Optimisation problems given a point inside a given angle

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.

Problems:

Find $M, N$ so that the area of triangle $MON$ is minimised.
Find $M, N$ so that the perimeter of triangle $MON$ is minimised.
Find $M, N$ so that the length $MN$ is minimised.
Find $M, N$ so that $MP.PN$ is minimised.
Find $M, N$ so that $1/MP + 1/PN$ is maximised.
Find $M, N$ so that $OM^p.ON^q$ is minimised (p, q > 0).
Find $M, N$ so that $OA = OB$ and $AP + PB$ is minimised.
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:

[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).
[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 $P$ .
[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 $OP$ .
[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 $P$ ).
[to maximise $1/MP + 1/PN$ ] Choose $MN$ perpendicular to $OP$ .
[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:
[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 $OA=OB$ .
[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!

Reference

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

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[...] 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 [...]

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