Pentagonal Recursion

Let the vertices of the exterior pentagon be $ABCDE$ , and the vertices of the interior pentagon be $A'B'C'D'E'$ , with $A'$ being the vertex directly opposite, or furthest from, $A$ . It is easy to calculate that each interior angle of a pentagon $=\frac{180^\circ \times 3}{5}=108^\circ$ , and by isosceles triangles, $\angle AEB = \angle CED = \frac{180^\circ-108^\circ}{2} = 36^\circ$ , and hence $\angle BEC = 36^\circ$ as well. Next, it clear by properties of (AAA) that $\triangle AEC$ is similar to $\triangle E'A'C$ is similar to $\triangle C'D'A$ . Thus, the required ratio, $\frac{AE}{C'D'} = \frac{AC}{AC'} = \frac{\sin\angle AC'C}{\sin\angle ACC'}$ . Simple calculations yield that $\angle ACC' = \frac{36^\circ}{2}=18^\circ$ and $\angle AC'C = 180^\circ - 36^\circ - 18^\circ = 126^\circ$ . Thus the required ratio is $\frac{\sin126^\circ}{\sin18^\circ} = \frac{\sin54^\circ}{\sin18^\circ}$ . Either by memorizing the value of $\sin18^\circ$ or by the derivation below, we have that $\sin18^\circ = \frac{\sqrt{5}-1}{4}$ , and then by the triple angle formula we have $\sin54^\circ = 3\sin18^\circ-4\sin^318^\circ$ , and solving, we easily obtain $\frac{\sin54^\circ}{\sin18^\circ}=\frac{3+\sqrt{5}}{2}$ . Thus the answer is 10.

Derivation of $\sin18^\circ$ : $\cos 18^\circ = \sin72^\circ=2\sin36^\circ \cos 36^\circ = 4\sin18^\circ \cos 18^\circ (1-2\sin^2 18^\circ)$ $\Rightarrow 1 = 4\sin18^\circ (1-2\sin^2 18^\circ)$ . Letting $\sin18^\circ = x$ , we have the cubic equation $8x^3 - 4x + 1 = (2x-1)(4x^2+2x-1)=0$ . Solving, we know the roots are $\frac{1}{2}, \frac{\pm \sqrt{5}-1}{4}$ Obviously $\sin18^\circ \neq \frac{1}{2}$ since $\sin30^\circ = \frac{1}{2}$ , and $\sin$ is increasing on $[0^\circ,90^\circ]$ , and the only other positive root is $\frac{\sqrt{5}-1}{4}$ , so that must be the value of $\sin18^\circ$ .

Let ABCDE be a regular pentagon where naming is done in clockwise direction. PQRST be the pentagon within it labelled clockwise such that AC and BE intersect at P. Let each side be equal to $a$ and each angle is obviously $108 ^ \circ$ . In isosceles triangle ABC, $\angle BAC = 36 ^ \circ$ . In isosceles triangle BCD $\angle DBC = 36 ^ \circ \Rightarrow \angle ABQ = 72 ^ \circ$ . So in triangle AQB by angle sum property , $\angle AQB = 72 ^ \circ$ . Hence AQB is isosceles $\Rightarrow AB = AQ = a$ . Now we are interested in finding $\frac {AB}{PQ} = \frac {AB}{AQ - AP} = \frac {a}{a - AP}$ . Again in isosceles triangle AEB $\angle ABE = 36 ^ \circ$ . So in APB $\angle APB = 108 ^ \circ$ . Now, in APB and ABC, $\angle APB = \angle ABC = 108 ^ \circ$ and $\angle A$ is common. So, APB and ABC are similar. So $( \frac {AP}{AB} )^2 = \frac {ar.(APB)}{ar.(ABC)}$ . Applying sine rule the equations reduce to $( \frac {AP}{AB} )^2 = \frac {AP \times AB}{AC \times AB}$ $\Rightarrow (AB)^2 = AP \times AC \Rightarrow a^2 = AP \times (a + AP)$ $\Rightarrow AP^2 + a AP + a^2 = 0$ . Solving this equation for AP we get $AP = \frac{-a \pm \sqrt{a^2-4(-a^2)}}{2}$ $= \frac {- a + a \sqrt {5}}{2} OR \frac {- a - a \sqrt {5}}{2}$ but the latter is not valid because AP is positive as it is length. So, $AP = \frac {- a + a \sqrt {5}}{2}$ . Therefore the required ratio is $\frac {a}{a - AP} = \frac {a}{a - \frac {- a + a \sqrt {5}}{2}}$ $= \frac {2}{3 - \sqrt {5}} = \frac {3 + \sqrt {5}}{2}$ by rationalizing the denominator. This is in the required form with a=3, b=5, c=2. Therefore 10 is the required answer.

