Lanjutan 2 Contoh Soal Persiapan untuk Numerasi AN BK 2022

$\LARGE\colorbox{yellow}{CONTOH SOAL 21}$.

$\begin{array}{ll}\\ &\textrm{Nilai eksak dari}\: \: \sin 36^{\circ}\: \: \: \textrm{adalah}\: ....\\ &\begin{array}{|l|}\hline \bullet \quad \sin 2\theta =2\sin \theta \cos \theta\\ \bullet \quad \cos 2\theta =2\cos ^{2}\theta -1\\ \bullet \quad \sin ^{2}\theta +\cos ^{2}\theta =1 \\\hline \end{array}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}&&&\textrm{d}.&\displaystyle \frac{\sqrt{5}-1}{4}\\ \textrm{b}.&\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}&&\quad &\textrm{e}.&\displaystyle \frac{\sqrt{5}-1}{2}\\ \textrm{c}.&\displaystyle \displaystyle \frac{\sqrt{5}+1}{4} \end{array} \end{array}$.

$\LARGE\colorbox{silver}{SOLUSI SOAL 21}$.

Perhatikanlah ilustrasi segitiga berikut ini

$.\qquad\begin{aligned}&\textrm{Perhatikan bahwa}\: \: \color{red}\bigtriangleup ABC\: \: \color{black}\textrm{sama kaki}\\ &\textrm{dengan}\: \: AD=DC=CB=1,\: AC=x\\ &\textrm{Diketahui pula}\: \: CD\: \: \textrm{adalah garis bagi}\\ &\textrm{serta}\: \: ABC\: \: \textrm{sebangun}\: \: \bigtriangleup BCD\\ &\textrm{akibatnya}:\\ &\color{red}\textrm{perbandingan sisi yang bersesuaian}\\ &\color{red}\textrm{akan sama},\: \: \color{black}\textrm{maka}\\ &\displaystyle \frac{AB}{BC}=\displaystyle \frac{BC}{AB-AD}\\ &\Leftrightarrow \displaystyle \frac{x}{1}=\frac{1}{x-1}\\ &\Leftrightarrow x(x-1)=1\\ &\Leftrightarrow x^{2}-x-1=0\\ &\Leftrightarrow x=\displaystyle \frac{1\pm \sqrt{5}}{2}\\ &\textrm{akibatnya}\: \: AB=AC=\displaystyle \frac{1+\sqrt{5}}{2}\\ &\textrm{Selanjutnya gunakan}\: \: \color{blue}\textrm{aturan sinus}\\ &\displaystyle \frac{AB}{\sin \angle C}=\frac{BC}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{AB}{BC}=\frac{\sin \angle C}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{ \left (\displaystyle \frac{1+\sqrt{5} }{2} \right )}{1}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=\displaystyle \frac{2\sin 36^{\circ}\cos 36^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=2\cos 36^{\circ}\\ &\Leftrightarrow \cos 36^{\circ}=\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{4}\\ &\textrm{Dari fakta di atas kita akan dengan}\\ &\textrm{mudah menentukan nilai sinusnya}\\ &\textrm{yaitu dengan menggunakan}\\ &\color{purple}\textrm{identitas trigonometri berikut}:\\ &\sin ^{2}36^{\circ}+\cos ^{2}36^{\circ}=1\\ &\Leftrightarrow \sin ^{2}36^{\circ}=1-\cos ^{2}36^{\circ}\\ &\Leftrightarrow \sin 36^{\circ}=\sqrt{1-\cos ^{2}36^{\circ}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\left ( \displaystyle \frac{1+\sqrt{5}}{4} \right )^{2}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\displaystyle \frac{6+2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{\displaystyle \frac{10-2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}} \end{aligned}$.

Untuk menjawab soal no.22,23,24, 25 dan 26 perlu diketahui bahwa

$\left \lfloor x \right \rfloor=\textrm{bilangan bulat terbesar}\leq x$.

$\left \lfloor 3,14 \right \rfloor=3, \left \lfloor -7,2 \right \rfloor=-8,\: \: \left \lfloor 2 \right \rfloor=2$.

$\LARGE\colorbox{yellow}{CONTOH SOAL 22}$.

$\begin{aligned}&\textrm{Jika}\: \: \left \lfloor a \right \rfloor\times a=68\: \: \textrm{dan}\: \: \left \lfloor b \right \rfloor\times b=109,\\ &\textrm{maka nilai dari}\: \: \left \lfloor a \right \rfloor\times \left \lfloor b \right \rfloor-\left \lfloor a+b \right \rfloor=\cdots  \end{aligned}$.

$\LARGE\colorbox{silver}{SOLUSI SOAL 22}$.

