6
$\begingroup$

From the Iranian Geometry Olympiad, 2017:

In the regular pentagon $ABCDE$, the perpendicular at $C$ to $CD$ meets $AB$ at $F$. Prove that $AE + AF = BE$. Construction: https://www.geogebra.org/calculator/bnmgctmk

I can't seem to make much headway on this problem. You could probably use trigonometry to find the length of $BE$, but I am guessing there is a much easier (and elegant) solution that is eluding me.

$\endgroup$
1
  • 4
    $\begingroup$ Have you seen this on the Art of Problem Solving? $\endgroup$
    – PinkyWay
    Jun 16, 2020 at 10:44

6 Answers 6

3
$\begingroup$

Extend the segment $CF$ to meet the line $AE$ at $G$. Easy angle chase we see that $AG = AF$ so we need to prove $EG = BE (= CE)$ and this is true since $\angle CEB = \angle BEG (= 36^{\circ})$ and $CG\bot BE$.

![enter image description here

$\endgroup$
2
  • $\begingroup$ Which tool did you use for drawing this? $\endgroup$ Jun 18, 2020 at 9:18
  • 1
    $\begingroup$ Geogebra............ $\endgroup$
    – nonuser
    Jun 18, 2020 at 10:58
2
$\begingroup$

enter image description here

Note

$$\angle BAG = \angle ABG = 36,\>\>\>\>\> \angle EAG = \angle AGE = 72$$

and the triangles CBF and CGF are congruent, which leads to

$$\angle AFG = \angle AGF = 72$$

So, the triangles AEG, AGB and AFG are all isosceles, which yield

$$ BG = AG = AF,\>\>\>\>\>EG = EA$$

Thus,

$$BE = BG + GE = AF + AE$$

$\endgroup$
1
  • $\begingroup$ Which tool do you use to draw figures like this? $\endgroup$ Jun 18, 2020 at 9:18
1
$\begingroup$

RegularPentagon

The answer described below does not use numerical values of angles. A couple of auxiliary lines is needed to be drawn in order to facilitate the proof. One of them is the line joining the two vertices $A$ and $C$, which intersects $BE$ at $G$. The other is $FG$.

Let the length of a side of the pentagon be $a$. Using the properties of a regular pentagon, we can state that $BE$ is parallel to $CD$, while $AC$ is parallel to $DE$. Thos makes $CDEG$ a parallelogram. However, because $CD=DE=a$ (two side of the pentagon) , $CDEG$ is an oblique equilateral parallelogram called a rhombus. Therefore, we have, $EG=GC= a$. Since $AE$ is also a sides of the pentagon, we can state, $$EG=AE \tag{1}$$

Furthermore, $CB=CG$, which confirms that $BCG$ is an isosceles triangle. Since $CD$ is parallel to $BE$, $CF$ is the perpendicular bisector of $BG$. Therefore, $BGF$ is also an isosceles triangle. Due to the prevalent symmetry of a regular pentagon, $AG=BG$, which makes $ABG$ is an isosceles triangle as well.

Let $\measuredangle GAB=\phi$ and $\measuredangle FGA=\psi$. Since $ABG$ is am isosceles triangle, we have $\measuredangle ABG=\phi$. Since $BGF$ is an isosceles triangle, $\measuredangle BGF=\phi$ as well. Consequently, $\measuredangle AFG$, which is one of the exterior angle the triangle $BGF$, is equal to $2\phi$. Now, by considering the sum of the three angles of the triangle $AFG$, we can write, $$\measuredangle GAF + \measuredangle AFG + \measuredangle FGA = 180^o \quad\rightarrow\quad 3\phi+\psi=180^o \tag{2}$$

Since $AB$ and $EA$ are two adjacent sides of the pentagon, $ABE$ is an isosceles triangle. Therefore, $\measuredangle BEA=\measuredangle ABE=\phi$. According to equation (1), $AGE$ is an isosceles triangle, which means $\measuredangle AGE = 90^o-\frac{\phi}{2}$. Now, we know three angles, which sum up to $180^o$, i.e., $$\measuredangle BGF + \measuredangle FGA + \measuredangle AGE=\phi+\psi+90^o-\frac{\phi}{2}=180^o \quad\rightarrow\quad \phi+2\psi=180^o \tag{3}$$

By subtracting equations (2) from (3), we can obtain the following relationship between $2\phi$ and $\psi$. $$\psi=2\phi$$

Therefore, $AFG$ is an isosceles triangle, where $AF=AG$. Bu we are already aware that $AG=BG$, because $BGA$ is an isosceles triangle. This means, $$BG=AF \tag{4}.$$

Now, we can prove the required relation using the equations (1) and (4) as shown below. $$EG+BG=AE+AF \quad\rightarrow\quad BE= AE+AF$$

$\endgroup$
2
  • $\begingroup$ Which tool did you use in the construction of this figure? $\endgroup$ Jun 18, 2020 at 9:19
  • 1
    $\begingroup$ I drew the figure in Microsoft Visio $\endgroup$
    – YNK
    Jun 19, 2020 at 9:29
0
$\begingroup$

Algebraic solution: Let the side length of the pentagon be 1. Let $x=\frac{\pi}5$. Then the internal angles of the pentagon are $3x$ and we have $\measuredangle CBE=2x$, $\measuredangle EBA=x$

Then: $$ BE=1+2\cos 2x,\quad AE+AF=2-\frac{\cos 2x}{\cos x}. $$ Hence: $$\begin{align} (BE-AE-AF)\cos x&=2\cos2x\,\cos x+\cos2x-\cos x\\ &=(\cos x+\cos3x)+\cos2x-\cos x\\ &=\cos3x+\cos2x=0. \end{align}$$

$\endgroup$
0
$\begingroup$

Let the side of the pentagon be $1$ and the point of intersection of $CF$ and $BE$ be $P$. We have $\hat{EPC}=90$ and $\hat{ABE}=\hat{BEC}=36$.

Then $$BE=2\cos36 = x = EC$$ $$\implies EP = x\cos 36\implies BP=x(1-\cos36)$$

Now, $$\frac{BP}{BF}=\cos36\implies BF=\frac{x(1-\cos36)}{\cos36}$$ $$\implies AF=1-\frac{x(1-\cos36)}{\cos36}$$ $$\implies AE+AF=2-\frac{x(1-\cos36)}{\cos36}=\frac{2\cos36-x+x\cos36}{\cos36}$$

But since $2\cos36 = x$, we have $AE+AF=x=BE$ QED

$\endgroup$
0
$\begingroup$

The diagonals of a regular polygon divide its vertex angles into equal parts. So we fold along line $\overline{EB}$:

enter image description here

From the angle equality noted above, the fold-projected point $A'$ lies on line $\overline{EC}$, and $\overline{A'G}$ which is perpendicular to $\overline{CF'}$ also bisects $\angle CA'F'$. That forces $A'F'=A'C$. Then

$EC=EA'+A'C=EA'+A'F'=EA+AF$

and from the regularity of the pentagon we also have $EC=EB$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .