The Inverse Sum Theorem

This post talks about the 'Inverse Sum theorem' which is at the heart of the video for my SoME2 submission.

Consider two right angled triangles $OAC$ and $OBC$ having a common side $OC=1$ and right angled at $O$. Let $F$ and $E$ be the foot of altitude of the hypotenuse of the respective triangles. Let the line connecting $E$ and $F$ meet the $x$-axis at $D$. We claim that

$$\frac{1}{OD}=\frac{1}{OA}+\frac{1}{OB}$$

We give three proofs of this result with Coordinate geometry.

Co-ordinate Geometry

Let $\angle OCA=\alpha$. Because $OC=1$, we can immediately see that $OA=\tan \alpha$, $CA=\sec \alpha$ and $OF=\sin \alpha$. As $\angle AOF=\alpha$, the co-ordinates of point $F$ can be seen to be $F \equiv (\sin \alpha \cos \alpha, \sin^2 \alpha)$.

Similarly, if we have $\angle OCB=\beta$, then we have $E \equiv (\sin\beta \cos\beta, \sin^2\beta)$

Then the equation of line $EF$ is

$$\frac{y-\sin^2\beta}{\sin^2\alpha-\sin^2\beta}=\frac{x-\sin\beta\cos\beta}{\sin\alpha\cos\alpha-\sin\beta\cos\beta}$$

$D$ is the point on where this line meets the $x$-axis. Therefore, using $y=0$ in the above equation and solving for $x$ gives us $OD$. After some algebraic manipulations, this gives us 

$$x=\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}$$

Therefore,

$$\displaystyle\frac{1}{OD}=\frac{1}{\tan\alpha}+\frac{1}{\tan\beta}=\frac{1}{OA}+\frac{1}{OB}$$

Trigonometry

Using the generalized Angled Bisector theorem in triangle $OEB$,

$$\frac{OD}{DB}=\frac{OE\sin\angle OED}{BE\sin\angle BED}$$

Using the law of Sines in triangle $OEB$, we get $OE\sin\angle EOB=BE\sin\angle EBO$. Therefore,

$$\frac{OD}{DB}=\frac{\sin\angle EBO}{\sin\angle EOB}\frac{\sin\angle OED}{\sin\angle BED}$$

But $\angle EOB$ and $\angle EBO$ are complementary. So are $\angle BED$ and $\angle OED$. Therefore,

$$\frac{OD}{DB}=\frac{\tan\angle EBO}{\tan\angle BED}=\frac{OC}{OB}\frac{1}{\tan\angle BED}$$

If we now draw a circle with $OC$ as diameter, we know that $E$ and $F$ must lie on this circle because of Thales theorem. Also, $OEF$ and $OCF$ must be equal because they are angles subtended by the chord $OF$. Therefore,

$$\angle BED=90^\circ - \angle OEF=90^\circ-\angle OCF=\angle OAC$$

Therefore,

$$\frac{OD}{DB}=\frac{OC}{OB}\frac{OA}{OC}\implies \frac{OD}{OB-OD}=\frac{OA}{OB}$$

Solving for $OD$, we get the desired result.

Circle Inversion

Using the diagram from above, we see that $OD^2=DF\cdot DE$ because of the tangent-secant theorem. If we imagine a circle $c$ with center $D$ and $OD$ as radius, then the above relation tells us that $E$ and $F$ are inverses of each other w.r.t circle $c$.

Here comes the nice part. Consider two points $U$ and $V$ which are reflections of each other w.r.t a line $l$. Let's say $U$, $V$ and $l$ are all reflected w.r.t a line $m$ to give $U'$, $V'$ and $l'$. Now it should be easy to see that $U'$ and $V'$ are reflections of each other w.r.t line $l'$. The amazing thing here is that this relation holds in case of Circle inversion as well.

Now imagine a circle $c'$ with $C$ as center and $CO$ as radius. Point $O$ is invariant as under an inversion w.r.t $c'$. Points $F$ and $A$ become inverses of each other under $c'$ whereas $E$ and $B$ become inverses. Circle $c$ we described above is invariant as it is orthogonal to Circle $c'$.

We know that $E$ and $F$ are inverses w.r.t $c$. Therefore, $A$ (inverse of $F$ w.r.t $c'$) and $B$ (inverse of $E$ w.r.t $c'$) must be inverses of each other w.r.t $c$ (which is its own inverse under an inversion in $c'$).

The cross ratio $\lambda$ of any four points $A$, $B$, $C$ and $D$ is defined as

$$\lambda=\frac{AB}{AC}\frac{CD}{BD}$$

If one of these points is at infinity, then the terms containing that point disappear from the definition. It is amazing that Cross ratio is invariant under Circle inversion.

The Cross ratio of $O$, $D$, $A$ and $B$ is then

$$\lambda=\frac{OD}{OA}\frac{AB}{DB}$$

The inverses of these points under an inversion w.r.t $c$ become $O$, $\infty$, $B$ and $A$. Therefore, the cross ratio after inversion is

$$\lambda=\frac{BA}{OB}$$

Because cross ratio is invariant under inversion. we have,

$$\frac{OD}{OA}\frac{AB}{DB}=\frac{BA}{OB} \implies \frac{OD}{DB}=\frac{OA}{OB}$$

which is the same relation we had in the previous case.

Interestingly, we also see that $OD^2=DA\cdot DB$ because $A$ and $B$ are inverses of each other w.r.t a circle with radius $OD$.

Until then

Yours Aye

Me