heureka

The numbers 1,2,3,4,5,6,7,8,9 are arranged in a list so that each number is
either greater than all the numbers that come before it
or is less than all the numbers that come before it.
For example, 4,5,6,3,2,7,1,8,9 is one such list:
notice that (for instance) the 6 is greater than all the numbers that come before it,
and the 2 is less than all the numbers that come before it.
How many such lists of the numbers 1,2,3,4,5,6,7,8,9 are possible?

\(\begin{array}{|r|r|} \hline 1.) & 123456789 \\ 2.) & 213456789 \\ 3.) & 231456789 \\ 4.) & 234156789 \\ 5.) & 234516789 \\ 6.) & 234561789 \\ 7.) & 234567189 \\ 8.) & 234567819 \\ 9.) & 234567891 \\ 10.) & 321456789 \\ 11.) & 324156789 \\ 12.) & 324516789 \\ 13.) & 324561789 \\ 14.) & 324567189 \\ 15.) & 324567819 \\ 16.) & 324567891 \\ 17.) & 342156789 \\ 18.) & 342516789 \\ 19.) & 342561789 \\ 20.) & 342567189 \\ 21.) & 342567819 \\ 22.) & 342567891 \\ 23.) & 345216789 \\ 24.) & 345261789 \\ 25.) & 345267189 \\ 26.) & 345267819 \\ 27.) & 345267891 \\ 28.) & 345621789 \\ 29.) & 345627189 \\ 30.) & 345627819 \\ 31.) & 345627891 \\ 32.) & 345672189 \\ 33.) & 345672819 \\ 34.) & 345672891 \\ 35.) & 345678219 \\ 36.) & 345678291 \\ 37.) & 345678921 \\ 38.) & 432156789 \\ 39.) & 432516789 \\ 40.) & 432561789 \\ 41.) & 432567189 \\ 42.) & 432567819 \\ 43.) & 432567891 \\ 44.) & 435216789 \\ 45.) & 435261789 \\ 46.) & 435267189 \\ 47.) & 435267819 \\ 48.) & 435267891 \\ 49.) & 435621789 \\ 50.) & 435627189 \\ 51.) & 435627819 \\ 52.) & 435627891 \\ 53.) & 435672189 \\ 54.) & 435672819 \\ 55.) & 435672891 \\ 56.) & 435678219 \\ 57.) & 435678291 \\ 58.) & 435678921 \\ 59.) & 453216789 \\ 60.) & 453261789 \\ 61.) & 453267189 \\ 62.) & 453267819 \\ 63.) & 453267891 \\ 64.) & 453621789 \\ 65.) & 453627189 \\ 66.) & 453627819 \\ 67.) & 453627891 \\ 68.) & 453672189 \\ 69.) & 453672819 \\ 70.) & 453672891 \\ 71.) & 453678219 \\ 72.) & 453678291 \\ 73.) & 453678921 \\ 74.) & 456321789 \\ 75.) & 456327189 \\ 76.) & 456327819 \\ 77.) & 456327891 \\ 78.) & 456372189 \\ 79.) & 456372819 \\ 80.) & 456372891 \\ 81.) & 456378219 \\ 82.) & 456378291 \\ 83.) & 456378921 \\ 84.) & 456732189 \\ 85.) & 456732819 \\ 86.) & 456732891 \\ 87.) & 456738219 \\ 88.) & 456738291 \\ 89.) & 456738921 \\ 90.) & 456783219 \\ 91.) & 456783291 \\ 92.) & 456783921 \\ 93.) & 456789321 \\ 94.) & 543216789 \\ 95.) & 543261789 \\ 96.) & 543267189 \\ 97.) & 543267819 \\ 98.) & 543267891 \\ 99.) & 543621789 \\ 100.) & 543627189 \\ 101.) & 543627819 \\ 102.) & 543627891 \\ 103.) & 543672189 \\ 104.) & 543672819 \\ 105.) & 543672891 \\ 106.) & 543678219 \\ 107.) & 543678291 \\ 108.) & 543678921 \\ 109.) & 546321789 \\ 110.) & 546327189 \\ 111.) & 546327819 \\ 112.) & 546327891 \\ 113.) & 546372189 \\ 114.) & 546372819 \\ 115.) & 546372891 \\ 116.) & 546378219 \\ 117.) & 546378291 \\ 118.) & 546378921 \\ 119.) & 546732189 \\ 120.) & 