Find area of Pentagram : Regular Pentagon if $AP : PQ = m:1$

Question

While I was doing an Olympiad geometry sum in Sri Lanka I found this question

A pentagram is a regular pentagon with its sides extended to their point of intersection. In the pentagram ABCDE shown below PQRST is a regular pentagon.

If AP:PQ = m:1 then what is the ratio ,the area of the pentagram : the area of the pentagon ?

There are 5 answers:-

The above is the image of the question

My Attempt:-

I drew SP, and SQ so I got congruent $\triangle PTS$ and $\triangle SQR$

And also I found that $\triangle ATP$,$\triangle BPQ$,$\triangle CQR$,$\triangle DSR$,$\triangle ETS$,$\triangle SPQ$ are congruent

And also I drew vertical heights h in $\triangle SRQ$ and H in $\triangle SPQ$

Then I took AP as $lm$ and PQ as $l$

I find the value of H and h from $l$ and $m$ but the simplifying of the answer is very hard to find the ratio

So anyone could help me with this question

Thank you

Answer 1

As mentioned by @user10354138 in comments, this is the most elegant answer.

Alternatively you can use the following method (which is a little bit lengthy).

Let $\angle PQS$ be $\theta$. (I will use $\text p$ to imply pentagon and $\triangle$ to represent all triangles those are congruent with $\triangle BPQ$). So we get $$\triangle=\frac 12\cdot m\cdot \sin \theta\implies \sin\theta=\frac{2\triangle}{m}$$ Remove $\triangle QRS$ from its position and paste it on $\triangle BFQ$. Now $BTSF$ is a parallelogram and its area is equal to $\text{pentagon}+\triangle$.$$\text p+\triangle=(m+1)\cdot 1 \cdot\sin\theta$$ $$\text p+\triangle=(m+1)\cdot \frac{2\triangle}{m}$$ $$\implies\text p=\frac{(m+2)\triangle}{m}$$

Now what we have to calculate is $$\frac{\text{pentagram}}{\text{pentagon}}=\frac {\text p+5\triangle}{\text p}$$ Can you do that?

Answer 2

Let the area of $\triangle APT$ be $A$ i.e. $[APT]=A$. Notice that $\triangle APT$ and $\triangle PQT$ have the same height. So $$[APT]:[PQT]=m:1\implies [PQT]=\frac{[APT]}{m}=\frac Am.$$ Now notice that $\triangle PQT$ and $\triangle RTS$ are congruent (why?). Also $\triangle QRT$ and $\triangle QRC$ are congruent because $\angle TQR=\angle TRQ=\angle CQR=\angle CRQ=72^\circ$ and they have the same side $QR$. In other words, they have the same area.

Now $[PAT]=[BPQ]=[QRC]=[RSD]=[SET]=A$ (by symmetry) and $[PQT]=[RTS]=\frac Am$. So area of the pentagon is $[QTR]+2[PQT]=A+\frac{2A}m$ and area of the pentagram is $A+\frac{2A}m+5A=6A+\frac{2A}m$.

Hence, $\text{area of the pentagram:area of pentagon}=6A+\frac{2A}m:A+\frac{2A}m=6m+2:m+2.$