Points $ P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7} $ are placed correspondingly on sides $ BC, CA, AB, BC, CA, AB, BC $ of triangle $ ABC $ and thease angles are equal:
$ P_{1}P_{2}C = AP_{2}P_{3} = P_{3}P_{4}B = CP_{4}P_{5} = P_{5}P_{6}A = BP_{6}P_{7} = 60^{o} $
Prove that
$ P_{1} = P_{7} $
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¬[ƒ(Gabriel)³²¹º]¼+½=¾
- Messaggi: 849
- Iscritto il: 22 ott 2006, 14:36
- Località: Carrara/Pisa
here a horrible trigonometric solution:
call $ \overline{CP_1} = x $ so for some law of sines
$ \displaystyle \begin{matrix} . \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \overline{CP_7} = a\ -\end{matrix} $$ \displaystyle \frac{\left ( c - \frac{\left ( b - \frac{\left ( a- \frac{\left ( c - \frac{\left ( b - \frac{x \cdot \sin (60+\gamma)}{\sin60} \right ) \cdot \sin60}{\sin (60+\alpha)} \right ) \cdot \sin (60+\beta)}{\sin60} \right ) \cdot \sin60}{\sin (60+\gamma)}\right ) \cdot \sin (60 + \alpha)}{\sin60}\right ) \cdot \sin60}{\sin (60 + \beta)} $
denote $ \delta = 60 + \alpha\ $, $ \ \theta = 60 + \beta \ $ and $ \ \sigma = 60 + \gamma $
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = a [\sin{\theta} \sin{\sigma} - \sin{\delta}\sin60] + b [\sin{\delta} \sin{\sigma} - $$ \displaystyle \sin{\theta}\sin60 ] + c [ \sin{\delta} \sin{\theta} - \sin{\sigma}\sin60 ] + x \cdot \sin{\theta} \sin{\sigma} $
but $ \ \ \sin{\theta}\sin{\sigma} - \sin{\delta}\sin60 = $$ \frac{1}{2}[\cos(\beta - \gamma) - \cos (60 + \alpha) - \cos {\alpha} + \cos (120 + \alpha)] = $$ \frac{1}{2}\cos(\beta - \gamma) - \cos {\alpha} $ and equally for the others, so
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = $$ a[\frac{1}{2}\cos(\beta - \gamma) - \cos {\alpha}] + $$ b[\frac{1}{2}\cos(\gamma - \alpha) - \cos {\beta}] + $$ c [\frac{1}{2}\cos(\beta - \alpha) - \cos {\gamma}] $$ + x \cdot \sin{\theta} \sin{\sigma} $
but $ \frac{1}{2}\cdot a \cdot \cos(\beta - \gamma) = R \cdot \cos(\beta - \gamma)\sin {\alpha} $ and equally for the others, where R is the circumradius of ABC
and $ a \cos {\alpha} = R \sin {2 \alpha} $ and equally for the others, then
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = $$ R[\cos(\beta - \gamma)\sin {\alpha} - \sin {2 \alpha} + $$ \cos(\gamma - \alpha)\sin{\beta} - \sin {2\beta} + $$ \cos(\beta - \alpha)\sin{\gamma} - \sin {2\gamma}] $$ + x \cdot \sin{\theta} \sin{\sigma} $
but $ \cos(\beta - \gamma)\sin {\alpha} = \sin(\beta + \gamma)\cos(\beta - \gamma) = \frac{1}{2}\sin{2\beta}+\frac{1}{2}\sin{2\gamma} $ and equally for the others, so
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = $$ R[\frac{1}{2}\sin{2\beta}+\frac{1}{2}\sin{2\gamma} - \sin {2 \alpha} + $$ \frac{1}{2}\sin{2\alpha}+\frac{1}{2}\sin{2\gamma} - \sin {2\beta} + $$ \frac{1}{2}\sin{2\beta}+\frac{1}{2}\sin{2\alpha} - \sin {2\gamma}] $$ + x \cdot \sin{\theta} \sin{\sigma} \ \Longleftrightarrow $