$\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\left \lfloor a \right \rfloor a=68\qquad \textrm{dan}\quad \left \lfloor b \right \rfloor b=109\\ &\Leftrightarrow \left \lfloor a \right \rfloor^{2}\leq \left \lfloor a \right \rfloor a\: \: \textrm{dan}\: \:  \left \lfloor b \right \rfloor^{2}\leq \left \lfloor b \right \rfloor b\\ &\Leftrightarrow \left \lfloor a \right \rfloor^{2}\leq 68\: \: \quad \textrm{dan}\: \: \left \lfloor b \right \rfloor^{2}\leq 109\\ &\Leftrightarrow \left \lfloor a \right \rfloor= 8\: \: \qquad \textrm{dan}\: \: \left \lfloor a \right \rfloor=10\\ &\textrm{Selanjutnya}\\ &\bullet \quad \left \lfloor a \right \rfloor a=68\Rightarrow  8a=68\Rightarrow a=\color{red}8,5\\ &\bullet \quad \left \lfloor b \right \rfloor b=109\Rightarrow  10b=109\Rightarrow b=\color{red}10,9\\ &\textrm{Sehingga nilai}\\ &\left \lfloor a \right \rfloor\times \left \lfloor b \right \rfloor-\left \lfloor a+ b \right \rfloor\\ &=8\times 10-\left \lfloor 8,5+ 10,9 \right \rfloor\\ &=80-\left \lfloor 19,4 \right \rfloor\\ &=80-19\\ &=\color{red}61 \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL 23}$.

$\begin{aligned}&\textrm{Jika}\: \: a\left \lfloor a \right \rfloor=17\: \: \textrm{dan}\: \: b\left \lfloor b \right \rfloor =11,\: \: \textrm{maka}\\ &\textrm{nilai dari}\: \: a-b=\cdots  \end{aligned}$.

$\LARGE\colorbox{silver}{SOLUSI SOAL 23}$.

$\begin{aligned}&\textrm{Perhatikan bahwa}\\ &a\left \lfloor a \right \rfloor =17\qquad \textrm{dan}\quad b\left \lfloor b \right \rfloor =11\\ &\Leftrightarrow \left \lfloor a \right \rfloor^{2}\leq a\left \lfloor a \right \rfloor \: \: \textrm{dan}\: \:  \left \lfloor b \right \rfloor^{2}\leq b\left \lfloor b \right \rfloor \\ &\Leftrightarrow \left \lfloor a \right \rfloor^{2}\leq 17\: \: \quad \textrm{dan}\: \: \left \lfloor b \right \rfloor^{2}\leq 11\\ &\Leftrightarrow \left \lfloor a \right \rfloor= 4\: \: \qquad \textrm{dan}\: \: \left \lfloor a \right \rfloor=3\\ &\textrm{Selanjutnya}\\ &\bullet \quad a\left \lfloor a \right \rfloor =17\Rightarrow  4a=17\Rightarrow a=\color{red}\displaystyle \frac{17}{4}\\ &\bullet \quad b\left \lfloor b \right \rfloor =11\Rightarrow  3b=11\Rightarrow b=\color{red}\displaystyle \frac{11}{3}\\ &\textrm{Sehingga nilai}\\ &a-b=\displaystyle \frac{17}{4}-\frac{11}{3}\\ &=\displaystyle \frac{51-44}{12}\\ &=\color{red}\displaystyle \frac{7}{12} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL 24}$.

$\begin{aligned}&\textrm{Jika}\: \: \left \lfloor x \right \rfloor +\left \lfloor y \right \rfloor +y=43,8\quad \textrm{dan}\\ &x+y-\left \lfloor x \right \rfloor=18,4\: ,\: \textrm{maka nilai}\\ &\textrm{dari}\quad 10(x+y)=\cdots   \end{aligned}$.

$\LARGE\colorbox{silver}{SOLUSI SOAL 24}$.

