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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.

 Dec 14, 2019
 #1
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                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

 

 

 

cool cool cool

 Dec 15, 2019

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