Concursului SARAGHIN 2013 prin corespondenta! (in engleza)
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Concursului SARAGHIN 2013 prin corespondenta! (in engleza)
Gasiti enunturile problemelor in limba engleza, in attachment!
Concursul fiind in desfasurare va rog sa nu postati solutiile acestor probleme inaintea datei de 1 aprilie 2013.
Pana atunci va puteti incerca fortele si poate va inscrieti la acest concurs si le trimiteti la adresa indicata... SUCCES!
Concursul fiind in desfasurare va rog sa nu postati solutiile acestor probleme inaintea datei de 1 aprilie 2013.
Pana atunci va puteti incerca fortele si poate va inscrieti la acest concurs si le trimiteti la adresa indicata... SUCCES!
- Fişiere ataşate
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- 09_zaochn-e(2013)(en).zip
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Imi cer scuze daca intrebarea nu prea are legatura cu matematica.Am vazut ca in material,la fiecare problema este specificata una din clasele 8-11.Daca sunt in clasa a 9-a,problemele pe care ar trebui sa incerc sa le rezolv sunt notate cu clasa a-8-a?
Don't wish it were easier, wish you were better
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Fiecare elev poate participa numai la clasa in care este, deci cei din clasa a 9-a la clasa a 9-a! Chiar daca sunt neconcordante intre programele noastre si cele ale rusilor. Poti rezolva orice problema si cele pentru clasele mai mici, cat si cele pentru clasele mai mari; insa la calculul punctajului obtinut, nu vor fi luate in considerare problemele care se adreseaza claselor mai mici. Evident problemele pentru care se refera la mai multe clase, de ex: 8-9, 9-11, 10-11 vor fi punctate pentru oricare dintre clasele mentionate...
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- Mesaje: 201
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Va multumesc pentru raspuns!
Don't wish it were easier, wish you were better
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- Mesaje: 1493
- Membru din: Mar Oct 26, 2010 9:21 pm
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Data limita pana la care se puteau trimite solutiile a EXPIRAT la 1 aprilie 2013!, asa ca puteti DEJA posta pe forum SOLUTIILE si COMENTARIILE voastre la aceste PROBLEME!!!
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Problem 1:
Let M be the midpoint of BC; since $AM\parallel DE$, we have $m(\widehat{MAD})=m(\widehat{ADE})$, but $m(\widehat{ADE})= m(\widehat{EAD})$; in conclusion AD is the bisector of $\angle BAM$, so m$(\widehat{CAD})=\frac{3}{4}\cdot m(\widehat{BAC})$.
Best regards,
sunken rock
Let M be the midpoint of BC; since $AM\parallel DE$, we have $m(\widehat{MAD})=m(\widehat{ADE})$, but $m(\widehat{ADE})= m(\widehat{EAD})$; in conclusion AD is the bisector of $\angle BAM$, so m$(\widehat{CAD})=\frac{3}{4}\cdot m(\widehat{BAC})$.
Best regards,
sunken rock
A blind man sees the details better.
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Problem 2:
Clearly $ABA_1B_1$ is an isosceles trapezoid and $AB_1=B_1A_1=A_1B$; construct the equilateral triangle $O_1A_1B_1, O_1$ being inside the triangle $B_1A_1C$; $m(\widehat{CB_1O_1})=20^\circ$, so $m(\widehat{O_1AC})=10^\circ=m(\widehat{O_1CA})$; consequently $CO_1=AO_1$ and $O_1\equiv O$.
Best regards,
sunken rock
Clearly $ABA_1B_1$ is an isosceles trapezoid and $AB_1=B_1A_1=A_1B$; construct the equilateral triangle $O_1A_1B_1, O_1$ being inside the triangle $B_1A_1C$; $m(\widehat{CB_1O_1})=20^\circ$, so $m(\widehat{O_1AC})=10^\circ=m(\widehat{O_1CA})$; consequently $CO_1=AO_1$ and $O_1\equiv O$.
Best regards,
sunken rock
A blind man sees the details better.
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Problem 20:
Easily, $C_2$ is Miquel point of $\triangle ABC$ with points $A_1, B_1,C_1$, consequently $\widehat{BC_1C}=\widehat{BB_1C}=\widehat{AC_1C_2}$, and $C_1C_2$ is the symmetrical of $CC_1$ across $AB$, hence it passes through the reflection of $C$ across $AB$.
Best regards,
sunken rock
Easily, $C_2$ is Miquel point of $\triangle ABC$ with points $A_1, B_1,C_1$, consequently $\widehat{BC_1C}=\widehat{BB_1C}=\widehat{AC_1C_2}$, and $C_1C_2$ is the symmetrical of $CC_1$ across $AB$, hence it passes through the reflection of $C$ across $AB$.
Best regards,
sunken rock
A blind man sees the details better.
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Problem 18:
By power of point, $BM^2=BX\cdot BY, CN^2=CY\cdot CX$ and dividing side by side: $\frac{BM^2}{CN^2}=\frac{BX\cdot BY}{CY\cdot CX}\ (\ *\ )$, but $\triangle ABM\sim\triangle ACN$, and $\frac{BM}{CN}=\frac{AB}{AC}$; by Steiner theorem, from $(*)$ we get the required angular relation.
Best regards,
sunken rock
By power of point, $BM^2=BX\cdot BY, CN^2=CY\cdot CX$ and dividing side by side: $\frac{BM^2}{CN^2}=\frac{BX\cdot BY}{CY\cdot CX}\ (\ *\ )$, but $\triangle ABM\sim\triangle ACN$, and $\frac{BM}{CN}=\frac{AB}{AC}$; by Steiner theorem, from $(*)$ we get the required angular relation.
Best regards,
sunken rock
A blind man sees the details better.
- sunken rock
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Re: Concursului SARAGHIN 2013 prin corespondenta! (in englez
Problem 6:
Sketch: Show easily that $BCYX$ is an isosceles trapezoid.
Best regards,
sunken rock
Sketch: Show easily that $BCYX$ is an isosceles trapezoid.
Best regards,
sunken rock
A blind man sees the details better.