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Source: Argentina MO 2008

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#1 h Nov 19, 2008, 11:25 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle, and $P$ a point on the internal bisector of angle $\angle BAC$ . It is known that $\angle ABP=30^{\circ}$ and $CP=BC$ . Find the measure of the angle $\angle APC$ .

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#2 h Nov 21, 2008, 5:26 PM • 1 Y

Y by Adventure10

Let
$X\in AB, PX\perp AB$
$Z\in PB, PZ = PX$
$Y\in AC, PY\perp AC$

$\angle XPZ = 90^0 - \angle XBP = 90^0 - 30^0 = 60^0$
$PX = PZ\implies \Delta PXZ$ is equilateral.
$\angle BXZ = 90^0 - \angle PXZ = 90^0 - 60^0 = 30^0 = \angle XBZ$
$\implies ZX = ZB$
By $AAS\implies \Delta AXP\equiv \Delta AYP$
$PX = PY$
$PY = PX = PZ = ZX = BZ$
$CZ\perp PB$
By $RHS\implies \Delta CPZ\equiv \Delta CPY$
$\angle APC = \angle APY + \angle YPC = \frac {1}{2}\angle XPY + \frac {1}{2}\angle YPZ = \frac {1}{2}(\angle XPY + \angle YPZ) =$
$= \frac {1}{2}(360^0 - \angle XPZ) = \frac {1}{2}(360^0 - 60^0) = \boxed{150^0}$

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#3 h Nov 21, 2008, 9:53 PM • 2 Y

Y by Adventure10, Mango247

Let the intersection of angle bisector of angle A and BC be K. Extend BP to meet AC at M. Extend CP to meet AB at L. Let angle BAK = angle CAK = $x$ . Since triangle BPC is isosceles let angle BPC = angle PBC = $y$ .
Simple angle chasing yields angle ACL = $y-30-2x$ . Also angle BCP = $180- 2y$ and Angle APC is equal to $210 + x - y$ . Now we applying Trignometric form of Ceva's theorem to cevians AK, CL and BM we get
$(sinx/sinx)(sin30/siny)(sin(180- 2y)/sin(y - 30 -2x)) = 1$
Simplifying we get $sin( y- 30 - 2x) = cosy (i)$ . From triangle BPC we know both $y$ and $y - 30- 2x$ are acute. From $(i)$ we now get $y- 30 - 2x = 90 - y$ .
From this we get $y - x = 60$ . But angle APC = $210 + x - y = 210 -( y- x)$ .
But $y - x = 60$ . This implies angle APC = $210 - 60 = 150$
$QED$

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#4 h Nov 22, 2008, 12:21 PM • 1 Y

Y by Adventure10

I will use Akashnil picture:) $PX = PY$ so $BP = 2PC*sin\angle PCY$ let $\angle PBC = x$ so from triangle BPC by law of sines( $CB = CP$ ) we get $BP = 2cosx*PC$ or $cosx = sin\angle PCY$ and $\angle XBP < \frac {\pi}{2}$ so $\angle PCY = \frac {\pi}{2} - x$
$\angle ABC + \angle BAC + \angle BCA = \pi$ or $\angle BAC = 2x - \frac {2\pi}{3}$ so $\angle PAC = x - \frac {\pi}{3}$ and from triangle APC $\angle APC = \frac {5\pi}{6}$

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