x_{3},...,x_{n}$ dan  $y_{1},y_{2},y_{3},...,y_{n}$, maka akan berlaku

$\begin{aligned}&(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}+...+x_{n}y_{n})^{\color{red}2}\\ &\leq (x_{1}^{\color{red}2}+x_{2}^{\color{red}2}+x_{3}^{\color{red}2}+...+x_{n}^{\color{red}2})(y_{1}^{\color{red}2}+y_{2}^{\color{red}2}+y_{3}^{\color{red}2}+...+y_{n}^{\color{red}2}) \end{aligned}$.

Jika dituliskan dengan notasi sigma adalah:

$\left ( \displaystyle \sum_{i=1}^{n}x_{i}y_{i} \right )^{\color{red}2}\leq \left ( \displaystyle \sum_{i=1}^{n}x_{i}^{\color{red}2} \right )\left ( \displaystyle \sum_{i=1}^{n}y_{i}^{\color{red}2} \right )$.

$\begin{aligned}&\color{blue}\textbf{Bukti}\: \: \color{black}\textrm{dari ketaksamaan ini adalah}:\\ &\textrm{Diberikan}\\&\alpha _{1},\alpha _{2},\alpha _{3},\cdots ,\alpha _{n}\: \: \textrm{dan}\: \: \beta_{1}+\beta_{2}+\beta _{3}+\cdots +\beta _{n}\\ &\textrm{Untuk setiap bilangan real}\: \: x\: ,\: \textrm{pada polinom kuadrat}\\ &\textrm{dapat berlaku}\\ &f(x)=\displaystyle \sum_{i=1}^{n}\left ( \alpha x+\beta  \right )^{2}\geq 0\\ &\Leftrightarrow \left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )x^{2}+2\left (\displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right )x+\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\geq 0\\ &\textrm{akibatnya diskriminan (D)}\leq 0,\: \textrm{maka}\\&D=b^{2}-4ac\leq 0\\ &\Leftrightarrow \left ( 2\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right ) \right )^{2}-4\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\leq 0\\ &\Leftrightarrow 4\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right )\leq 4\left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\\ &\Leftrightarrow \left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}\beta _{i} \right )\leq \left ( \displaystyle \sum_{i=1}^{n}\alpha _{i}^{2} \right )\left ( \displaystyle \sum_{i=1}^{n}\beta _{i}^{2} \right )\qquad \blacksquare   \end{aligned}$.

Ada hal yang sangat menarik ketika substitusi bentuk  $x_{i}=\displaystyle \frac{p_{i}}{\sqrt{q_{i}}}$  dan  $y_{i}=\sqrt{q_{i}}$, yaitu:

$\begin{aligned}&\color{blue}\textrm{Perhatikan bentuk berikut}\\ &(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}+...+x_{n}y_{n})^{\color{red}2}\\ &\leq (x_{1}^{\color{red}2}+x_{2}^{\color{red}2}+x_{3}^{\color{red}2}+...+x_{n}^{\color{red}2})(y_{1}^{\color{red}2}+y_{2}^{\color{red}2}+y_{3}^{\color{red}2}+...+y_{n}^{\color{red}2})\\ \end{aligned}$.

$\begin{aligned} &\textrm{dengan substitusi}\\ &\begin{cases} x_{i} &= \displaystyle \frac{p_{i}}{\sqrt{q_{i}}} \\ y_{i} &= \sqrt{q_{i}}  \end{cases}\\ &\textrm{maka menjadi bentuk}\\ \end{aligned}$.

$\begin{aligned} &(p_{1}+p_{2}+p_{3}+...+p_{n})^{2}\\ &\leq \left ( \displaystyle \frac{p_{1}^{2}}{q_{1}}+\frac{p_{2}^{2}}{q_{2}}+\frac{p_{3}^{2}}{q_{3}}+...+\frac{p_{n}^{2}}{q_{n}} \right )\left ( q_{1}+q_{2}+q_{3}+...+q_{n} \right )\\ &\Leftrightarrow \displaystyle \frac{(p_{1}+p_{2}+p_{3}+...+p_{n})^{2}}{\left ( q_{1}+q_{2}+q_{3}+...+q_{n} \right )}\leq \left ( \displaystyle \frac{p_{1}^{2}}{q_{1}}+\frac{p_{2}^{2}}{q_{2}}+\frac{p_{3}^{2}}{q_{3}}+...+\frac{p_{n}^{2}}{q_{n}} \right ) \end{aligned}$.

Bentuk di atas selanjutnya lebih dikenal dengan sebutan Cauchy-Schwarz Engel, karena bentuk ketaksamaan ini dipopulerkan oleh Arthur Engel dan biasa disebut dengan sebutan CS-Engel.

Perhatikan lemma berikut berkaitan dengan ketaksamaan CS-Engel di atas.

Jika diberikan $a,b$ bilangan real dan $x,y$ real positif akan ditunjukkan berlaku

$\displaystyle \frac{a^{2}}{x}+\frac{b^{2}}{y}\geq \displaystyle \frac{(a+b)^{2}}{x+y}$.

$\begin{aligned}&\color{blue}\textbf{Bukti}\\ &\textrm{Perhatikan bahwa}\: \: (ay-bx)^{2}\geq 0\\ &\Leftrightarrow a^{2}y^{2}+b^{2}y^{2}-2abxy\geq 0\\ &\Leftrightarrow a^{2}y^{2}+b^{2}y^{2}\geq 2abxy\\ &\Leftrightarrow \color{red}a^{2}xy\color{black}+a^{2}y^{2}+b^{2}y^{2}+\color{red}b^{2}xy\color{black}\geq \color{red}a^{2}xy+\color{black}2abxy+\color{red}b^{2}xy\\ &\Leftrightarrow (a^{2}y+b^{2}x)(x+y)\geq (a^{2}+2ab+b^{2})xy\\ &\Leftrightarrow \displaystyle \frac{(a^{2}y+b^{2}x)}{xy}\geq \displaystyle \frac{(a^{2}+2ab+b^{2})}{(x+y)}\\ &\Leftrightarrow \displaystyle \frac{a^{2}}{x}+\frac{b^{2}}{y}\geq \displaystyle \frac{(a+b)^{2}}{x+y}\qquad \blacksquare  \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: x,y\: \: \textrm{bilangan positif, tunjukkan}\\ &x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\\\\ &\textbf{Bukti}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &(x-y)^{2}\geq 0\Leftrightarrow x^{2}-2xy+y^{2}\geq 0\\ &\Leftrightarrow x^{2}+y^{2}\geq 2xy\\ &\Leftrightarrow 2x^{2}+2y^{2}\geq x^{2}+y^{2}+2xy\\ &\Leftrightarrow 2(x^{2}+y^{2})\geq (x+y)^{2}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare  \end{aligned}\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}}{2}}\geq \displaystyle \frac{x+y}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}}{2} \right )\geq \left ( \displaystyle \frac{x+y}{2} \right )^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}}{2} \right )\geq \displaystyle \frac{(x+y)^{2}}{4}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare  \end{aligned}\\  &\color{blue}\textrm{Alternatif 3}\\ &\color{purple}\textrm{Dengan ketaksamaan CS-Engel}\\ &\textrm{kita dapat peroleh}\\ &\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{1}\geq \displaystyle \frac{(x+y)^{2}}{2}\\ &\Leftrightarrow x^{2}+y^{2}\geq \displaystyle \frac{(x+y)^{2}}{2}\qquad \blacksquare  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Perhatikan bahwa}\: \: (a-b)^{2}\geq 0\\  &(a-c)^{2}\geq 0,\: \: \textrm{dan}\: \: (b-c)^{2}\geq 0\\ &\textrm{adalah benar, maka}\\ &(a-b)^{2}=a^{2}-2ab+b^{2}\geq 0\\ &\Leftrightarrow a^{2}+b^{2}\geq 2ab\: .....(1)\\ &\textrm{Dengan cara yang kurang lebih sama}\\ &\textrm{akan didapatkan}\\ &\bullet \quad a^{2}+c^{2}\geq 2ac\: .....(2)\\ &\bullet \quad b^{2}+c^{2}\geq 2bc\: .....(1)\\ &\textrm{Jika ketaksamaan}\quad (1),(2), \& \: (3)\: \: \textrm{dijumlahkan}\\ &\textrm{akan didapatkan bentuk}\\ &2a^{2}+2b^{2}+2c^{2}\geq 2ab+2ac+2bc\\ &\Leftrightarrow \: a^{2}+b^{2}+c^{2}\geq ab+ac+bc\quad \blacksquare\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan ketaksamaan}\: \: \color{red}\textbf{Cauchy-Schwarz}\\ &(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+a^{2})\geq (ab+bc+ca)^{2}\\ &\Leftrightarrow a^{2}+b^{2}+c^{2}\geq ab+ac+bc\qquad \blacksquare   \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Buktikan bahwa setiap bilangan real}\\ &\textrm{positif}\: \: a,\: b\: \: \textrm{dan}\: \: c\: \: \textrm{berlaku}\\ & a^{2}+b^{2}+c^{2}\geq 3(ab+ac+bc)\\\\ &\textbf{Bukti}\\  &\begin{aligned}&\textrm{Lihat jawaban no.2 di atas, yaitu}\\ &\color{red}a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\ &\textrm{karena nilai dari}\: \: \: (a+b+c)^{2}\\ &= a^{2}+b^{2}+c^{2}+2(ab+ac+bc)\\ &\textrm{maka nilai}\\ &a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+ac+bc)\\ &\textrm{Sehingga nilai untuk}\\ &\color{red}a^{2}+b^{2}+c^{2}\geq ab+ac+bc\: \: \: \color{black}\textrm{menjadi}\\ &(a+b+c)^{2}-2(ab+ac+bc)\geq ab+ac+bc\\ &\Leftrightarrow (a+b+c)^{2}\geq 3(ab+ac+bc)\qquad \blacksquare      \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: x,y,z\: \: \textrm{adalah bilangan real }\\ &\textrm{positif. Tunjukkan bahwa}\\ &\qquad\qquad  (x+y+z)^{2}\leq 3(x^{2}+y^{2}+z^{2})\\\\ &\textbf{Bukti}\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\color{red}\textrm{Dengan ketaksamaan Cauchy-Schwarz}\\ &\textrm{Kita dapat peroleh}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{untuk}\: \: \color{red}a=b=c=1,\: \color{black}\textrm{akan diperoleh}\\ &(ax+by+cz)^{2}\leq (1^{2}+1^{2}+1^{2})(x^{2}+y^{2}+z^{2})\\ &\Leftrightarrow (ax+by+cz)^{2}\leq (3)(x^{2}+y^{2}+z^{2})\quad \blacksquare \\ \end{aligned} \end{array}$.

