What's the measure of the segment $PC$ in the pentagon below?

Question

For reference: On the arc $AB$ of the circumscribed circumference a regular pentagon $ABCDE$ becomes a point $P$; if $AP + BP = 8; PD = 12$ and $PE = 11$ calculate $PC$ (Answer:$9$)

My progress:

$AP+PB =8\\ PBDE: PB.DE+PE.BD =BE.PD\\ PB.l+11l \varphi=l\varphi.12\therefore \boxed{PB = \varphi}\\ PEDC: PE.CD+DE.PC=PD.CE\implies\\ 11l+PCl = 12.\varphi\therefore \boxed{PC=12 \frac{\varphi}{l}-11}\\ APDE:\boxed{PA = 11\varphi - 12}\\ PAEB: \boxed{PA = \frac{11}{\varphi} - 1}\\ ACPB: \boxed{PA+PB\varphi = PC}$

...???

It doesn't reach the result...I think there is something wrong with the statement... I couldn't do it in geogebra with this data either.

Answer 1

First, let me show another way to prove that the given data are inconsistent.

Say the length of a side of the pentagon is $x$. Then the length of the diagonal is $\varphi x$ where $\varphi$ is golden ratio.

Apply Ptolemy's theorem for quadrilateral $APBD$, considering $AP+BP=8$.

$$AP\cdot\varphi x+BP\cdot\varphi x=12\cdot x\implies \varphi=\frac32$$

This is clearly wrong and thus the problem statement is wrong.

So let me give the correction.

If we take $AP+BP+DP=20$, this problem is equivalent to your last problem, (Disclosure: the accepted answer is mine) and you will get the answer $9$ as expected (Verified).

What was wrong in the original problem is that in the scaled figure with $PE=11$ and $PC=9$, $PD$ is approximately equal to $12.3$ and $AP$, $BP$ are approximately $5.5$ and $2.2$ respectively. Therefore probably the author of the problem have rounded them to the nearest integer values. Although those three lengths add up to $20$ nicely.

Answer 2

The question seems incorrect as the given measurements are inconsistent. First, $PC$ is fixed given the lengths of $PD$ and $PE$ but it is not $9$. If we assume side of the pentagon is $a$ and $PC = x$, then applying Ptolemy's theorem in $PCDE$ -

$ \displaystyle 11 a + a \cdot x = 12 \cdot \frac{ (1 + \sqrt 5) a}{2} = (6 + 6 \sqrt 5) a$

$\implies x + 11 = 6 + 6 \sqrt 5 ~ $ or $~ x = 6 \sqrt 5 - 5 \ne 9$

Now if you go by $AP + BP = 8$ and apply Ptolemy's theorem in $APBE$ and $APCD$, you obtain a different value of $PC$.