Let the vertices of the bigger pentagon be A, B, C, D and E in anti-clockwise direction and the vertices of the smaller pentagon be F, G, H, I and J in anti-clockwise direction such that Points A, F, G, C are collinear.

Let AE = b, AJ = a and FJ = 1. We are trying to find $\frac {b} {1}$ = b.

We can see triangle AFJ is similar to triangle ACD as they are both $36 ^ \circ$ , $72 ^ \circ$ , $72 ^ \circ$ triangles. We get $\frac {a} {1}$ = $\frac {2a+1} {b}$ (1).

We also can see triangle ABF is similar to triangle EBA as they are both $36 ^ \circ$ , $36 ^ \circ$ , $108 ^ \circ$ triangles. We get $\frac {b} {2a+1}$ = $\frac {a} {b}$ (2).

From the above two equalities, we can get $\frac {a} {b}$ = $\frac {1} {a}$ , so $b = \sqrt {a}$

Substituting b for $\sqrt {a}$ in (2), we get $\frac {b} {2 \cdot \sqrt {b} + 1}$ = $\frac {\sqrt {b}} {b}$ .

By cross multiplying, we get $b^2$ = $2b + \sqrt {b}$ . Let k be $\sqrt {b}$ , we get $k^4$ = $2k^2 + k$ . By factorisation the equation, we get $k(k+1)( k^2 -k -1)$ =0. Obviously, k can't be negative or 0. By solving $k^2 -k -1=0$ , we get $k = \frac{1 \pm \sqrt{1^2+4}}{2}$ . Since k can't be negative or 0, $k= \frac{1 + \sqrt{5}}{2}$ .

Then, $k^2 = b = \frac{3 + \sqrt{5}}{2}$ . The required sum will be 3+5+2=10.

Call the larger pentagon $ABCDE$ and the smaller pentagon $FGHIJ$ , with F closest to AB, G closest to BC, and so on.

An internal angle of a regular pentagon measures $\frac{180 \cdot (5-2)}{5}=108^\circ$ , so $\angle ABC=\angle FGH=\angle CGB$ . $\bigtriangleup ABC$ and $\bigtriangleup CGB$ are isosceles, so $\angle BAC=\angle GBC=(180-108)/2=36^\circ$ . By AA Similarity, $\bigtriangleup ABC \sim \bigtriangleup CGB$ . Hence,

$\frac{AB}{AF+FG+GC}=\frac{GC}{CB}.AB=CB,AF=GC$ , so $\frac{AB}{2AF+FG}=\frac{AF}{AB}$ (*).

$\angle AGB=180-\angle CGB=72^\circ$ and $\angle ABG=180- \angle BAC - \angle AGB=72^\circ$ . Thus, $AB=AF+FG$ or $AF=AB-FG$ . Substituting this in $*$ and simplifying yields $AB^2-(3FG)(AB)+FG^2=0$ . We solve for $AB$ using quadratic formula, then divide by $FG$ to get $\frac{AB}{FG}=\frac{3+\sqrt5}{2}$ . (We disregard the other root since $\frac{AB}{FG}>1$ .)

Thus, the answer is $3+5+2=10$ .

This question definitely requires a pen and a paper in handy.

Let the pentagon so formed be ABCDE. Let AC meet BE at P and BD at Q. Let AD meet BE at T and CE at S. Let BD and CE meet at R.

So we know that all angles of a regular polygon of n sides are $\frac{180(n-2)}{n}$ .

This, for a pentagon, will be equal to $108^\circ$ .

