Fie \( ABCD \) un patrat cu \( AB=a, \{O\}=AC \cap BD \) si \( M,N,P,Q,R,S,T \) mijloacele laturilor \( AD,OB,OC,OD,OA,DC,BC \). Daca \( \{E\}=QP \cap SR \), \( \{F\}=MN \cap SR \), \( \{G\}=MN \cap TR \), \( \{H\}=PN \cap TR \), [/tex]\( \{I\}=QP \cap SO \), \( \{K\}=HO \cap SR \), \( \{L\}=BO \cap TR \), atunci aratati ca:
a) \( HO \perp RS \)
b) \( IL \perp RT \)
c) Aria lui \( EFGHP \) este \( \frac{3a^2}{20} \)
Jianu Ovidiu, G.M. Constanta 2001
Problema gazeta constanta 2001
Moderators: Laurian Filip, Filip Chindea, maky, Cosmin Pohoata, Virgil Nicula
a) Fie \( OX\perp NP \Rightarrow OX=\frac{a}{4} \) (se arata imediat), fie \( RN\cap SO=\{V\} \) si fie \( RN\cap BC=\{W\}\Rightarrow \Delta RNH\sim \Delta RWT \) \( \Rightarrow \frac{RN}{RW}=\frac{NH}{WT}\Rightarrow NH=\frac{RN\cdot WT}{RW}= \)\( \frac{\frac{a}{2}\cdot \frac{a}{4}}{\frac{3a}{4}}=\frac{a}{6} \)
\( NX=NH+HX\Rightarrow HX=NX-NH=\frac{a}{4}-\frac{a}{6}=\frac{a}{12} \)
In \( \Delta HOX , \widehat{HXO}=90^{o} \Rightarrow HO^2=HX^2+OX^2 \Rightarrow HO=\frac{a\sqrt{10}}{12} \)
In \( \Delta RVS , \widehat{RVS}=90^{o} \Rightarrow \) prin Pitagora \( RS=\frac{a\sqrt{10}}{4} \) .
\( \Rightarrow \frac{HX}{RV}=\frac{OX}{OV}=\frac{OH}{RS} \Leftrightarrow \) \( \frac{\frac{a}{12}}{\frac{a}{4}}=\frac{\frac{a}{4}}{\frac{3a}{4}}= \)\( \frac{\frac{a\sqrt{10}}{12}}{\frac{a\sqrt{10}}{4}}=\frac{1}{3}\Rightarrow \) \( \Delta RVS\sim \Delta HOX \Rightarrow m(\widehat{OSK})=m(\widehat{HOX}) \), dar \(
OS\parallel HX \Rightarrow \widehat{KOS}=\widehat{OHX} \) (corespondente) \( \Rightarrow \) \( \Delta HOX \sim \Delta KOS \Rightarrow m(\widehat{OKS})=m(\widehat{OHX})=90^{o} \)
\( NX=NH+HX\Rightarrow HX=NX-NH=\frac{a}{4}-\frac{a}{6}=\frac{a}{12} \)
In \( \Delta HOX , \widehat{HXO}=90^{o} \Rightarrow HO^2=HX^2+OX^2 \Rightarrow HO=\frac{a\sqrt{10}}{12} \)
In \( \Delta RVS , \widehat{RVS}=90^{o} \Rightarrow \) prin Pitagora \( RS=\frac{a\sqrt{10}}{4} \) .