546732819 \\ 121.) & 546732891 \\ 122.) & 546738219 \\ 123.) & 546738291 \\ 124.) & 546738921 \\ 125.) & 546783219 \\ 126.) & 546783291 \\ 127.) & 546783921 \\ 128.) & 546789321 \\ 129.) & 564321789 \\ 130.) & 564327189 \\ 131.) & 564327819 \\ 132.) & 564327891 \\ 133.) & 564372189 \\ 134.) & 564372819 \\ 135.) & 564372891 \\ 136.) & 564378219 \\ 137.) & 564378291 \\ 138.) & 564378921 \\ 139.) & 564732189 \\ 140.) & 564732819 \\ 141.) & 564732891 \\ 142.) & 564738219 \\ 143.) & 564738291 \\ 144.) & 564738921 \\ 145.) & 564783219 \\ 146.) & 564783291 \\ 147.) & 564783921 \\ 148.) & 564789321 \\ 149.) & 567432189 \\ 150.) & 567432819 \\ 151.) & 567432891 \\ 152.) & 567438219 \\ 153.) & 567438291 \\ 154.) & 567438921 \\ 155.) & 567483219 \\ 156.) & 567483291 \\ 157.) & 567483921 \\ 158.) & 567489321 \\ 159.) & 567843219 \\ 160.) & 567843291 \\ 161.) & 567843921 \\ 162.) & 567849321 \\ 163.) & 567894321 \\ 164.) & 654321789 \\ 165.) & 654327189 \\ 166.) & 654327819 \\ 167.) & 654327891 \\ 168.) & 654372189 \\ 169.) & 654372819 \\ 170.) & 654372891 \\ 171.) & 654378219 \\ 172.) & 654378291 \\ 173.) & 654378921 \\ 174.) & 654732189 \\ 175.) & 654732819 \\ 176.) & 654732891 \\ 177.) & 654738219 \\ 178.) & 654738291 \\ 179.) & 654738921 \\ 180.) & 654783219 \\ 181.) & 654783291 \\ 182.) & 654783921 \\ 183.) & 654789321 \\ 184.) & 657432189 \\ 185.) & 657432819 \\ 186.) & 657432891 \\ 187.) & 657438219 \\ 188.) & 657438291 \\ 189.) & 657438921 \\ 190.) & 657483219 \\ 191.) & 657483291 \\ 192.) & 657483921 \\ 193.) & 657489321 \\ 194.) & 657843219 \\ 195.) & 657843291 \\ 196.) & 657843921 \\ 197.) & 657849321 \\ 198.) & 657894321 \\ 199.) & 675432189 \\ 200.) & 675432819 \\ 201.) & 675432891 \\ 202.) & 675438219 \\ 203.) & 675438291 \\ 204.) & 675438921 \\ 205.) & 675483219 \\ 206.) & 675483291 \\ 207.) & 675483921 \\ 208.) & 675489321 \\ 209.) & 675843219 \\ 210.) & 675843291 \\ 211.) & 675843921 \\ 212.) & 675849321 \\ 213.) & 675894321 \\ 214.) & 678543219 \\ 215.) & 678543291 \\ 216.) & 678543921 \\ 217.) & 678549321 \\ 218.) & 678594321 \\ 219.) & 678954321 \\ 220.) & 765432189 \\ 221.) & 765432819 \\ 222.) & 765432891 \\ 223.) & 765438219 \\ 224.) & 765438291 \\ 225.) & 765438921 \\ 226.) & 765483219 \\ 227.) & 765483291 \\ 228.) & 765483921 \\ 229.) & 765489321 \\ 230.) & 765843219 \\ 231.) & 765843291 \\ 232.) & 765843921 \\ 233.) & 765849321 \\ 234.) & 765894321 \\ 235.) & 768543219 \\ 236.) & 768543291 \\ 237.) & 768543921 \\ 238.) & 768549321 \\ 239.) & 768594321 \\ 240.) & 768954321 \\ 241.) & 786543219 \\ 242.) & 786543291 \\ 243.) & 786543921 \\ 244.) & 786549321 \\ 245.) & 786594321 \\ 246.) & 786954321 \\ 247.) & 789654321 \\ 248.) & 876543219 \\ 249.) & 876543291 \\ 250.) & 876543921 \\ 251.) & 876549321 \\ 252.) & 876594321 \\ 253.) & 876954321 \\ 254.) & 879654321 \\ 255.) & 897654321 \\ 256.) & 987654321 \\ \hline \end{array}\)

