heureka

Find positive integers a,b and c such that
$\dfrac{3}{11} = \frac {1} {a+ \frac1{b+ \frac1c}}$.

$\begin{array}{|r|rcl|rcl|} \hline \dfrac{3}{11} = \frac {1} {a+ \frac1{b+ \frac1c}} &\dfrac{11}{3} &=& a+ \frac1{b+ \frac1c} \\\\ &{\color{red}3}+\dfrac{2}{3} &=& {\color{red}a}+ \frac1{b+ \frac1c}~\Rightarrow \mathbf{a=3} \\\\ &\dfrac{2}{3} &=& \frac1{b+ \frac1c} &\dfrac{3}{2} &=& b+ \frac1c \\\\ &&&&{\color{red}1}+\dfrac{1}{2} &=& {\color{red}b}+ \frac1c ~\Rightarrow \mathbf{b=1} \\\\ &&&&\dfrac{1}{2} &=& \dfrac1c \\\\ &&&& \mathbf{2} &=& \mathbf{c} \\ \hline \end{array}$

$\dfrac{3}{11} = \frac {1} {3+ \frac1{1+ \frac12}}.$

If we write $\sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) }$
in the form $\dfrac{(a*\sqrt{2} + b*\sqrt{3})}{c}$ such that a, b, and c

are positive integers and c is as small as possible,
then what is a + b + c?

$\begin{array}{|rcll|} \hline && \sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) } * \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (2*\sqrt{2} - 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (2*\sqrt{2} + 3*\sqrt{3})(2*\sqrt{2} - 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (4*2-9*3) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ -19 } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{-(2*\sqrt{2} - 3*\sqrt{3})}{ 19 } \\\\ &=& \dfrac{19*(\sqrt{2} + \sqrt{3}) -(2*\sqrt{2} - 3*\sqrt{3})}{ 19 } \\\\ &=& \dfrac{19*\sqrt{2} + 19*\sqrt{3} -2*\sqrt{2} + 3*\sqrt{3}}{ 19 } \\\\ &=& \dfrac{{\color{red}17}*\sqrt{2} + {\color{red}22}*\sqrt{3}}{ {\color{red}19} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a+b+c &=& 17+22+19 \\ \mathbf{a+b+c} &=& \mathbf{58} \\ \hline \end{array}$

If

$d-6c=4$ and $2d-9c=20$, find the value of $\dfrac{d}{c}$.

$\begin{array}{|rcll|} \hline 2d-9c &=& 20 \quad | \quad :c \\\\ \mathbf{\dfrac{2d}{c}-9} &=& \mathbf{\dfrac{20}{c}} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline d-6c &=& 4 \quad | \quad :c \\\\ \dfrac{d}{c}-6 &=& \dfrac{4}{c} \quad | \quad *5 \\\\ \mathbf{\dfrac{5d}{c}-30} &=& \mathbf{\dfrac{20}{c}} \qquad (2) \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (2)=(1): & \dfrac{5d}{c}-30 &=& \dfrac{2d}{c}-9 \\\\ & \dfrac{3d}{c} &=& 30 -9 \\\\ & \dfrac{3d}{c} &=& 21 \quad | \quad :3 \\\\ & \mathbf{\dfrac{d}{c}} &=& \mathbf{7} \\ \hline \end{array}$

Given that x satisfies $x^2 - 5x + 1 = 0$,
find the value of $x^4 +\dfrac{1}{ x^4 }$.

$\begin{array}{|rcll|} \hline \mathbf{x^2 - 5x + 1} &=& \mathbf{0} \\\\ x_{1,2} &=& \dfrac{5\pm\sqrt{25-4}}{2} \\ x_{1,2} &=& \dfrac{5\pm\sqrt{21}}{2} \\ \hline x_{1,2}^2 &=& \left( \dfrac{5\pm\sqrt{21}}{2} \right)^2 \\ x_{1,2}^2 &=& \dfrac{123\pm 5\sqrt{21}}{2}\\ \hline x_{1,2}^4 &=& \left( \dfrac{123\pm 5\sqrt{21}}{2} \right)^2 \\ x_{1,2}^4 &=& \dfrac{527\pm 115\sqrt{21}}{2} \\ \hline \end{array}$

$x_1:$

$\begin{array}{|rcll|} \hline x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2}{527+115\sqrt{21}} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2}{(527+115\sqrt{21})}*\dfrac{(527-115\sqrt{21})}{(527-115\sqrt{21})} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2(527-115\sqrt{21})}{4} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{(527-115\sqrt{21})}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}+527-115\sqrt{21}}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{2*527}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& 527 \\ \hline \end{array}$

$x_2:$

$\begin{array}{|rcll|} \hline x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2}{527-115\sqrt{21}} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2}{(527-115\sqrt{21})}*\dfrac{(527+115\sqrt{21})}{(527+115\sqrt{21})} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2(527+115\sqrt{21})}{4} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{(527+115\sqrt{21})}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}+527+115\sqrt{21}}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{2*527}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& 527 \\ \hline \end{array}$

$x^4 +\dfrac{1}{ x^4 } =527$

As shown in the diagram,$~BD/DC=2,~CE/EA=3$,  and $AF/FB=4$.
Find $[DEF]/[ABC]$.

