Fie \( VABC \) un tetraedru si \( M \) un punct variabil pe arcul mic \( AB \) de pe cercul \( \mathcal{C}\left(O,\ R\right) \) circumscris triunghiului \( ABC \). Fie \( VD,\ VE,\ VF \) distantele de la \( V \) la \( MC,\ MA,\ MB \) respectiv. Demonstrati ca \( \max\left(VD^{2}+VE^{2}+VF^{2}\right)=3\left(VO^{2}+\frac{R^{2}}{2}\right) \).
Virginia si Vasile Tica, lista scurta, 2007
Valoare maxima a sumei
Moderators: Bogdan Posa, Laurian Filip
-
Claudiu Mindrila
- Fermat
- Posts: 520
- Joined: Mon Oct 01, 2007 2:25 pm
- Location: Targoviste
- Contact:
Valoare maxima a sumei
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
-
Andi Brojbeanu
- Bernoulli
- Posts: 294
- Joined: Sun Mar 22, 2009 6:31 pm
- Location: Targoviste (Dambovita)
In enuntul original, \( VABC \) este priamida triunghiulara regulata. In urma rezolvarii, se obtine ca suma \( VD^2+VE^2+VF^2 \) este egala cu \( 3(VO^2+\frac{R^2}{2}) \).
Evident, \( VO\perp (ABC) \), deci triunghiurile \( VOD, VOE, VOF \) sunt dreptunghice si \( VD^2=VO^2+OD^2 \) si analoagele.
Atunci \( VD^2+VE^2+VF^2=3VO^2+OD^2+OE^2+OF^2=3(VO^2+\frac{R^2}{2})\Leftrightarrow OD^2+OE^2+OF^2=\frac{3}{2}R^2=\frac{3}{2}OM^2\Leftrightarrow \)
\( (\frac{OD}{OM})^2+(\frac{OE}{OM})^2+(\frac{OF}{OM})^2=\frac{3}{2}\Leftrightarrow cos^2(\angle{MOE})+cos^2(\angle{MOF})+cos^2(\angle{MOD})=\frac{3}{2} \).
Evident, triunghiurile \( AOM, AOB \) si \( AOC \) sunt isoscele. Notam cu \( x \) masura unghiului \( \angle {MOF} \).
Atunci \( m(\angle{MOB})=2x \), \( m(MOE)=\frac{m(\angle{AOM})}{2}=\frac{m(\angle{AOB)}-m(\angle{MOB})}{2}=\frac{120-2x}{2}=60-x \) si \( m(\angle{MOD})=\frac{m(\angle{MOC)}}{2}=\frac{m(\angle{BOC})+m(\angle{MOB})}{2}=\frac{120+2x}{2}=60+x \). (In functie de pozitia punctului \( M \), masurile celor trei unghiuri se pot schimba intre ele, dar vor avea masurile de \( x, 60+x \) si \( 60-x \))
In continuare, problema se poate rezolva in doua moduri:
Solutia 1 (la nivelul clasei a VII-a):
Fie \( \bigtriangleup{ABC} \) un triunghi echilateral. Luam \( M \) un punct pe \( (BC) \) astfel incat \( m(\angle{MBC})=x(x<60\textdegree) \). Avem \( m(\angle{MBA})=60-x \) si \( m(\angle{AMB})=60+x \).
Ducem \( ME\perp BC, MF\perp AB \) si \( BP\perp AC \).
Atunci, \( cos^2x+cos^2(60-x)+cos^2(60+x)=(\frac{BE}{BM})^2+(\frac{BF}{BM})^2+(\frac{PM}{BM})^2=\frac{1}{BM^2}(BE^2+BF^2+MP^2) \).
Notam cu \( a \) latura \( \bigtriangleup{ABC} \) si cu \( b \) segmentul \( [MP] \).
