Problema de "slicing" (100,50,30) (own)

Marius Stănean
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Problema de "slicing" (100,50,30) (own)

Mesaj de Marius Stănean »

Fie $\triangle ABC,\,m(\angle A)=50^\circ,m(\angle C)=30^\circ$. Punctele $D \in (AB),\,E \in (AC)$ astfel incat $m(\angle ACD)=10^\circ,\,m(\angle ABE)=20^\circ$. Sa se afle $m(\angle{ADE})$.
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bilalkaan
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Re: Problema de "slicing" (100,50,30)

Mesaj de bilalkaan »

Construim bisectoarea unghiului $\angle BCD$ si notam cu $O$ intersectia $BE$ si $CD$ si vom avea ca triunghiul $COB$ este isoscel.Fie $Y$ si $X$ intersectia bisectoare unghiului $\angle BCD$ cu segmentele $AB$ si $BE$. Din calculul de unghiuri ne va rezulta ca triunghiul $CEY$ este isoscel si rezulta ca masura unghiului $ADE$ este $60^\circ$. :D
Marius Stănean
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Re: Problema de "slicing" (100,50,30)

Mesaj de Marius Stănean »

Felicitari Bilal Kaan, o solutie frumoasa. :)
bilalkaan scrie:Construim bisectoarea unghiului $\angle BCD$ si notam cu $O$ intersectia $BE$ si $CD$ si vom avea ca triunghiul $COB$ este isoscel.Fie $Y$ si $X$ intersectia bisectoare unghiului $\angle BCD$ cu segmentele $AB$ si $BE$. Din calculul de unghiuri ne va rezulta ca triunghiul $CEY$ este isoscel si rezulta ca masura unghiului $ADE$ este $60^\circ$. :D
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pr_50_30_bilalkaan.png (9.09 KiB) Vizualizat de 4714 ori
Deci patrulaterul $BCEY$ este inscriptibil $(\angle ECY=\angle EBY=20^\circ)\Rightarrow$ $\angle CEY=180^\circ-\angle B=80^\circ \Rightarrow$ $\angle CYE=180^\circ-\angle CEY-\angle ECY=80^\circ\Rightarrow \triangle CEY$ este isoscel.
Deci $CY=CE\Rightarrow \triangle CDY\equiv \triangle CDE \;(L.U.L.)$ $\Rightarrow \angle ADE=2\cdot\angle DYE=2\cdot\angle C=60^\circ.$


Solutia mea se bazeaza pe construirea poligonului regulat cu 18 laturi.
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pr_50_30_d.png (12.01 KiB) Vizualizat de 4719 ori
$DE\parallel BF \Leftrightarrow \dfrac{AD}{AB}=\dfrac{AE}{AF}\Leftrightarrow\ldots \Leftrightarrow$ $\dfrac{\sin 10^\circ\cdot \sin 100^\circ \cdot\sin 50^\circ \cdot\sin 70^\circ \cdot\sin 60^\circ }{\sin 50^\circ \cdot\sin 60^\circ \cdot\sin 20^\circ \cdot\sin 30^\circ \cdot\sin 70^\circ}=1\Leftrightarrow$ $\dfrac{\sin 10^\circ \cdot\sin 100^\circ}{\sin 20^\circ \cdot\sin 30^\circ}=1\Leftrightarrow \dfrac{2\sin 10^\circ \cdot\sin 100^\circ}{2\sin 10^\circ \cdot\cos 10^\circ }=1\Leftrightarrow\dfrac{\cos (-10^\circ)}{ \cos 10^\circ }=1.$
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sunken rock
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Re: Problema de "slicing" (100,50,30) (own)

Mesaj de sunken rock »

Take $F$ as intersection of $BE$ with the symmetrical of $CD$ w.r.t. $AC$; since $\angle DCF=\angle DBF=20^\circ, CBDF$ is cyclic; with $\angle CDF=80^\circ$, we get $\Delta CDF$ isosceles and, by symmetry $\angle FDE=\angle DFE$, but $\angle DFE=\angle DCB=20^\circ$. With $\angle ADF=\angle BCF=40^\circ$ we get $\angle ADE=60^\circ$.

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sunken rock
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Virgil Nicula
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Re: Problema de "slicing" (100,50,30) (own)

Mesaj de Virgil Nicula »

PP. Fie $\triangle ABC,\,m(\angle A)=50^\circ,m(\angle C)=30^\circ$. Punctele $D \in (AB),\,E \in (AC)$ astfel incat $m(\angle ACD)=10^\circ,\,m(\angle ABE)=20^\circ$. Sa se afle $m(\angle{ADE})$.

Demonstratie. Notam $m\left(\widehat{CDE}\right)=x$ si $m\left(\widehat{BED}\right)=y$ , unde $x+y=100^{\circ}$ . In patrulaterul convex $BDEC$ aplicam relatia trigonometrica cunoscuta:

$\sin\widehat{EBD}\sin\widehat{CDE}\sin\widehat{BEC}\sin\widehat{DCB}=$ $\sin\widehat{CBE}\sin\widehat{BDC}\sin\widehat{DEB}\sin\widehat{ECD}\iff$ $\sin 20^{\circ}\sin x\sin 70^{\circ}\sin 20^{\circ}=\sin 80^{\circ}\sin 60^{\circ}\sin y\sin 10^{\circ}$ $\iff$

$\sin 20^{\circ}\sin x\cos 20^{\circ}\sin 20^{\circ}=\cos 10^{\circ}\sin 60^{\circ}\sin y\sin 10^{\circ}\iff$ $\sin 40^{\circ}\sin x\sin 20^{\circ}=\sin 60^{\circ}\sin y\sin 20^{\circ}\iff$ $\boxed{\sin 40^{\circ}\sin x=\cos 30^{\circ}\cos (10^{\circ} -x)}$

$\iff$ $\cos (40^{\circ} -x)-\cos (40^{\circ}+x)=\cos (40^{\circ}-x)+\cos (20^{\circ}+x)\iff$ $\cos (140^{\circ}-x)=\cos (20^{\circ}+x)\iff$ $140^{\circ} -x=20^{\circ}+x\iff x=60^{\circ}$ .
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sunken rock
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Re: Problema de "slicing" (100,50,30) (own)

Mesaj de sunken rock »

Take $O$ the circumcenter of $\triangle BCE$; it belongs to $CD$. Because $\angle BEO=\angle BDO=60^\circ$, $BDEO$ is cyclic, thus $\angle CDE=\angle ODE=60^\circ$, making $\angle ADE=60^\circ$.

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sunken rock
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sunken rock
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Re: Problema de "slicing" (100,50,30) (own)

Mesaj de sunken rock »

Take ${F}\in BE\cap CD$ and $G$ reflection of $F$ in $BC$; $G$ lies onto $AB$. Now take ${H}\in GF\cap AC$. Since $\angle DGH=\angle DCH=10^\circ$ we get $GDHC$ cyclic hence, simple angle inspection shows $DHEF$ cyclic as well, with $\angle EDH=10^\circ$, finally giving $\angle EDF=60^\circ\iff\angle ADE=60^\circ$.

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sunken rock
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