So $\angle B = 108^\circ$ . (1)

In triangle BCD, we have

BC=CD and $\angle BCD = 108 ^ \circ$

So we get, $\angle CBD = \angle CDB = 36 ^\circ$ . (2)

Similarly, $\angle EBA = 36^\circ$ . (3)

So by using (1), (2) and (3), we obtain $\angle DBE = \angle B - \angle CBD - \angle EBA = 36^\circ$ .

So every angle of the pentagon is trisected by the diagonals, and each of the angle is equal to $36^\circ$ .

Now, in triangle CDS,

$\angle CSD$ = $180^\circ - \angle DCS -\angle CDS$ = $(180-36-72)^\circ$ = $72^\circ$

So CD = CS or, CD = CR+RS which means that CR = CD-RS..........(i)

Also, DR bisects $\angle CDS$

So, we have $\frac{CD}{DS}$ = $\frac{CR}{RS}$

by symmetry of the pentagon, we see that CR = DS

$\frac{CD}{CR}$ = $\frac{CR}{RS}$

which means that $CR^2$ = $CD*RS$

by using (i), we get

$(CD-RS)^2$ = $CD*RS$

$CD^2 +RS^2 -2*CD*RS$ = $CD*RS$

$CD^2 + RS^2 - 3*CD*RS$ = 0

Dividing both sides by $RS^2$ , we get

$\frac{CD^2}{RS^2} -\frac{3*CD}{RS} +1$ = 0

or, $z^2-3z+1$ = 0, where z = $\frac{CD}{RS}$

or, z= $\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}$ = $\frac{3\pm\sqrt5}{2}$

observing that z must be greater than 1, we see that

z= $\frac{3+\sqrt5}{2}$

So, from the question, we get

a=3 b=5 c=2

Therefore, $a+b+c$ = 10

1. Draw a regular pentagon, and draw in all the diagonals to reveal the inscribed pentagon.

2. Using triangle angle identities (all angles must add up to 180 degrees), determine the angles between all line segments.

3. Bisect each isoceles triangle to get two right triangles, and use sine and cosine rules to get the ratios between line segment lengths.

4. Simplify the ratios to get the final ratio between the length of a side of the outer pentagon $S$ , and the inscribed pentagon $s$ . This ratio should be $\frac{S}{s} = \frac{\cos 36^\circ}{\sin 18^\circ}$

5. According to trig identities $\cos 36^\circ = \frac{1 + \sqrt{5}}{4}$ , and $\sin 18^\circ = \frac{-1 + \sqrt{5}}{4}$

6. Simplify $\frac{S}{s} = \frac{\frac{1 + \sqrt{5}}{4}}{\frac{-1 + \sqrt{5}}{4}}$ and rationalize the root.

7. You should get $\frac{S}{s} = \frac{3 + \sqrt{5}}{2}$

8. $a + b + c = 3 + 5 + 2 = 10$

Let the vertices of the larger pentagon be $A, B, C, D$ and $E$ in order. Let the center of the pentagon be $O$ . Let the vertices of the smaller pentagon be $A’B’C’D’E’$ such that $A$ is opposite of $A’$ , $B$ is opposite of $B’$ and so on. By symmetry, $A A’$ is perpendicular to $DC$ , hence it passes through $O$ and the same can be said for $BB’$ , $CC’$ , etc. Also, the center of the smaller pentagon is likewise $O$ . Since the pentagon is regular, we have $BE \parallel CD$ , so triangles $AD'C'$ and $ACD$ are similar. Then, the ratio of sides is $\frac { CD } { C’D’ } = \frac { AD} {AC’}$ .

Since $ABC'$ and $ADB$ are $72^\circ-36^\circ-72^\circ$ triangles, by sine rule we have $\frac {AD}{AB} = \frac {\sin 72^\circ} { \sin 36^\circ}$ and $\frac {AB}{AC'} = \frac {\sin 72^\circ}{\sin 36^\circ}$ , thus $\frac {AD}{AC'} = ( \frac {\sin 72^\circ}{\sin 36^\circ} )^2 = (2 \cos 36^\circ)^2 = ( \frac {1 +\sqrt{5} } {2} )^2 = \frac {3+\sqrt{5} } {2}$ . Hence, $a + b + c = 3 + 5 + 2 = 10$ .