$\begin{array}{rcll} \text{ Let } BC&=&a\\ \text{ Let } BD&=&\frac{2}{3}a \\ \text{ Let } DC&=&\frac{1}{3}a \\\\ \text{ Let } CA&=&b \\ \text{ Let } CE&=&\frac{3}{4}b \\ \text{ Let } EA&=&\frac{1}{4}b \\\\ \text{ Let } AB&=&c \\ \text{ Let } AF&=&\frac{4}{5}c \\ \text{ Let } FB&=&\frac{1}{5}c \\ \end{array}$

$\begin{array}{rcll} \text{ Let } \angle CAB&=& A\\ \text{ Let } \angle ABC&=& B\\ \text{ Let } \angle BCA&=& C\\ \end{array}$

$\begin{array}{|rcll|} \hline 2*[ABC] &=& bc\sin(A) \\\\ 2*[AFE] &=& \frac{1}{4}b*\frac{4}{5}c \sin(A) \\ 2*[AFE] &=& \frac{1}{5}bc \sin(A) \\ 2*5*[AFE] &=& bc \sin(A) \\ \hline bc\sin(A) = 2*[ABC] &=& 2*5*[AFE] \\ 2*[ABC] &=& 2*5*[AFE] \\ [ABC] &=& 5*[AFE] \\ \mathbf{[AFE]} &=& \mathbf{\frac{1}{5}[ABC]} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline 2*[ABC] &=& ca\sin(B) \\\\ 2*[FBD] &=& \frac{1}{5}c*\frac{2}{3}a \sin(B) \\ 2*[FBD] &=& \frac{2}{15}ca \sin(B) \\ 2*\frac{15}{2}*[FBD] &=& ca \sin(B) \\ \hline ca \sin(B) = 2*[ABC]&=&2*\frac{15}{2}[FBD] \\ 2*[ABC]&=&2*\frac{15}{2}[FBD] \\ [ABC]&=&\frac{15}{2}[FBD] \\ \mathbf{[FBD]} &=& \mathbf{\frac{2}{15}[ABC]} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline 2*[ABC] &=& ab\sin(C) \\\\ 2*[EDC] &=& \frac{1}{3}a*\frac{3}{4}b \sin(C) \\ 2*[EDC] &=& \frac{1}{4}ab \sin(C) \\ 2*4*[EDC] &=& ab \sin(C) \\ \hline ab \sin(C) = 2*[ABC] &=& 2*4*[EDC] \\ 2*[ABC] &=& 2*4*[EDC] \\ [ABC] &=& 4*[EDC] \\ \mathbf{[EDC]} &=& \mathbf{\frac{1}{4}[ABC]} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline [DEF] + [AFE]+[FBD]+[EDC] &=& [ABC] \\ [DEF] + \frac{1}{5}[ABC]+\frac{2}{15}[ABC]+\frac{1}{4}[ABC] &=& [ABC] \\ [DEF] + [ABC] \left(\frac{1}{5}+\frac{2}{15}+\frac{1}{4} \right) &=& [ABC] \\ [DEF] + \frac{7}{12}[ABC] &=& [ABC] \\ [DEF] &=& [ABC]- \frac{7}{12}[ABC] \\ [DEF] &=& [ABC] \left(1- \frac{7}{12} \right) \\ [DEF] &=& \frac{5}{12}[ABC] \\ \mathbf{\frac{[DEF]}{[ABC]}} &=& \mathbf{\frac{5}{12}} \\ \hline \end{array}$

help

(b)

For some positive integer $n$ the expansion of $(1 + x)^n$
has three consecutive coefficients $a,~b,~c$ that satisfy
$a:b:c = 1:7:35$.
What must $n$ be?

$\text{Let $a=\dbinom{n}{k - 1}$ } \\ \text{Let $b=\dbinom{n}{k}$ } \\ \text{Let $c=\dbinom{n}{k+1}$ }$

