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October 16, 2011

Optimisation problems given a point inside a given angle

Filed under: mathematics — ckrao @ 10:59 pm

Given a point inside an angle, it easy to come up with many minimisation problems to solve. Here are some mostly taken from [1], with the answers given below, but proofs are left as exercises to the interested reader. 🙂

Let XOY be the angle, P the point inside the angle and MN be a variable straight line through P with M on OX and N on OY. Let A, B be variable points on OX, OY respectively.


  1. Find M, N so that the area of triangle MON is minimised.
  2. Find M, N so that the perimeter of triangle MON is minimised.
  3. Find M, N so that the length MN is minimised.
  4. Find M, N so that MP.PN is minimised.
  5. Find M, N so that 1/MP + 1/PN is maximised.
  6. Find M, N so that OM^p.ON^q is minimised (p, q > 0).
  7. Find M, N so that OA = OB and AP + PB is minimised.
  8. Find M, N so that the perimeter of triangle PAB is minimised.

Do try at least one of these before reading below for the answers!

Answers without proof:

  1. [to minimise the area of MON] Choose MN so that MP = PN (extend OP to twice its length, then complete the parallelogram with P as centre and MN as diagonal).
  2. [to minimise the perimeter of MON] Construct a circle tangent to the angle and through P, then MN is its tangent through P. The circle is formed by first choosing any point on the angle bisector and drawing any initial circle tangent to XOY (its radius can be found by dropping a perpendicular to OX). This initial circle can be scaled up or down by drawing parallel lines so that the final circle passes through P.
  3. [to minimise the length of MN] The solution is Philo’s line, mentioned in a previous blog entry. The points M and N cannot be found by straight edge and compass, but should be chosen so that they are equidistant to the midpoint of OP.
  4. [to minimise MP.PN] Choose MN so that OMN is isosceles (construct the angle bisector of angle XOY, then draw the perpendicular to this line through P).
  5. [to maximise 1/MP + 1/PN] Choose MN perpendicular to OP.
  6. [to minimise OM^p.ON^q] Choose M, N so that MP/PN = q/p. Given the lengths p, q drawn from P this can be done via parallel lines and similar triangles as shown:
  7. [to minimise AP + PB so that OA = OB] Rotate P by \angle XOY about O to P'. Then A is found by the intersection of MN and PP' and B is constructed so that OA=OB.
  8. [to minimise the perimeter of triangle PAB] Reflect P in OX and OY. Then A and B are where the line joining these reflected images meets OX and OY. If \angle XOY \geq 90^{\circ} then A=B=O.

So given the same setup we have at least 8 different problems with 8 different constructions and answers!


[1] T. Andreescu, O. Mushkarov, L. Stoyanov, Geometric Problems on Maxima and Minima, Birkhäuser, 2006.


1 Comment »

  1. […] or minimise various functions on the plane. This is on a similar theme to an earlier post on optimisation problems given a point inside a given angle. Many of the results are proved in […]

    Pingback by Extremal points of a triangle « Chaitanya's Random Pages — January 9, 2012 @ 11:41 am | Reply

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