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Today I've encountered a question like the following I am adding a picture because I have to;

enter image description here

The question paragraph says;

$\text{Given} \quad |OF|=6 \quad \text{and} \quad |BF|=4$

What is $|CH|=x$

My Attempts;

I have noticed that the diameter $r=10$ (1)

I have drawn a line from $C$ to $O$ which also is $r$ (2)

I have written $|HO|=\sqrt{100-x^2}$ but couldn't go further,

What do you suggest?

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  • $\begingroup$ How about $\angle EOF$? $\endgroup$
    – Math Lover
    Sep 19, 2017 at 15:30
  • $\begingroup$ Do you imply that it is $90^\circ$? It didn't work $8$ is the wrong answer:( $\endgroup$
    – Deniz Tuna Yalçın
    Sep 19, 2017 at 15:32
  • $\begingroup$ Do you mean the diameter or the radius is 10? $\endgroup$
    – Seyed
    Sep 19, 2017 at 15:39
  • $\begingroup$ $|OB|=10$ so $|OC|=10$ too, was the diameter $2r$ or $r$ I probably remembered it wrongly. $\endgroup$
    – Deniz Tuna Yalçın
    Sep 19, 2017 at 15:40
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    $\begingroup$ Something more is needed here. Is it also given that EC=ED=EF, as marked in the figure? $\endgroup$
    – Edward Porcella
    Sep 19, 2017 at 16:06

2 Answers 2

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Let $\alpha=\angle HFC$ and $a=EF$, so that $x=2a\sin\alpha$.

From the cosine law applied to triangle $OFC$ we get: $$ OF^2+FC^2-2OF\cdot FC\cos\alpha=OC^2, \quad\hbox{that is:}\quad a^2-6a\cos\alpha=16. $$

From the cosine law applied to triangle $OFD$ (notice that $FD=\sqrt2a$) we get: $$ OF^2+FD^2-2OF\cdot FD\cos(\pi/4+\alpha)=OD^2, \quad\hbox{that is:}\quad a^2-6a\cos\alpha+6a\sin\alpha=32. $$ Substituting here our previous result we thus get $6a\sin\alpha=16$, that is $x=2a\sin\alpha={16\over3}$.

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  • $\begingroup$ Is it ODF instead of EFD and I got lost in The part where you said $\cos(\pi/4+\alpha)$ why added $\alpha$. Excuse me for my ignorance and many thanks for this elegant solution:)) $\endgroup$
    – Deniz Tuna Yalçın
    Sep 19, 2017 at 16:38
  • $\begingroup$ @Deniz: $\angle DFE=\pi/4$ from triangle $EFD$ $\endgroup$
    – Vasili
    Sep 19, 2017 at 16:48
  • $\begingroup$ $\angle OFD=\angle OFE+\angle EFD=\alpha+\pi/4$. And then you need of course the cosine addition formula. $\endgroup$
    – Intelligenti pauca
    Sep 19, 2017 at 16:56
  • $\begingroup$ I know the cosine sum formula, but I carelessly thought that $\pi/4=90^\circ$ I then realised it was $45^\circ$, thank you:) $\endgroup$
    – Deniz Tuna Yalçın
    Sep 19, 2017 at 17:04
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This is only a partial solution, because I get the results using analytic geometry

Anyway I hope it can be useful

enter image description here

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  • $\begingroup$ Which software do you use to draw these figures? $\endgroup$
    – Math Lover
    Sep 19, 2017 at 17:01
  • $\begingroup$ @MathLover GeoGebra at geogebra.org It is much much more than a tool for drawing! $\endgroup$
    – Raffaele
    Sep 19, 2017 at 17:08
  • $\begingroup$ Thanks @Raffaele. $\endgroup$
    – Math Lover
    Sep 19, 2017 at 17:10

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