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The alternative vertices of two regular octagons are joined to form another octagon. Find the ratio of their areas.

I used the direct formulas for area of regular polygon given by $$\Delta=\frac{a^2n}{4\tan(\frac{180^{\circ}}{n})}$$ where $n$ is number of sides of regular polygon and $a$ is side length.

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Now from the figure $AB=a=BC$ and $\angle{ABC}=135^{\circ}$ and by cosine rule we get

$$AC=a\sqrt{2+\sqrt{2}}$$

But since the new octagon is $IJKLMNOP$ how can we find the length $IJ$

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    $\begingroup$ BI = BJ and IJ = $\sqrt2$BI. So AC=AI+IJ+JC=(2+ $\sqrt2$)IJ. $\endgroup$
    – fleablood
    Commented Feb 18, 2016 at 5:42
  • $\begingroup$ how can we prove mathematically that $BI=BJ$ and $\angle{IBJ}=90^{\circ}$ $\endgroup$ Commented Feb 18, 2016 at 6:03
  • $\begingroup$ It's a regular polygon. All the angles and sides are congruent. Just pick the congruent triangle to congruent triangle. By symmetry one can't be longer than other because they we're all constructed by the same method... but more in next comment. $\endgroup$
    – fleablood
    Commented Feb 18, 2016 at 6:32
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    $\begingroup$ Okay, HA = AB = BC etc. as are the vertices angles so by congruent parts of congruent triangles those square sides are all equal. Angle ABI and JBC etc. are therefore equal. Subtract them from the vertices angles and thus all the square corners are equal. So they are all right angles. Triangles ABI and JBC and all the rest are congruent by ASA. SO BI = BJ. $\endgroup$
    – fleablood
    Commented Feb 18, 2016 at 6:39
  • $\begingroup$ You don't need any formula for the area of any polygon. All you need is to find the ratio $(AB)^2/(IJ)^2$. By similarity of the regular octagons, that's the answer. $\endgroup$
    – David K
    Commented Feb 18, 2016 at 7:14

1 Answer 1

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Let $R, r$ be the radius of the circumscribed circle of $ABCDEFGH$ and the radius of the inscribed circle of $IJKLMNOP$ respectively. Then

Area of $\displaystyle ABCDEFGH=8\times \frac{1}{2} R^{2}\sin 45^{\circ}=2\sqrt{2}R^{2}$

Note that $IJ=2r\tan 13.5^{\circ}$

Area of $IJKLMNOP=8\times r^{2}\tan 13.5^{\circ}=8(\sqrt{2}-1)r^{2}$

Also note that $\displaystyle \frac{r}{R}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

The required ratio $\displaystyle \,= \frac{2\sqrt{2}R^{2}}{8(\sqrt{2}-1)r^{2}}= \frac{\sqrt{2}}{2(\sqrt{2}-1)}= \frac{\sqrt{2}(\sqrt{2}+1)}{2}= 1+\frac{\sqrt{2}}{2}$

P.S.: See the link here for the proof of $\tan 13.5^{\circ}=\sqrt{2}-1$. Of course, it also can be verified by Pythagoras' Theorem in $\Delta BIJ$ together with $AI=BI=BJ=CJ$.

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