Fie prisma triunghiulara regulata ABCA'B'C' si \( P\in(AC) \) oarecare. Fie \( Q\in(AP) \) astfel incat \( m\angle{[(BPB^{\prime}),(BQB^{\prime})]}=m\angle{[(BQB^{\prime}),(BAB^{\prime})]} \). Notam \( \{T\}=(BQB^{\prime})\cap [(AB^{\prime}C^{\prime})\cap (ABC)] \) .
Demonstrati ca \( S_{\triangle BPB^{\prime}}=S_{\triangle CPC^{\prime}}+S_{\triangle ATA^{\prime}} \).
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Relatie cu arii (in spatiu) - relatie cu segmente (in plan)
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Marius Mainea
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Din ipoteza, deducem ca planul \( (BQB^{\prime}) \) este planul bisector al unghiului diedru format de planele \( (BPB^{\prime}) \) si \( (BAB^{\prime}) \), deci \( (AQ \) este bisectoarea unghiului \( \widehat{ABP} \).
Deoarece \( \{T\}=(BQB^{\prime})\cap [(AB^{\prime}C^{\prime})\cap (ABC)] \), rezulta ca \( T \) este punctul in care dreapta de intersectie a planelor \( (AB^{\prime}C^{\prime}) \) si \( (ABC) \) intersecteaza planul \( (BQB^{\prime}) \).
Dar, cum planele \( (AB^{\prime}C^{\prime}) \) si \( (ABC) \) contin dreptele paralele \( B^{\prime}C^{\prime} \) si \( BC \), dreapta de intersectie a celor doua plane va fi paralela dusa prin \( A \) la \( BC \). Punctul \( T \) va fi punctul de intersectie al dreptei \( BQ \) cu aceasta paralela.
Deoarece \( ABCA^{\prime}B^{\prime}C^{\prime} \) este prisma triunghiulara regulata, rezulta ca \( m(\angle{ACB})=60\textdegree \), \( AA^{\prime}=BB^{\prime}=CC^{\prime} \) si \( AA^{\prime}, BB^{\prime}, CC^{\prime}\perp (ABC) \), deci \( AA^{\prime}\perp AT, BB^{\prime}\perp BP, CC^{\prime}\perp CP \).
Atunci, \( S_{\triangle BPB^{\prime}}=S_{\triangle CPC^{\prime}}+S_{\triangle ATA^{\prime}}\Leftrightarrow \frac{BB^{\prime}\cdot BP}{2}=\frac{AA^{\prime}\cdot AT}{2}+\frac{CC^{\prime}\cdot PC}{2}\Leftrightarrow BP=AT+PC \).
Rezolvarea problemei se reduce la demonstrarea unei relatii cu segmente in plan.
Facem urmatoarele notatii: \( PC=x, AB=AC=BC=l, BP=a \).
Ne propunem sa il scriem si pe \( AT \) in functie de \( a, x \) si \( l \).
Din teorema bisectoarei in \( \bigtriangleup{ABP} \), avem:\( \frac{AQ}{PQ}=\frac{AP}{BP}\Rightarrow AQ=\frac{l(l-x-AQ)}{a}\Rightarrow AQ=\frac{l(l-x)}{a}-\frac{l\cdot AQ}{a}\Rightarrow AQ(1+\frac{l}{a})=\frac{l(l-x)}{a}\Rightarrow AQ=\frac{l(l-x)}{a}\cdot \frac{a}{l+a}=\frac{l(l-x)}{l+a} \).
\( AT\parallel BC\Rightarrow \bigtriangleup{ATQ}\sim \bigtriangleup{CBQ}\Rightarrow \frac{AT}{BC}=\frac{AQ}{QC}\Rightarrow AT=\frac{BC\cdot AQ}{AC-AQ}=\frac{l\cdot\frac{l(l-x)}{l+a}}{l-\frac{l(l-x)}{l+a}}=\frac{l^2(l-x)}{l(l+a)-l(l-x)}=\frac{l^2(l-x)}{l^2+al-l^2+xl}=\frac{l^2(l-x)}{al+xl}=\frac{l^2(l-x)}{l(x+a)}=\frac{l(l-x)}{x+a} \).
Atunci relatia \( BP=AT+PC \) se mai scrie:\( a=\frac{l(l-x)}{x+a}+x\Leftrightarrow a(x+a)=l(l-x)+x(x+a)\Leftrightarrow ax+a^2=l^2-xl+x^2+ax\Leftrightarrow a^2=l^2+x^2-lx\Leftrightarrow a=\sqrt{x^2+l^2-xl} \).
