Geometrie

[alex2008]

Fie dreptunghiul \( ABCD \) in care \( AB=BC\sqrt{2} \) si punctul \( M \) care apartine semicercului de diametru \( AB \), situat in acelasi semiplan cu dreapta \( CD \). Notam \( F\in DM\cap AB \) si \( G\in CM\cap AB \). Sa se arate relatia \( AG^2+BF^2=AB^2 \).

Geometrie Plana (sintetica, vectoriala, analitica), Virgil Nicula

[Virgil Nicula]

OFF-topic. Aceasta frumoasa problema apartine lui Fermat (1601-1665), unic cu acest nume in matematica ! Deci are o vechime de aprox. patru secole (prin adaos !) si a aparut in numeroase culegeri de probleme si reviste, mai vechi sau mai recente.

ON-topic. Va recomand sa folositi geometria analitica (dupa parerea mea, cea mai scurta demonstratie !) alegand dreapta \( AB \) ca axa \( Ox \) si mediatoarea laturii \( [AB] \) ca axa \( Oy \) astfel incat \( B(1,0) \) si \( C(1,\sqrt 2)\ . \)

In solutia mea metrica am ajuns in final la identitatea (pentru \( a=b=1 \) si \( c=-\sqrt 2 \) ) , unde \( \phi=m\ (\angle BOM) \) :

\( \underline{\overline{\left\|\ (a+b\cdot\cos\phi +c\cdot\sin\phi )^2+(a-b\cdot\cos\phi +c\cdot\sin\phi )^2=2(c+a\cdot\sin\phi )^2+2\left(a^2+b^2-c^2\right)\cdot\cos^2\phi\ \right\|}} \) .

[Virgil Nicula]

Fie dreptunghiul \( ABCD \) in care \( AB=BC\sqrt{2} \) si punctul \( M \) care apartine semicercului de diametru \( AB \),

situat in acelasi semiplan cu dreapta \( CD \). Notam \( F\in DM\cap AB \) si \( G\in CM\cap AB \).

Sa se arate relatia \( AG^2+BF^2=AB^2 \) (Geometrie Plana - sintetica, vectoriala, analitica, Virgil Nicula)

Dem. Presupunem fara a restrange generalitatea ca \( AB=CD=2 \) si \( AD=BC=\sqrt 2\ . \) Notam

\( \left\|\ \begin{array}{c}
m(\widehat {BOM})=2x\\\\\\\\
m(\widehat {BCM})=u\\\\\\\\
m(\widehat {ADM})=v\end{array}\ \right\|\ . \) Se observa ca \( \left\|\ \begin{array}{c}
MA=2\cos x\\\\\\\\
MB=2\sin x\end{array}\ \right\| \) si aplicam teorema sinusurilor in triunghiurile :

\( \odot\ \ \triangle\ ADM\ :\ \frac {MA}{\sin v}=\frac {AD}{\cos (x-v)}\ \Longrightarrow\ \frac {2\cos x}{\sin v}=\frac {\sqrt 2}{\cos (x-v)}\ \Longrightarrow\ \sqrt 2\cos x=\frac {\tan v}{\cos x+\sin x\tan v}\ \Longrightarrow \)

\( \frac {AF}{AD}=\tan v=\frac {\sqrt 2\cos^2x}{1-\sqrt 2\cos x\sin x}\ \Longrightarrow\ AF=\frac {2\cos^2x}{1-\sqrt 2\cos x\sin x}\ \Longrightarrow\ \overline{\underline{\left\|\ FB=\frac {2\sin^2x-\sqrt 2\sin 2x}{1-\sqrt 2\sin x\cos x}\ \right\|}} \) .

\( \odot\ \ \triangle\ BCM\ :\ \frac {MB}{\sin u}=\frac {BC}{\sin (x+u)}\ \Longrightarrow\ \frac {2\sin x}{\sin u}=\frac {\sqrt 2}{\sin (x+u)}\ \Longrightarrow\ \sqrt 2\sin x=\frac {\tan u}{\sin x+\cos x\tan u}\ \Longrightarrow \)

\( \frac {BG}{BC}=\tan u=\frac {\sqrt 2\sin^2x}{1-\sqrt 2\cos x\sin x}\ \Longrightarrow\ BG=\frac {2\sin^2x}{1-\sqrt 2\cos x\sin x}\ \Longrightarrow\ \overline{\underline{\left\|\ GA=\frac {\sqrt 2\sin 2x-2\cos^2x}{1-\sqrt 2\sin x\cos x}\ \right\|}} \) .

Deci \( \overline{\underline{\left\|\ FB^2+GA^2=AB^2\ \right\|}}\ \Longleftrightarrow\ \left(2\sin^2x-\sqrt 2\sin 2x\right)^2+\left(2\cos^2x-\sqrt 2\sin 2x\right)^2=4\left(1-\sqrt 2\sin x\cos x\right)^2 \)

adica \( \left(1-\cos 2x-\sqrt 2\sin 2x\right)^2+\left(1+\cos 2x-\sqrt 2\sin 2x\right)^2=\left(2-\sqrt 2\sin 2x\right)^2\ . \) Notam \( 2x=y\ . \) Prin urmare,

\( \left(1-\cos y-\sqrt 2\sin y\right)^2+\left(1+\cos y-\sqrt 2\sin y\right)^2=\left(2-\sqrt 2\sin y\right)^2\ \Longleftrightarrow \)

\( 2+2\cos^2y+4\sin^2y-4\sqrt 2\sin y=4-4\sqrt 2\sin y+2\sin^2y\ \Longleftrightarrow\ \underline{\overline{\left\|\ \cos^2y+\sin^2y=1\ \right\|}} \) , evident.