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Author Topic: Bine v-am gasit !  (Read 2535 times)
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February 08, 2014, 09:09:25 AM
 #1

Bine v-am gasit !
Sunt adminul site-ului :

http://www.howtoearnbtc.com

Ma bucur ca sunt printre voi.

Doresc multa bafta celor care indragesc BITCOIN ! Kiss
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February 08, 2014, 11:27:46 AM
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Bine ai venit, sedere placuta si succes cu proiectul/proiectele tale Smiley.

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February 09, 2014, 12:11:10 PM
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Multumesc la fel !
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December 27, 2014, 09:07:18 PM
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Bine v-am gasit !
Sunt adminul site-ului :

http://www.howtoearnbtc.com

Ma bucur ca sunt printre voi.

Doresc multa bafta celor care indragesc BITCOIN ! Kiss

Fain algoritm la joculetul ala... nu reusesc sa-l bat Sad

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December 27, 2014, 09:21:20 PM
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M-am gandit ca o sa fie usor dar deh... Undecided e chiar imposibil  Cry

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December 27, 2014, 09:32:07 PM
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L-am batut si am primit un MARE...NIMIC! deci care e treaba cu jocul ?  Cheesy

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December 27, 2014, 09:53:41 PM
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L-am batut si am primit un MARE...NIMIC! deci care e treaba cu jocul ?  Cheesy



Capeti experienta in X si O  ..  Cheesy
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December 28, 2014, 12:39:53 PM
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L-am batut si am primit un MARE...NIMIC! deci care e treaba cu jocul ?  Cheesy



Cum ai reusit sa-l bati? Ce tactica ai folosit?

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December 28, 2014, 03:43:07 PM
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L-am batut si am primit un MARE...NIMIC! deci care e treaba cu jocul ?  Cheesy



Cum ai reusit sa-l bati? Ce tactica ai folosit?

Ceva simplu..  Cheesy

Daca o dreapta d este perpendiculara pe un plan \alpha si prin piciorul ei trece o dreapta a, continuta in plan, care este perpendiculara pe o alta dreapta b continuta in plan, atunci o dreapta c care uneste orice punct M al dreptei d cu intersectia P a celor doua drepte a si b, este perpendiculara pe cea de-a treia latura.

Cum aplicam Teorema celor trei perpendiculare

d\perp\alpha    \\a\subset\alpha, O\in a

a\perp b, b\subset\alpha, a\cap b=\left\{P\right\}, M\in D

\Rightarrow MP\perp b
Acum cele doua reciproce sunt foarte importante deoarece  putem afla distanta de la un punct la altul sau distanta de la un punct la un plan.

 Reciprocele teoremei  celor trei perpendiculare

R.T.3\perp 1

Cum aplicam prima reciproca a celor trei perpendiculare

 

d\perp \alpha, d\cap\alpha=\left\{O\right\}    a\subset\alpha, O\in a, b\subset\alpha , a\cap b=\left\{P\right\},    M\in d, MP\perp b\Rightarrow a\perp b

R.T.3\perp 2

Cum aplicam Reciproca a doua a celor trei perpendiculare

d\perp a, d\cap a=\left\{O\right\}, a\subset\alpha
a\perp b, a, b\subset\alpha, a\cap b=\left\{P\right\}, M\in d,

 

MP\perp b\Rightarrow d\perp \alpha
 

Rezolvam probleme in care aplicam teorema celor trei perpendiculare

1) Pe planul triunghiului isoscel ABC  cu AB=AC=20 cm  si BC=32 cm se ridica perpendiculara AP, cu AP=12\sqrt{3} cm. Aflati:

a) distanta de la punctul P la dreapta BC

b) distanta de la  punctul A la planul  (PBC).

cum aplicam teorema celor trei perpendiculare

 

Stim ca

AP\perp\left(ABC\right)
Construim AD\perp BC, deci prin piciorul dreptei BC trece o dreapta perpendiculara pe o alta dreapta, atunci rezulta ca AD\perp BC

AP\perp\left(ABC\right)
AD\perp BC, BC\subset \left(ABC\right), AD\cap BC=\left\{P\right\}\Rightarrow AD\perp BC

Am aplicat Teorema celor trei perpendiculare si astfel am gasit ca d\left(    A, BC\right)=AD.

Acum aflam valoarea numerica a distantei

Cum Ad este inaltime, stim ca intr-un triunghi isoscel mediana, mediatoarea, bisectoarea si inaltimea coincid, deci observam  ca AD  este si mediana, astfel BD=\frac{BC}{2}\Rightarrow BD=\frac{32}{2}\Rightarrow BD=16 cm, acum aplicam teorema lui Pitagora in triunghiul ABD pentru a afla AD

AD^{2}=AB^{2}-BD^{2}\Rightarrow AD^{2}=400-256\Rightarrow AD^{2}=144\Rightarrow AD=\sqrt{144}\Rightarrow AD=12 cm.

Acum aplicam Teorema lui Pitagora in triunghiul PAD

PD^{2}=AP^{2}+AD^{2}\Rightarrow PD^{2}=\left(12\sqrt{3}\right)^{2}+12^{2}\Rightarrow PD^{2}=144\cdot 3+144\Rightarrow PD^{2}=144\left(3+1\right)\Rightarrow PD^{2}=144\cdot 4\Rightarrow PD=\sqrt{144\cdot 4}\Rightarrow PD=12\cdot 2\Rightarrow PD=24 cm.

b)d\left(A, \left(PBC\right)\right)=

CUM CALCULAM DISTANTA DE LA UN PUNCT LA UN PLAN

 

Daca AE\perp PD, PD\subset \left(PDC\right), rezulta cu cea de doua reciproca a teoremei celor trei perpendiculare ca AE\perp \left(PBC\right), deci trebuie sa aflam pe AE, cum stim ca triunghiul PAD este dreptunghic in A, aplicam teorema inaltimii

AD=\frac{c_{1}\cdot c_{2}}{ipotenuza}\Rightarrow AD=\frac{12\sqrt{3}\cdot 12}{24}\Rightarrow AD=\frac{144\sqrt{3}}{24}=6\sqrt{3}.

Deci important sa intelegem atat teorema celor trei perpendiculare, dar si reciprocele teoremei celor trei perpendiculare pentru a putea reusii in X si O
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