$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\color{purple}\textrm{Dengan ketaksamaan CS-Engel}\\  &\textrm{Kita dapat peroleh}\\  &\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{1}+\frac{z^{2}}{1}\geq \displaystyle \frac{(x+y+z)^{2}}{1+1+1}\\ &\Leftrightarrow x^{2}+y^{2}+z^{2}\geq \displaystyle \frac{(x+y+z)^{2}}{3}\\ &\Leftrightarrow 3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2},\\ &\quad \textrm{atau}\\ &\Leftrightarrow (x+y+z)^{2}\leq 3(x^{2}+y^{2}+z^{2})\qquad \blacksquare  \end{aligned}$.

$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 3}\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}+z^{2}}{3}}\geq \displaystyle \frac{x+y+z}{3}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}+z^{2}}{3} \right )\geq \left ( \displaystyle \frac{x+y+z}{3} \right )^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}+y^{2}+z^{2}}{3} \right )\geq \displaystyle \frac{(x+y+z)^{2}}{9}\\ &\Leftrightarrow x^{2}+y^{2}+z^{2}\geq \displaystyle \frac{(x+y+z)^{2}}{3}\\ &\Leftrightarrow 3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2}\quad \textrm{atau}\\ &\Leftrightarrow  (x+y+z)^{2}\leq 3(x^{2}+y^{2}+z^{2}) \qquad \blacksquare  \end{aligned}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: x,y,z\: \: \textrm{adalah bilangan real positif}\\ &\textrm{dengan}\: \: x^{2}+y^{2}+z^{2}=27.\: \textrm{Tunjukkan}\\ & \textrm{bahwa}\: \: x^{3}+y^{3}+z^{3}\geq 81\\\\ &\textbf{Bukti}\\ &\color{red}\textrm{Pada contoh soal no.2 terdapat}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{ganti mengganti}\: \:  a=b=c=1,\: \textrm{maka menjadi}\\ &(x+y+z)^{2}\leq (1^{2}+1^{2}+1^{2})(x^{2}+y^{2}+z^{2})\\ &\Leftrightarrow (x+y+z)^{2}\leq (3)(x^{2}+y^{2}+z^{2})\: \color{red}........(1)\\ &\textrm{Selanjutnya dengan mengganti dengan}\\ &x^{.^{\frac{3}{2}}},y^{.^{\frac{3}{2}}},z^{.^{\frac{3}{2}}}\: \: \textrm{dan}\: \: x^{.^{\frac{1}{2}}},y^{.^{\frac{1}{2}}},z^{.^{\frac{1}{2}}},\: \textrm{pada}\\ &(ax+by+cz)^{2}\leq (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})\\ &\textrm{Kita akan dapatkan}\\ &(x^{2}+y^{2}+z^{2})\leq (x^{3}+y^{3}+z^{3})(x+y+z)\: \color{red}........(2)\\ &\textrm{Jika masing-masing ruas dikuadratkan, maka}\\ &(x^{2}+y^{2}+z^{2})^{4}\leq (x^{3}+y^{3}+z^{3})^{2}(x+y+z)^{2}\\ &\Leftrightarrow (x^{2}+y^{2}+z^{2})^{4} \leq 3(x^{3}+y^{3}+z^{3})^{2}\left ( x^{2}+y^{2}+z^{2} \right )\\ &\Leftrightarrow (x^{2}+y^{2}+z^{2})^{3} \leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow (27)^{3}\leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow 3^{9}\leq 3(x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow 3^{8}\leq (x^{3}+y^{3}+z^{3})^{2}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})^{2}\geq 3^{8}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})\geq 3^{.^{\frac{8}{2}}}\\ &\Leftrightarrow (x^{3}+y^{3}+z^{3})\geq 3^{4}=81\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Untuk}\: \: a,b,c,d\: \: \textrm{adalah bilangan real }\\ &\textrm{positif, tunjukkan bahwa}\\ &\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \displaystyle \frac{64}{a+b+c+d}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\color{red}\textrm{Dengan ketaksamaan CS-Engel}\\ &(a+b+c+d)\left (\displaystyle \frac{1^{2}}{a}+\displaystyle \frac{1^{2}}{b}+\displaystyle \frac{2^{2}}{c}+\displaystyle \frac{4^{2}}{d}  \right )\geq (1+1+2+4)^{2}\\ &\Leftrightarrow \displaystyle \frac{1^{2}}{a}+\displaystyle \frac{1^{2}}{b}+\displaystyle \frac{2^{2}}{c}+\displaystyle \frac{4^{2}}{d}\geq \displaystyle \frac{(1+1+2+4)^{2}}{a+b+c+d}\\ &\Leftrightarrow \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \displaystyle \frac{64}{a+b+c+d}\quad \blacksquare  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: a,b\: \: \textrm{bilangan real positif}\\ &\textrm{dan}\: \: a+b=1.\: \textrm{Tunjukkan bahwa}\\ &\qquad  \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left ( b+\displaystyle \frac{1}{b} \right )^{2}\geq \displaystyle \frac{25}{2}\\\\  &\textbf{Bukti}\\   &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: a+b=1\\ &\color{red}\textrm{Dengan QM-AM kita akan mendapatkan}\\ &\sqrt{\displaystyle \frac{x^{2}+y^{2}}{2}}\geq \displaystyle \frac{x+y}{2}\Leftrightarrow \displaystyle \frac{x^{2}+y^{2}}{2}\geq \left ( \displaystyle \frac{x+y}{2} \right )^{2}\\ &\textrm{Kita misalkan}\: \: \color{purple}\begin{cases} x & =a+\displaystyle \frac{1}{a} \\\\  y & =b+\displaystyle \frac{1}{b}  \end{cases}\\ &\textrm{maka}\\ &\begin{aligned}\displaystyle \frac{\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}}{2}&\geq \left ( \displaystyle \frac{a+\displaystyle \frac{1}{a}+b+\displaystyle \frac{1}{b}}{2} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( a+b+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+ \displaystyle \frac{a+b}{ab}\right )^{2}\\ &\geq \displaystyle \frac{1}{4}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2} \end{aligned}\\ &\color{red}\textrm{Dari GM-AM kita akan mendapatkan}\\ &\sqrt{ab}\leq \displaystyle \frac{a+b}{2}\Leftrightarrow ab\leq \left ( \displaystyle \frac{a+b}{2} \right )^{2}\\ &\Leftrightarrow ab\leq \left ( \displaystyle \frac{1}{2} \right )^{2}\Leftrightarrow ab\leq \displaystyle \frac{1}{4} \\ &\textrm{Sehingga}\\ &\begin{aligned}\left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}&\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+ \displaystyle \frac{1}{\displaystyle \frac{1}{4}}\right )^{2}\\ &\geq \displaystyle \frac{1}{2}\left ( 1+4 \right )^{2}\\ &\geq \displaystyle \frac{1}{2}\times 25\\ &\geq \displaystyle \frac{25}{2}\qquad \blacksquare  \end{aligned}  \end{aligned}  \end{array}$.

$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 2}\\ &\color{red}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &1=a+b\geq 2\sqrt{ab}\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\Leftrightarrow 2\leq \displaystyle \frac{1}{\sqrt{ab}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{ab}}\geq 2\Leftrightarrow \displaystyle \frac{1}{ab}\geq 4\\ &\color{red}\textrm{Perhatikan soal, dengan QM-AM}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\begin{aligned}\displaystyle \frac{\left ( a+\frac{1}{a} \right )^{2}+\left (b+\frac{1}{b}  \right )^{2}}{2}&\geq \left ( \displaystyle \frac{a+\frac{1}{a}+b+\frac{1}{b}}{2} \right )^{2}\\ \left ( a+\displaystyle \frac{1}{a} \right )^{2}+\left (b+\displaystyle \frac{1}{b}  \right )^{2}&\geq \displaystyle \frac{1}{2}\left ( a+b+\displaystyle \frac{1}{a}+\frac{1}{b} \right )^{2}\\ &=\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{a+b}{ab} \right )^{2}\\ &=\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &=\displaystyle \frac{1}{2}(1+4)^{2}\\ &= \displaystyle \frac{1}{2}\times 25\\ &\geq \displaystyle \frac{25}{2}\qquad \blacksquare    \end{aligned}    \end{aligned}$.

$\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Alternatif 3}\\ &\color{red}\textrm{Dengan AM-GM kita akan mendapatkan}\\ &1=a+b\geq 2\sqrt{ab}\Leftrightarrow \displaystyle \frac{1}{2}\geq \sqrt{ab}\Leftrightarrow 2\leq \displaystyle \frac{1}{\sqrt{ab}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{ab}}\geq 2\Leftrightarrow \displaystyle \frac{1}{ab}\geq 4\\ &\color{red}\textrm{Perhatikan soal, dengan CS-Engel}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\begin{aligned} &(1+1)\left (\displaystyle \frac{\left ( a+\frac{1}{a} \right )^{2}}{1} +\displaystyle \frac{\left ( b+\frac{1}{b} \right )^{2}}{1} \right )\geq \left (a+\frac{1}{a}+b+\frac{1}{b}  \right )^{2}\\ &\Leftrightarrow 2\left (\left (\displaystyle a+\frac{1}{a}  \right)^{2} +\left (b+\frac{1}{b}  \right )^{2} \right )\geq \left (a+b+\frac{1}{a}+\frac{1}{b}  \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq \displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{a+b}{ab} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq\displaystyle \frac{1}{2}\left ( 1+\displaystyle \frac{1}{ab} \right )^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq\displaystyle \frac{1}{2}(1+4)^{2}\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq \displaystyle \frac{1}{2}\times 25\\ &\Leftrightarrow \left (\displaystyle a+\frac{1}{a}  \right )^{2} +\left (b+\frac{1}{b}  \right )^{2}\geq \displaystyle \frac{25}{2}\qquad \blacksquare    \end{aligned}      \end{aligned}$.