There are 256 lists of the numbers 1,2,3,4,5,6,7,8,9 possible.

In the coordinate plane, let F = (5,0).

Let P be a point, and let Q be the projection of the point P onto the line \(x=\dfrac{16}{5}\).

The point traces a curve in the plane, so that  \(\dfrac{PF}{PQ} = \dfrac{5}{4}\) for all points on the curve.

Find the equation of this curve.

(Enter it in standard form.) 

The equation of this curve is a hyperbola : \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\)

The focus is the Point \(F = (5,0) =(c,0)\)  so \(c = 5 \).

\(\mathbf{a=\ ?}\)

\(\begin{array}{|rcll|} \hline \dfrac{PF}{PQ} = \dfrac{c}{a} &=& \dfrac{5}{4} \\\\ \dfrac{c}{a} &=& \dfrac{5}{4} \quad | \quad c = 5 \\\\ \dfrac{5}{a} &=& \dfrac{5}{4} \\\\ \dfrac{1}{a} &=& \dfrac{1}{4} \\\\ \mathbf{a} &\mathbf{=}& \mathbf{4} \\ \hline \end{array} \)

\(\mathbf{b=\ ?}\)

\(\begin{array}{|rcll|} \hline b^2 &=& c^2-a^2 \quad | \quad c = 5,~ a= 4 \\ b^2 &=& 5^2-4^2 \\ b^2 &=&25-16 \\ b^2 &=& 9 \\ \mathbf{b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}\)

check:
\(\begin{array}{rcl} \dfrac{16}{5} &=& \dfrac{a^2}{c} \\\\ &=& \dfrac{4^2}{5} \\\\ &=& \dfrac{16}{5}\ \checkmark \\ \end{array}\)

\(\text{Let $ P=(x_p,y_p) $ }\)

The equation of this curve is : \(\mathbf{\dfrac{x_p^2}{4^2} - \dfrac{y_p^2}{3^2} = 1} \)

circle:

\(|z-z_0|=1\\ if~z=x+yi ~and~z_0=1+2i\\ then~(x-1)^2+(y-2)^2=1\\ or~x^2+y^2-2x-4y+4=0\)

Formula:

\(z\overline{z}+pz+q\overline{z}+r=0\\ if~z=x+yi~and~\overline{z}=x-yi\\~and~z_0=1+2i\\ then~x^2+y^2+(p+q)x+(pi-qi)y+r=0\)

p,q = ?

\(p+q=-2\\\ (p-q)i=-4\\ r=4\\ \Rightarrow \\ q=-1-2i\qquad p=-1+2i\qquad r=4\)

(p,q,r) = ( -1+2i, -1-2i, 4 )

(a+b)^2=a^2+b^2+2ab

=193+2*84

=361

a+b=19

If I have a list of 5 numbers each of them following another [1,2,3,4,5] and I want to multiply all of them together. 

How do I write (1*2)+(1*3)+(1*4)+(1*5)+(2*3)+(2*4)+(2*5)+(3*4)+(3*5)+(4*5)

But with 4000 numbers in the list instead of 5 without having to write every instance out?

Is it possible to make a rule and calculate it in a common calculator? 

