I noticed that many people have been posting problems involving Descartes' Four Circle Theorem, but this site has no documentation on it (especially the proof of it). It seems that all the proofs presented online are by Coxeter, which is short but non-elementary. I will attempt to prove the theorem from scratch and see where I get in the next few days. For the time being, here is Coxeter's proof.

**Interesting Trivia**

Descartes found the theorem without proof and sent it to Princess Elizabeth of Bohemia in 1643. The theorem was first proven by Jakob Steiner nearly 200 years later in 1826. Interestingly, this theorem was known to the Japanese around 1830 in Hashimoto Masataka's book *Sanpo Tenzan Shogakusho* (筭法點竄初學抄).

**Key Equations**

\[{r}_{4}=\frac{{r}_{1}{r}_{2}{r}_{3}}{{r}_{1}{r}_{2}+{r}_{1}{r}_{3}+{r}_{2}{r}_{3}+2\sqrt{{r}_{1}{r}_{2}{r}_{3}({r}_{1}+{r}_{2}+{r}_{3})}}\]

\[{R}_{4}=\frac{-{r}_{1}{r}_{2}{r}_{3}}{{r}_{1}{r}_{2}+{r}_{1}{r}_{3}+{r}_{2}{r}_{3}-2\sqrt{{r}_{1}{r}_{2}{r}_{3}({r}_{1}+{r}_{2}+{r}_{3})}}\]

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## Comments

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TopNewestYou can help the community by contributing a wiki article on it! Just click on 'post something' and start from there.

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This should be helpful.

Cheers!

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Exactly. @Steven Zheng Unfortunately, some people don't take the time to look at notes. So putting it in a wiki will help people find this theorem easier and quicker. Also, the wiki is pretty much permanent and can be edited and revised as well.

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Check my proof of Desecartes Theorom, in the Desecartes Theorom wiki.

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Gogeometry right!!!!!!!!!!!!

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Could you help add to the Descartes Circle Theorem Wiki? Thanks

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Sorry, I have been very busy lately with school and extracurricular activities. I've tried using Heron's formula to calculate the radius of the incircle, but to no avail. However I am writing notes on quantum mechanics. Perhaps those can be turned into wikis.

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It would also be nice to see proofs of formulas relating to excircles of triangles because I can't them anywhere!

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sighs... I still don't understand the proof...Log in to reply

me neither

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