11
$\begingroup$

A student sent me a question and I am stucked. I'd like to use classical geometry, instead of hard trigonometry if possible (can be trigonometric, but I'd like to not use scientific calculator; the only answer I got I need this), but I am in circles. The pentagon $ABCDE$ is regular. The answer is $48^\circ$ (I've constructed in Geogebra). Thank you in advance!

enter image description here

$\endgroup$
5
  • 1
    $\begingroup$ It can't be a constant number if you move a line CD. Otherwise, it should have a fixed position. I guess this pentagon is regular, isn't it? $\endgroup$ Feb 7 '20 at 14:10
  • 1
    $\begingroup$ @mathfux It does say so in the title line, though not in the question itself. $\endgroup$ Feb 7 '20 at 14:11
  • $\begingroup$ Now it's clear. $\endgroup$ Feb 7 '20 at 14:12
  • 1
    $\begingroup$ This was a really cool problem. Thanks for posting it! +1 $\endgroup$
    – user729424
    Feb 9 '20 at 13:59
  • $\begingroup$ @AndrewOstergaard, thank you! $\endgroup$ Feb 9 '20 at 14:06
10
+50
$\begingroup$

Let $G$ be reflection of $A$ about $EF$. Clearly $EG=AE=ED$ and $\angle GED = \angle AED - \angle AEF - \angle FEG = 108^\circ - 24^\circ -24^\circ=60^\circ$. Hence $GED$ is an equilateral triangle. So $GD=DE=DC$ and $\angle GDC=48^\circ$. Hence $\angle DCG=\angle CGD=66^\circ$, $\angle DGE=60^\circ$, and $\angle EGF =\angle FAE = 54^\circ$. So angles $CGD, DGE, EGF$ sum up to $180^\circ$. So $G$ lies on $FC$. So $\angle DCF = 66^\circ$ and by symmetry $\angle FDC=66^\circ$. It follows that $\angle CFD=48^\circ$.

$\endgroup$
5
  • $\begingroup$ Fantastic! Many thanks. $\endgroup$ Feb 9 '20 at 2:12
  • 2
    $\begingroup$ Your solution is beautiful! Thanks for posting it! +1 $\endgroup$
    – user729424
    Feb 9 '20 at 2:23
  • $\begingroup$ I hope you will take my imitation as sincere flattery: How can I solve this geometry problem without trigonometry? $\endgroup$ Apr 7 '20 at 23:01
  • $\begingroup$ Very elegant and creative Solution. I have a question, if we change the angle 24° for 42°, will this process work. I have make some attempts and seems that this will be a different way. $\endgroup$ Apr 15 '20 at 23:11
  • $\begingroup$ @PaúlAguilar This approach certainly would not work for your problem since the position of point $F$ would be different than in the OP's problem. Another construction would be needed. $\endgroup$
    – timon92
    Apr 16 '20 at 2:23
3
$\begingroup$

Without loss of generality, let the sides of the pentagon be $1$. Also let $ \mid BF \mid =x$ and $ \mid CF \mid =y$.

enter image description here

Apply the sine rule to $ABF$ \begin{eqnarray*} \frac{x}{ \sin(54)} =\frac{1}{\sin(102)}. \end{eqnarray*} Next apply the cosine rule to $BCF$ \begin{eqnarray*} y^2=x^2+1-2\cos(84). \end{eqnarray*} Cosine rule again, this time on $CDF$ \begin{eqnarray*} 1= 2y^2-2y^2\cos(\theta). \end{eqnarray*} Plug that into your casio (other brands of calculator are available) & you get $ \theta= \color{red}{48^{\circ}}$.

With such a neat final answer you certainly get the feeling there could be a much more elegant method ?

$\endgroup$
1
  • $\begingroup$ Many thanks for attention! My initial answer was similar to yours, but I've used sine rule more times. In fact, yours is better than mine (cleaner). But so I though as you... Is there a more elegant method? I am happy to say yes now, did you see the answer of timon92? Best regards. $\endgroup$ Feb 11 '20 at 14:39
2
$\begingroup$

Consider the point T such that TE = SE = SB Be the angle SET = 60 So the triangle SET is equal to time. On the other hand CTD = TED = ASE So the angle specified in the figure is proven (because I didn't have the right to post the photo) We know STC = 96 and because ST = CT So CST = SCT = 42 So SCD = SDC = 66 and so on CSD = 48

$\endgroup$
3
  • 1
    $\begingroup$ Welcome to math SE. Have a look at mathjax for your mathematical expressions. $\endgroup$ Feb 10 '20 at 20:12
  • $\begingroup$ Many thanks for attention! I am not sure that I can follow your thought. Did you mean $F $ by $S $? Maybe a figure would be fine if possible. Thank you so much. $\endgroup$ Feb 11 '20 at 1:03
  • $\begingroup$ Post a question ... you will be able to attach a link to a photo ... we can copy that & edit it into your answer ... I need to see the diagram, I am not understanding your solution at the moment. $\endgroup$ Feb 11 '20 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.