\( \Rightarrow \frac{HX}{RV}=\frac{OX}{OV}=\frac{OH}{RS} \Leftrightarrow \) \( \frac{\frac{a}{12}}{\frac{a}{4}}=\frac{\frac{a}{4}}{\frac{3a}{4}}= \)\( \frac{\frac{a\sqrt{10}}{12}}{\frac{a\sqrt{10}}{4}}=\frac{1}{3}\Rightarrow \) \( \Delta RVS\sim \Delta HOX \Rightarrow m(\widehat{OSK})=m(\widehat{HOX}) \), dar \(
OS\parallel HX \Rightarrow \widehat{KOS}=\widehat{OHX} \) (corespondente) \( \Rightarrow \) \( \Delta HOX \sim \Delta KOS \Rightarrow m(\widehat{OKS})=m(\widehat{OHX})=90^{o} \)
. A snake that slithers on the ground can only dream of flying through the air.
b) Se vede ca \( L \) este mijlocul lui \( RT \).
Fie \( IY\perp BC \Rightarrow \) in \( \Delta IYT \) prin Pitagora \( IT=\frac{a\sqrt{5}}{4} \)
\( RQ\perp IQ, \Rightarrow \) in \( \Delta IQR \) prin Pitagora \( IR=\frac{a\sqrt{5}}{4} \)
Deci \( \Delta IRT \) este isoscel, dar intr-un triunghi isoscel mediana care pleca din varf este si inaltime \( \Rightarrow IL\perp RT \).
Fie \( IY\perp BC \Rightarrow \) in \( \Delta IYT \) prin Pitagora \( IT=\frac{a\sqrt{5}}{4} \)
\( RQ\perp IQ, \Rightarrow \) in \( \Delta IQR \) prin Pitagora \( IR=\frac{a\sqrt{5}}{4} \)
Deci \( \Delta IRT \) este isoscel, dar intr-un triunghi isoscel mediana care pleca din varf este si inaltime \( \Rightarrow IL\perp RT \).
. A snake that slithers on the ground can only dream of flying through the air.
c) Fie \( RQ\cap MN=\{J\} \) , \( RQ\cap CD=\{U\} \) si \( UZ\perp RS \)
\( \Delta RJN \equiv \Delta RNH \Rightarrow RJ=\frac{a}{6} \)
In \( \Delta RUS , UZ=\frac{RU\cdot US}{RS}=\frac{3a\sqrt{10}}{40} \) . Se arata imediat ca \( FJ \parallel UZ \) deci \( \Delta RFJ\sim \Delta RUZ\Rightarrow \frac{RF}{RZ}=\frac{RJ}{RU} \) . Mai intai prin Pitagora \( RZ=\frac{9a\sqrt{10}}{40}\Rightarrow RF=\frac{a\sqrt{10}}{20}\Rightarrow \)
\( FN=\frac{3a\sqrt{10}}{20}\Rightarrow A_{RFN}=\frac{15a^2}{400} \)
\( A_{EFGHP}=A_{RNPQ}-A_{RQE}-A_{RFN}-A_{GHN}=\frac{a^2}{4}= \)\( \frac{a^2}{24}-\frac{15a^2}{400}-\frac{a^2}{48}=\frac{3a^2}{20} \)
\( \Delta RJN \equiv \Delta RNH \Rightarrow RJ=\frac{a}{6} \)
In \( \Delta RUS , UZ=\frac{RU\cdot US}{RS}=\frac{3a\sqrt{10}}{40} \) . Se arata imediat ca \( FJ \parallel UZ \) deci \( \Delta RFJ\sim \Delta RUZ\Rightarrow \frac{RF}{RZ}=\frac{RJ}{RU} \) . Mai intai prin Pitagora \( RZ=\frac{9a\sqrt{10}}{40}\Rightarrow RF=\frac{a\sqrt{10}}{20}\Rightarrow \)
\( FN=\frac{3a\sqrt{10}}{20}\Rightarrow A_{RFN}=\frac{15a^2}{400} \)
\( A_{EFGHP}=A_{RNPQ}-A_{RQE}-A_{RFN}-A_{GHN}=\frac{a^2}{4}= \)\( \frac{a^2}{24}-\frac{15a^2}{400}-\frac{a^2}{48}=\frac{3a^2}{20} \)
. A snake that slithers on the ground can only dream of flying through the air.