Fie patru puncte necoplanare \( A, B, C, D \) astfel incat \( DA\perp (ABC) \), \( m(\angle{BAC})=120\textdegree \) si punctul \( M\in (BC) \) cu proprietatea ca \( 2BM=MC \).
Demonstrati ca \( DM\perp AC\Leftrightarrow AB=AC \).
Constantin Barascu.
Concursul Nicolae Paun editia 2009 subiectul III
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Andi Brojbeanu
- Bernoulli
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Andi Brojbeanu
- Bernoulli
- Posts: 294
- Joined: Sun Mar 22, 2009 6:31 pm
- Location: Targoviste (Dambovita)
\( "\Leftarrow" \):\( AB=AC\Rightarrow m(\angle{ABC})=30\textdegree \).
Daca \( BM=a \), atunci \( AB=AC=a\sqrt{3} \) si \( AM=\sqrt{AB^2+BM^2-2AB\cdot BM\cdot cos(30\textdegree)}=\sqrt{a^2+3a^2-a\cdot a\sqrt{3}\cdot\sqrt{3}}=\sqrt{a^2+3a^2-3a^2}=\sqrt{a^2}=a \).
In \( \bigtriangleup{AMC} \), avem \( MC^2=AM^2+AC^2 \)(\( 4a^2=a^2+3a^2 \)), deci triunghiul este dreptunghic, adica \( MA\perp AC\Rightarrow AC\perp MA \).
Din \( DA\perp (ABC)\Rightarrow DA\perp AC\Rightarrow AC\perp DA \).
Din \( AC\perp DA \) si \( AC\perp MA \Rightarrow AC\perp (DAM)\Rightarrow AC\perp DM\Rightarrow DM\perp AC \).
\( "\Rightarrow" \): \( DM\perp AC\Rightarrow AC\perp DM \).
Din \( DA\perp (BAC)\Rightarrow DA\perp AC\Rightarrow AC\perp DA \).
Din \( AC\perp DM \) si \( AC\perp DA\Rightarrow AC\perp (DAM)\Rightarrow AC\perp MA \).
Fie \( N \) mijlocul lui \( [MC] \).Notam \( BM=MN=NC=a, AB=c, AC=b \). Evident, \( AN=\frac{MC}{2}=a \).
In \( \bigtriangleup{BAN} \), din teorema medianei obtinem ca:\( AM^2=\frac{2(AB^2+AN^2)-BN^2}{4}=\frac{2(c^2+a^2)-(2a)^2}{4}=\frac{2c^2+2a^2-4a^2}{4}=\frac{2c^2-2a^2}{4}=\frac{c^2-a^2}{2} \).
Din teorema lui Pitagora in \( \bigtriangleup{MAC} \), obtinem ca \( AC^2=MC^2-AM^2=4a^2-\frac{c^2-a^2}{2}=\frac{8a^2-c^2+a^2}{2}=\frac{9a^2-c^2}{2} \).
Dar \( AC^2=b^2 \). Rezulta ca \( \frac{9a^2-c^2}{2}=b^2 \), deci \( 9a^2=2b^2+c^2 \).
Din teorema cosinuslui in \( \bigtriangleup{ABC} \) avem ca \( BC^2=AB^2+AC^2-2AB\cdot AC\cdot cos(\angle{BAC}) \), deci \( 9a^2=b^2+c^2-2\cdot b\cdot c\cdot (-\frac{1}{2})=b^2+c^2+bc \).
Prin egalarea ultimelor doua relatii obtinute, obtinem ca \( 2b^2+c^2=b^2+c^2+bc \), adica \( b^2=bc \) sau \( b=c \). Deci, \( AB=AC \).
Daca \( BM=a \), atunci \( AB=AC=a\sqrt{3} \) si \( AM=\sqrt{AB^2+BM^2-2AB\cdot BM\cdot cos(30\textdegree)}=\sqrt{a^2+3a^2-a\cdot a\sqrt{3}\cdot\sqrt{3}}=\sqrt{a^2+3a^2-3a^2}=\sqrt{a^2}=a \).
In \( \bigtriangleup{AMC} \), avem \( MC^2=AM^2+AC^2 \)(\( 4a^2=a^2+3a^2 \)), deci triunghiul este dreptunghic, adica \( MA\perp AC\Rightarrow AC\perp MA \).
Din \( DA\perp (ABC)\Rightarrow DA\perp AC\Rightarrow AC\perp DA \).
Din \( AC\perp DA \) si \( AC\perp MA \Rightarrow AC\perp (DAM)\Rightarrow AC\perp DM\Rightarrow DM\perp AC \).
\( "\Rightarrow" \): \( DM\perp AC\Rightarrow AC\perp DM \).
Din \( DA\perp (BAC)\Rightarrow DA\perp AC\Rightarrow AC\perp DA \).
Din \( AC\perp DM \) si \( AC\perp DA\Rightarrow AC\perp (DAM)\Rightarrow AC\perp MA \).
Fie \( N \) mijlocul lui \( [MC] \).Notam \( BM=MN=NC=a, AB=c, AC=b \). Evident, \( AN=\frac{MC}{2}=a \).
In \( \bigtriangleup{BAN} \), din teorema medianei obtinem ca:\( AM^2=\frac{2(AB^2+AN^2)-BN^2}{4}=\frac{2(c^2+a^2)-(2a)^2}{4}=\frac{2c^2+2a^2-4a^2}{4}=\frac{2c^2-2a^2}{4}=\frac{c^2-a^2}{2} \).
Din teorema lui Pitagora in \( \bigtriangleup{MAC} \), obtinem ca \( AC^2=MC^2-AM^2=4a^2-\frac{c^2-a^2}{2}=\frac{8a^2-c^2+a^2}{2}=\frac{9a^2-c^2}{2} \).
Dar \( AC^2=b^2 \). Rezulta ca \( \frac{9a^2-c^2}{2}=b^2 \), deci \( 9a^2=2b^2+c^2 \).
Din teorema cosinuslui in \( \bigtriangleup{ABC} \) avem ca \( BC^2=AB^2+AC^2-2AB\cdot AC\cdot cos(\angle{BAC}) \), deci \( 9a^2=b^2+c^2-2\cdot b\cdot c\cdot (-\frac{1}{2})=b^2+c^2+bc \).
Prin egalarea ultimelor doua relatii obtinute, obtinem ca \( 2b^2+c^2=b^2+c^2+bc \), adica \( b^2=bc \) sau \( b=c \). Deci, \( AB=AC \).