2019 AIME II Problems/Problem 13

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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 1

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 denote the radius of the circle, O be the center of the circle. Then: \[r^2= 1^2 + \left(\sqrt{2}+1\right)^2= 4+2\sqrt{2}.\]

Now, we need to find the "D" shape, the small area enclosed by one side of the octagon and 18 of the circumference of the circle: \[D= \frac{1}{8} \pi r^2 - [A_1 A_2 O]=\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\]

Let PU be the height of A1A2P, PV be the height of A3A4P, PW be the height of A6A7P. From the 17 and 19 condition we have \[\triangle P A_1 A_2= \frac{\pi r^2}{7} - D= \frac{1}{7} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)\] \[\triangle P A_3 A_4= \frac{\pi r^2}{9} - D= \frac{1}{9} \pi \left(4+2\sqrt{2}\right)-\left(\frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right)\right)\] which gives PU=(1718)π(4+22)+2+1 and PV=(1918)π(4+22)+2+1.

Now, let A1A2 intersects A3A4 at X, A1A2 intersects A6A7 at Y,A6A7 intersects A3A4 at Z.

Clearly, XYZ is an isosceles right triangle, with right angle at X and the height with regard to which shall be 3+22. Now PU2+PV2+PW=3+22 which gives:

PW=3+22PU2PV2=3+2212((1718)π(4+22)+2+1+(1918)π(4+22)+2+1)=1+212(17+1914)π(4+22)

Now, we have the area for D and the area for PA6A7, so we add them together:

Target Area=18π(4+22)(2+1)+(1+2)12(17+1914)π(4+22)=(1812(17+1914))Total Area

The answer should therefore be 1822(16631664)=182504. The answer is 504.

SpecialBeing2017

Solution 2

Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is horizontal (and therefore $A_3A_4$ is vertical). Note that the area bounded by $\overline{A_iA_j}$ and the arc $\widehat{A_iA_j}$ is fixed, so we only need to consider the relevant triangles.

[asy] size(7cm); draw(Circle((0,0),1));  pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red);  dot(P);  for(int i=0; i<8; ++i) {   draw(dir(22.5+45i)--dir(67.5+45i));   draw((0,0)--dir(22.5+45i),gray+dashed); }  draw(dir(135)--dir(-45),blue+linewidth(1));  label("$P$", P, dir(-75));    label("$A_1$", dir(112.5), dir(112.5)); label("$A_2$", dir(112.5-45), dir(112.5-45)); label("$A_3$", dir(112.5-90), dir(112.5-90)); label("$A_4$", dir(112.5-135), dir(112.5-135)); label("$A_5$", dir(112.5-180), dir(112.5-180)); label("$A_6$", dir(112.5-225), dir(112.5-225)); label("$A_7$", dir(112.5-270), dir(112.5-270)); label("$A_8$", dir(112.5-315), dir(112.5-315));  dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]

Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\tfrac{1}{7}-\tfrac{1}{8}=\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\tfrac{1}{7}$. Similarly, $P$ was moved right by $\tfrac{1}{8}-\tfrac{1}{9}=\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\tfrac{1}{9}$. This means that $P$ has coordinates $(\tfrac{1}{72},-\tfrac{1}{56})$.

Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\tfrac{\sqrt{2}}{2}(\tfrac{1}{56}-\tfrac{1}{72})=\tfrac{\sqrt{2}}{504}$.

Remembering the definition of our unit, this yields a final area of \[\frac{1}{8}-\frac{\sqrt{2}}{\boxed{504}}.\]

-Archeon

Video Solution by MOP 2024

https://youtube.com/watch?v=CHJ15nlpZZk

~r00tsOfUnity

Video Solution by On the Spot STEM

https://youtu.be/B_Drjjn0vv0

See Also

2019 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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