A Geometric Optimisation problem

Consider two circles $A(B)$ and $C(B)$ intersecting at $B$ and $D$, and a line passing through $D$. Let this line intersect the circle $A(B)$ at $F$ and the circle $C(B)$ at $G$.

The question of maximising the arithmetic mean of $DF$ and $DG$ is well known. It is also known that Euclidean construction of the line that maximises the Harmonic mean of $DF$ and $DG$ is impossible. We already saw the construction of the line that maximises the Geometric mean of $DF$ and $DG$ in this blog.

In this post we are interested in constructing the line such that the Quadratic mean of $DF$ and $DG$ is maximised. In fact, we solve a slightly more general problem. For a given $\theta$, we construct the line such that $\sqrt{DF^2+DG^2-2 \cdot DF \cdot DG \cdot \cos \theta}$ is maximised.

We first use some auxiliary points and lines that helps us in the final solution. For a given $\theta$, construct points $D'$ and $D'_1$ such that $\angle DAD'=\angle DCD'_1=\theta$. Construct the circle that passes through $D$, $B$ and $D'_1$.

For any point $E$ on this circle, let the line $ED'$ intersect circle $A(B)$ at $F$. Let $G$ be likewise on circle $C(B)$. Construct line $FG$. We first show that the line $FG$ passes through $D$.

We know that $\angle FD'B+\angle FDB=\pi$ because they are opposite angles in a cyclic quadrilateral. But $\angle FD'B=\angle ED'B=\pi-\angle ED'_1B$ because $\angle ED'B$ and $\angle ED'_1B$ are opposite angles of a cyclic quadrilateral. Now, $\angle FD'B=\angle BD'_1G=\angle BDG$.

Therefore, $\angle BDG + \angle FDB=\pi$ which shows that $F$, $D$ and $G$ are collinear.

Now, by construction, $\theta / 2=\angle D'BD=\angle D'_1BD$ and therefore, $\angle D'BD'_1=\theta$. This shows that $\angle D'ED'_1=\pi-\theta$.

We know that $\angle DGD'_1=\angle DBD'_1=\theta/2$ because they are angles in a same segment of a circle. Because $\angle DAD'=\theta$, we can see that $\angle DFD'=\pi-\theta/2$, and therefore $\angle DFE=\theta/2$.

Thus we have shown that $\triangle FEG$ is isosceles with $FE=EG$. Therefore, we can quickly see that $FG=2EF\cos(\theta/2)$.

Using Stewart's theorem on this triangle for the cevian DE, we have,

$EF^2=DE^2+DF\cdot DG$

Multiplying by $(2\cos(\theta/2))^2$ on both sides and simplifying, we get

$(2\cos(\theta/2))^2DE^2=FG^2-2DF\cdot DG \cdot 2\cos^2(\theta/2)$

Using the fact that $FG=DF+DG$ and $\cos\theta=2\cos^2(\theta/2)-1$,

$(2\cos(\theta/2))^2DE^2=DF^2+DG^2-2\cdot DF \cdot DG \cdot \cos\theta$

As $\theta$ is given, the above expression shows that maximising the RHS amounts to maximising $DE$. But this can be easily achieved. We just draw the line joining $D$ and center of circle $D'BD'_1$. The point at which this line intersects the circle gives us $E$ from which we can trivially construct the line $FG$ using what we've described before.

Hope you enjoyed this. See ya soon.

Until then, Yours Aye

Me