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Let ABCDE be a regular pentagon. If $\overline{BF} = \overline{BC}$, calculate $\alpha$.

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Using some trigonometry, it's a pretty simple exercise as you can reduce your problem to: $\dfrac{\sin(66°)}{\sin(42°+\frac{\alpha}{2})} = \dfrac{\sin(108°-\alpha)}{\sin(36°+\frac{\alpha}{2})}$.

Sadly, I've been having some problem to prove it geometrically. Any ideas? Thanks in advance.

Feripinho
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  • Essentially the same as https://math.stackexchange.com/questions/3627671/another-angle-inside-a-pentagon – timon92 Oct 12 '22 at 19:33

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Let's construct equilateral triangle $ABF$.

We have that $\angle{EAF}=48^\circ$, $\angle{AEF}=\frac{1}{2}(180^\circ-48^\circ)=66^\circ$.

Thus, $\angle{DEF}=108^\circ-66^\circ=42^\circ$ and $\alpha = \angle{EAF} = \angle{FBC}=48^\circ$

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Vasili
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  • Or just $\alpha=108-60=48$. I see why $BF=BA$, but why is the whole triangle equilateral? – TheBestMagician Oct 08 '22 at 16:14
  • @TheBestMagician The solution is, of course, correct. A "backward" one. You must construct the triangle and notice that the angles are equivalent. So, it's the same construction, then you use $\alpha = 108 - 60 = 48$. The only thing is why would ya do it. – Feripinho Oct 09 '22 at 00:24
  • I’m asking why $BFA$ is equilateral @Feripinho – TheBestMagician Oct 09 '22 at 03:29
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    @TheBestMagician, I think Vasili's diagram will be clearer if the point F is replaced by F'. That is he draw equilateral triangle BAF' as shown. He deduces that $\angle DEF'=42^o$ and $\angle CBF'=48^o$. From $ BF'=BC$ and $\angle DEF'=42^o$, one can see that F and F' coincide. Hence $\alpha= \angle CBF =\angle CBF'=48^o$ – Li Kwok Keung Oct 09 '22 at 06:35
  • @TheBestMagician: Should have replaced $F$ by $F'$. The triangle is equilateral by construction, we just show that $F'$ will coincide with $F$. – Vasili Oct 09 '22 at 11:14
  • Thanks, I was being silly – TheBestMagician Oct 10 '22 at 13:44
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angle alpha in pentagon Arguing in reverse:

Given the regular pentagon $ABCDE$, draw circles with radii $BC=AE=AB$ about centers $B$ and $A$, intersecting at $F$, and join $EF$, $FA$, and $FB$.

Since $\triangle ABF$ is equilateral, then$$\angle CBF=(108^o-60^o=48^o$$And since $\triangle EFA$ is isosceles, and $\angle FAE=\angle CBF=48^o$, then$$\angle AEF=\angle EFA=66^o$$making$$\angle DEF=(108^o-66^o=42^o$$Conversely then, given $\angle DEF=42^o$ and $BF=BC$, it follows that $\triangle ABF$ is equilateral and $\angle CBF=48^o$.

Edward Porcella
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The other answers refer to an equilateral triangle being present in thos construction. Here that advance knowledge is not required.

Draw diagonal $\overline{BE}$, thus completing $\triangle BEF$. From the regularity of the pentagon we know that $\angle BEA$ measures $72°$ and $\angle FEA$ is given to measure $42°$, so the difference $\angle BEF$ measures $30°$. Also the diagonL/side ratio of the pentagon is given by $2\sin54°$, where $\overline{BE}$ is a diagonal and $\overline{FB$ is given to be congruent to the sides.

So the Law of Sines applied to $\triangle BEF$ gives

$\dfrac{\sin\angle EFB}{\sin\angle BEF}=\dfrac{BE}{FB}$

$\dfrac{\sin\angle EFB}{\sin30°}=2\sin54°.$

Plugging $\sin30°=1/2$ then gives

$\sin\angle EFB=\sin 54°,$

but observe that this means $\angle EFB$ could measure either $54°$ or its supplement $126°$. With $F$ lying inside the pentagon and $FB$ congruent to the sides of this pentagon, we must have $|\angle EFB|>72°$ so we select

$\angle EFB|=(180-54)°=126°.$

That leaves $|\angle FBE|=(180-30-126)°=24°$ from which

$|\angle FBC|=(72-24)°=48°.$

We also have

$|\angle ABF|=(108-48)°=60°,$

proving that $\triangle ABF$ is equilateral indeed.

Oscar Lanzi
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