$\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad \left \lfloor x \right \rfloor +\left \lfloor y \right \rfloor +y=43,8\quad \textrm{dan}\\ &\bullet \quad x+y-\left \lfloor x \right \rfloor=18,4\\ &\textrm{Sekarang misalkan untuk}\\ &\bullet \quad a\leq x< a+1\Rightarrow \left \lfloor x \right \rfloor=\color{red}a\\ &\bullet \quad b\leq y\leq b+1\Rightarrow \left \lfloor y \right \rfloor=\color{red}b\\ &\textrm{Selanjutnya untuk}\: \: x\: \: \textrm{dan}\: \: y\\ &\textrm{dapat kita nyatakan dengan}\\ &\bullet\quad x=a+m,\: \: \textrm{dengan}\: \: 0\leq m< 1\\ &\bullet\quad y=b+n,\: \: \: \: \textrm{dengan}\: \: \: 0\leq n< 1\\ &\begin{aligned}&\textbf{Untuk persamaan pertama}\\ &\left \lfloor x \right \rfloor +\left \lfloor y \right \rfloor +\quad y=43,8\\ &\Leftrightarrow a+b\quad+ b+n=43,8\\ &\Leftrightarrow a+2b+n=43,8\\ &\textrm{didapatkan}\: \: a+2b=43,\: \: \textrm{dan}\: \: n=0,8\\ &\textbf{Untuk persamaan kedua}\\ &x+y\quad-\quad\left \lfloor x \right \rfloor =18,4\\ &\Leftrightarrow a+m+ b+n-a=18,4\\ &\Leftrightarrow b+m+n\qquad=18,4\\ &\Leftrightarrow b+m+0,8\quad=18,4\\ &\Leftrightarrow b+m=17,6\\ &\textrm{didapatkan}\quad b=17,\: \: \textrm{dan}\: \: m=0,6\\ &\textbf{Selanjutnya perhatikan bahwa}\\ &a+2b=43\Leftrightarrow a+2.(17)=43\\ &\Leftrightarrow a+34=43\Leftrightarrow a=43-34=9\\ &\textbf{Sehingga kita akan mendapatkan nilai}\\ &\bullet \quad x=a+m=9+0,6=9,6\\ &\bullet \quad y=b+n=17+0,8=17,8\\ &\textrm{Jadi},\: 10(x+y)=10(9,6+17,8)=\color{red}274   \end{aligned} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL 25}$.

$\begin{aligned}&\textrm{Jika}\: \: \left \lfloor 3x+y \right \rfloor=12\: \: \textrm{dan}\: \: \left \lfloor x+3y \right \rfloor=14,\\ &\textrm{maka nilai}\: \: \left \lfloor x+y \right \rfloor=\cdots  \end{aligned}$.

$\LARGE\colorbox{silver}{SOLUSI SOAL 25}$.

$\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad\left \lfloor 3x+y \right \rfloor=12\Leftrightarrow \color{blue}12\leq 3x+y< 13\color{black}\cdots (1)\\ &\bullet \quad\left \lfloor x+3y \right \rfloor=14\Leftrightarrow \color{blue}14\leq x+3y< 15\color{black}\cdots (2)\\ &\textrm{Jika kedua ketaksamaan dijumlahkan, maka}\\ &26\leq 4x+4y< 28\Leftrightarrow \displaystyle \frac{26}{4}\leq x+y< \displaystyle \frac{28}{4}\\ &\Leftrightarrow 6\displaystyle \frac{1}{2}\leq x+y< 7\\ &\textrm{Jadi},\: \left \lfloor x+y \right \rfloor=\color{red}6 \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL 26}$.

$\begin{aligned}&\textrm{Jika}\: \: \left \lfloor \displaystyle \frac{3x-7}{5} \right \rfloor=\displaystyle \frac{x}{3}\: ,\: \textrm{maka untuk}\: \:x\: \: \textrm{bulat terkecil}=\cdots  \end{aligned}$.

$\LARGE\colorbox{silver}{SOLUSI SOAL 26}$.

$\begin{aligned}&\textrm{Diketahui bahwa}\\ &\left \lfloor \displaystyle \frac{3x-7}{5} \right \rfloor=\displaystyle \frac{x}{3}\\ &\Leftrightarrow \displaystyle \frac{x}{3}\leq \displaystyle \frac{3x-7}{5}< \displaystyle \frac{x}{3}+1\\ &\Leftrightarrow \displaystyle \frac{x}{3}-\color{red}\frac{x}{3}\color{black}\leq \displaystyle \frac{3x-7}{5}-\color{red}\frac{x}{3}\color{black}< \displaystyle \frac{x}{3}-\color{red}\frac{x}{3}\color{black}+1\\ &\Leftrightarrow 0\leq \displaystyle \frac{4x-21}{15}< 1\\ &\Leftrightarrow 0\leq 4x-21< 15\\ &\Leftrightarrow 0+\color{red}21\color{black}\leq 4x-21+\color{red}21\color{black}< 15+\color{red}21\color{black}\\ &\Leftrightarrow 21\leq 4x< 36\\ &\Leftrightarrow \displaystyle \frac{21}{4}\leq x< 9\\ &\textrm{Jadi}, \: x=\color{red}6 \end{aligned}$.

DAFTAR PUSTAKA

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3. Mustofa, O. 2016. Olimpiyatlarina Hazirlik 1 Temel Bilgiler-1.Ankara
4. Tim Matematika. 2007. Program Pembinaan Kompetensi Siswa Bidang Matematika Tahap I. Bandung: LPPM ITB