angles wanted, BP = AC + AB, <BAP=63^o (2001 Argentina OMA L2 p2)

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angles wanted, BP = AC + AB, <BAP=63^o (2001 Argentina OMA L2 p2)

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#1 h Jun 10, 2020, 6:25 AM

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In triangle $ABC$ , which has $\angle BAC = 63^o$ , the bisector of angle $\angle BAC$ was drawn. Let $\ell$ be the line that passes through $A$ and is perpendicular to this bisector. If line $\ell$ intersects line $BC$ at $P$ such that $BP = AC + AB$ , find the measures of the angles of triangle $ABC$ . Give all the possibilities.

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#2 h Jun 11, 2020, 2:01 PM

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Choose $AB=1$ , let $\widehat{ABC}=x$ .

Rule of sines in $\triangle ABC\ :\ \frac{1}{\sin(117^{\circ}}=\frac{AC}{\sin x}=\frac{BC}{\sin 63^{\circ}}$ .
Rule of sines in $\triangle ABP\ :\ \frac{BP}{\sin(90^{\circ}+\frac{63^{\circ}}{2})}=\frac{1}{\sin(90^{\circ}-\frac{63^{\circ}}{2}-x)}$ .

$\frac{1+AC}{\sin(90^{\circ}+\frac{63^{\circ}}{2})}=\frac{1}{\cos(\frac{63^{\circ}}{2}+x)}$ .
$[1+\frac{\sin x}{\sin(63^{\circ}+x)}]\cos(\frac{63^{\circ}}{2}+x)=\cos \frac{63^{\circ}}{2}$ .
$[\sin(63^{\circ}+x)+\sin x]\cos(\frac{63^{\circ}}{2}+x)=\cos \frac{63^{\circ}}{2}\sin(63^{\circ}+x)$ .

$2\sin(\frac{63^{\circ}}{2}+x)\cos \frac{63^{\circ}}{2}\cos(\frac{63^{\circ}}{2}+x)=\cos \frac{63^{\circ}}{2}\sin(63^{\circ}+x)$ .
$2\sin(\frac{63^{\circ}}{2}+x)\cos(\frac{63^{\circ}}{2}+x)=\sin(63^{\circ}+x)$ .
$\sin(63^{\circ}+2x)=\sin(63^{\circ}+x)$ .

$x=18^{\circ}$ .

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