Dar, teorema cosinusului in \( \bigtriangleup{BPC} \), obtinem ca: \( BP=a=\sqrt{PC^2+BC^2-2PC\cdot BC\cdot cos(\angle{ACB})}=\sqrt{x^2+l^2-2\cdot x\cdot l\cdot \frac{1}{2}}=\sqrt{x^2+l^2-xl} \).
\( \mathcal{q.e.d.} \)
Deoarece \( \{T\}=(BQB^{\prime})\cap [(AB^{\prime}C^{\prime})\cap (ABC)] \), rezulta ca \( T \) este punctul in care dreapta de intersectie a planelor \( (AB^{\prime}C^{\prime}) \) si \( (ABC) \) intersecteaza planul \( (BQB^{\prime}) \).
Dar, cum planele \( (AB^{\prime}C^{\prime}) \) si \( (ABC) \) contin dreptele paralele \( B^{\prime}C^{\prime} \) si \( BC \), dreapta de intersectie a celor doua plane va fi paralela dusa prin \( A \) la \( BC \). Punctul \( T \) va fi punctul de intersectie al dreptei \( BQ \) cu aceasta paralela.
Deoarece \( ABCA^{\prime}B^{\prime}C^{\prime} \) este prisma triunghiulara regulata, rezulta ca \( m(\angle{ACB})=60\textdegree \), \( AA^{\prime}=BB^{\prime}=CC^{\prime} \) si \( AA^{\prime}, BB^{\prime}, CC^{\prime}\perp (ABC) \), deci \( AA^{\prime}\perp AT, BB^{\prime}\perp BP, CC^{\prime}\perp CP \).
Atunci, \( S_{\triangle BPB^{\prime}}=S_{\triangle CPC^{\prime}}+S_{\triangle ATA^{\prime}}\Leftrightarrow \frac{BB^{\prime}\cdot BP}{2}=\frac{AA^{\prime}\cdot AT}{2}+\frac{CC^{\prime}\cdot PC}{2}\Leftrightarrow BP=AT+PC \).
Rezolvarea problemei se reduce la demonstrarea unei relatii cu segmente in plan.
Facem urmatoarele notatii: \( PC=x, AB=AC=BC=l, BP=a \).
Ne propunem sa il scriem si pe \( AT \) in functie de \( a, x \) si \( l \).
Din teorema bisectoarei in \( \bigtriangleup{ABP} \), avem:\( \frac{AQ}{PQ}=\frac{AP}{BP}\Rightarrow AQ=\frac{l(l-x-AQ)}{a}\Rightarrow AQ=\frac{l(l-x)}{a}-\frac{l\cdot AQ}{a}\Rightarrow AQ(1+\frac{l}{a})=\frac{l(l-x)}{a}\Rightarrow AQ=\frac{l(l-x)}{a}\cdot \frac{a}{l+a}=\frac{l(l-x)}{l+a} \).
\( AT\parallel BC\Rightarrow \bigtriangleup{ATQ}\sim \bigtriangleup{CBQ}\Rightarrow \frac{AT}{BC}=\frac{AQ}{QC}\Rightarrow AT=\frac{BC\cdot AQ}{AC-AQ}=\frac{l\cdot\frac{l(l-x)}{l+a}}{l-\frac{l(l-x)}{l+a}}=\frac{l^2(l-x)}{l(l+a)-l(l-x)}=\frac{l^2(l-x)}{l^2+al-l^2+xl}=\frac{l^2(l-x)}{al+xl}=\frac{l^2(l-x)}{l(x+a)}=\frac{l(l-x)}{x+a} \).
Atunci relatia \( BP=AT+PC \) se mai scrie:\( a=\frac{l(l-x)}{x+a}+x\Leftrightarrow a(x+a)=l(l-x)+x(x+a)\Leftrightarrow ax+a^2=l^2-xl+x^2+ax\Leftrightarrow a^2=l^2+x^2-lx\Leftrightarrow a=\sqrt{x^2+l^2-xl} \).
Dar, teorema cosinusului in \( \bigtriangleup{BPC} \), obtinem ca: \( BP=a=\sqrt{PC^2+BC^2-2PC\cdot BC\cdot cos(\angle{ACB})}=\sqrt{x^2+l^2-2\cdot x\cdot l\cdot \frac{1}{2}}=\sqrt{x^2+l^2-xl} \).
\( \mathcal{q.e.d.} \)