Atunci, \( BE=BC-EC=BC-MC\cdot cos(\angle{ACB})=BC-\frac{1}{2}\cdot (PC-PM)=a-\frac{1}{2}\cdot ({\frac{a}{2}-x)=\frac{3a}{4}+\frac{x}{2} \);
\( BF=AB-AF=AB-AM\cdot cos(\angle{BAC})=AB-\frac{1}{2}\cdot (AP+PM)=a-\frac{1}{2}\cdot (\frac{a}{2}+x)=\frac{3a}{4}-\frac{x}{2} \);
\( BM^2=BP^2+MP^2=(\frac{a\sqrt{3}}{2})^2+x^2=\frac{3a^2}{4}+x^2 \).
Atunci, \( \frac{1}{BM^2}(BE^2+BF^2+MP^2)=\frac{1}{\frac{3a^2}{4}+x^2}[(\frac{3a}{4}+\frac{x}{2})^2+(\frac{3a}{4}-\frac{x}{2})^2+x^2]=\frac{1}{\frac{3a^2}{4}+x^2}(\frac{9a^2}{16}+\frac{x^2}{4}+2\cdot \frac{3a}{4}\cdot \frac{x}{2}+\frac{9a^2}{16}+\frac{x^2}{4}-2\cdot \frac{3a}{4}\cdot \frac{x}{2}+x^2)=\frac{1}{\frac{3a^2}{4}+x^2}\cdot \frac{3}{2}\cdot (\frac{3a^2}{4}+x^2)=\frac{3}{2} \).
Solutia 2 (la nivelul clasei a IX-a):
Se cunoaste ca: \( cos^2x=\frac{1+cos2x}{2} \) si \( cos x+cos y=2\cdot cos(\frac{x+y}{2})\cdot cos(\frac{x-y}{2}) \).
Atunci, \( cos^2x+cos^2(60-x)+cos^2(60+x)=\frac{1+cos2x}{2}+\frac{1+cos 2(60-x)}{2}+\frac{1+cos 2(60+x)}{2}=\frac{3}{2}+ \)
\( +\frac{cos 2x+cos(120-2x)+cos(120+2x)}{2}=\frac{3}{2}+\frac{cos2x+2\cdot cos(\frac{120-2x+120+2x}{2})\cdot cos(\frac{(120+2x-(120-2x))}{2})}{2}=\frac{3}{2}+\frac{cos2x+2\cdot cos(120\textdegree)\cdot cos 2x}{2}=\frac{3}{2}+\frac{cos2x+2\cdot \frac{-1}{2}\cdot cos2x}{2}=\frac{3}{2}+\frac{cos2x-cos2x}{2}=\frac{3}{2} \).
\( \mathcal{q.e.d.} \)
Evident, \( VO\perp (ABC) \), deci triunghiurile \( VOD, VOE, VOF \) sunt dreptunghice si \( VD^2=VO^2+OD^2 \) si analoagele.
Atunci \( VD^2+VE^2+VF^2=3VO^2+OD^2+OE^2+OF^2=3(VO^2+\frac{R^2}{2})\Leftrightarrow OD^2+OE^2+OF^2=\frac{3}{2}R^2=\frac{3}{2}OM^2\Leftrightarrow \)
\( (\frac{OD}{OM})^2+(\frac{OE}{OM})^2+(\frac{OF}{OM})^2=\frac{3}{2}\Leftrightarrow cos^2(\angle{MOE})+cos^2(\angle{MOF})+cos^2(\angle{MOD})=\frac{3}{2} \).
Evident, triunghiurile \( AOM, AOB \) si \( AOC \) sunt isoscele. Notam cu \( x \) masura unghiului \( \angle {MOF} \).
