Crawford Lynch
10/26/2024 · High School

equivale a \( 2 \sin 2 x=1 \); il problema ha due soluzioni: \( x=\frac{\pi}{12} \) v \( x=\frac{5 \pi}{12} \) wol Data una semicirconferenza di diametro \( \overline{A B}=2 r \), determina su di essa un punto \( P \) in modo che, detto \( Q \) il puy win cui la bisettrice di \( B \widehat{A} P \) interseca la semicirconferenza, risulti \( \overline{A P}+\overline{P Q}+\overline{Q B}=3 r \).

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I due punti \( P \) che soddisfano la condizione sono \( x = \frac{\pi}{12} \) e \( x = \frac{5\pi}{12} \).

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