GM 2/2009
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
GM 2/2009
Fie un triunghi ABC, D este punctul de tangenta al cercului inscris cu latura (BC) si \( R_1,R_2 \) razele cercurilor inscrise in triunghiurile ABD, respectiv ACD. Aratati ca \( sqrt{\frac{R_1}{R_2}}+sqrt{\frac{R_2}{R_1}}=\frac{2}{\sin ADB} \).
S.S.
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Lema. Intr-un patrulater convex \( ABCD \) consideram cercurile inscrise \( w_a \), \( w_b \), \( w_c \), \( w_d \) in triunghiurile \( BCD \), \( CDA \), \( DAB \), \( ABC \) respectiv. Atunci \( ABCD \) este circumscriptibil \( \Longleftrightarrow \) cercurile \( w_a \), \( w_c \) sunt tangente (in acelasi punct al diagonalei \( BD \))\( \Longleftrightarrow \) cercurile \( w_b \), \( w_d \) sunt tangente (in acelasi punct al diagonalei \( AC \)).
Dem. Notam \( AB=a \) , \( BC=b \) , \( CD=c \) , \( DA=d \) , \( AC=e \) , \( BD=f \) , punctele de tangenta \( T_a \) , \( T_c \) ale cercurilor \( w_a \) , \( w_c \)
respectiv cu diagonala \( [BD] \) si punctele de tangenta \( T_b \) , \( T_d \) ale cercurilor \( w_b \) , \( w_d \) respectiv cu diagonala \( [AC] \) . Se observa ca
\( \left\|\ \begin{array}{c}
2\cdot BT_a=b+f-c\\\\\\\\
2\cdot BT_c=a+f-d\end{array}\ \right\| \) si \( \left\|\ \begin{array}{c}
2\cdot AT_b=d+e-c\\\\\\\\
2\cdot AT_d=a+e-b\end{array}\ \right\| \) . Asadar, patrulaterul \( ABCD \) este circumscriptibil \( \ \Longleftrightarrow\ a+c=b+d\ \Longleftrightarrow \)
\( BT_a=BT_c\ \Longleftrightarrow\ T_a\equiv T_c\ \Longleftrightarrow\ w_a\ ,\ w_c \) sunt tangente \( \ \Longleftrightarrow\ AT_b=AT_d\ \Longleftrightarrow\ T_b\equiv T_d\ \Longleftrightarrow\ w_b\ ,\ w_d \) sunt tangente.
Revenim la problema propusa. Triunghiul \( ABC \) poate fi considerat ca un patrulater circumscriptibl degenerat (la limita !) \( ABDC \) , unde
\( D \) apartine cercului inscris acestuia. Deci \( DB+b=DC+c \) . In virtutea lemei precedente, cercurile inscrise \( w_c=C(I_1,r_1) \) ,
\( w_b=C(I_2,r_2) \) in triunghiurile \( ABD \) , \( ACD \) sunt tangente, mai exact sunt tangente cevienei \( AD \) in acelasi punct \( S \) , adica
\( S\in AD\cap w_c\cap w_b \) . Notam \( X\in w_c\cap BC \) , \( Y\in w_b\cap BC \) si \( T\in I_1I_2\cap BC \) . Si acum sa privim la trapezul dreptunghic \( I_1XYI_2 \) .
Observam ca \( \left\|\ \begin{array}{c}
S\in I_1I_2\ ;\ I_1I_2\perp AD\\\\\\\\
I_1I_2=r_1+r_2\ ;\ XY=2\sqrt {r_1r_2}\\\\\\\\
XY=I_1I_2\cdot\cos\widehat {STC}\\\\\\\\
m(\widehat {STC})+m(\widehat {ADB})=90^{\circ}\end{array}\ \right\|\ \Longrightarrow\ 2\sqrt {r_1r_2}=\left(r_1+r_2\right)\cdot \sin \widehat {ADB}\ \Longrightarrow\ \sqrt {\frac {r_1}{r_2}}+\sqrt {\frac {r_2}{r_1}}=\frac {2}{\sin\widehat {ADB}} \) .
Dem. Notam \( AB=a \) , \( BC=b \) , \( CD=c \) , \( DA=d \) , \( AC=e \) , \( BD=f \) , punctele de tangenta \( T_a \) , \( T_c \) ale cercurilor \( w_a \) , \( w_c \)
respectiv cu diagonala \( [BD] \) si punctele de tangenta \( T_b \) , \( T_d \) ale cercurilor \( w_b \) , \( w_d \) respectiv cu diagonala \( [AC] \) . Se observa ca
\( \left\|\ \begin{array}{c}
2\cdot BT_a=b+f-c\\\\\\\\
2\cdot BT_c=a+f-d\end{array}\ \right\| \) si \( \left\|\ \begin{array}{c}
2\cdot AT_b=d+e-c\\\\\\\\
2\cdot AT_d=a+e-b\end{array}\ \right\| \) . Asadar, patrulaterul \( ABCD \) este circumscriptibil \( \ \Longleftrightarrow\ a+c=b+d\ \Longleftrightarrow \)
\( BT_a=BT_c\ \Longleftrightarrow\ T_a\equiv T_c\ \Longleftrightarrow\ w_a\ ,\ w_c \) sunt tangente \( \ \Longleftrightarrow\ AT_b=AT_d\ \Longleftrightarrow\ T_b\equiv T_d\ \Longleftrightarrow\ w_b\ ,\ w_d \) sunt tangente.
Revenim la problema propusa. Triunghiul \( ABC \) poate fi considerat ca un patrulater circumscriptibl degenerat (la limita !) \( ABDC \) , unde
\( D \) apartine cercului inscris acestuia. Deci \( DB+b=DC+c \) . In virtutea lemei precedente, cercurile inscrise \( w_c=C(I_1,r_1) \) ,
\( w_b=C(I_2,r_2) \) in triunghiurile \( ABD \) , \( ACD \) sunt tangente, mai exact sunt tangente cevienei \( AD \) in acelasi punct \( S \) , adica
\( S\in AD\cap w_c\cap w_b \) . Notam \( X\in w_c\cap BC \) , \( Y\in w_b\cap BC \) si \( T\in I_1I_2\cap BC \) . Si acum sa privim la trapezul dreptunghic \( I_1XYI_2 \) .
Observam ca \( \left\|\ \begin{array}{c}
S\in I_1I_2\ ;\ I_1I_2\perp AD\\\\\\\\
I_1I_2=r_1+r_2\ ;\ XY=2\sqrt {r_1r_2}\\\\\\\\
XY=I_1I_2\cdot\cos\widehat {STC}\\\\\\\\
m(\widehat {STC})+m(\widehat {ADB})=90^{\circ}\end{array}\ \right\|\ \Longrightarrow\ 2\sqrt {r_1r_2}=\left(r_1+r_2\right)\cdot \sin \widehat {ADB}\ \Longrightarrow\ \sqrt {\frac {r_1}{r_2}}+\sqrt {\frac {r_2}{r_1}}=\frac {2}{\sin\widehat {ADB}} \) .