What is the maximum area of an octagon inscribed in a circle?

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    2017
In summary, an octagon inscribed in a circle is a regular octagon whose vertices lie on the circumference of a circle, with sides tangent to the circle. The maximum area of this octagon can be calculated by dividing the circle into 8 equal parts and using the formula A = (2 + √2) * r^2. The radius of the circle is equal to the distance from the center to any vertex of the octagon. The maximum area of an octagon inscribed in a circle is always the same, regardless of the size of the circle, and can never be greater than the area of the circle itself.
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Ackbach
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Here is this week's POTW:

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The octagon $P_1P_2P_3P_4P_5P_6P_7P_8$ is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon $P_1P_3P_5P_7$ is a square of area 5, and the polygon $P_2P_4P_6P_8$ is a rectangle of area 4, find the maximum possible area of the octagon.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 261 - May 01, 2017

This was Problem A-3 in the 2000 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows (extra kudos for terrific tikz drawing):

[TIKZ][scale=0.75]
\coordinate [label=above: $P_1$] (P1) at (90:5cm) ;
\coordinate [label=above right: $P_2$] (P2) at (33.13:5cm) ;
\coordinate [label=right: $P_3$] (P3) at (0:5cm) ;
\coordinate [label=below right: $P_4$] (P4) at (340:5cm) ;
\coordinate [label=below: $P_5$] (P5) at (270:5cm) ;
\coordinate [label=below left: $P_6$] (P6) at (213.13:5cm) ;
\coordinate [label=left: $P_7$] (P7) at (180:5cm) ;
\coordinate [label=above left: $P_8$] (P8) at (160:5cm) ;
\coordinate [label=left: $O$] (O) at (0,0) ;
\draw (P1) -- (P3) -- (P5) -- (P7) -- cycle ;
\draw (P2) -- (P4) -- (P6) -- (P8) -- cycle ;
\draw (P1) -- (P2) -- (P3) -- (P4) -- (P5) -- (P6) -- (P7) -- (P8) -- cycle ;
\draw (O) circle (5cm) ;
\fill (O) circle (2pt);
\draw[dashed] (P3) -- (O) -- (P2) ;
\draw[dashed] (P1) -- (O) -- (P4) ;
\draw[dashed] (O) -- (P5) ;
\draw (1,0.3) node {$\theta$} ;
\draw (1,-0.15) node {$\phi$} ;
[/TIKZ]​
The square $P_1P_3P_5P_7$ has area $5$, so it has side $\sqrt5$ and diagonal $\sqrt{10}$. The diagonal of the square is the diameter of the circle, which therefore has radius $\sqrt{5/2}.$

If the sides of the rectangle $P_2P_4P_6P_8$ are $a$ and $b$, then $ab=4$ and (by Pythagoras) $a^2+b^2 = 10.$ Therefore the sides are $\sqrt2$ and $2\sqrt2.$ It follows that $\tan(\angle P_2P_6P_4) = \frac12.$

Let $\alpha = \angle P_2OP_4$. Then $\alpha = 2(\angle P_2P_6P_4).$ So $\tan\frac\alpha2 = \frac1 2$, from which $\sin\frac\alpha2 = \frac1 {\sqrt5}$ and $\cos\frac\alpha2 = \frac2{\sqrt5}.$

Let $\theta = \angle P_2OP_3$ and $\phi = \angle P_3OP_4$, as in the diagram, and notice that $\theta+\phi = \alpha.$

The area $| P_2OP_3|$ of the triangle $ P_2OP_3$ is $\frac54\sin\theta$ (remembering that the radius of the circle is $\sqrt{5/2}$). Similarly $| P_3OP_4| = \frac54\sin\phi.$ Also, $|P_1OP_2| = \frac54\sin\bigl(\frac\pi2 - \theta\bigr) = \frac54\cos\theta$, and $|P_4OP_5| = \frac54\cos\phi.$

The sum of the areas of those four triangles is the area of the right half of the octagon. The left half of the octagon is congruent to the right half and therefore has the same area. So the are of the whole octagon is $\frac52\bigl(\sin\theta + \sin\phi + \cos\theta+ \cos\phi\bigr).$

Now apply the "sum-to-product" trig formulas to write that as $$5\Bigl(\sin\tfrac{\theta+\phi}2\cos\tfrac{\theta-\phi}2 + \cos\tfrac{\theta+\phi}2\cos\tfrac{\theta-\phi}2\Bigr) = 5\bigl(\sin\tfrac\alpha2 + \cos\tfrac\alpha2\bigr)\cos\tfrac{\theta-\phi}2 = 5\bigl(\tfrac1{\sqrt5} + \tfrac2{\sqrt5}\bigr)\cos\tfrac{\theta-\phi}2 = 3\sqrt5\cos\tfrac{\theta-\phi}2.$$ That is obviously maximised when $\theta=\phi$ so that $\cos\tfrac{\theta-\phi}2 = 1$. In conclusion, the maximum area of the octagon is $3\sqrt5$.
 

FAQ: What is the maximum area of an octagon inscribed in a circle?

What is the definition of an octagon inscribed in a circle?

An octagon inscribed in a circle is a regular octagon whose vertices lie on the circumference of a circle, and the sides of the octagon are tangent to the circle.

How is the maximum area of an octagon inscribed in a circle calculated?

The maximum area of an octagon inscribed in a circle is calculated by dividing the circle into 8 equal parts, and then using the formula A = (2 + √2) * r2, where r is the radius of the circle.

What is the relationship between the radius of the circle and the side length of the inscribed octagon?

The radius of the circle is equal to the distance from the center of the circle to any vertex of the octagon. This distance is also equal to the length of the octagon's side.

Is the maximum area of an octagon inscribed in a circle always the same?

Yes, the maximum area of an octagon inscribed in a circle is always the same. It does not depend on the size of the circle, as long as the octagon is inscribed in it.

Can the maximum area of an octagon inscribed in a circle be greater than the area of the circle?

No, the maximum area of an octagon inscribed in a circle can never be greater than the area of the circle. The area of the circle is always greater than the area of the inscribed octagon.

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