$\begin{array}{|rcll|} \hline \dfrac{b}{a} = \dfrac{7}{1} &=& \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} \\\\ \dfrac{7}{1} &=& \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} \\\\ \dfrac{7}{1} &=& \dfrac{n-k+1}{k} \\\\ 7k &=& n-k+1 \\ 8k &=& n+1 \\ \mathbf{n} &=& \mathbf{8k-1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \dbinom{n}{k + 1} &=& \dfrac{n!}{(k+1)!\Big( n-(k+1)\Big)!} \\\\ &=& \dfrac{n!}{(k+1)!\Big( n-k-1)\Big)!} \\\\ && \boxed{ (k+1)!=k!(k+1)} \\\\ &=& \dfrac{n!}{k!(k+1)( n-k-1)!} \\\\ && \boxed{ ( n-k-1)!(n-k)=(n-k)! \\ ( n-k-1)!=\dfrac{(n-k)!}{n-k} } \\\\ &=& \dfrac{n!*(n-k)}{k!(k+1)(n-k)!} \\\\ &=& \dfrac{n!}{k!(n-k)! } *\dfrac{(n-k)}{(k+1)} \\\\ && \boxed{ \dbinom{n}{k}=\dfrac{n!}{k!(n-k)! } } \\\\ &=& \dbinom{n}{k} *\dfrac{(n-k)}{(k+1)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \dfrac{c}{b} = \dfrac{35}{7} &=& \dfrac{\dbinom{n}{k+1}}{\dbinom{n}{k}} \\\\ \dfrac{35}{7} &=& \dfrac{\dbinom{n}{k} *\dfrac{(n-k)}{(k+1)}}{\dbinom{n}{k}} \\\\ \dfrac{35}{7} &=&\dfrac{(n-k)}{(k+1)} \\\\ 35(k+1) &=&7(n-k) \\ 42k &=& 7n-35 \\ \mathbf{k} &=& \mathbf{ \dfrac{7n-35}{42}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n &=& 8k-1 \quad | \quad \mathbf{k=\dfrac{7n-35}{42}} \\\\ n &=& 8\left(\dfrac{7n-35}{42}\right)-1 \\\\ n &=& \dfrac{4(7n-35)}{21}-1 \\\\ n &=& \dfrac{4(7n-35)-21}{21} \\\\ 21n &=&4(7n-35)-21\\ 21n &=&28n-140-21\\ 7n &=& 161 \quad | \quad :7 \\ \mathbf{n} &=& \mathbf{23} \\ \hline k &=& \dfrac{7n-35}{42} \\\\ k &=& \dfrac{7*23-35}{42} \\\\ \mathbf{k} &=& \mathbf{3} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a:b:c =\dbinom{n}{k - 1}:\dbinom{n}{k}:\dbinom{n}{k + 1} &=& 1:7:35 \\\\ \dbinom{n}{k - 1}:\dbinom{n}{k}:\dbinom{n}{k + 1} &=& 1:7:35 \\\\ \dbinom{23}{3 - 1}:\dbinom{23}{3}:\dbinom{23}{3 + 1} &=& 1:7:35 \\\\ \dbinom{23}{2}:\dbinom{23}{3}:\dbinom{23}{4} &=& 1:7:35 \\\\ 253:1771:8855 &=& 1:7:35 \\ \hline \end{array}$

(a) Simplify

$\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}$.

$\begin{array}{|rcll|} \hline \dbinom{n}{k - 1} &=& \dfrac{n!}{(k-1)!\Big( n-(k-1)\Big)!} \\\\ &=& \dfrac{n!}{(k-1)!\Big( n-k+1)\Big)!} \\\\ && \boxed{ (k-1)!k=k!\\ (k-1)! =\dfrac {k!}{k} } \\\\ &=& \dfrac{n!*k}{k!\Big( n-k+1)\Big)!} \\\\ && \boxed{ \Big( n-k+1)\Big)!=(n-k)!(n-k+1) } \\\\ &=& \dfrac{n!*k}{k!(n-k)!(n-k+1) } \\\\ &=& \dfrac{n!}{k!(n-k)! } *\dfrac{k}{n-k+1} \\\\ && \boxed{ \dbinom{n}{k}=\dfrac{n!}{k!(n-k)! } } \\\\ &=& \dbinom{n}{k} *\dfrac{k}{n-k+1} \\\\ \hline \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} &=& \dfrac{\dbinom{n}{k}}{\dbinom{n}{k} *\dfrac{k}{n-k+1}} \\\\ \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} &=& \dfrac{1}{\dfrac{k}{n-k+1}} \\\\ \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} &=& \dfrac{n-k+1}{k}\\ \hline \end{array}$

Suppose we have $PQ=6$, $QR=7$, and $PR=9$, as in the picture below:

Find the length of the median from $R$ to $\overline{PQ}$.

cos Rule:

$\begin{array}{|rcll|} \hline 9^2&=&(2*3)^2+7^2-2*2*3*7*\cos(Q) \\ 9^2&=&6^2+7^2-2*2*3*7*\cos(Q) \\ 2*2*3*7*\cos(Q) &=& 6^2+7^2-9^2 \\ 2*2*3*7*\cos(Q) &=& 4 \quad | \quad :2 \\ 2*3*7*\cos(Q) &=& 2 \qquad (1) \\ \hline \end{array}$

cos Rule:

$\begin{array}{|rcll|} \hline d^2&=&3^2+7^2-2*3*7*\cos(Q) \\ 2*3*7*\cos(Q) &=& 3^2+7^2-d^2 \\ 2*3*7*\cos(Q) &=& 58-d^2 \qquad (2) \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline (1)=(2):& 2 &=& 58-d^2 \\ & d^2&=& 58-2 \\ & d^2&=& 56 \\ & d^2&=& 4*14 \\ & \mathbf{d}&=&\mathbf{ 2\sqrt{14} } \\ \hline \end{array}$

Sorry, There is a typo!

$\mathbf{PM} = \mathbf{ 2\sqrt{37} }$