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Find-the-point-in-interior-of-a-convex-quadrilateral-such-that-the-sum-of-its-distances-to-the-4-vertices-is-minimal-Find-the-point-in-interior-of-a-convex-quadrilateral-such-that-the-sum-of-its-dis

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Question Number 17373 by mrW1 last updated on 04/Jul/17
Find the point in interior of a convex quadrilateral such that the sum of its distances to the 4 vertices is minimal. Find the point in interior of a convex quadrilateral such that the sum of its distances to the 4 sides is minimal.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex} \\ $$$$\mathrm{quadrilateral}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{4}\:\mathrm{vertices}\:\mathrm{is}\:\mathrm{minimal}. \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex} \\ $$$$\mathrm{quadrilateral}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{4}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{minimal}. \\ $$
Answered by ajfour last updated on 05/Jul/17
let x_A be the distance of the point (P) from vertex A and so on .. let L=x_A +x_C +x_B +x_D x_A +x_C ≥ AC x_B +x_D ≥ BD so L_(minimum) = AC+BD so point P is the intersection of the diagonals of the quadrilateral. when sum of the distances of a point from the four sides is minimum the point is probably at the corner of the quadrilateral having the greatest angle (cannot prove yet).
$$\mathrm{let}\:\mathrm{x}_{\mathrm{A}} \mathrm{be}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{P}\right) \\ $$$$\mathrm{from}\:\mathrm{vertex}\:\mathrm{A}\:\mathrm{and}\:\mathrm{so}\:\mathrm{on}\:.. \\ $$$$\:\mathrm{let}\:\mathrm{L}=\mathrm{x}_{\mathrm{A}} +\mathrm{x}_{\mathrm{C}} +\mathrm{x}_{\mathrm{B}} +\mathrm{x}_{\mathrm{D}} \\ $$$$\:\:\:\:\:\mathrm{x}_{\mathrm{A}} +\mathrm{x}_{\mathrm{C}} \:\geqslant\:\mathrm{AC} \\ $$$$\:\:\:\:\:\mathrm{x}_{\mathrm{B}} +\mathrm{x}_{\mathrm{D}} \:\geqslant\:\mathrm{BD} \\ $$$$\mathrm{so}\:\mathrm{L}_{\mathrm{minimum}} =\:\mathrm{AC}+\mathrm{BD} \\ $$$$\mathrm{so}\:\mathrm{point}\:\mathrm{P}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{diagonals}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadrilateral}. \\ $$$$ \\ $$$$\mathrm{when}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{four}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{minimum} \\ $$$$\mathrm{the}\:\mathrm{point}\:\mathrm{is}\:\mathrm{probably}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{corner}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadrilateral}\:\mathrm{having} \\ $$$$\mathrm{the}\:\mathrm{greatest}\:\mathrm{angle}\:\left(\mathrm{cannot}\:\mathrm{prove}\:\mathrm{yet}\right). \\ $$
Commented by mrW1 last updated on 05/Jul/17
answer to part 1 is correct. answer to part 2 is to check. please consider different cases.
$$\mathrm{answer}\:\mathrm{to}\:\mathrm{part}\:\mathrm{1}\:\mathrm{is}\:\mathrm{correct}. \\ $$$$\mathrm{answer}\:\mathrm{to}\:\mathrm{part}\:\mathrm{2}\:\mathrm{is}\:\mathrm{to}\:\mathrm{check}.\:\mathrm{please} \\ $$$$\mathrm{consider}\:\mathrm{different}\:\mathrm{cases}.\: \\ $$
Answered by mrW1 last updated on 05/Jul/17
To part 1: The answer is as given by ajfour the intersection point of the diagonals. To part 2: case 1: both pairs of opposite sides are not parallel, see diagram. Point A is the point whose sum of distances to the sides is minimal.
$$\mathrm{To}\:\mathrm{part}\:\mathrm{1}: \\ $$$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{as}\:\mathrm{given}\:\mathrm{by}\:\mathrm{ajfour}\:\mathrm{the} \\ $$$$\mathrm{intersection}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagonals}. \\ $$$$ \\ $$$$\mathrm{To}\:\mathrm{part}\:\mathrm{2}: \\ $$$$\mathrm{case}\:\mathrm{1}: \\ $$$$\mathrm{both}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{parallel},\:\mathrm{see}\:\mathrm{diagram}. \\ $$$$\mathrm{Point}\:\mathrm{A}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{whose}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{minimal}. \\ $$
Commented by mrW1 last updated on 05/Jul/17
Commented by mrW1 last updated on 05/Jul/17
case 2: one pair of opposite sides is parallel. The point which is closest to K is the solution, here it is point A.
$$\mathrm{case}\:\mathrm{2}: \\ $$$$\mathrm{one}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{parallel}. \\ $$$$\mathrm{The}\:\mathrm{point}\:\mathrm{which}\:\mathrm{is}\:\mathrm{closest}\:\mathrm{to}\:\mathrm{K}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{solution},\:\mathrm{here}\:\mathrm{it}\:\mathrm{is}\:\mathrm{point}\:\mathrm{A}. \\ $$
Commented by mrW1 last updated on 05/Jul/17
Commented by mrW1 last updated on 05/Jul/17
case 3: both pairs of opposite sides are parallel. The sum of distances from every point to the sides is constant.
$$\mathrm{case}\:\mathrm{3}: \\ $$$$\mathrm{both}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{parallel}. \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{distances}\:\mathrm{from}\:\mathrm{every}\:\mathrm{point} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{constant}. \\ $$
Commented by mrW1 last updated on 05/Jul/17

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