If the length of one of the diagonals of a regular pentagon is $d,$ how can we represent the perimeter of the pentagon in terms of $d$?

  • $\begingroup$ have you tried drawing a diagram and seeing what happens? $\endgroup$ – The Integrator Jun 1 '18 at 15:56
  • $\begingroup$ What have you attempted? $\endgroup$ – Andrew Li Jun 1 '18 at 15:56
  • $\begingroup$ In the multiple choices that I had, all of them had either a sine 36 or a sine 54. So I tried law of sines, but I ended up with 2 sines, but all the options had only one sine. Sorry if my question is too trivial. $\endgroup$ – Nithya S Jun 1 '18 at 15:58
  • $\begingroup$ What do you mean by 2 sines? Also you can easily take out the value of sine 36 or sine 54 $\endgroup$ – Love Invariants Jun 1 '18 at 16:03
  • $\begingroup$ For example, one of the answer was P = 5d/sin 36. They have left it at that not taking the value of sin 36. $\endgroup$ – Nithya S Jun 1 '18 at 16:04

As the perpendicular bisectors of the sides of the regular pentagon are its axes of symmetry, a fast angles computation shows the angle made by a side and the adjacent diagonal is equal to π/5, so we deduce that, denoting $s$ the length of a side, $d$ the length of a diagonal $$\frac d2=s\cos\frac\pi 5$$ and the perimeter $p$ is $$p=5s=\frac{5d}{2\cos\frac\pi 5}=\frac{5(\sqrt 5-1)d}{2}.$$

  • $\begingroup$ it should be $\sec{\frac\pi5}$. please correct it $\endgroup$ – The Integrator Jun 1 '18 at 17:07
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    $\begingroup$ @TheIntegrator: Oops! I should not compute directly on screen… Fixed. Thanks for pointing it! $\endgroup$ – Bernard Jun 1 '18 at 17:53
  • $\begingroup$ thanks for the edit , was bugging me watching that XD $\endgroup$ – The Integrator Jun 1 '18 at 17:54
  • $\begingroup$ How you evaluated $\cos\frac\pi5$ is not clear from this. $\endgroup$ – Michael Hardy Jun 1 '18 at 18:09

In Euclid you have a theorem that considers two angles subtended by an arc of a circle: one with its vertex at the center of the circle, and the other with its vertex and any point on the circle – any point at all other circle; it doesn't matter which one. The theorem says the measure of the latter angle is exactly half that of the former.

Circumscribe a circle around the regular pentagon. Look at the point on the circle that is an endpoint of the diagonal you are considering. That will be the vertex of several angles: (1) the angle between the tangent line to the circle and one of the sides; (2) the angle between that side and one of the diagonals of the pentagon; (3) the angle between that diagonal and the next diagonal; (4) the angle between that last diagonal and the next side of the pentagon; (5) the angle between that last side and the tangent line. The five arcs of the circle are equally long; therefore the measures of these five angles are equal; therefore each is $180^\circ/5 = 36^\circ.$

Those two diagonals and one side have three angles adding up to $180^\circ$ and two of them have equal measures and the third is $36^\circ;$ therefore the other two are $72^\circ.$

So how long is the short side of that isosceles triangle as a function of the length of one of the diagonals? Call the length of the diagonal $d$ and the length of the side we seek $s.$

If you bisect one of those $72^\circ$ angles you cut off two other triangles, the smaller of which has the same three angles and hence the same shape as the whole triangle, with two sides of length $s.$ The third side of that small triangle must have length $\dfrac s d\cdot s,$ by similarity. The larger of those two triangles cut off by bisection has a side of length $s.$ That side of length $s,$ plus the aforementioned side of length $\dfrac s d \cdot s,$ make up a diagonal of length $d.$ Therefore $$ s + \frac s d \cdot s = d $$ $$ \frac s d + \left( \frac s d\right)^2 = 1 $$ Solving this quadratic equation for $s/d$ yields $$ \frac s d = \frac{\sqrt 5 -1} 2. $$ Hence $$ s = \frac {d(\sqrt 5 -1 )} 2 $$ and the circumference of the pentagon is therefore $$ \frac{5d(\sqrt 5 -1)} 2. $$


Drawing the diagram gives us the answer;

enter image description here

$\beta $ and $\gamma$ can be found as shown below;

let $\beta =x$

in $\triangle$ABE , ${AB} = AE\implies \angle{ ABE} = \angle AEB = 108 - x$

$2(108-x)+108 = 180$

$x = 72^\circ$

In $\triangle BED$, $EB =BD = d$ and $\angle BED =\angle BDE = 72^\circ$

$\implies \gamma= 180 - 2(72) = 36^\circ$

Now applying the sine rule gives us;


where $s= ED$

$s= d\cdot \dfrac{\sin(36)}{\sin(72)}$

$s= d\cdot \dfrac{\sin(36)}{2\sin(36)\cos(36)}$

$s= \dfrac{d}2\cdot \sec(36)$

So the total perimeter is $p = \dfrac{5d}2\sec(36^\circ)$

  • $\begingroup$ Thank you very much. That was so clear. $\endgroup$ – Nithya S Jun 2 '18 at 0:16
  • $\begingroup$ @NithyaS if you found that helpful , consider voting and accepting the answer as the correct one. $\endgroup$ – The Integrator Jun 2 '18 at 7:41

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