In triangle ABC, angle A is 45 degrees and angle C is 30 degrees. If altitude BH intersects median AM at P, then AP:PM = 1:k. Find k.
B
M
P
A H D C
Since angle BAC = 45 and angle BHA = 90 .....then angle ABH = 45
So....BH = AH
And since angle BCH = 30 and angle BHC = 90....then angle HBC = 60
So...HC = BH * sqrt (3)
BC = 2BH
AH + HC = BH ( 1 + sqrt (3)) = AC
Let MD be perpendicular to AC
Triangles BHC and MDC are similar
MC/ BC = DC/ HC
1/2 = DC / [BH sqrt(3)]
DC = sqrt (3)/2) BH
And MD = (1/2)BH
So AD = AC - DC = BH + sqrt(3)BH - sqrt (3)/2 BH = BH [ 1 + sqrt(3)/2 ]
So AM = sqrt ( AD^2 + MD^2) = BH * sqrt ( 7/4 + sqrt (3) + 1/4) =
BH* sqrt ( 2 + sqrt (3) )
And triangles APH and AMD are similar
So
AH / AD = AP / AM
BH / [ BH (1 + sqrt (3)/2) ) ] = AP/AM
1 / ( 1 + sqrt (3)/2) = AP / AM
AM = AP (1 + sqrt (3)/2)
So
PM = AM - AP
PM = AP [ ( 1 + sqrt (3)/2) - 1 ] = AP* sqrt (3)/2
So
AP : PM = AP / [AP * sqrt (3)/2) ] = 1 / [sqrt (3)/2)]
So k = sqrt (3)/2