$ \Longleftrightarrow \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = x \cdot \sin{\theta} \sin{\sigma} \ \Longleftrightarrow\ $$ CP_7 = x \Longleftrightarrow \boxed{P_1 \equiv P_7} $
call $ \overline{CP_1} = x $ so for some law of sines
$ \displaystyle \begin{matrix} . \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \overline{CP_7} = a\ -\end{matrix} $$ \displaystyle \frac{\left ( c - \frac{\left ( b - \frac{\left ( a- \frac{\left ( c - \frac{\left ( b - \frac{x \cdot \sin (60+\gamma)}{\sin60} \right ) \cdot \sin60}{\sin (60+\alpha)} \right ) \cdot \sin (60+\beta)}{\sin60} \right ) \cdot \sin60}{\sin (60+\gamma)}\right ) \cdot \sin (60 + \alpha)}{\sin60}\right ) \cdot \sin60}{\sin (60 + \beta)} $
denote $ \delta = 60 + \alpha\ $, $ \ \theta = 60 + \beta \ $ and $ \ \sigma = 60 + \gamma $
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = a [\sin{\theta} \sin{\sigma} - \sin{\delta}\sin60] + b [\sin{\delta} \sin{\sigma} - $$ \displaystyle \sin{\theta}\sin60 ] + c [ \sin{\delta} \sin{\theta} - \sin{\sigma}\sin60 ] + x \cdot \sin{\theta} \sin{\sigma} $
but $ \ \ \sin{\theta}\sin{\sigma} - \sin{\delta}\sin60 = $$ \frac{1}{2}[\cos(\beta - \gamma) - \cos (60 + \alpha) - \cos {\alpha} + \cos (120 + \alpha)] = $$ \frac{1}{2}\cos(\beta - \gamma) - \cos {\alpha} $ and equally for the others, so
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = $$ a[\frac{1}{2}\cos(\beta - \gamma) - \cos {\alpha}] + $$ b[\frac{1}{2}\cos(\gamma - \alpha) - \cos {\beta}] + $$ c [\frac{1}{2}\cos(\beta - \alpha) - \cos {\gamma}] $$ + x \cdot \sin{\theta} \sin{\sigma} $
but $ \frac{1}{2}\cdot a \cdot \cos(\beta - \gamma) = R \cdot \cos(\beta - \gamma)\sin {\alpha} $ and equally for the others, where R is the circumradius of ABC
and $ a \cos {\alpha} = R \sin {2 \alpha} $ and equally for the others, then
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = $$ R[\cos(\beta - \gamma)\sin {\alpha} - \sin {2 \alpha} + $$ \cos(\gamma - \alpha)\sin{\beta} - \sin {2\beta} + $$ \cos(\beta - \alpha)\sin{\gamma} - \sin {2\gamma}] $$ + x \cdot \sin{\theta} \sin{\sigma} $
but $ \cos(\beta - \gamma)\sin {\alpha} = \sin(\beta + \gamma)\cos(\beta - \gamma) = \frac{1}{2}\sin{2\beta}+\frac{1}{2}\sin{2\gamma} $ and equally for the others, so
$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = $$ R[\frac{1}{2}\sin{2\beta}+\frac{1}{2}\sin{2\gamma} - \sin {2 \alpha} + $$ \frac{1}{2}\sin{2\alpha}+\frac{1}{2}\sin{2\gamma} - \sin {2\beta} + $$ \frac{1}{2}\sin{2\beta}+\frac{1}{2}\sin{2\alpha} - \sin {2\gamma}] $$ + x \cdot \sin{\theta} \sin{\sigma} \ \Longleftrightarrow $
$ \Longleftrightarrow \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} = x \cdot \sin{\theta} \sin{\sigma} \ \Longleftrightarrow\ $$ CP_7 = x \Longleftrightarrow \boxed{P_1 \equiv P_7} $