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan positif, buktikan}\\ &\textrm{a}.\quad 2(a^{2}+b^{2})\geq (a+b)^{2}\\ &\textrm{b}.\quad 4(a^{3}+b^{3})\geq (a+b)^{3}\\ &\textrm{c}.\quad 8(a^{4}+b^{4})\geq (a+b)^{4}\\ &\textrm{d}.\quad 16(a^{5}+b^{5})\geq (a+b)^{5}\\ &\textrm{e}.\quad 32(a^{6}+b^{6})\geq (a+b)^{6}\\ &\textrm{f}.\quad 64(a^{7}+b^{7})\geq (a+b)^{7}\\ &\textrm{g}.\quad 128(a^{8}+b^{8})\geq (a+b)^{8}\\\\ &\textbf{Bukti}:\\  &\textrm{Akan ditunjukkan bukti poin 8.c saja}\\ &\textrm{untuk poin yang lain, silahkan pembaca}\\ &\textrm{sekalian untuk dibuktikan sendiri sebagai}\\ &\textrm{bahan latihan mandiri}.\\ &\textrm{Adapun bukti poin 8.c adalah sebagaimana}\\ &\textrm{berikut ini}\\ &\begin{aligned} &\color{red}\textrm{Dengan ketaksamaan CS-Engel}\: \color{black}\textrm{akan}\\ &\textrm{didapatkan}\\ &\bullet \: \: (1+1)(a^{4}+b^{4})\geq (a^{2}+b^{2})^{2}\\ &\: \quad\Leftrightarrow 2\left (a^{4}+b^{4}  \right )\geq (a^{2}+b^{2})^{2}\: \color{red}..........(1)\\ &\bullet \: \: (1+1)(a^{2}+b^{2})\geq (a+b)^{2}\\ &\: \quad\Leftrightarrow 2\left (a^{2}+b^{2}  \right )\geq \displaystyle (a+b)^{2},\quad (\textrm{kuadratkan})\\ &\: \quad \Leftrightarrow 4\displaystyle \left (a^{2}+b^{2}  \right )^{2}\geq (a+b)^{4}\\ &\: \quad \Leftrightarrow \left (a^{2}+b^{2}  \right )^{2}\geq \displaystyle \frac{(a+b)^{4}}{4}\: \color{red}...........(2)\\ &\textrm{Dari (1) dan (2) didapatkan hubungan}\\ &2\left (a^{4}+b^{4}  \right )\geq \left (a^{2}+b^{2}  \right )^{2}\geq \displaystyle \frac{(a+b)^{4}}{4}\\ &\Leftrightarrow 8\left ( a^{4}+b^{4} \right )\geq 4\left ( a^{2}+b^{2} \right )^{2}\geq (a+b)^{4}\\ &\Leftrightarrow 8\left ( a^{4}+b^{4} \right )\geq (a+b)^{4}\qquad \blacksquare   \end{aligned}    \end{array}$.

$.\quad\qquad\begin{aligned}&\color{red}\textrm{Mengenal penulisan pola}\: \textbf{Siklik dan Simetri}\\ &\textrm{Misal untuk}\: \: n=3,\: \: \textrm{pada penulisan unsur}\\ &x,y,\: \: \textrm{dan}\: \: z,\: \textrm{maka}\\ &\begin{array}{|l|l|}\hline \textbf{Pola Siklik}&\textbf{Pola Simetri}\\\hline\begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}=x^{2}+y^{2}+z^{2}\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}\displaystyle\sum_{\textrm{sym}}^{.}x^{2}&=x^{2}+x^{2}\\ &+y^{2}+y^{2}\\ \\ &+z^{2}+z^{2}\\ \\ &=2\left (x^{2}+y^{2}+z^{2}  \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{3}=x^{3}+y^{3}+z^{3}\end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{3}=2\left (x^{3}+y^{3}+z^{3}  \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y=x^{2}y+y^{2}z+z^{2}x\\ &\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{2}y&=x^{2}y+x^{2}z\\ &+y^{2}x+y^{2}z\\\\ &+z^{2}x+z^{2}y \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}xyz&=xyz+yzx+zxy\\ &=3xyz \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}xyz&=xyz+xzy+\cdots \\ &=6xyz \end{aligned}  \\\hline \end{array} \end{aligned}$.

$\begin{array}{ll}\\ 9.&(\textbf{IMO 1995})\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{bilangan-bilangan real positif}\\ &\textrm{dengan}\: \: abc=1,\: \: \textrm{maka tunjukkan bahwa}\\ &\displaystyle \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\geq \displaystyle \frac{3}{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Misalkan}\: \: x=\displaystyle \frac{1}{a},\: y=\displaystyle \frac{1}{b},\: \: \textrm{dan}\: \: z=\displaystyle \frac{1}{c},\\ & \textrm{maka}\\ &\displaystyle \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\\ &=\displaystyle \frac{x^{3}yz}{y+z}+ \frac{y^{3}xz}{x+z}+ \frac{z^{3}xy}{x+y},\: \: \textrm{karena}\: xyz=1\\ &=\displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\\ &\textrm{Dengan ketaksamaan}\: \textbf{Cauchy-Schwarz}\\ &\left ( 2\displaystyle \sum_{\textrm{siklik}}^{.}y+z \right )\left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq (x+y+z)^{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{(x+y+z)^{2}}{2(x+y+z)}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{(a+b+c)}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{3\sqrt[3]{xyz}}{2}\\ &\Leftrightarrow \left ( \displaystyle \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right )\geq \displaystyle \frac{3}{2}\qquad \blacksquare  \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Tung. K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.
2. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Lanjutan Materi 1 Segitiga dan Ketaksamaan QM-AM-GM-HM

1. Identitas Trigonometri dalam Segitiga

Perhatikan segitiga sembarang berikut

$\begin{aligned}&\textrm{Dalam Sebuah segitiga akan berlaku}\\ &\textrm{identitas berikut}\\ &1.\quad \sin \alpha +\sin \beta +\sin \gamma =4\cos \displaystyle \displaystyle \frac{\alpha }{2}\cos \displaystyle \frac{\beta }{2}\cos \displaystyle \frac{\gamma }{2}\\ &2.\quad \cos \alpha +\cos \beta +\cos \gamma =4\sin \displaystyle \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}+1\\ &3.\quad \tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \tan \beta \tan \gamma \\ &4.\quad \sin^{2} \alpha +\sin^{2} \beta +\sin^{2} \gamma =2\cos \alpha \cos \beta \cos \gamma +2\\ &5.\quad \sin 2\alpha +\sin 2\beta +\sin 2\gamma =4\sin \alpha \sin \beta \sin \gamma \\ &6.\quad \cot \displaystyle \frac{\alpha }{2}+\cot \frac{\beta}{2}+\cot \displaystyle \frac{\gamma }{2}=\cot \displaystyle \frac{\alpha }{2}+\cot \frac{\beta}{2}+\cot \displaystyle \frac{\gamma }{2}\\ &7.\quad \cot \alpha \cot \beta +\cot \alpha \cot \gamma + \cot \beta \cot \gamma =1 \end{aligned}$.

2. Segitiga dan Pertidaksamaan Segitiga

Pada sebuah segitiga pengklasifikasiannya dapat berdasarkan berdasarkan panjang sisinya ataupun jenis sudut-sudutnya. Berikut untuk klasifikasi berdasarkan panjang sisinya

$\begin{aligned} &\textrm{a}.\quad \textrm{sembarang}\\ &\textrm{b}.\quad \textrm{samakaki}\\ &\textrm{c}.\quad \textrm{samasisi}\\ \end{aligned}$.

Dan berdasarkan jenis sudutnya sebuah segitiga dapat dikategorikan dengan

$\begin{aligned} &\textrm{a}.\quad \textrm{lancip}\\ &\textrm{b}.\quad \textrm{siku-siku}\\ &\textrm{c}.\quad \textrm{tumpul}\\ \end{aligned}$.

Adapun berkaitan dengan segmen garis yang akan menjadi penyusun sebuah segitiga, maka sebuah segitiga hanya bisa dibuat dari ketiga segmen garis yang mana segemen garis yang terpanjang akan selalu lebih pendek dari pada jumlah panjang kedua segmen garis yang lainnya atau segmen garis yang terpendek akan selalu lebih panjang dari pada selisih panjang dari kedua segemn garis yang lainnya. Sifat tersebut lazim dinamkan dengan ketidaksamaan dalam segitiga.

$\begin{aligned} &\textbf{Pertidaksamaan dalam Segitiga}\\ &\begin{cases} a+b &>c \\ a+c &>b \\ b+c &>a \end{cases}\\ &\textbf{atau}\\ &\begin{cases} \left | a-b \right | &<c \\  \left | a-c \right | &<b \\  \left | b-c \right | &<a \end{cases} \end{aligned}$. 

3. Ketaksamaan (Inequality)

Di sini yang akan dibahas adalah beberapa ketaksamaan secara umum yang tentunya sebagian berlaku pada segitiga untuk membantu para siswa dijenjang SMP atau SMA atau sederajat juga menjadi pengingat buat penulis sendiri, karena materi ini hampir menuntut daya nalar yang lebih dengan prasyarat telah terbiasa dengan soal-soal semisal aljabar dan trigonometri.

3. 1 Ketaksamaan QM-AM-GM-HM

Dalam setiap soal yang melibatkan ketaksamaan biasanya muncul dalam soal berkategori KSN (Kompetisi Sains Nasional) baik tingkat kabupaten, provinsi bahkan nasional maupun juga KSM (Kompetisi Sains Madrasah) dengan jenjang yang sama serta soal-soal dengan kategori kompetisi yang semisal. Model penyelesaian yang digunakan hampir sering akan melibatkan penggunaan ketaksamaan metode ini, yaitu QM-AM-GM-HM. Pada beberapa contoh soal di bawah dapat Anda cermati tentang penggunaan penyelesaian cara ini demikian pula pada halaman-halaman berikutnya pada blog ini akan dibahas beberapa soal dan diselesaikan dengan cara ini.

$\begin{aligned}&\textbf{Pada segitiga akan berlaku}\\ &\bullet \: \textrm{Quadratic Mean}=QM=\sqrt{\displaystyle \frac{a^{2}+b^{2}+c^{2}}{3}}\\ &\bullet \:  \textrm{Arithmetic Mean}=AM= \frac{a+b+c}{3}\\ &\bullet \: \textrm{Geometric Mean}=GM=\sqrt[3]{abc}\\ &\bullet \: \textrm{Harmonic Mean}=HM=\displaystyle \frac{3}{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\\ \end{aligned}$.

Dengan kata lain QM adalah rataan kuadratik, AM adalah rataan aritmetik, dan GM adalah rataan geometri, serta HM rataan harmoni dan besarnya $QM\geq AM\geq GM\geq HM$.

$\begin{aligned}&\textrm{Misalkan diberikan}\: \: x_{1},x_{2},x_{3},\cdots ,x_{n}\\ & \textrm{bilangan real positif, maka hubungan }\\ &\textrm{ketaksamaan}\: \: \color{red}\textrm{QM-AM-GM-HM}\\ & \color{black}\textrm{dapat dituliskan}\\ &\sqrt{\displaystyle \frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\cdots +x_{n}^{2}}{n}}\qquad\qquad \color{red}(\textrm{QM})\\ &\geq \displaystyle \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}\: \: \: \quad\quad\quad \color{red}(\textrm{AM})\\ &\geq \sqrt[n]{x_{1}.x_{2}.x_{3}\cdots x_{n}}\: \: \: \: \qquad\qquad\qquad \color{red}(\textrm{GM})\\ &\geq \displaystyle \frac{n}{\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots +\frac{1}{x_{n}}}\quad\quad \color{red}(\textrm{HM}) \end{aligned}$.

Anda juga bisa klik di sini untuk QM, AM, GM, dan HM.

$\begin{aligned}&\textrm{Misalkan untuk}\: \: \color{blue}n=2\\ &\textrm{maka hubungan ketaksamaannya adalah}\\ &\sqrt{\displaystyle \frac{x_{1}^{2}+x_{2}^{2}}{2}}\geq \displaystyle \frac{x_{1}+x_{2}}{2}\geq \sqrt{x_{1}x_{2}}\geq \displaystyle \frac{2}{\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}} \end{aligned}$.

$\begin{aligned}&\textrm{Dan misalkan untuk}\: \: \color{blue}n=3\\ &\textrm{maka hubungan ketaksamaannya adalah}\\ &\sqrt{\displaystyle \frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{3}}\geq \displaystyle \frac{x_{1}+x_{2}+x_{3}}{3}\geq \sqrt[3]{x_{1}x_{2}x_{3}}\geq \displaystyle \frac{3}{\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}} \end{aligned}$.

Demikian seterusnya.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\sin 2\alpha +\sin 2\beta +\sin 2\gamma =4\sin \alpha \sin \beta \sin \gamma \\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ &\textrm{Maka}\\ &\color{red}\sin 2\alpha +\sin 2\beta +\sin 2\gamma \\ &=2\sin \left ( \displaystyle \frac{2\alpha +2\beta }{2} \right )\cos \left ( \displaystyle \frac{2\alpha -2\beta }{2} \right )+2\sin \gamma \cos \gamma \\ &=2\sin (\alpha +\beta )\cos (\alpha -\beta )+2\sin \gamma \cos \gamma \\ &=2\sin \gamma \cos (\alpha -\beta )+2\sin \gamma \cos \gamma \\ &=2\sin \gamma \left (\cos (\alpha -\beta )+ \cos \gamma  \right ) \\ &=2\sin \gamma \left ( \cos (\alpha -\beta )-\cos (\alpha +\beta ) \right )\\ &=2\sin \gamma \left ( -2\sin \alpha \sin (-\beta ) \right )\\ &=4\sin \gamma \sin \alpha \sin \beta \\ &=4 \sin \alpha \sin \beta \sin \gamma \qquad \blacksquare \\    \end{aligned}  \end{array}$.

Catatan: 

Kotak persegi kecil hitam diletakkan diakhir pembuktian menunjukkan pembuktian telah dianggap cukup dan memenuhi

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{adalah bilangan real}\\ &\textrm{positif, tunjukkan bahwa}\: \: \displaystyle \frac{a}{b}+\frac{b}{a}\geq 2\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Dengan AM-GM}\\ &\displaystyle \frac{\displaystyle \frac{a}{b}+\frac{b}{a}}{2}\geq \sqrt{\displaystyle \frac{a}{b}.\frac{b}{a}}\\ &\Leftrightarrow \: \displaystyle \frac{\displaystyle \frac{a}{b}+\frac{b}{a}}{2}\geq 1\\ &\Leftrightarrow \: \displaystyle \frac{a}{b}+\frac{b}{a}\geq 2\qquad \blacksquare  \end{aligned}\\ &\color{red}\textrm{Alternatif 2}\\ &\begin{aligned}&\left (  \sqrt{\displaystyle \frac{a}{b}}-\sqrt{\displaystyle \frac{b}{a}}\right )^{2}\geq 0\\ &\Leftrightarrow \: \displaystyle \frac{a}{b}-2+\displaystyle \frac{b}{a}\geq 0\\ &\Leftrightarrow \: \displaystyle \frac{a}{b}+\frac{b}{a}\geq 2\qquad \blacksquare  \end{aligned}\\ \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Untuk sembarang}\: \: x>0\\ &\textrm{Tunjukkan bahwa}\: \: \displaystyle \frac{x^{2}}{1+x^{4}}\leq \displaystyle \frac{1}{2}\\\\ &\textbf{Bukti}:\\ &\color{red}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: \: \left (x^{2}-1 \right )^{2}\geq 0\\ &\textrm{maka}\\ &\left (x^{2}-1 \right )^{2}=x^{4}-2x^{2}+1\geq 0\\ &\Leftrightarrow \: x^{4}+1\geq 2x^{2}\\ &\Leftrightarrow \: 1\geq \displaystyle \frac{2x^{2}}{x^{4}+1}\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\geq \displaystyle \frac{x^{2}}{x^{2}+1}\: \: \textrm{atau}\\ &\Leftrightarrow \: \displaystyle \frac{x^{2}}{x^{4}+1}\leq \displaystyle \frac{1}{2}\qquad \blacksquare   \end{aligned} \\ &\color{red}\textrm{Alternatif 2}\\ &\begin{aligned} &\textrm{Dengan AM-GM}\\ &\displaystyle \frac{1+x^{4}}{2}\geq \sqrt{1\times x^{4}}\\ &\Leftrightarrow \: \displaystyle \frac{1+x^{4}}{2}\geq x^{2}\\ &\Leftrightarrow \: \displaystyle \frac{1+x^{4}}{x^{2}}\geq  2\\ &\Leftrightarrow \: \displaystyle \frac{x^{2}}{1+x^{4}}\leq \displaystyle \frac{1}{2} \qquad \blacksquare \\  \end{aligned} \\ &\color{red}\textrm{Alternatif 3}\\ &\begin{aligned} &\displaystyle \frac{x^{2}}{1+x^{4}}=\displaystyle \frac{1}{\displaystyle \frac{1}{x^{2}}+x^{2}}\\ &\textrm{Dengan AM-GM}\\ &\displaystyle \frac{\displaystyle \frac{1}{x^{2}}+x^{2}}{2}\geq \sqrt{\displaystyle \frac{1}{x^{2}}\times x^{2}}\\ &\Leftrightarrow \: \displaystyle \frac{\displaystyle \frac{1}{x^{2}}+x^{2}}{2}\geq 1\\ &\Leftrightarrow \: \displaystyle \frac{1}{\displaystyle \frac{\displaystyle \frac{1}{x^{2}}+x^{2}}{2}}\leq 1\\ &\Leftrightarrow \: \displaystyle \frac{2}{\displaystyle \frac{\displaystyle \frac{1}{x^{2}}+x^{2}}{2}}\leq 1\\ &\Leftrightarrow \: \displaystyle \frac{1}{\displaystyle \frac{\displaystyle \frac{1}{x^{2}}+x^{2}}{2}}\leq \displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \displaystyle \frac{1}{\displaystyle \frac{\displaystyle \frac{1}{x^{2}}+x^{2}}{2}}=\displaystyle \frac{x^{2}}{1+x^{4}}\leq \displaystyle \frac{1}{2}\qquad \blacksquare \\  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah sisi segitiga yang}\\  &\textrm{memenuhi}\: \: a+b+c=1,\: \textrm{tunjukkan}\\ &\textrm{bahwa}\: \: ab+ac+bc\leq \displaystyle \frac{1}{2}\\\\  &\textbf{Bukti}:\\   &\begin{aligned}&(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\\ &\Leftrightarrow \: 1=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\\ &\Leftrightarrow \: 2ab+2ac+2bc=1-\left (a^{2}+b^{2}+c^{2}  \right )\\ &\Leftrightarrow \: ab+bac+bc=\displaystyle \frac{1}{2}-\displaystyle \frac{1}{2}\left ( a^{2}+b^{2}+c^{2} \right )\\ &\Leftrightarrow \: ab+bac+bc\leq \displaystyle \frac{1}{2}\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah bilangan real positif}\\  &\textrm{dengan}\: \: a+b+c=1,\: \textrm{tunjukkan}\\ &\textrm{bahwa}\: \: \left ( \displaystyle \frac{1}{a}-1 \right )\left ( \displaystyle \frac{1}{b}-1 \right )\left ( \displaystyle \frac{1}{c}-1 \right )\geq 8\\\\  &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Telah diketahui bahwa}:\: a+b+c=1 \\ &\textrm{Dengan ketaksamaan AM-GM kita}\\ &\textrm{bisa mendapatkan}\\ &\bullet \: \: \displaystyle \frac{a+b}{2}\geq \sqrt{ab}\Leftrightarrow a+b\geq 2\sqrt{ab}\\ &\bullet \: \: \displaystyle \frac{b+c}{2}\geq \sqrt{bc}\Leftrightarrow b+c\geq 2\sqrt{bc}\\ &\bullet \: \: \displaystyle \frac{c+a}{2}\geq \sqrt{ca}\Leftrightarrow c+a\geq 2\sqrt{ca}\\ &\textrm{Selanjutnya kembali kepersoalan, yaitu}:\\ &\begin{aligned} \left ( \displaystyle \frac{1}{a}-1 \right )&\left ( \displaystyle \frac{1}{b}-1 \right )\left ( \displaystyle \frac{1}{c}-1 \right )\\ &= \left ( \displaystyle \frac{1-a}{a} \right )\left ( \displaystyle \frac{1-b}{b} \right )\left ( \displaystyle \frac{1-c}{c} \right )\\ &= \left ( \displaystyle \frac{b+c}{a} \right )\left ( \displaystyle \frac{a+c}{b} \right )\left ( \displaystyle \frac{a+b}{c} \right )\\ &\geq \left (\displaystyle \frac{2\sqrt{bc}}{a}  \right )\left ( \displaystyle \frac{2\sqrt{ac}}{b} \right )\left ( \displaystyle \frac{2\sqrt{ab}}{c} \right )\\ &\geq \left (\displaystyle \frac{8\sqrt{(abc)^{2}}}{abc}  \right )\\ &\geq \displaystyle \frac{8abc}{abc}\\ &\geq 8\qquad \blacksquare  \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah bilangan real positif}\\  &\textrm{dengan}\: \: a+b+c=1,\: \textrm{tunjukkan}\\ &\textrm{bahwa}\: \: \left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq 64\\\\  &\textbf{Bukti}:\\ &\begin{aligned} &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\\ &=\displaystyle \frac{1}{abc}+\left (\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}  \right )+\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}  \right )+1\\ &\textrm{Dengan AM-GM kita mendapatkan}\\ &\bullet \: \: \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\sqrt[3]{\displaystyle \frac{1}{abc}}\\ &\bullet \: \: \displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}\\ &\textrm{Kita tulis sintak prosesnya di atas}\\ &=1+\left (\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}  \right )+\left (\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}  \right )+\displaystyle \frac{1}{abc}\\ &\geq 1+3\sqrt[3]{\displaystyle \frac{1}{abc}}+3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}+\sqrt[3]{\displaystyle \frac{1}{(abc)^{3}}}\\ &\geq 1+3\sqrt[3]{\displaystyle \frac{1}{abc}}+3\sqrt[3]{\displaystyle \frac{1}{(abc)^{2}}}+\sqrt[3]{\displaystyle \frac{1}{(abc)^{3}}}\\ &=\left ( 1+\displaystyle \frac{1}{\sqrt[3]{abc}} \right )^{3}\\ &\textrm{Karena}\: \: \sqrt[3]{abc}\leq \displaystyle \frac{a+b+c}{3}=\displaystyle \frac{1}{3},\: \textrm{maka}\\ &\left ( \displaystyle \frac{1}{a}+1 \right )\left ( \displaystyle \frac{1}{b}+1 \right )\left ( \displaystyle \frac{1}{c}+1 \right )\geq \left ( 1+\displaystyle \frac{1}{\sqrt[3]{abc}} \right )^{3}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq  \left ( 1+\displaystyle \frac{1}{\left (\frac{1}{3}  \right )} \right )^{3}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq \left ( 1+3 \right )^{4}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq  4^{4}\\ &\: \qquad\qquad\qquad\qquad\qquad\quad\quad \geq 64\qquad \blacksquare  \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Matematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
2. Budhi, W.S. 2014. Matematika 4: Bahan Ajar Persiapan Menuju Olimpiade Matematika Sain Nasional/Internasional SMA. Jakarta: TRISULA ADISAKTI.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Lanjutan Luas Segitiga 2

D. 3 Aturan Sinus

Perhatikan ilustasi berikut

$\Large\begin{array}{|c|}\hline \displaystyle \frac{a}{\sin A}= \frac{b}{\sin B}=\frac{c}{\sin C}=2R\\\hline \end{array}$.

$\begin{aligned}1.\quad a&=b.\displaystyle \frac{\sin \angle A}{\sin \angle B}=c.\displaystyle \frac{\sin \angle A}{\sin \angle C}=2R\sin \angle A\\ 2.\quad b&=c.\displaystyle \frac{\sin \angle B}{\sin \angle C}=a.\displaystyle \frac{\sin \angle B}{\sin \angle A}=2R\sin \angle B\\ 3.\quad c&=a.\displaystyle \frac{\sin \angle C}{\sin \angle A}=b.\displaystyle \frac{\sin \angle C}{\sin \angle B}=2R\sin \angle C \end{aligned}$.

Sehingga luas segitiga dapat dituliskan sebagai berikut:

$\begin{aligned} 1.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}a\left ( a.\displaystyle \frac{\sin \angle B}{\sin \angle A} \right )\sin \angle C\\ &=\displaystyle \frac{1}{2}a^{2}\displaystyle \frac{\sin \angle B\sin \angle C}{\sin \angle A}\\ 2.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}bc\sin \angle A\\ &=\displaystyle \frac{1}{2}b\left ( b.\displaystyle \frac{\sin \angle C}{\sin \angle B} \right )\sin \angle A\\ &=\displaystyle \frac{1}{2}b^{2}\displaystyle \frac{\sin \angle B\sin \angle A}{\sin \angle B}\\ 3.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ac\sin \angle B\\ &=\displaystyle \frac{1}{2}\left ( c.\displaystyle \frac{\sin \angle A}{\sin \angle C} \right )c\sin \angle B\\ &=\displaystyle \frac{1}{2}c^{2}\displaystyle \frac{\sin \angle A\sin \angle B}{\sin \angle C} \end{aligned}$.

D. 4 Luas segitiga sama sisi

$\begin{aligned}L_{\bigtriangleup }ABC&=\displaystyle \frac{1}{2}ab\sin \angle C,\quad a=b=c\\ &\qquad\quad\quad \textrm{dan}\: \: \angle A=\angle B\angle C=60^{\circ}\\ &=\displaystyle \frac{1}{2}a.a\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}a^{2}\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\color{red}\displaystyle \frac{1}{4}a^{2}\sqrt{3} \end{aligned}$.

D. 5 Lingkaran Luar Segitiga

Perhatikan lagi lingkaran luar segitiga di atas, dari sana kita akan mendapatkan rumus luas segitiga yang dapat kita munculkan harga R nya, yaitu:

$\begin{aligned}1.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}(2R\sin \angle A)(2R\sin \angle B)\sin \angle C\\ &=2R^{2}\sin \angle A\sin \angle B\sin \angle C\\ 2.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}ab\left ( \displaystyle \frac{c}{2R} \right )\\ &=\color{red}\displaystyle \frac{abc}{4R} \end{aligned}$.

D. 6 Lingkaran dalam segitiga

Perhatikanlah gambar berikut

$\begin{aligned}\textrm{Diketahu}&\textrm{i}\\ L_{\bigtriangleup }AOB&=\displaystyle \frac{1}{2}(AB)(OD)=\color{red}\displaystyle \frac{1}{2}cr\\ L_{\bigtriangleup }AOC&=\displaystyle \frac{1}{2}(AC)(OF)=\color{red}\displaystyle \frac{1}{2}br\\ L_{\bigtriangleup }BOC&=\displaystyle \frac{1}{2}(BC)(OE)=\color{red}\displaystyle \frac{1}{2}ar\\ \textrm{Sehingga}&\\ L_{\bigtriangleup }ABC&=\left [ ABC \right ]\\ &=\color{red}\displaystyle \frac{1}{2}ar+\displaystyle \frac{1}{2}br+\displaystyle \frac{1}{2}cr\\ &=\displaystyle \frac{1}{2}r(a+b+c)\\ &=\displaystyle \frac{1}{2}r(2s)\\ &=rs \end{aligned}$.

D. 7 Lingkaran singgung segitiga

Sebagai ilustrasinya adalah gambar berikut

$\begin{aligned}&\textrm{Diketahui}\\ &DO=EO=FO=r_{a}\\ &\textrm{maka}\\ &1.\quad L_{\bigtriangleup}ABO=\displaystyle \frac{1}{2}(AB)(OD)=\displaystyle \frac{1}{2}cr_{a}\\ &2.\quad L_{\bigtriangleup}ACO=\displaystyle \frac{1}{2}(AC)(OE)=\displaystyle \frac{1}{2}br_{a}\\ &3.\quad L_{\bigtriangleup}BCO=\displaystyle \frac{1}{2}(BC)(OF)=\displaystyle \frac{1}{2}ar_{a} \end{aligned}$.

$\begin{aligned} \textrm{Sehingga}&\\ L_{\bigtriangleup }ABC&=\left [ ABC \right ]\\ &=\left [ ACO \right ]+\left [ ABO \right ]-\left [ BCO \right ]\\ &=\color{red}\displaystyle \frac{1}{2}br_{a}+\displaystyle \frac{1}{2}cr_{a}-\displaystyle \frac{1}{2}ar_{a}\\ &=\displaystyle \frac{1}{2}r_{a}(b+c-a)\\ &=\displaystyle \frac{1}{2}r_{a}(a+b+c-2a)\\ &=\displaystyle \frac{1}{2}r_{a}(2s-2a)\\ &=\color{red}r_{a}(s-a) \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diberikan sembarang}\: \: \bigtriangleup ABC\: .\: \textrm{Jika}\: \: r\\ & \textrm{merupakan jari-jari lingkaran singgung }\\ &\textrm{dalam pada}\: \: \bigtriangleup ABC\: \: \textrm{dan}\: \: r_{a},\: r_{b},\: r_{c}\\ &\textrm{adalah jari-jari singgung luar pad}\: \: \bigtriangleup ABC\\ &\textrm{tunjukkan bahwa}:\: \displaystyle \frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{1}{r}\\\\ &\textbf{Bukti}:\\ &\begin{aligned} \textrm{Diketahu}&\textrm{i}\\ L_{\bigtriangleup }ABC&=\color{red}r_{a}(s-a),\: \color{black}\Rightarrow r_{a}=\displaystyle \frac{\left [ ABC \right ]}{s-a}\\ L_{\bigtriangleup }ABC&=\color{red}r_{b}(s-b),\: \color{black}\Rightarrow r_{b}=\displaystyle \frac{\left [ ABC \right ]}{s-b}\\ L_{\bigtriangleup }ABC&=\color{red}r_{c}(s-c),\: \color{black}\Rightarrow r_{c}=\displaystyle \frac{\left [ ABC \right ]}{s-c}\\ \textrm{maka}\: \quad&\\ \displaystyle \frac{1}{r_{a}}+\frac{1}{r_{b}}&+\frac{1}{r_{c}}\\ &=\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-a}}+\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-b}}+\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-c}}\\ &=\displaystyle \frac{s-a}{\left [ ABC \right ]}+\displaystyle \frac{s-b}{\left [ ABC \right ]}+\displaystyle \frac{s-c}{\left [ ABC \right ]}\\ &=\displaystyle \frac{s-a+s-b+s-c}{\left [ ABC \right ]}\\ &=\displaystyle \frac{3s-(a+b+c)}{\left [ ABC \right ]}\\ &=\displaystyle \frac{3s-2s}{\left [ ABC \right ]}\\ &=\displaystyle \frac{s}{\left [ ABC \right ]}\\ &=\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s}}\\ &=\color{blue}\displaystyle \frac{1}{r}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

Lanjutan Luas Segitiga 1

C. Luas Segitiga dengan Integral Tentu

Misalkan suatu garis linear di sekitar  sumbu-X yang tidak sejajar dan sumbu-X itu sendiri yang membatasi suatu daerah di antara keduanya serta dibatasi pula oleh 2 garis yang sejajar dengan sumbu Y yang keduanya tidak berimpit, maka hasil dari proses integral tentu ini akan menghasilkan luas segitiga.

Secara rumus integral tentu untuk model di atas adalah:

$\begin{aligned}&\textrm{Integral Tentu}\\ &=\displaystyle \int_{p}^{q}f(x)dx=\displaystyle \left [ F(x) \right ]_{p}^{q}\\ &=F(x)|_{p}^{q}=F(q)-F(p) \end{aligned}$.

$\begin{aligned}&\textrm{Diketahui}\\ &y=\color{red}m\color{black}x,\quad \textrm{dengan}\: \: m=\displaystyle \frac{b}{a} ,\\ &\textrm{sehingga}\\ &y=\displaystyle \frac{b}{a}x\: \: \textrm{atau}\: \: f(x)=\displaystyle \frac{b}{a}x\\ &\textrm{maka}\\ &L_{area}=\displaystyle \int_{a}^{b}f(x)\: dx\\ &\: \qquad =\displaystyle \int_{a}^{b}\displaystyle \frac{b}{a}x\: dx\\ &\: \qquad =\displaystyle \frac{b}{a}\displaystyle \int_{a}^{b}x\: dx\\ &\: \qquad =\displaystyle \frac{b}{a}\left ( \displaystyle \frac{1}{2}x^{2} \right )\begin{matrix} a\\ |\\ 0 \end{matrix}\\ &\: \qquad =\displaystyle \frac{b}{a}\left ( \displaystyle \frac{1}{2}(a)^{2} \right )-\displaystyle \frac{b}{a}\left ( \displaystyle \frac{1}{2}(0)^{2} \right )\\ &\: \qquad =\displaystyle \frac{b}{a}\left ( \displaystyle \frac{1}{2}(a)^{2} \right )-0\\ &\: \qquad =\displaystyle \frac{b\times a^{2}}{2a}\\ &\: \qquad =\displaystyle \frac{a\times b}{2}\\ &\: \qquad =\color{red}\displaystyle \frac{1}{2}\times ab \end{aligned}$.

D. Luas segitiga dengan Trigonometri

D. 1 Segitiga siku-siku

Perhatikan ilustrasi segitiga siku-siku berikut

$\begin{aligned}&1.\quad \sin \angle B=\displaystyle \frac{b}{a}\: \: \: \textrm{dan}\: \: \cos \angle C=\displaystyle \frac{b}{a}\\ &\qquad \textrm{sehingga}\: \: b=a\sin \angle B=a\cos \angle C\\ &2.\quad \sin \angle C=\displaystyle \frac{c}{a}\: \: \: \textrm{dan}\: \: \cos \angle B=\displaystyle \frac{c}{a}\\ &\qquad \textrm{sehingga}\: \: c=a\sin \angle C=a\cos \angle B\\ &3.\quad \tan \angle B=\displaystyle \frac{b}{c}\: \: \: \textrm{dan}\: \: \cot \angle C=\displaystyle \frac{b}{c}\\ &\qquad \textrm{sehingga}\: \: b=c\tan \angle B=c\cot \angle C\\ &4.\quad \tan \angle C=\displaystyle \frac{c}{b}\: \: \: \textrm{dan}\: \: \cot \angle B=\displaystyle \frac{c}{b}\\ &\qquad \textrm{sehingga}\: \: c=b\tan \angle C=b\cot \angle B\\ &5.\quad \sin \angle B=\displaystyle \frac{h}{c}\: \: \: \textrm{dan}\: \: \sin \angle C=\displaystyle \frac{h}{b}\\ &\qquad \textrm{sehingga}\: \: h=c\sin \angle B=b\sin \angle C\\ \end{aligned}$.

Dari fakta-fakta di atas dapat ditunjukkan beberapa rumus segitiga, yaitu:

$\begin{array}{|c|c|l|}\hline 1.&L_{\bigtriangleup }&\displaystyle \frac{1}{2}bc\\\hline &&\begin{aligned}&\displaystyle \frac{1}{2}\left ( a\sin \angle B \right )\left ( a\sin \angle C \right )\\ &=\displaystyle \frac{1}{2}a^{2}\sin \angle B\sin \angle C \end{aligned}\\ &&\begin{aligned}&\displaystyle \frac{1}{2}\left ( a\cos \angle C \right )\left ( a\cos \angle B \right )\\ &=\displaystyle \frac{1}{2}a^{2}\cos \angle B\cos \angle C \end{aligned}\\ &&\begin{aligned}&\displaystyle \frac{1}{2}(b)\left ( b\tan \angle C \right )\\ &=\displaystyle \frac{1}{2}b^{2}\tan \angle C \end{aligned}\\ &&\begin{aligned}&\displaystyle \frac{1}{2}(c\tan \angle B)(c)\\ &=\displaystyle \frac{1}{2}c^{2}\tan \angle B \end{aligned}\\ &&\begin{aligned}&\displaystyle \frac{1}{2}(b)(b\cot \angle B)\\ &=\displaystyle \frac{1}{2}b^{2}\cot \angle B \end{aligned}\\ &&\begin{aligned}&\displaystyle \frac{1}{2}(c\cot \angle C)(c)\\ &=\displaystyle \frac{1}{2}c^{2}\cot \angle C \end{aligned}\\\hline 2.&L_{\bigtriangleup } &\displaystyle \frac{1}{2}ha\\\hline &&\begin{aligned}&\displaystyle \frac{1}{2}(c\sin \angle B)(a)\\ &=\displaystyle \frac{1}{2}ac\sin \angle B \end{aligned}\\ &&\begin{aligned}&\displaystyle \frac{1}{2}(b\sin \angle C)(a)\\ &=\displaystyle \frac{1}{2}ab\sin \angle C \end{aligned}\\\hline \end{array}$.

D. 2 Segitiga tidak siku-siku

Perhatikan ilustrasi segitiga tidak siku-siku berikut

Jika dari tiap titik sudut ditarik garis tinggi sampai memotong sisi di depannya, misal titik A, maka garis tingginya sebagaimana gambar berikut:

$\begin{aligned}1.\quad L_{\bigtriangleup }&=\displaystyle \frac{1}{2}h_{A}.a\\ &=\displaystyle \frac{1}{2}c\sin \angle B.(a)=\displaystyle \frac{1}{2}ac\sin \angle B\\ L_{\bigtriangleup }&=\displaystyle \frac{1}{2}h_{A}.a\\ &=\displaystyle \frac{1}{2}b\sin \angle C.(a)=\displaystyle \frac{1}{2}ab\sin \angle C\\ 2.\quad L_{\bigtriangleup }&=\displaystyle \frac{1}{2}h_{B}.b\\ &=\displaystyle \frac{1}{2}c\sin \angle A.(b)=\displaystyle \frac{1}{2}bc\sin \angle A\\ L_{\bigtriangleup }&=\displaystyle \frac{1}{2}h_{B}.b\\ &=\displaystyle \frac{1}{2}a\sin \angle C.(b)=\displaystyle \frac{1}{2}ab\sin \angle C\\ 3.\quad L_{\bigtriangleup }&=\displaystyle \frac{1}{2}h_{C}.c\\ &=\displaystyle \frac{1}{2}a\sin \angle B.(c)=\displaystyle \frac{1}{2}ac\sin \angle B\\ L_{\bigtriangleup }&=\displaystyle \frac{1}{2}h_{C}.c\\ &=\displaystyle \frac{1}{2}b\sin \angle A.(c)=\displaystyle \frac{1}{2}bc\sin \angle A \end{aligned}$.

Jika diringkas menjadi

$\begin{aligned}\textbf{Luas}\: \triangle \: ABC&=\frac{1}{2}bc.\sin \angle A\\ &=\frac{1}{2}ac.\sin \angle B\\ &=\frac{1}{2}ab.\sin \angle C \end{aligned}$.

Selingan-Luas Segitiga

A. Heron's Formula

Bukti Luas Segitiga dengan sisi a, b, dan c

$\begin{aligned}& \textrm{Bagaimana membuktikan luas suatu}\\ &\textrm{segitiga jika diketahui sisinya}\: a,b\: \textrm{dan}\: c\\ &\textrm{berupa rumus}\\ &L_{\bigtriangleup }=\left [ ABC \right ]=\sqrt{s(s-a)(s-b)(s-c)}\\ & \textrm{dengan}\\ &\qquad s=\displaystyle \frac{1}{2}(a+b+c) \end{aligned}$.

Berikut akan dipaparkan buktinya

$\begin{aligned}\displaystyle \textrm{L}{\bigtriangleup }\textrm{ABC}&=\frac{1}{2}bc\sin\angle A\\ &=\displaystyle \frac{1}{2}bc\sqrt{\sin ^{2}\angle A}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( \sin ^{2}\angle A \right )}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( 1-\cos ^{2}\angle A \right )},\\ &\textrm{ingat bahwa};\: \cos \angle A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( 1-\left ( \frac{b^{2}+c^{2}-a^{2}}{2bc} \right )^{2} \right )}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}-\left ( \frac{b^{2}+c^{2}-a^{2}}{2} \right )^{2}}\\ &=\displaystyle \frac{1}{2}\sqrt{\frac{4b^{2}c^{2}-\left ( b^{2}+c^{2}-a^{2} \right )^{2}}{4}}\\ &=\displaystyle \frac{1}{2}.\frac{1}{2}\sqrt{\left ( 2bc \right )^{2}-\left ( b^{2}+c^{2}-a^{2} \right )^{2}}\\ &=\displaystyle \frac{1}{4}\sqrt{\left ( 2bc+b^{2}+c^{2}-a^{2} \right )\left (2bc-b^{2}-c^{2}+a^{2} \right )}\\ &=\frac{1}{4}\sqrt{\left \{ \left ( b+c \right )^{2}-a^{2} \right \}\left \{ a^{2}-\left ( b-c \right )^{2} \right \}}\\ &=\displaystyle \frac{1}{4}\sqrt{\left ( b+c+a \right )\left ( b+c-a \right )\left ( a+b-c \right )\left ( a-b+c \right )},\\ &\textrm{dengan mengingat bahwa}\: 2s=a+b+c\\ &=\displaystyle \frac{1}{4}\sqrt{(2s)(2s-2a)(2s-2b)(2s-2c)}\\ &=\displaystyle \frac{1}{4}.\sqrt{16.s(s-a)(s-b)(s-c)}\\ &=\displaystyle \frac{1}{4}.4\sqrt{s(s-a)(s-b)(s-c)}\\ \textrm{L}\bigtriangleup \textrm{ABC}&=\sqrt{s(s-a)(s-b)(s-c)}\quad \blacksquare \end{aligned}$.

Rumus di atas lebih dikenal dengan istilah rumus Heron lihat Heron's formula di sini.

Sumber tulisan lagi di antara silahkan kunjungi di sini.

B. Luas segitiga

Luas segitiga juga bisa kita dapatkan dengan dengan simulasi berikut

$\begin{aligned}1.\quad &\textrm{Buat suatu segitiga bebas, misal} \end{aligned}$.

$\begin{aligned}2.\quad &\textrm{Perhatikan bahwa luas segitiga sama}\\ &\textrm{dengan luas dua kali persegi panjang}\\ &\textrm{yang terbentuk, perhatikan gambar berikut} \end{aligned}$.

dengan rincian

$\begin{aligned}3.\quad &\textrm{Panjang dari persegi panjang adalah}\\ &\textrm{setengah dari alas segitiga dan lebar }\\ &\textrm{persegi panjang adalah setengah dari}\\ &\textrm{tinggi segitiga} \end{aligned}$.

$.\qquad\begin{aligned} \textrm{Luas}\: &\textrm{segitiga}\\ &=2\times \left ( \textrm{luas persegi panjang} \right )\\ &=2\times \left ( \textrm{panjang}\times \textrm{lebar} \right )\\ &=2\times \left ( \displaystyle \frac{1}{2}a \right )\times \left ( \displaystyle \frac{1}{2}t \right )\\ &=\displaystyle \frac{1}{2}\times a\times t \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui segitiga ABC dengan}\: \: AB=21\: \: cm\\ & AC=10\: \: cm,\: \: \textrm{dan}\: \: BC=17\: \: cm.\: \textrm{Tentukan}\\ &\textrm{segitiga yang ditarik dari titik}\: \: C\\\\ &\textbf{Jawab}:\\ &\textrm{Sebelumnya perhatikan ilustrasi berikut ini} \end{array}$.

$.\: \qquad\begin{aligned}L_{\bigtriangleup }&=\left [ ABC \right ]=\textrm{Luas segitiga}\\ &=\sqrt{s(s-a)(s-b)(s-c)}\\ &\quad \textrm{dengan}\quad s=\displaystyle \frac{1}{2}(a+b+c)\\ &\quad \textrm{sehingga}\: s=\displaystyle \frac{1}{2}(21+10+17)=\color{blue}24\\ \textrm{ma}&\textrm{ka}\\ L_{\bigtriangleup }&=\sqrt{24(24-17)(24-10)(24-21)}\\ &=\sqrt{24(7)(14)(3)}=\sqrt{4.6.7.2.7.3}\\ &=\sqrt{4.49.36}=2.7.6=84\\ L_{\bigtriangleup }&=\displaystyle \frac{1}{2}\left ( \textrm{alas di depan sudut} \right )\times t\\ 84&=\displaystyle \frac{1}{2}.21\times t\\ t&=\color{red}8 \end{aligned}$.

$\begin{array}{ll}\\ 2.&(\textbf{OSK 2008})\textrm{Segitiga sama kaki dengan}\\ & AB=AC\: \: \textrm{dengan keliling}\: \: 32.\: \textrm{Jika panjang}\\ &\textrm{garis tinggi dari titik}\: \: A\: \: \textrm{adalah}\: \: 8,\: \textrm{maka}\\ &\textrm{panjang}\: \: AC\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 9\displaystyle \frac{1}{3}\\ &\textrm{b}.\quad 10\\ &\textrm{c}.\quad 10\displaystyle \frac{2}{3}\\ &\textrm{d}.\quad 11\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad 12\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut ini} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Diketahui bahwa}\: \: AB=AC=x\\ &\textrm{dengan keliling}\: \bigtriangleup ABC=32\\ &\textrm{maka}\\ &\color{blue}x+x+BC=32\\ &\Leftrightarrow BC=32-2x,\: \textrm{sedangkan}\\ &\qquad DC=16-x\\ &\textrm{Diketahui pula garis tinggi dari}\\ &\textrm{titik}\: \: A=8, \\ &\textrm{perhatikan}\: \: \bigtriangleup ADC\\ &\textrm{dengan}\: \: AC^{2}=AD^{2}+DC^{2}\\ &\Leftrightarrow x^{2}=8^{2}+(16-x)^{2}\\ &\Leftrightarrow x^{2}=8^{2}+256-32x+x^{2}\\ &\Leftrightarrow 32x=320\\ &\Leftrightarrow x=\color{red}10 \end{aligned}$.

DAFTRA PUSTAKA

1. Sembiring, S., Sukino. 2020. Super Master KSN Matematika SMA/MA. Bandung: YRAMA WIDYA.
2. Sobel, M.A., Maletsky, E.M. 2004. Mengajar Matematika: Sebuah Buku Sumber Alat Peraga, Aktivitas, dan Strategi untuk Guru Matematika SD, SMP, SMA. ed. ke-3. Alih Bahasa: Suyono. Jakarta: ERLANGGA.

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas XII

  

Jika jarak titik A ke C adalah 8 cm, 

maka jarak titik A ke D adalah ... cm

Jawab : b

Pembahasan diserahkan kepada pembaca yang budiman

Jika jarak K ke Q adalah 9 cm,

maka jarak titik K ke L adalah ... cm

Jawab : b

Pembahasan juga diserahkan kepada pembaca yang budiman

Jika panjang rusuk tegaknya adalah 13 cm, 

serta AB = 8 cm, dan BC = 6 cm, 

maka jarak T ke bidang ABCD (Titik tengah bidang ABCD)

Jawab : e

Pembahasan juga diserahkan kepada pembaca yang budiman

Jika panjang AB = 4 BC = 2 CG = 8 cm,

maka panjang AG adalah ... . cm

Jawab : a

$\begin{aligned}&\textrm{Misalkan}\\ &\textrm{Panjang diagonal sisi}\: \: =s\sqrt{2}=4\: \textrm{cm}\\ &\color{blue}\textbf{Alternatif 1}\\ &s\sqrt{2}=4\\ &\Leftrightarrow s=\displaystyle \frac{4}{\sqrt{2}}\\ &\Leftrightarrow s=\displaystyle \frac{4}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{4}{\sqrt{4}}\sqrt{2}=\displaystyle \frac{4}{2}\sqrt{2}=2\sqrt{2}\\ &\textrm{Jadi, panjang sisinya}\: \: \color{red}2\sqrt{2}\\ &\color{blue}\textbf{Alternatif 2}\\ &\textrm{dikuadratkan kedua ruasnya, yaitu}:\\ &(s\sqrt{2})^{2}=(4)^{2}\Leftrightarrow s^{2}.2=18\\ &\Leftrightarrow s^{2}=\displaystyle \frac{16}{2}=8\\ &\Leftrightarrow s=\sqrt{8}=\sqrt{4.2}=\sqrt{4}.\sqrt{2}=2\sqrt{2}\\ &\textrm{Jadi, panjang sisinya}\: \: \color{red}2\sqrt{2} \end{aligned}$.

