A geometry problem by rajdeep brahma
Geometry
Level
4
Let the sides of a triangle be 8,10,12.Let the altitude through A be p, through B be q, through C be r and x be the distance of vertex A, y be the distance of vertex B, z be the distance of vertex C to the orthocenter in the acute angled triangle .Find px+qy+rz-96.
The answer is 58.
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1 solution
Let D,E,F be the feet of the altitudes from A,B,C respectively. Let H be the orthocenter of triangle ABC. Then triangle ACD is similar to triangle AHE. So p.x =AD.AH=AC.AE= AE.b. Similarly ABD is similar to triangle AHF , so p. x=AD.AH=AB.AF=AB.c. Therefore p.x= 2 A E . b + A F . c .....Similarly q.y= 2 B F . c + B D . a ,r.z= 2 C D . a + C E . b .Now dude it is a cake walk.....add all of them to get px+qy+rz= 2 a ∗ a + b ∗ b + c ∗ c =154(as per given values of a,b,c).