A nice Geometric Optimisation Problem - Part 2

This post picks up where we left earlier.

In addition to the centre of circumcircle of $\triangle KDL$, the midpoint of the segment $KL$ also lies in the circle. In fact, this can be proved more easily.

Let $A$ and $C$ be the centers of the respective circles. It can be easily seen that $2\angle CAD=\angle BAD=2\angle BQD=2\angle BKD$ which means $\angle CAD=\angle BKD$. Similarly, we can show that $\angle ACD=\angle BLD$. Hence, we see that $\triangle ACD \sim \triangle KLD$.

If we let $O$ be the midpoint of $AC$ and $P$ be the midpoint of $KL$, then by the similarity of the triangle we established before, we have $\angle DOC=\angle DPL=\angle DPB$. Because $\angle DOC$ is constant, we see that $P$ subtends a constant angle with segment $BD$ and hence lies on a circle containing $B$ and $D$.

Also, note that $\angle BOD=2\angle DOC=2\angle DPB=2\angle DSB$ where $S$ is as shown below. Because the angle subtended by $O$ with segment $BD$ is twice that of $P$, we know that $O$ is the center of circle containing $P$.

As both $J$ and $O$ lie on the perpendicular bisector of $AC$ at the product-maximising-position, segment $DT$ (where $T$ is shown below) is parallel to $AC$. Also, $\angle BDT$ is right angled which makes $BT$ the diameter of $P$'s locus.

Because $\angle NDT$ is right angle, we see that $NT$, which is the perpendicular bisector of $KL$ at the product-maximising position, is the diameter of $\triangle KLD$'s circumcircle.

We know that $\angle KTN=\angle KDN$ and $\angle LDN=\angle LTN$ (angles subtended by the same chord in a circle), we see that $\angle KDB=\angle LDB$.

Therefore, we finally see that at the product-maximising position, $ND$ (or) $BD$ should be the angle bisector $\angle KDB$ which is the relation we were looking for in the previous post.

Hope you enjoyed this. See ya later.

Until then
Yours Aye
Me

A nice Geometric Optimisation Problem - Part I

Consider two circles of different radii intersecting at $B$ and $D$. If we now draw a line $KL$ through $B$ as shown in the following figure, the question of determining the position at which the sum of $BK$ and $BL$ maximised is one of the famous non trivial problems in Geometric Optimisation.

We can solve this by noting that $\triangle KDL$ remains self similar. Because both $\angle BKD$ and $\angle BLD$ are angles subtended by the same chord $BD$ in the respective circles, the angles in the triangle remain constant irrespective of the position of the line $KL$. 

 Therefore $KL$ will be maximised when $DK$ (or $DL$) is as large as possible which happens when they are the respective diameters of the given circles. We could also see that, at this position $KL$ and $BD$ will be perpendicular to each other.

A natural follow up question would be find the position of $KL$ such that the product of $BK$ and $BL$ is maximised. Find a geometric construction of the position along with a geometric proof for the same proved more challenging that I had anticipated. That proof will be the subject of this post.

Before I start, I also want to point out that the proof here may seem unnecessarily complicated. But if anyone can share the shorter and elegant geometric proof for this problem, I would be really grateful.

To begin with, if we construct the circle passing through $K$, $D$ and $L$, and extend the segment $BD$ so that it intersects the circumcircle at $N$, then by the power-of-the-point theorem, we know that

$BK \cdot BL = BD \cdot BN$

Therefore, maximising the product of $BK$ and $BL$ is the same as maximising $BD \cdot BN$. But $BD$ is a constant independent of the position of the segment $KL$. Hence, the problem we are after reduces to that of maximising the segment $BN$ or $DN$.

To maximise $DN$, we turn our attention to the locus of the centre of the circumcircle, say $J$ (Take a wild guess as to what the locus might be before continuing).

Let $A$ and $C$ be the centres of the respective circles and let the line $AC$ meet the circles at $Q$ and $R$ as shown below.

Then, for any general position of $KL$, $\angle CAD=2\angle CQD=\angle BQD=\angle BKD$

Also, because $LD$ is a chord on both the circumcircle and (one of) the given circle, it's perpendicular bisector pass through both $C$ and $J$. Therefore, $\angle BKD=\angle LKD=\angle DJC$.

We've seen that the segment $CD$ subtends the same angle at both $A$ and $J$ in any general position. Therefore, $J$ lies on the circle containing the points $A$, $D$ and $C$ i.e. the circumcircle of $\triangle ADC$.

It should be easy to see that $DN$ (which is a chord on the circumcircle of $\triangle KDL$) is maximised when the projection of $DJ$ along the line $BD$ is at its highest. This happens when $J$ is at the 'top' of the circle i.e. at the intersection of the $\triangle ACD$'s circumcircle and the perpendicular bisector of $AC$.

Now that we know the position of $J$, we can draw a circle with $J$ as centre and $JD$ as radius. Then, the intersection of this circle with the given circles gives us points $K$ and $L$ such that the product $BK \cdot BL$ is maximised.

Even though we have now shown the geometric construction of $KL$ to maximize the product, it's not as clean as the statement '$KL$ must be perpendicular to $BD$' we had in the maximising-the-sum case.

Do we even have such a relation in the product case? We'll see about that in the next post.

Hope you enjoyed this. See ya later.

Until then
Yours Aye
Me