Table:

\(\begin{array}{|r|r|r|r|r|r} \hline \text{multiply} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & \color{red}1 & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline 2 & \color{grey}2 & \color{red}4 & \mathbf{6} & \mathbf{8} & \mathbf{10} \\ \hline 3 & \color{grey}3 & \color{grey}6 & \color{red}9 & \mathbf{12} & \mathbf{15} \\ \hline 4 & \color{grey}4 & \color{grey}8 & \color{grey}12 & \color{red}16 & \mathbf{20} \\ \hline 5 & \color{grey}5 & \color{grey}10 & \color{grey}15 & \color{grey}20 & \color{red}25 \\ \hline sum & \color{blue}15 & \color{blue}30 & \color{blue}45 & \color{blue}60 & \color{blue}75 & \color{blue}\text{arithmetical sequence} \\ \hline \end{array} \)

\(\text{Let the sum of all numbers in the table $\mathbf{S} = \color{blue}{15+30+45+60+75} = \color{blue}225 $} \\ \begin{array}{rcll} \text{Let the sum we are looking for } \mathbf{s=} \\ && (1*2)+(1*3)+(1*4)+(1*5)\\ && +(2*3)+(2*4)+(2*5) \\ && +(3*4)+(3*5) \\ && +(4*5) \\ &=& \mathbf{2+3+4+5} \\ && \mathbf{+6+8+10} \\ && \mathbf{+12+15} \\ && \mathbf{+20} \\ &=& 85\\ \end{array} \\ \begin{array}{rcll} \text{Let the sum of all square numbers in the diagonal of the table $\mathbf{q} =$} \color{red}{1+4+9+16+25}=\color{red}55 \\ \end{array} \)

So we see the formula: \(\boxed{S = 2s + q}\)  or \(\boxed{s = \dfrac{1}{2}\cdot (S-q) }\)

\(\mathbf{ S=\ ?}\)

\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)d \quad | \quad d=a_1 \\ &=& a_1+(n-1)a_1 \\ &=& a_1+ na_1-a_1 \\ &=& na_1 \quad | \quad a_1=(1+n) \dfrac{n}{2} \\ &=& (1+n) \dfrac{n^2}{2} \\\\ S &=& (a_1 + a_n)\cdot \dfrac{n}{2} \\ &=& \left((1+n) \dfrac{n}{2} +(1+n) \dfrac{n^2}{2} \right)\cdot \dfrac{n}{2} \\ &=& \Big((1+n) +(1+n) n \Big)\cdot \dfrac{n^2}{4} \\ &=& (n^2+2n+1 )\cdot \dfrac{n^2}{4} \\ &=& (n+1)^2\cdot \dfrac{n^2}{4} \\ \mathbf{S} & \mathbf{=} & \mathbf{\left[\dfrac{(n+1)\cdot n}{2}\right]^2} \\ \hline \end{array}\)

\(\mathbf{ q=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{q} & \mathbf{=} & \mathbf{\dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3}} \quad | \quad \text{from the formula collection} \\ \hline \end{array} \)

\(\mathbf{ s=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{ \dfrac{1}{2}\cdot (S-q) } \\ &=& \dfrac{1}{2}\cdot \left(\left[\dfrac{(n+1)\cdot n}{2}\right]^2 - \dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3} \right) \\ &=& \dfrac{(n+1)\cdot n}{4}\cdot \left[ \dfrac{(n+1)\cdot n}{2} - \dfrac{(2n+1)}{3} \right] \\ &=& \dfrac{(n+1)\cdot n}{24}\cdot \Big[ 3n(n+1) - 2(2n+1) \Big] \\ &=& \dfrac{(n+1)\cdot n\cdot ( 3n^2-n-2)}{24} \\ &=& \dfrac{(n+1)\cdot n\cdot (3n+2)\cdot (n-1)}{24} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(n-1)\cdot n \cdot (n+1)\cdot (3n+2)}{24} } \\ \hline \end{array}\)

Example: \([1,2,3,4,5]\)

So \(n = 5 \)