Atunci \( m(\angle{MOB})=2x \), \( m(MOE)=\frac{m(\angle{AOM})}{2}=\frac{m(\angle{AOB)}-m(\angle{MOB})}{2}=\frac{120-2x}{2}=60-x \) si \( m(\angle{MOD})=\frac{m(\angle{MOC)}}{2}=\frac{m(\angle{BOC})+m(\angle{MOB})}{2}=\frac{120+2x}{2}=60+x \). (In functie de pozitia punctului \( M \), masurile celor trei unghiuri se pot schimba intre ele, dar vor avea masurile de \( x, 60+x \) si \( 60-x \))
In continuare, problema se poate rezolva in doua moduri:
Solutia 1 (la nivelul clasei a VII-a):
Fie \( \bigtriangleup{ABC} \) un triunghi echilateral. Luam \( M \) un punct pe \( (BC) \) astfel incat \( m(\angle{MBC})=x(x<60\textdegree) \). Avem \( m(\angle{MBA})=60-x \) si \( m(\angle{AMB})=60+x \).
Ducem \( ME\perp BC, MF\perp AB \) si \( BP\perp AC \).
Atunci, \( cos^2x+cos^2(60-x)+cos^2(60+x)=(\frac{BE}{BM})^2+(\frac{BF}{BM})^2+(\frac{PM}{BM})^2=\frac{1}{BM^2}(BE^2+BF^2+MP^2) \).
Notam cu \( a \) latura \( \bigtriangleup{ABC} \) si cu \( b \) segmentul \( [MP] \).
Atunci, \( BE=BC-EC=BC-MC\cdot cos(\angle{ACB})=BC-\frac{1}{2}\cdot (PC-PM)=a-\frac{1}{2}\cdot ({\frac{a}{2}-x)=\frac{3a}{4}+\frac{x}{2} \);
\( BF=AB-AF=AB-AM\cdot cos(\angle{BAC})=AB-\frac{1}{2}\cdot (AP+PM)=a-\frac{1}{2}\cdot (\frac{a}{2}+x)=\frac{3a}{4}-\frac{x}{2} \);
\( BM^2=BP^2+MP^2=(\frac{a\sqrt{3}}{2})^2+x^2=\frac{3a^2}{4}+x^2 \).
Atunci, \( \frac{1}{BM^2}(BE^2+BF^2+MP^2)=\frac{1}{\frac{3a^2}{4}+x^2}[(\frac{3a}{4}+\frac{x}{2})^2+(\frac{3a}{4}-\frac{x}{2})^2+x^2]=\frac{1}{\frac{3a^2}{4}+x^2}(\frac{9a^2}{16}+\frac{x^2}{4}+2\cdot \frac{3a}{4}\cdot \frac{x}{2}+\frac{9a^2}{16}+\frac{x^2}{4}-2\cdot \frac{3a}{4}\cdot \frac{x}{2}+x^2)=\frac{1}{\frac{3a^2}{4}+x^2}\cdot \frac{3}{2}\cdot (\frac{3a^2}{4}+x^2)=\frac{3}{2} \).
Solutia 2 (la nivelul clasei a IX-a):
Se cunoaste ca: \( cos^2x=\frac{1+cos2x}{2} \) si \( cos x+cos y=2\cdot cos(\frac{x+y}{2})\cdot cos(\frac{x-y}{2}) \).
Atunci, \( cos^2x+cos^2(60-x)+cos^2(60+x)=\frac{1+cos2x}{2}+\frac{1+cos 2(60-x)}{2}+\frac{1+cos 2(60+x)}{2}=\frac{3}{2}+ \)
\( +\frac{cos 2x+cos(120-2x)+cos(120+2x)}{2}=\frac{3}{2}+\frac{cos2x+2\cdot cos(\frac{120-2x+120+2x}{2})\cdot cos(\frac{(120+2x-(120-2x))}{2})}{2}=\frac{3}{2}+\frac{cos2x+2\cdot cos(120\textdegree)\cdot cos 2x}{2}=\frac{3}{2}+\frac{cos2x+2\cdot \frac{-1}{2}\cdot cos2x}{2}=\frac{3}{2}+\frac{cos2x-cos2x}{2}=\frac{3}{2} \).
\( \mathcal{q.e.d.} \)