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. đź™‚

Let be the angle, the point inside the angle and be a variable straight line through with on and on . Let be variable points on respectively.

**Problems**:

- Find so that the area of triangle is minimised.
- Find so that the perimeter of triangle is minimised.
- Find so that the length is minimised.
- Find so that is minimised.
- Find so that is maximised.
- Find so that is minimised (p, q > 0).
- Find so that and is minimised.
- Find so that the perimeter of triangle is minimised.

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

**Answers without proof**:

- [to minimise the area of ] Choose so that (extend to twice its length, then complete the parallelogram with as centre and as diagonal).

- [to minimise the perimeter of ] Construct a circle tangent to the angle and through , then is its tangent through . The circle is formed by first choosing any point on the angle bisector and drawing any initial circle tangent to (its radius can be found by dropping a perpendicular to ). This initial circle can be scaled up or down by drawing parallel lines so that the final circle passes through .

- [to minimise the length of ] The solution is Philo’s line, mentioned in a previous blog entry. The points and cannot be found by straight edge and compass, but should be chosen so that they are equidistant to the midpoint of .

- [to minimise ] Choose so that is isosceles (construct the angle bisector of angle , then draw the perpendicular to this line through ).

- [to maximise ] Choose perpendicular to .

- [to minimise ] Choose so that . Given the lengths drawn from this can be done via parallel lines and similar triangles as shown:

- [to minimise so that ] Rotate by about to . Then is found by the intersection of and and is constructed so that .

- [to minimise the perimeter of triangle ] Reflect in and . Then and are where the line joining these reflected images meets and . If then .

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

**Reference**

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

[…] 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 givenÂ angle. Many of the results are proved in […]

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