\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(5-1)\cdot 5 \cdot (5+1)\cdot (3\cdot 5+2)}{24} } \\ &=& \dfrac{4\cdot 5 \cdot 6\cdot 17}{24} \\ &=& 5 \cdot 17 \\ &\mathbf{=}& \mathbf{85}\ \checkmark\\ \hline \end{array}\)

Example:\( [1,2,3,4,5, \ldots ,4000]\)

So \(n = 4000\)

\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(4000-1)\cdot 4000 \cdot (4000+1)\cdot (3\cdot 4000+2)}{24} } \\ &=& \dfrac{3999\cdot 4000 \cdot 4001\cdot 12002 }{24} \\ &=& 1333\cdot 1000 \cdot 4001\cdot 6001 \\ &\mathbf{=}& \mathbf{32\ 005\ 331\ 333\ 000} \\ \hline \end{array} \)

How many zeroes are in the solution of 65^20-65^16

\(\begin{array}{|rcll|} \hline && 65^{20}-65^{16} \\ &=& 65^{16}(65^4-1) \\ &=& 65^{16}(65^2-1)(65^2+1) \\ &=& (5\cdot 13)^{16}(4224)(4226) \quad \text{Decomposition into prime factors} \\ &=& (5\cdot 13)^{16}(2^7\cdot 3 \cdot 11)(2\cdot 2113 ) \quad 10~\text{is the factor of $5$ and $2$ } \\ &=& 2^8\cdot 5^{16}\ldots \quad \text{The minimum of $16$ and $8$ is $8$ is the number of zeroes in the product. } \\ \hline \end{array} \)

In the solution of \(\mathbf{65^{20}-65^{16}}\) are \(\mathbf{8}\) zeroes.

Proof  \(65^{20}-65^{16} = 1\ 812\ 454\ 482\ 098\ 982\ 309\ 886\ 289\ 062\ 500\ 000\ 000\)

Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2.

Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3.

Find \[PA_1^2 + PA_2^2 + \dots + PA_{11}^2.\]

\(\text{Let $A_n=\dbinom{x_a}{y_a}$, $~x_a =2\cos\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big)$, $~y_a =2\sin\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big) $, $~ n=1,2,\ldots 11$ }\\ \text{Let $P=\dbinom{x_p}{y_p}$, $~x_p =3\cos(\beta)$, $~y_p =3\sin(\beta)$ } \\ \text{Let $PA^2 = (x_p-x_a)^2 + (y_p-y_a)^2$}\)

\(\begin{array}{|rcll|} \hline PA_1^2 &=& \left[ 3\cos(\beta) -2\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_2^2 &=& \left[ 3\cos(\beta) -2\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_3^2 &=& \left[ 3\cos(\beta) -2\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ \ldots \\\\ PA_{11}^2 &=& \left[ 3\cos(\beta) -2\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right)\right]\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && PA_1^2 + PA_2^2 + PA_3^2 + \dots + PA_{11}^2 \\\\ &=& 11\cdot 13 \\ && -12\cdot \Big\{ \cos(\beta) \Big(\underbrace{ \cos(0\cdot \dfrac{360^{\circ}}{11}) + \cos(1\cdot \dfrac{360^{\circ}}{11}) + \cos(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \cos(10\cdot \dfrac{360^{\circ}}{11}) }_{=0} \Big) \\ && +\sin(\beta) \Big(\underbrace{ \sin(0\cdot \dfrac{360^{\circ}}{11}) + \sin(1\cdot \dfrac{360^{\circ}}{11}) + \sin(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \sin(10\cdot \dfrac{360^{\circ}}{11}) \Big) }_{=0} \Big\} \\\\ &=& 11\cdot 13 -12\cdot \Big( \cos(\beta)\cdot 0 +\sin(\beta)\cdot 0 \Big) \\ &=& 11\cdot 13 -12\cdot 0 \\ &=& 11\cdot 13 \\ &\mathbf{=}& \mathbf{143} \\ \hline \end{array}\)

\(PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \mathbf{143}\)

z=-i^4 kann das jemand in polarkoordinaten umwandeln danke!

