# Cassini's Ovals and Geometric Optimization

### What is this about?

7 July 2013, Created with GeoGebra

### Problem

Elsewhere we considered a problem from the 2009 Asian Pacific Mathematical Olympiad:

Consider all the triangles ABC which have a fixed base BC and whose altitude from A is a constant h. For which of these triangles is the product of its altitudes a maximum?

It was noticed in the comments by Maurizio Frigeni that the solution was incomplete. Below we have a second look at the problem.

### Hint

The problem is of geometric optimization. As is often the case, it is useful to consider level curves of the function involved.

### Solution

Assume the two given points are $A(-a,0)$ and $B(a,0),$ for a parameter $a.$ It is helpful to find the loci of a point $P(x,y)$ the product of whose distances to $A$ and $B$ is constant, say, $b^2$:

$[(x+a)^2+y^2][(x-a)^2+y^2]=b^4.$

The so defined curves are known as *Cassini's ovals* and their shape depends on the ratio $e = b/a.$ A point on one of Cassini's oval will solve the given problem if and only if the given line $r$ is tangent to the oval at that point.

Here's how Cassini's ovals look like:

For $e = b/a \gt 1$ the curve is indeed a closed loop. When $e \lt 1$ the "oval" consists of two disjoint pieces. Finally, when $e=1$ the curve is intersects itself and looks like the symbol of infinity, $\infty.$ This one is known as the *Lemniscate of Bernoulli*.

For a fixed $a,$ Cassini's ovals cover the plane and do not intersect. Some of them are convex, others are not. In the given problem, $r\parallel AB.$ It may therefore happen that $r$ is tangent to a non-convex oval:

and then the olympiad problem would have two solutions. When does this happen? The answer to that is when $e \lt \sqrt{2},$ which corresponds to the case where $\mbox{dist}(A,BC)\lt BC/2,$ as was claimed by Maurizio Frigeni.

The emergence of the circle $x^2+y^2=a^2$ through the foci $B$ and $C$ is explained by implicit differentiation of the equation of the ovals. Assuming $y'=0,$ i.e., the points where the tangent to the curve is horizontal leads exactly to that circle. Before you differentiate it's worth to convert the equation to a different form:

$(x^2+y^2+a^2)^2-4a^2x^2=b^4.$

In his comment Maurizio asked about the case where $r \not\parallel BC.$

In this case, unless $r$ passes through the origin, the solution is unique:

If $r$ passes through the origin, there are two solutions when $e\lt 1:$

### Acknowledgment

The graphics of the family Cassini's ovals are courtesy of wikipedia and wikimedia commons.

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