asinus

\(Evaluate\\ a^3- \dfrac{1}{a^3}\\ if\\ a^2 - a - 1 = 0.\\ a=\frac{1}{2}\pm\sqrt{\frac{1}{4}+1}=\frac{1}{2}\pm\sqrt{\frac{1+4}{4}}=\frac{1}{2}(1\pm \sqrt{5})\\ \color{blue}a\in\{1.618, -0.618\}\)

\( a^3- \dfrac{1}{a^3}\\ =\dfrac{a^6-1}{a^3}\\ =\dfrac{\dfrac{1}{64}(1\pm\sqrt{5})^6-1}{\dfrac{1}{8}(1\pm \sqrt{5})^3}\\ =\dfrac{1}{8}(1\pm\sqrt{5})^3-\dfrac{1}{\dfrac{1}{8}(1\pm \sqrt{5})^3}\\ \color{blue} a^3- \dfrac{1}{a^3}=4\)

Let the distance on the base line, from the right edge to the strip, be x.

Then the following applies:

\( A_{rectangle}=1\cdot\sqrt{8^2+x^2}+8x=10\cdot 8\\ \sqrt{8^2+x^2}=80-8x\\ 8^2+x^2=6400-1280x+64x^2\\ 63x^2-1280x+6336=0 \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}=\frac{640-8\sqrt{163}}{63}\\ x=8.5375\\ A_{strip}=1\cdot \sqrt{8.5375^2+8^2}\\ \color{blue}A_{strip}=11.700\)

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\(\angle TOP=\angle UOP=45^{\circ}\\ \angle TOU=\angle TOP+\angle UOP=2\cdot 45^{\circ}\\ \color{blue}\angle TOU=90^\circ\)

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\(h^2=1^2-(2r-2)^2=r^2-(2-r)^2\\ 1-(4r^2-8r+4)=r^2-(4-4r+r^2)\\ 1-4r^2+8r-4=r^2-4+4r-r^2\\ 4r^2-4r-1=0\\ r^2-r-0.25=0\\ \)

\(r=0.5\pm\sqrt{0.25+0.25}\\ r\in\{1.207,-0.207\}\\ \color{blue}r=0.5+\sqrt{0.5}=1.207\\ h^2=1^2-(2r-2)^2\\ h^2=1-(1+\sqrt{2}-2)^2\\ h^2=1-(\sqrt{2}-1)^2\\ h^2=1-(2-2\sqrt{2}+1)\\ h=\sqrt{2\sqrt{2}-2}\\ \color{blue}h=\overline{CH}=0.91\)

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From the center M of the circles, draw rays to the corner E and to the center S of the side belonging to the corner. The result is a right-angled triangle M, E, S with sides s/2, r and R.

Then:

\(R^2=(\frac{s}{2})^2+r^2\\r=\sqrt{R^2-(\frac{s}{2})^2}\\ \frac{s}{2}=R\cdot sin(60^{\circ})\)

\(r=\sqrt{R^2-R^2\cdot sin^260^{\circ}}\\ r=R\cdot \sqrt{1-sin^260^{\circ}}\\ r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)

or very simply:

In the right triangle SME, angle SME=60°.

So is:

\(r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)

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Ist a Element der negativen reellen Zahlen, ist a < x < 0,

ist a Element der positiven reellen Zahlen, ist 0 < x < a.

Ich behaupte, meine Aussage ist richtig.

Ist x = a  ist die Ungleichung nicht erfüllt.

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Ich setze beide Diagonalen gleich:

\(2(R+\overline{VC})=2(R+r+\overline{MB})\\ 2(R+R\sqrt{2})=2(R+r+r\sqrt{2})\\ 2R+2\sqrt{2}R=2R+2r+2\sqrt{2}r\\ 2\sqrt{2}R=r(2+2\sqrt{2})\\ r=R\cdot \frac{2\sqrt{2}}{2+2\sqrt{2}}\cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}\\ r=R\cdot \frac{4\sqrt{2}-4\cdot 2}{4-4\cdot 2}=R\cdot \frac{4\sqrt{2}-4\cdot 2}{-4}\)

\(\color{blue}r=R\cdot(2-\sqrt{2})\\ q.e.d\)

\(R(1+2\sqrt{2})=r(2+2\sqrt{2})\\ r=R\cdot \frac{1+2\sqrt{2}}{2+2\sqrt{2}}\cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}} =R\cdot \frac{2-2\sqrt{2}+4 \sqrt{2}-8}{4-8}\\ r=R\cdot \frac{-6+2\sqrt{2}}{-4}\\ \color{blue}r=\frac{R}{2}\cdot (3-\sqrt{2})\)

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Leider auch falsch! Bin etwas aus der Übung.

\(2R+2\sqrt{2}R=R+2r+2\sqrt{2}r\\ R+2\sqrt{2}R=2r+2\sqrt{2}r\\ R(1+2\sqrt{2})=r(2+2\sqrt{2})\\ \color{blue}r=0,792893218813\cdot R\\ 2-\sqrt{2}=0,585786...\\ \color{blue } !\ r\neq (2-\sqrt{2})R\ !\)

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Da habe ich mich leider verrechnet. Sorry!