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I have noticed problems when posting on Math StackExchange recently. Mathjax, whether posted by me or by others comes out garbled. Usually, all is well when I refresh the download. Most recently, while I can read others' postings just fine, my own come out as a dreadful mess, and refreshing doesn't work. I am well aware that a missing or surplus dollar sign can wreak havoc, but I have checked carefully, and I don't think that any remaining typos would cause this degree of corruption.

I have tried using a different browser, to no avail. Does anyone have an idea what might be causing this? I copy, below, my latest attempt at posting.

We may conveniently partition $\angle BAC$ as $\angle BAP=18^\circ+\alpha$ and $\angle CAP=30^\circ-\alpha\$, where $-18^\circ<\alpha<30^\circ$. Comparing the triangles $BAP$ and $CAP$, we note that $AP$ is common and $|BP|=|CP|$ (since $\triangle BPC$ is isosceles). Hence, applying the sine rule in a similar way to each triangle yields $\sin30^\circ\sin(30^\circ-\alpha)=\sin54^\circ\sin(18^\circ+\alpha)$, which may be written $$\cos60^\circ\cos(60^\circ+\alpha)=\cos36^\circ\cos(72^\circ-\alpha).\qquad\qquad(1)$$ It is well known that $\cos36^\circ=\frac14(\surd5+1)$ and $\cos72^\circ=\frac14(\surd5-1)$. (A proof is appended.) Hence $\cos36^\circ\cos72^\circ=\frac14=\cos^260^\circ$, and therefore $\alpha=0$ is a solution to eqn $1$. Also, the LHS of eqn $1$ is decreasing while the RHS is increasing in the given range for $\alpha$. Hence the solution $\alpha=0$ is unique.

Appendix$\quad$ A property of the cosine is that $$\cos(3\times72^\circ)=-\cos36^\circ=\cos(2\times72^\circ).$$ From the identities $\cos2\theta=2\cos^2\theta-1$ and $\cos3\theta=4\cos^3\theta-3\cos\theta\$, it follows that $x=\cos72^\circ$ is a solution of the cubic $4x^3-3x=2x^2-1$, or $$(x-1)(4x^2+2x-1)=0.$$ Since $0<\cos72^\circ<1$, the linear factor may be ignored, and the required value is the positive root of the quadratic: $\cos72^\circ=\frac14(\surd5-1)$. The value $\cos36^\circ=\frac14(\surd5+1)$ follows from the initially mentioned properties.

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    $\begingroup$ Look at the spot where the rendering stops working. You have finished a math section with \$, which renders a dollar sign, rather than ending the math. Delete the `\`. $\endgroup$
    – Xander Henderson Mod
    May 26, 2023 at 18:54

1 Answer 1

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In the example you posted here on meta, you have "$\backslash\$$" or "\\$" in several places. For example,"\$\angle CAP=30^\circ-\alpha\$" does work. But if you write "\$\angle CAP=30^\circ-\alpha\\$" then the MathJax did not end there and it does not render correctly: $\angle CAP=30^\circ-\alpha\$ The text here until the next dollar will still be considered as part of the MathJax formula$

Let me try to copy-paste your text and remove all instances of \\\$.


We may conveniently partition $\angle BAC$ as $\angle BAP=18^\circ+\alpha$ and $\angle CAP=30^\circ-\alpha$, where $-18^\circ<\alpha<30^\circ$. Comparing the triangles $BAP$ and $CAP$, we note that $AP$ is common and $|BP|=|CP|$ (since $\triangle BPC$ is isosceles). Hence, applying the sine rule in a similar way to each triangle yields $\sin30^\circ\sin(30^\circ-\alpha)=\sin54^\circ\sin(18^\circ+\alpha)$, which may be written $$\cos60^\circ\cos(60^\circ+\alpha)=\cos36^\circ\cos(72^\circ-\alpha).\qquad\qquad(1)$$ It is well known that $\cos36^\circ=\frac14(\surd5+1)$ and $\cos72^\circ=\frac14(\surd5-1)$. (A proof is appended.) Hence $\cos36^\circ\cos72^\circ=\frac14=\cos^260^\circ$, and therefore $\alpha=0$ is a solution to eqn $1$. Also, the LHS of eqn $1$ is decreasing while the RHS is increasing in the given range for $\alpha$. Hence the solution $\alpha=0$ is unique.

Appendix$\quad$ A property of the cosine is that $$\cos(3\times72^\circ)=-\cos36^\circ=\cos(2\times72^\circ).$$ From the identities $\cos2\theta=2\cos^2\theta-1$ and $\cos3\theta=4\cos^3\theta-3\cos\theta$, it follows that $x=\cos72^\circ$ is a solution of the cubic $4x^3-3x=2x^2-1$, or $$(x-1)(4x^2+2x-1)=0.$$ Since $0<\cos72^\circ<1$, the linear factor may be ignored, and the required value is the positive root of the quadratic: $\cos72^\circ=\frac14(\surd5-1)$. The value $\cos36^\circ=\frac14(\surd5+1)$ follows from the initially mentioned properties.


Interestingly, I am not sure how to best type backslash followed by a dollar. When I try just to enclose it by backticks, I get this: \\\$. It seems to behave in the same way when used as code (after four spaces):

$\angle CAP=30^\circ-\alpha\\\$
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  • $\begingroup$ Thank you very much, Martin. It seems that I fumble-fingered the backslash key instead of the shift key; but these backslashes don't show up, either in the typed code or in the output on the site. That's something I need to watch out for in future! $\endgroup$ May 26, 2023 at 19:06
  • $\begingroup$ I still don't understand what's going on. The copy of my post in your answer looks fine after your editing out the pesky backslashes. But when I copy and paste the code for it on the intended web page, the same old garbage appears. Help! $\endgroup$ May 26, 2023 at 19:23
  • $\begingroup$ @JohnBentin I have edited your answer. You can see what changes I made in the revision history. $\endgroup$ May 26, 2023 at 19:32
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    $\begingroup$ Thanks again, Martin. I don't know how those two rogue backslashes crept in there after I copied your perfectly rendering code. Anyway, I am very happy with the result! $\endgroup$ May 26, 2023 at 19:39
  • $\begingroup$ Regarding the aside at the end of your post, I had a conversation with Calvin Khor about the bug in the rendering of "backslash+dollar" beginning here: math.meta.stackexchange.com/questions/34574/… $\endgroup$ May 27, 2023 at 11:28

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