# Chaitanya's Random Pages

## 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:

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!

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!

Reference

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