Problem: Segments AD=10,BE=6,CF=24 are drawn from the vertices of triangle ABC, each perpendicular to a straight line RS, not intersecting the triangle. Points D,E,F are the intersection points of RS with the perpendiculars. If x is the length of the perpendicular segment drawn to RS from the intersection point of the medians of the triangle, then x is:
Answer Choices:
A. 40/3
B. 16
C. 56/3
D. 80/3
E. undetermined
Solution:
Let M be the midpoint of CB and let G be the intersection point of the three medians. Draw MN perpendicular to RS. Then AD,x,BE,MN, and CF are parallel. MN is the median of trapezoid BEFC. β΄MN=21β(6+24)=15. In trapezoid ADNM draw AHβ₯MN and let AH intersect GK in J. Then HN=10 and MH=5 and JK=10 and GJ =32β(5). β΄x=GK=310β+10=340β.
or
Start with MN=15 and use the Principle of Weighted Means:
since the ratio AG:GM=2:1, we have GK=32β 15+1β 10β=340β.
or
Using vectors with O as origin, we first show that gΛβ=31β(aΛ+bΛ+cΛ) where gΛβ=OG,aΛ=OA,bΛ=OB,cΛ=OC. We have
Let unit vector i be in the direction RS and unit vector j be in the direction DA. Then gΛβ=ix+jy,aΛ=ix1β+10j.bΛ=ix2β+6j,cΛ=ix3β+24jβ΄jy=31β(10+6+24)jβ΄y=340β.
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