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 Internal Triangle Problem (Posted on 2007-02-02)

In triangle ABC, lines are drawn from the vertices to the first trisection points of the opposite sides, forming an internal triangle. What is the ratio of the area of the internal triangle to triangle ABC?

 No Solution Yet Submitted by David Johnson Rating: 4.0000 (1 votes)

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 Solution | Comment 1 of 4
`Let <PQ> denote the vector from point Pto point Q and [IJK...N] denote the areaof polygon IJK...N. Let ABC be the triangleand AA', BB', and CC' the cevians inquestion. Let A* be the intersection ofAA' and CC', B* be the intersection ofBB' and AA', and C* be the intersection ofCC' and BB'.`
`  <AA*> = x<AC'> + (1-x)<AC>              = x<AB>/3 + (1-x)<AC>            (1)`
`  <AA*> = <AC> + y<CA'> + (1-y)<CA>`
`        = 2y<AB>/3 + y<AC>/3             (2)`
`Since <AB> and <AC> are independentvectors, (1) and (2) imply`
`  y = 3/7`
`and therefore by (2)`
`  <AA*> = 2<AB>/7 + <AC>/7 `
`We can find the area of triangles byusing the cross product of vectors`
`  [ABC] = (1/2) |<AB> x <AC>|`
`  [AC'A*] = (1/2) |<AC'> x <AA*>|`
`          = (1/2) |<AB>/3 x (2<AB>/7 + <AC>/7)|`
`          = (1/2) |(<AB> x <AC>)/21|`
`Therefore, [AC'A*] = [ABC]/21 and by symmetry,`
`  [AC'A*] = [BA'B*] = [CB'C*]`
`Now we find the area of the quadrilaterals:`
`  [ABA'] = [AC'A*] + [C'BB*A*] + [BA'B*]`
`         = [C'BB*A*] + 2[ABC]/7 = [ABC]/3`
`Therefore, [C'BB*A*] = 5[ABC]/21 and by symmetry,`
`  [C'BB*A*] = [A'CC*B*] = [B'AA*C*]`
`Now we find the area of the internal triangle:`
`  [ABC] = [A*B*C*] + [AC'A*] + [BA'B*] + [CB'C*]`
`          + [C'BB*A*] + [A'CC*B*] + [B'AA*C*]`
`        = [A*B*C*] + 3[ABC]/21 + 15[ABC]/21`
`Therefore, [A*B*C*] = [ABC]/7`
`Note: The areas above were verified by Geometer's Sketchpad.`
` `

 Posted by Bractals on 2007-02-02 12:15:06

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