2019 AIME II Problems/Problem 13
Contents
Problem
Regular octagon is inscribed in a circle of area Point lies inside the circle so that the region bounded by and the minor arc of the circle has area while the region bounded by and the minor arc of the circle has area There is a positive integer such that the area of the region bounded by and the minor arc of the circle is equal to Find
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:
Now, we need to find the "D" shape, the small area enclosed by one side of the octagon and \(\frac{1}{8}\) of the circumference of the circle:
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 \(\frac{1}{7}\) and \(\frac{1}{9}\) condition we have which gives \(PU= \left(\frac{1}{7}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\) and \(PV= \left(\frac{1}{9}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\).
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 \(3+2\sqrt2\). Now \(\frac{PU}{\sqrt{2}} + \frac{PV}{\sqrt{2}} + PW = 3+2\sqrt2\) which gives:
\begin{align*} PW&= 3+2\sqrt2-\frac{PU}{\sqrt{2}} - \frac{PV}{\sqrt{2}} \\ &=3+2\sqrt{2}-\frac{1}{\sqrt{2}}\left(\left(\frac{1}{7}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1+\left(\frac{1}{9}-\frac{1}{8}\right) \pi \left(4+ 2\sqrt{2}\right) + \sqrt{2}+1\right)\\ &=1+\sqrt{2}- \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\pi\left(4+2\sqrt{2}\right) \end{align*}
Now, we have the area for \(D\) and the area for \(\triangle P A_6 A_7\), so we add them together:
\begin{align*} \text{Target Area} &= \frac{1}{8} \pi \left(4+2\sqrt{2}\right)- \left(\sqrt{2}+1\right) + \left(1+\sqrt{2}\right)- \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\pi\left(4+2\sqrt{2}\right)\\ &=\left(\frac{1}{8} - \frac{1}{\sqrt{2}}\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}\right)\right)\text{Total Area} \end{align*}
The answer should therefore be \(\frac{1}{8}- \frac{\sqrt{2}}{2}\left(\frac{16}{63}-\frac{16}{64}\right)=\frac{1}{8}- \frac{\sqrt{2}}{504}\). The answer is \(\boxed{504}\).
SpecialBeing2017
Solution 2
Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point 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 is horizontal (and therefore is vertical). Note that the area bounded by and the arc is fixed, so we only need to consider the relevant triangles.
Define one arbitrary unit as the distance that you need to move from to change the area of by . We can see that was moved down by units to make the area defined by , , and . Similarly, was moved right by to make the area defined by , , and . This means that has coordinates .
Now, we need to consider how this displacement in affected the area defined by , , and . This is equivalent to finding the shortest distance between and the blue line in the diagram (as and the blue line represents while is fixed). Using an isosceles right triangle, one can find the that shortest distance between and this line is .
Remembering the definition of our unit, this yields a final area of
-Archeon
Video Solution by MOP 2024
https://youtube.com/watch?v=CHJ15nlpZZk
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See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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