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$\underline{\text{Introduction}:-}$

I think the eyeball theorem is not really that popular and it also doesn't have a Wikipedia page. I have first found it on Quora, however I don't currently remember what the question was, so I cannot link it.

I have also found a question in MO

Searching the name on google also gives some proofs, but those proofs didn't really satisfied me. So I tried proving it on my own.

$\underline{\text{Eyeball Theorem}:-}$

$■\text{Statement}:$

enter image description here

$|GH|=|EF|$

$■\text{My Proof}:-$

enter image description here (I couldn't find a similar picture, so I just drew it quickly with a sketch pen. If you have problem understanding this picture, then please ask me)

From my picture,

$\angle AI_2P=\angle BI_1R=\angle ACB=\angle ADB=90°$

So,

$\begin{cases}\sin{\alpha}=\frac{RI_1}{BR}=\frac{AC}{AB}\\\sin{\theta}=\frac{PI_2}{AP}=\frac{BD}{AB}\end{cases}$

$\implies\begin{cases}\frac{RI_1}{r_1}=\frac{r_2}{AB}\\\frac{PI_2}{r_2}=\frac{r_1}{AB}\end{cases}$

$\implies\begin{cases}AB\cdot RI_1=r_1r_2\\AB\cdot PI_2=r_1r_2\end{cases}$

$\implies AB\cdot RI_1=AB\cdot PI_2$

$\implies 2RI_2=2PI_2$

$\implies PQ=RS$. $\blacksquare$

$《\bigstar》\text{Generalization}:-$

I think that the $2D$ case implies the $N-D$ case.($N\geq 2$)

Since, the $N$-cones from the centres of the two $N$-balls will create an $N-1$-ball in the places of intersection and since their radii are the same then their $N-1$ volume would be same too which is our concern here.

Correct me if I'm wrong.

$\underline{\text{My Question}:-}$

I would like to have my proof reviewed and I would also like to have more elegant proofs of this theorem.

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Your proof seems to be absolutely correct.

But, since you asked for elegancy, I think, it's nicer to not use trignometry in a geometry problem, especially when you can avoid it. Note that the $\sin \theta$ and the $\sin \alpha$ expressions that you invoked were mainly used to find $$\frac{RI_1}{r_1}=\frac{r_2}{AB}\\ \frac{PI_2}{r_2}=\frac{r_1}{AB}$$ which can be very easily and more elegantly derived from noticing the fact that $$\Delta I_2AP\sim \Delta DAB\\ \Delta I_1BR\sim \Delta CBA$$

Also, here's one more proof-

$\underline{\text{My Approach}:-}$

Let $CB$ and $DA$ intersect at $O$ and let $AQ$ and $BS$ extended intersect at $O^\prime$. Notice that $$\Delta COA\sim \Delta DOB$$ Also, because of symmetry, $$AB\perp OO^\prime$$ Using this, we can see that $$\Delta APQ\sim \Delta AOO^\prime\\ \Delta BRS\sim \Delta BOO^\prime$$ The rest is trivial and only requires some algebra with the sides of the similar triangles.

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  • $\begingroup$ Hey, I forgot to ask, where is $O^\prime$? $\endgroup$ Aug 26 at 3:49
  • $\begingroup$ @RounakSarkar $AQ$ and $BS$ extended meet at $O^\prime$ $\endgroup$ Aug 26 at 4:27
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In [1], Chapter 21 Section 2 (The eyeball theorem) Page 141 says

Much the same thing happens in a curious theorem discovered in the 1960s by the Peruvian geometer, Antonio Gutierrez.


The proof, given in Page 259 of [1], is presented below (adapted by me for this answer and more or less the same as OP's proof); the image referenced in this proof is the same as the one in question. Note: The same proof appears in [2] and [3]


Set $RI_1 = \frac{PS}{2}$ and $PI_2 = \frac{PQ}{2}$.

We only need to show $RI_1 = PI_2$.

Since $\triangle AI_2P \sim \triangle ADB$, $\frac{PI_2}{DB} = \frac{AP}{AB} \implies PI_2 = \frac{DB \: \times \: AP}{AB} = \frac{r_1 r_2}{AB}$.

Similarly, $RI_1 = \frac{AC \: \times \: RB}{AB} = \frac{r_2 r_1}{AB}$.

Hence, proved. The "lower half" is can also proven similarly to conclude $PQ = RS.$


[1] Acheson, David, The wonder book of geometry. A mathematical story, Oxford: Oxford University Press (ISBN 978-0-19-884638-3/hbk). 288 p. (2020). ZBL1460.00006.

[2] Konhauser, Joseph D. E.; Vellemen, Dan; Wagon, Stan, Which way did the bicycle go? The Mathematical Association of America. xv, 235 p. (1996). ZBL0860.00010.

[3] Alsina, Claudi; Nelsen, Roger B., Icons of mathematics. An exploration of twenty key images, Washington, DC: Mathematical Association of America (MAA). xvii, 327 p. (2011). ZBL1230.00001.

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