In triunghiul \( ABC \) inaltimea \( AD(d\in BC) \) si mediana \( AA\prime(A\prime\in BC) \) impart unghiul \( BAC \) in trei parti egale.
a) Determinati masurile unghiurilor triunghiului.
b) Fie \( P\in (AA\prime) \). Daca \( BP\cap AC=\{B\prime\} \)\( ,CP\cap AB=\{C\prime\} \)\( , B\prime C\prime\cap AA\prime=\{T\} \)\( , B\prime C\prime \cap AD=\{S\} \), aratati ca \( B\prime C\prime\parallel BC. \) Daca in plus \( [ST] \) este linie mijlocie in triunghiul\( ADA\prime \), determinati valoarea raportului \( \frac{PA\prime}{AA\prime} \).
Probleme date la olimpiade, RMT 1/1998
O.VII.30
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Andi Brojbeanu
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O.VII.30
Last edited by Andi Brojbeanu on Mon Apr 20, 2009 11:36 am, edited 1 time in total.
a) in\( \triangle ABM \),\( AD \) bisectoare si inaltime\( \Longrightarrow \) triunghiul este isoscel,\( BD=DM \).
Aplicand \( T.B \) in \( \triangle ADC \),obtinem\( \frac{AD}{AC}=\frac{1}{2} \)
\( sin(\angle C)=\frac{1}{2}\Longrightarrow m(\angle C)=30. \)Obtinem apoi \( m(\angle A)=90, m(\angle B)=60 \).
-am notat M mijlocul laturii BC.
b) din \( T.CEVA \) avem ca\( \frac{BC^\prime}{C^\prime A}\cdot \frac{AB^\prime}{B^\prime C}\cdot \frac{CM}{MB}=1 \)
dar \( CM=MB \) \( \Longrightarrow \) \( \frac{BC^\prime}{C^\prime A}\cdot \frac{AB^\prime}{B^\prime C}=1 \),adica \( \frac{AC^\prime}{C^\prime B}=\frac{AB^\prime}{B^\prime C} \) si din \( T.Th \) \( \Longrightarrow \) \( B^\prime C^\prime \parallel BC \).
-din \( TB^\prime \parallel BC\Longrightarrow \)\( m(\angle TB^\prime A)=30,m(\angle B^\prime AT)=30\Longrightarrow m(\angle ATB^\prime)=120,AT=TB^\prime=TM\Longrightarrow TB^\prime=TM, m(\angle B^\prime TM)=180-m(\angle ATB^\prime)=180-120=60\Longrightarrow \triangle TBM \) echilateral.
\( \triangle B^\prime TP\sim \triangle BMP\Longrightarrow\frac{B^\prime T}{BM}=\frac{TP}{MP}
\)
\( B^\prime T=\frac{AM}{2} \), \( BM=\frac{BC}{2} \), \( AM=\frac{BC}{2}\Longrightarrow B^\prime T=\frac{BM}{2} \)
\( \frac{TP}{MP}=\frac{1}{2}\Longrightarrow TP=\frac{MP}{2} \).
\( AM=2(TP+PM)=2(\frac{PM}{2}+PM)=PM+2PM=3PM\Longrightarrow \frac{PM}{AM}=\frac{1}{3} \).
Aplicand \( T.B \) in \( \triangle ADC \),obtinem\( \frac{AD}{AC}=\frac{1}{2} \)
\( sin(\angle C)=\frac{1}{2}\Longrightarrow m(\angle C)=30. \)Obtinem apoi \( m(\angle A)=90, m(\angle B)=60 \).
-am notat M mijlocul laturii BC.
b) din \( T.CEVA \) avem ca\( \frac{BC^\prime}{C^\prime A}\cdot \frac{AB^\prime}{B^\prime C}\cdot \frac{CM}{MB}=1 \)
dar \( CM=MB \) \( \Longrightarrow \) \( \frac{BC^\prime}{C^\prime A}\cdot \frac{AB^\prime}{B^\prime C}=1 \),adica \( \frac{AC^\prime}{C^\prime B}=\frac{AB^\prime}{B^\prime C} \) si din \( T.Th \) \( \Longrightarrow \) \( B^\prime C^\prime \parallel BC \).
-din \( TB^\prime \parallel BC\Longrightarrow \)\( m(\angle TB^\prime A)=30,m(\angle B^\prime AT)=30\Longrightarrow m(\angle ATB^\prime)=120,AT=TB^\prime=TM\Longrightarrow TB^\prime=TM, m(\angle B^\prime TM)=180-m(\angle ATB^\prime)=180-120=60\Longrightarrow \triangle TBM \) echilateral.
\( \triangle B^\prime TP\sim \triangle BMP\Longrightarrow\frac{B^\prime T}{BM}=\frac{TP}{MP}
\)
\( B^\prime T=\frac{AM}{2} \), \( BM=\frac{BC}{2} \), \( AM=\frac{BC}{2}\Longrightarrow B^\prime T=\frac{BM}{2} \)
\( \frac{TP}{MP}=\frac{1}{2}\Longrightarrow TP=\frac{MP}{2} \).
\( AM=2(TP+PM)=2(\frac{PM}{2}+PM)=PM+2PM=3PM\Longrightarrow \frac{PM}{AM}=\frac{1}{3} \).