another "strange" geometry problem of Bulgaria, but nice

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another "strange" geometry problem of Bulgaria, but nice

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Source: bmo 1988

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#1 h Apr 23, 2007, 7:37 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle and let $M,N,P$ be points on the line $BC$ such that $AM,AN,AP$ are the altitude, the angle bisector and the median of the triangle, respectively. It is known that

$\frac{[AMP]}{[ABC]}=\frac{1}{4}$ and $\frac{[ANP]}{[ABC]}=1-\frac{\sqrt{3}}{2}$ .
Find the angles of triangle $ABC$ .

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anonymouslonely

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anonymouslonely

#2 h Jul 25, 2011, 9:29 AM • 2 Y

Y by Adventure10, Mango247

aria(ABP)=1/2aria(ABC),aria(AMP)=1/4aria(ABC), so M is the midpoint of BP. this means the triangle ABP has AB=AP and this means the measure of the angle PAC is the difference between the angles B and C of the triangle ABC.

using trigonometry we obtain 3sinCcosB=sinBcosC where C<=B.

NP=BC(b-c)/2(b+c) and we have the ratio between the area of the triangle ANP and ABC,so we obtain b/c=sqrt3 and using he other relation we obtain C=30,B=60,A=90 and B=30,C=60,A=90 because we suppose C<=B.

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