What is the semiperimeter of the ABCDE pentagon?
By John Campbell •
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For reference: (answer $3R+r_1+r_2$)
My progress: I think it can be solved by the property below together with the poncelet theorem and algebraic manipulation of tangents but I couldn't..
2 Answers
$\begingroup$Hints :
- Perimeter of pentagon = Perimeter of square $AHNE$ $-$ $(FN+NG-FG)$
- $FN+NG-FG=2\times$ inradius of $\triangle FNG$
which is similar to other three right triangles in picture and hence, its inradius is easily found.
$\endgroup$ 4 $\begingroup$$$8R - 2R\times \frac{FG}{BC}$$ $$=8R - 2R \left( 1 - \frac{r_1}{R} - \frac{r_2}{R} \right)$$
Say, $s$ is the sub-perimeter.
As the circle with radius $R$ is the ex-circle of $\triangle ABG$ and $\triangle EDL$ and circles with $r_1$ and $r_2$ are incircles, we have
$HE=ES=r_2 , AQ=AH=r_1$
Therefore,
$2s=(BF+FD+BA+AE+ED)\\
2s=(2R+2R+(R+r_1)+(r_1+r_2)+(R+r_2))\\
2s=(6R+2r_1+2r_2)\\
\boxed{s=3R+r_1+r_2}$