$\begin{array}{ll}\\ 26.&\textrm{Perhatikan gambar berikut} \end{array}$.

$\: \qquad \begin{aligned}&\textrm{Pernyataan yang tepat adalah}\: ...\\ &\begin{array}{ll} \textrm{a}.&\textrm{Titik sudut}\: \: H\: \: \textrm{pada rusuk}\: \: AB\\ \textrm{b}.&\textrm{Titik sudut}\: \: C\: \: \textrm{pada rusuk}\: \: BD\\ \textrm{c}.&\textrm{Titik sudut}\: \: A\: \: \textrm{pada rusuk}\: \: DC\\ \textrm{d}.&\textrm{Titik sudut}\: \: D\: \: \textrm{pada rusuk}\: \: BC\\ \textrm{e}.&\color{red}\textrm{Titik sudut}\: \: G\: \: \textrm{pada rusuk}\: \: GF\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\textrm{Cukup jelas bahwa Titik}\: \: G\: \: \textrm{pada ruas}\\ &\textrm{garis}\: \: GF \end{aligned}$.

$\begin{array}{ll}\\ 27.&\textrm{Diketahui kubus dengan rusuk 10}\: \: cm.\\ &\textrm{Titik}\: \: P\: \: \textrm{dan}\: \: Q\: \: \textrm{berturut-turut adalah}\\ &\textrm{titik tengah}\: \: AB\: \: \textrm{dan}\: \: DC.\: \textrm{Titik}\: \: O\\ &\textrm{merupakan perpotongan diagonal bidang}\\ &\textrm{alas}\: \: ABCD.\: \textrm{Pernyataan yang tepat}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.&\textrm{Titik}\: \: P\: \: \textrm{terletak di luar garis}\: \: AB\\ \textrm{b}.&\textrm{Titik}\: \: P\: \: \textrm{terletak pada}\: \: DC\\ \textrm{c}.&\textrm{Titik}\: \: Q\: \: \textrm{terletak di luar garis}\: \: DC\\ \textrm{d}.&\color{red}\textrm{Titik}\: \: P\: \: \textrm{terletak pada garis}\: \: AB\\ \textrm{e}.&\color{red}\textrm{Titik}\: \: P\: \: \textrm{terletak pada garis}\: \: PQ\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d,e}\\ &\textrm{ada 2 jawaban yang tepat}\\ &\textrm{perhatikan ilustrasi berikut ini}\\ &\textrm{untuk lebih memahami jawaban di atas} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Pada kubus}\: \: ABCD.EFGH,\: \: ABCD\\ &\textrm{merupakan wakil bidang}\: \: \alpha \: \: \textrm{dan}\: \: BCGF\\ &\textrm{merupakan wakil dari bidang}\: \: \beta .\\ &\textrm{Pernyataan yang kurang tepat adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.&\textrm{Titik}\: \: A\: \: \textrm{terletak pada bidang}\: \: \alpha \: \: \textrm{saja}\\ \textrm{b}.&\textrm{Titik}\: \: D\: \: \textrm{terletak pada bidang}\: \: \alpha \: \: \textrm{saja}\\ \textrm{c}.&\textrm{Titik}\: \: F\: \: \textrm{terletak pada bidang}\: \: \beta \: \: \textrm{saja}\\ \textrm{d}.&\color{red}\textrm{Titik}\: \: B\: \: \textrm{terletak pada bidang}\: \: \alpha \: \: \textrm{saja}\\ \textrm{e}.&\textrm{Titik}\: \: G\: \: \textrm{terletak pada bidang}\: \: \beta \: \: \textrm{saja}\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\textrm{Titik}\: \: B\: \: \textrm{terletak pada bidang}\: \: \alpha \: \: \textrm{dan}\: \: \beta \\ &\textrm{sekaligus, perhatikan ilustrasi berikut ini}\\ &\textrm{untuk lebih memahami jawaban di atas} \end{array}$.

$\begin{array}{ll} 29.&\textrm{Diketahui panjang rusuk kubus}\: ABCD.EFGH\\ &\textrm{adalah}\: \: 6\: \: cm\: .\: \textrm{Jarak titik}\: \: E\: \: \textrm{ke bidang}\: \: BDG\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \sqrt{3}\: \: cm\\ &\textrm{b}.\quad 2\sqrt{3}\: \: cm\\ &\textrm{c}.\quad 3\sqrt{3}\: \: cm\\ &\textrm{d}.\quad \color{red}4\sqrt{3}\: \: cm\\ &\textrm{e}.\quad 6\sqrt{3}\: \: cm\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Perhatikanlah ilustrasi berikut ini} \end{array}$.

$.\qquad\begin{aligned}\textrm{Jelas}&\: \textrm{bahwa}\\ &\begin{cases} AB & =BC=CG=QQ'=6\: \: cm\: \: (\color{red}\textrm{dari soal}) \\ AC &=EG =6\sqrt{2}\: \: cm\: \: (\textrm{diagonal sisi kubus}) \\ EQ & =QG= \displaystyle \frac{1}{2}(\textbf{sisi})\sqrt{6}=3\sqrt{6}\: \: cm \end{cases}\\ \textrm{Perh}&\textrm{atikan}\: \: \bigtriangleup EQG\\ &\begin{aligned}&\textrm{Dengan perbandingan luas}\: \: \bigtriangleup EQG=\bigtriangleup EQG\\ &\displaystyle \frac{1}{2}\times QG\times EE'=\displaystyle \frac{1}{2}\times QQ'\times EG\\ &\displaystyle \frac{1}{2}\times \left ( 3\sqrt{6} \right )\times EE'=\displaystyle \frac{1}{2}\times 6\times 6\sqrt{2}\\ &\quad\quad\quad EE'= \displaystyle \frac{\displaystyle \frac{1}{2}\times \left ( 6\sqrt{2} \right )\times 6}{\displaystyle \frac{1}{2}\times \left ( 3\sqrt{6} \right )}\\ &\: \: \: \quad\quad\qquad =\color{blue}4\sqrt{3}\: \: \color{black}cm \end{aligned} \end{aligned}$.

$\begin{array}{ll} 30.&\textrm{Diketahui panjang rusuk kubus}\: ABCD.EFGH\\ &\textrm{adalah}\: \: 8\: \: cm\: .\: \textrm{Jarak titik}\: \: C\: \: \textrm{ke bidang}\: \: BDG\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad 4\sqrt{3}\: \: cm\\ &\textrm{b}.\quad 6\sqrt{3}\: \: cm\\ &\textrm{c}.\quad \color{red}\displaystyle \frac{8}{3}\sqrt{3}\: \: cm\\ &\textrm{d}.\quad 8\sqrt{3}\: \: cm\\ &\textrm{e}.\quad \displaystyle \frac{10}{3}\sqrt{3}\: \: cm\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\textrm{Perhatikanlah ilustrasi} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Jika gambarnya dipartisi lagi di bagian}\\ & \textrm{segitiga} \: \: GCG'\: \: \textrm{maka akan tampak}\\ & \textrm{seperti ilustrasi berikut} \end{aligned}$.

$.\qquad\begin{aligned}\textrm{Jelas}&\: \textrm{bahwa}\\ &\begin{cases} GC & =8\: \: cm\: \: (\color{red}\textrm{dari soal}) \\ BD &=AC =8\sqrt{2}\: \: cm\: \: (\textrm{diagonal sisi kubus}) \\ CG' & = \displaystyle \frac{1}{2}AC=4\sqrt{2}\: \: cm \end{cases}\\ \textrm{maka}&\: \textrm{dengan rumus Pythagoras dapat}\\ \textrm{panja}&\textrm{ng}\: \: GG',\: \textrm{yaitu}:\\ \left ( GG' \right )&^{2}=\left ( G'C \right )^{2}+CG^{2}\\ GG'&=\sqrt{\left ( G'C \right )^{2}+CG^{2}}=\sqrt{\left ( 4\sqrt{2} \right )^{2}+8^{2}}\\ &=\sqrt{32+64}=\sqrt{96}=\sqrt{16.6}=4\sqrt{6}\: \: cm\\ \textrm{Perha}&\textrm{tikan}\: \: \bigtriangleup GCG'\\ &\begin{aligned}&\textrm{Dengan perbandingan luas}\: \: \bigtriangleup GCG'=\bigtriangleup GCG'\\ &\displaystyle \frac{1}{2}\times CC'\times GG'=\displaystyle \frac{1}{2}\times CG'\times CG\\ &\displaystyle \frac{1}{2}\times CC'\times \left ( 4\sqrt{6} \right )=\displaystyle \frac{1}{2}\times \left ( 4\sqrt{2} \right )\times 8\\ &\quad\quad\quad CC'= \displaystyle \frac{\displaystyle \frac{1}{2}\times \left ( 4\sqrt{2} \right )\times 8}{\displaystyle \frac{1}{2}\times \left ( 4\sqrt{6} \right )}\\ &\: \: \: \quad\quad\qquad =\displaystyle \frac{8}{\sqrt{3}}\\ &\: \: \: \quad\quad\qquad =\displaystyle \frac{8}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &\: \: \: \quad\quad\qquad =\color{blue}\displaystyle \frac{8}{3}\sqrt{3}\: \: \color{black}cm \end{aligned} \end{aligned}$.