# Golden ratio as a property of conic section (is it known?)

I am looking for a proof of a discovery as follows:

Let $$ABC$$ be arbitrary triangle and $$(\Omega)$$ be an arbitrary circumconic of $$ABC$$ let $$A'B'C'$$ is its tangential triangle of $$ABC$$ respect to $$(\Omega)$$. Let $$BB'$$ meet $$AC$$ at $$D$$ and $$CC'$$ meet $$AB$$ at $$E$$, let $$DE$$ meet the circumconic at $$F$$. A line through $$F$$ and parallel to $$B'C'$$ meets $$AB$$, $$AC$$ at $$H$$, $$G$$ (see Figure) then: $$\frac{HG}{GF}=\frac{\sqrt{5}+1}{2}.$$

• I think 'amazing' doesn't belong in an MO question, at least not describing your own work. I have edited it out, but of course you may revert if you find it in keeping with MO norms. May 10, 2021 at 1:45
• @LSpice thank you very much, I open to your suggestions May 10, 2021 at 2:03
• An equivalent statement is $HG^2=FG\cdot FH.$ This may be easier to prove. May 10, 2021 at 3:47
• @brainjam If $\frac{HG}{GF}=\varphi$ $\Rightarrow$ $HG^2=FG.FH$ but $HG^2=FG.FH$ no $\Rightarrow$ $\frac{HG}{GF}=\varphi$ May 10, 2021 at 4:41
• This may be a consequence of Poncelet's Porism for $n=3$. May 10, 2021 at 5:23

Consider generalizing the last line to

Let $$I$$ be a point on $$B'C'$$, and let $$FI$$ meet $$AB$$, $$AC$$ at $$H$$, $$G$$.

Then the cross ratio of $$H,F,G,I$$ is the negative golden ratio: $$\frac{HG\cdot FI}{FG\cdot HI}=-\frac{1+\sqrt{5}}{2} = -\phi$$

The original version is the limit of the general version as $$I$$ goes to infinity.

But the general version is projectively invariant, so we can prove it using coordinates where $$A$$ is at vertical infinity, the conic is $$y=x^2$$, and the tangent at $$A$$ is the line at infinity.

Then the first coordinates are simple: \begin{align} A &= \infty(0,1)\\ B &= (b,\ b^2)\\ C &= (c,\ c^2)\\ A' &= ((b+c)/2,\ bc)\\ B' &= \infty(1,2c)\\ C' &= \infty(1,2b)\\ D &= (c,\ (b-c)^2+c^2)\\ E &= (b,\ (b-c)^2+b^2)\\ F &= ((1-\phi)b+\phi c,\ ((1-\phi)b+\phi c)^2)\\ \end{align} and we can stop computing coordinates there. The graphic below shows the case $$b=-1$$, $$c=2$$.

(There is also another solution for $$F$$, not mentioned in the original question, which leads to switching $$\phi$$ and $$1-\phi$$ in the above and the below.)

We want to evaluate the cross ratio $$(HG\cdot FI)/(FG \cdot HI)$$. Since $$I$$ is a point at infinity, the $$FI$$ and $$HI$$ terms will cancel. Then we can compute the directed quotient $$HG/FG$$ from the $$x-$$coordinates, which are $$c$$ and $$b$$ for $$G$$ and $$H$$. This gives us the desired result: $$\frac{HG\cdot FI}{FG \cdot HI}=\frac{HG}{FG}=\frac{b-c}{(b-c)(1-\phi)}=-\phi$$