\( 1. \) Pe segmentul \( (AB) \) se considera un punct \( P \). De aceeasi parte a dreptei \( AB \) se construiesc triunghiurile \( \Delta APC \) si \( \Delta APD \) astfel incat \( [PA] \equiv [PC] \), \( [PB] \equiv [PD] \) si \( m(\angle APC)=m(\angle BPD)=a \). Dreptele \( AD \) si \( BC \) se intersecteaza in \( Q \). Sa se determine \( m(\angle AQC) \) in functie de \( a \).
Maria Mihet, R.M.T. 2/2008
\( 2. \) Se considera triunghiul \( \Delta ABC \), dreptunghic in \( A \). Sa se arate ca unul din unghiurile \( B \) si \( C \) este egal cu \( 30^ \circ \) daca si numai daca \( AI=MI \) unde \( I \) este centrul cercului inscris triunghiului, iar \( M \) mijlocul ipotenuzei.
Mihail Mogosanu, R.M.T. 2/2008
Doua probleme interesante de geometrie
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Claudiu Mindrila
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Doua probleme interesante de geometrie
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Andi Brojbeanu
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Problema 1. Sunt 3 posibilitati:
1) \( a<90\textdegree. \) \( m(\widehat{APD})=m(\widehat{BPC})(=a+m(\widehat{CPD})) \), deci \( \bigtriangleup {APD}\equiv\bigtriangleup{CPB}(L.U.L) \). Obtinem ca \( m(\widehat{BCP})=m(\widehat{DAP}) \). Deci, \( m(\widehat{ACQ})=m(\widehat{QAB})+m(\widehat{QBA)}=m(\widehat{BCP})+m(\widehat{CBP})=m(\widehat{CPA})=a. \)
2) \( a=90\textdegree. \) Deoarece Q este punctul unde dreapta \( BC \) taie dreapta \( AD \), inseamna ca \( m(\widehat{BCQ})=180\textdegree \). Atunci \( m(\widehat{ACQ})=m(\widehat{BCQ})-m(\widehat{ACP})-m(\widehat{BCP})=180\textdegree-45\textdegree-45\textdegree=90\textdegree \).
3) \( a>90\textdegree. \) Insa \( m(\widehat{APD})=m(\widehat{CPB})(=a-m(\widehat{CPD})) \), rezulta \( \bigtriangleup{APD}\equiv\bigtriangleup{CPB}(L.U.L) \). Analog cazului anterior, \( m(\widehat{BCP})=m(\widehat{DAP}) \).
\( m(\widehat{ACQ})=180\textdegree-m(\widehat{QAB})-m(\widehat{QBA})=180\textdegree=m(\widehat{BCP})-m(\widehat{CPB})=m(\widehat{CPB})=180\textdegree-a \).
Problema 2. "\( \Rightarrow \)" Daca \( m(\widehat{B})=30\textdegree \), rezulta \( AC=\frac{BC}{2}=CM \). Deci, \( \bigtriangleup{AIC}\equiv\bigtriangleup{MIC}(L.U.L) \) rezulta \( AI=MI \).
"\( \Leftarrow \)" Luam \( D \) si \( E \) proiectiile lui\( I \) pe [AC], respectiv \( BC \). Evident, \( ID=IE \), deci \( \bigtriangleup{AID}\equiv \bigtriangleup{MIE}(L.U.L) \).
Daca \( AB\leq AC \), avem \( \widehat{IMP}\equiv\widehat{IMB}\equiv\widehat{IAN}\equiv\widehat{CAI}\equiv\widehat{BAI} \). Rezulta \( \bigtriangleup{AIB}\equiv\bigtriangleup{MIB}(U.L.U) \), adica \( AB=BM=\frac{BC}{2} \), de unde \( m(\widehat{C})=30\textdegree \).
Analog, pentru \( AB>AC \), \( AC=MC=\frac{BC}{2} \), rezulta \( m(\widehat{B})=30\textdegree. \)
1) \( a<90\textdegree. \) \( m(\widehat{APD})=m(\widehat{BPC})(=a+m(\widehat{CPD})) \), deci \( \bigtriangleup {APD}\equiv\bigtriangleup{CPB}(L.U.L) \). Obtinem ca \( m(\widehat{BCP})=m(\widehat{DAP}) \). Deci, \( m(\widehat{ACQ})=m(\widehat{QAB})+m(\widehat{QBA)}=m(\widehat{BCP})+m(\widehat{CBP})=m(\widehat{CPA})=a. \)
2) \( a=90\textdegree. \) Deoarece Q este punctul unde dreapta \( BC \) taie dreapta \( AD \), inseamna ca \( m(\widehat{BCQ})=180\textdegree \). Atunci \( m(\widehat{ACQ})=m(\widehat{BCQ})-m(\widehat{ACP})-m(\widehat{BCP})=180\textdegree-45\textdegree-45\textdegree=90\textdegree \).
3) \( a>90\textdegree. \) Insa \( m(\widehat{APD})=m(\widehat{CPB})(=a-m(\widehat{CPD})) \), rezulta \( \bigtriangleup{APD}\equiv\bigtriangleup{CPB}(L.U.L) \). Analog cazului anterior, \( m(\widehat{BCP})=m(\widehat{DAP}) \).
\( m(\widehat{ACQ})=180\textdegree-m(\widehat{QAB})-m(\widehat{QBA})=180\textdegree=m(\widehat{BCP})-m(\widehat{CPB})=m(\widehat{CPB})=180\textdegree-a \).
Problema 2. "\( \Rightarrow \)" Daca \( m(\widehat{B})=30\textdegree \), rezulta \( AC=\frac{BC}{2}=CM \). Deci, \( \bigtriangleup{AIC}\equiv\bigtriangleup{MIC}(L.U.L) \) rezulta \( AI=MI \).
"\( \Leftarrow \)" Luam \( D \) si \( E \) proiectiile lui\( I \) pe [AC], respectiv \( BC \). Evident, \( ID=IE \), deci \( \bigtriangleup{AID}\equiv \bigtriangleup{MIE}(L.U.L) \).
Daca \( AB\leq AC \), avem \( \widehat{IMP}\equiv\widehat{IMB}\equiv\widehat{IAN}\equiv\widehat{CAI}\equiv\widehat{BAI} \). Rezulta \( \bigtriangleup{AIB}\equiv\bigtriangleup{MIB}(U.L.U) \), adica \( AB=BM=\frac{BC}{2} \), de unde \( m(\widehat{C})=30\textdegree \).
Analog, pentru \( AB>AC \), \( AC=MC=\frac{BC}{2} \), rezulta \( m(\widehat{B})=30\textdegree. \)