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. ∴MN=21​(6+24)=15. In trapezoid ADNM draw AH⊥MN and let AH intersect GK in J. Then HN=10 and MH=5 and JK=10 and GJ =32​(5). ∴x=GK=310​+10=340​.
or
Start with MN=15 and use the Principle of Weighted Means:
since the ratio AG:GM=2:1, we have GK=32⋅15+1⋅10​=340​.
or
Using vectors with O as origin, we first show that gˉ​=31​(aˉ+bˉ+cˉ) where gˉ​=OG,aˉ=OA,bˉ=OB,cˉ=OC. We have
Let unit vector i be in the direction RS and unit vector j be in the direction DA. Then gˉ​=ix+jy,aˉ=ix1​+10j.bˉ=ix2​+6j,cˉ=ix3​+24j∴jy=31​(10+6+24)j∴y=340​.
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