2019 AIME II Problems/Problem 13

Problem

Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ and the minor arc $\widehat{A_3A_4}$ of the circle has area $\tfrac{1}{9}.$ There is a positive integer $n$ such that the area of the region bounded by $\overline{PA_6},\overline{PA_7},$ and the minor arc $\widehat{A_6A_7}$ of the circle is equal to $\tfrac{1}{8}-\tfrac{\sqrt2}{n}.$ Find $n.$

Solution

This problem is not difficult, but the calculation is tormenting.

The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, $1$ , and assume the side length of the octagon is $2$

Let $r$ denotes the radius of the circle, $O$ be the center of the circle.

$r^2= 1^2 + (\sqrt{2}+1)^2= 4+2\sqrt{2}$

Now, we need to find the "D"shape, the small area enclosed by one side of the octagon and 1/8 of the circumference of the circle

$D= \frac{1}{8} \pi r^2 - [A_1 A_2 O]=\frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1)$

Let $PU$ be the height of $\triangle A_1 A_2 P$ , $PV$ be the height of $\triangle A_3 A_4 P$ , $PW$ be the height of $\triangle A_6 A_7 P$ ,

From the 1/7 and 1/9 condition

we have

$\triangle P A_1 A_2= \frac{\pi r^2}{7} - D= \frac{1}{7} \pi (4+2\sqrt{2})-(\frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1))$

$\triangle P A_3 A_4= \frac{\pi r^2}{9} - D= \frac{1}{9} \pi (4+2\sqrt{2})-(\frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1))$

which gives

$PU= (\frac{1}{7}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1$

$PV= (\frac{1}{9}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1$

Now, let $A_1 A_2$ intersects $A_3 A_4$ at $X$ , $A_1 A_2$ intersects $A_6 A_7$ at $Y$ , $A_6 A_7$ intersects $A_3 A_4$ at $Z$

Clearly, $\triangle XYZ$ is an isosceles right triangle, with right angle at $X$

and the height with regard to which shall be $3+2\sqrt2$

That $\frac{PU}{\sqrt{2}} + \frac{PV}{\sqrt{2}} + PW = 3+2\sqrt2$ is a common sense

which gives $PW= 3+2\sqrt2-\frac{PU}{\sqrt{2}} - \frac{PV}{\sqrt{2}}$

$=3+2\sqrt{2}-\frac{1}{\sqrt{2}}((\frac{1}{7}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1+(\frac{1}{9}-\frac{1}{8}) \pi (4+ 2\sqrt{2}) + \sqrt{2}+1))$

$=1+\sqrt{2}- \frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4})\pi(4+2\sqrt{2})$

Now, we have the area for $D$ and the area for $\triangle P A_6 A_7$

we add them together

$Target Area= \frac{1}{8} \pi (4+2\sqrt{2})- (\sqrt{2}+1) + (1+\sqrt{2})- \frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4})\pi(4+2\sqrt{2})$

$=(\frac{1}{8} - \frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}))Total Area$

The answer should therefore be $\frac{1}{8}- \frac{\sqrt{2}}{2}(\frac{16}{63}-\frac{16}{64})=\frac{1}{8}- \frac{\sqrt{2}}{504}$

The final answer is, therefore, $\boxed{504}$

-By SpecialBeing2017

2019 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2019 AIME II Problems/Problem 13

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