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A triangle $ABC$ with centroid $G$ is such that a line $l$ passing through $G$ intersects $AB$, $BC$, and $AC$ at $H, I, J$, respectively. Show that out of the 3 distances $d(G, I), d(G, H), d(G, J)$, one is the harmonic mean of the others.

http://i.imgur.com/yf6zAlB.png

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  • $\begingroup$ I think you mean one of them is HALF of the harmonic mean of the others. Don't you? $\endgroup$ Commented Jan 9, 2015 at 13:18

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Let P, Q trisect BC so that BP=PQ=QC.

Then GP is parallel to AB and GQ parallel to AC

Thus GIP ~ HIB and GIQ ~ JIC.

Consider GI(1/GI + 1/GH + 1/GJ) = 1+ GI/GH + GI/GJ

= 1 + PI/PB + QI/QC

= 1 + PI/PB - QI/PB (since PB = -QC taking direction into account)

= 1 + PI/PB - (QP + PI)/PB

= 1 - QP/PB = 0

GI must be non-zero as the centroid cannot lie on any of the edges, so we can deduce that 1/GI + 1/GH + 1/GJ = 0

Which, taking directions into account can be arranged to give that one of the directions is HALF the harmonic mean of the other two.

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