Ich vermute die Aufgabe lautet: \(z = -(i^4)\)

\(\begin{array}{|rcll|} \hline & z &=& -(i^4) \\ & &=& -(i^2)(i^2) \quad & | \quad i^2 = -1 \\ & &=& -(-1)(-1) \\ & \mathbf{z} &\mathbf{=}& \mathbf{-1} \\ \text{ oder }& \mathbf{z} &\mathbf{=}& \mathbf{e^{i\pi}} \\ \text{ oder }& \mathbf{z} &\mathbf{=}& \mathbf{\cos(\pi)+i\sin(\pi)} \\ \hline \end{array}\)

Of all numbers between 1 and 15! (factorial),

how many of them are BOTH perfect squares and perfect cubes for the same number(such as 64)?

lcm(2,3) = 6  (least common multiple)

So the numbers that are going to be perfect squares and perfect cubes are precisely the 6th powers of the integers:
\(\begin{array}{rcll} 1^6 &=& 1 \\ 2^6 &=& 64 \\ 3^6 &=& 729 \\ 4^6 &=& 4096 \\ 5^6 &=& 15625 \\ ... \\ n^6 &<& 15! \\ \end{array}\)

So n are the numbers between 1 and 15! (factorial)

\(\begin{array}{|rcll|} \hline n^6 &<& 15! \\ n &<& \sqrt[6]{15!} \\ n &<& \sqrt[6]{1~307~674~368~000} \\ n &<& 104.57228596314317\ldots \\ \mathbf{n} & \mathbf{=} & \mathbf{104}\\ \hline \end{array}\)

104 of them are BOTH perfect  squares and perfect cubes.

In rhombus ABCD, points E,F,G, and H are the midpoints of lines AB,BC,CD, and DA respectively.
Quadrilateral EFGH has area 14 and perimeter 16.
Find the side length for rhombus ABCD.

\(\text{Let $EF$ = x } \\ \text{Let $EH$ = y } \\ \text{The side length for rhombus $=AB=BC=CD=DA=HF$ = s } \)

\(\begin{array}{|lrcll|} \hline 1. & xy&=&14 \\ & y&=& \dfrac{14}{x} \\\\ 2. & 2(x+y)&=& 16 \\ & \mathbf{x+y} & \mathbf{=} & \mathbf{8} \quad & | \quad y= \dfrac{14}{x}\\ & x+\dfrac{14}{x} &=& 8 \quad & | \quad \cdot x \\ & x^2+ 14 &=& 8x \\ & x^2-8x+ 14 &=& 0 \\ & x &=& \dfrac{8\pm \sqrt{64-4\cdot 14} }{2} \\ & x &=& \dfrac{8\pm \sqrt{4(16-14)} }{2} \\ & x &=& \dfrac{8\pm 2\sqrt{ 16-14 } }{2} \\ & x &=& \dfrac{8\pm 2\sqrt{ 2 } }{2} \\ & x & = & 4\pm \sqrt{ 2 } \\ & \mathbf{x} & \mathbf{=} & \mathbf{4- \sqrt{ 2 }} \\\\ & \mathbf{x+y} & \mathbf{=} & \mathbf{8} \quad & | \quad x=4-\sqrt{ 2 } \\ & 4-\sqrt{ 2 }+y &=& 8 \\ & y &=& 8-(4-\sqrt{ 2 }) \\ & y &=& 8-4+\sqrt{ 2 }) \\ & \mathbf{y} & \mathbf{=} & \mathbf{4+ \sqrt{ 2 }} \\ \hline \end{array} \)

\(\mathbf{s =\ ?}\)

\(\begin{array}{|rcll|} \hline s^2 &=& x^2+y^2 \quad & | \quad \mathbf{x= 4- \sqrt{ 2 }},~ \mathbf{y= 4+ \sqrt{ 2 }} \\ s^2 &=& (4-\sqrt{2})^2+(4+\sqrt{2})^2 \\ s^2 &=& 16-8\sqrt{2} + 2 + 16 +8\sqrt{2} + 2 \\ s^2 &=& 16 + 2 + 16 + 2 \\ s^2 &=& 36 \\ \mathbf{s} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

The side length for rhombus ABCD is 6