Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com