Difference between revisions of "2019 AIME II Problems/Problem 13"
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| + | ==Problem== | ||
| + | Regular octagon <math>A_1A_2A_3A_4A_5A_6A_7A_8</math> is inscribed in a circle of area <math>1.</math> Point <math>P</math> lies inside the circle so that the region bounded by <math>\overline{PA_1},\overline{PA_2},</math> and the minor arc <math>\widehat{A_1A_2}</math> of the circle has area <math>\tfrac{1}{7},</math> while the region bounded by <math>\overline{PA_3},\overline{PA_4},</math> and the minor arc <math>\widehat{A_3A_4}</math> of the circle has area <math>\tfrac{1}{9}.</math> There is a positive integer <math>n</math> such that the area of the region bounded by <math>\overline{PA_6},\overline{PA_7},</math> and the minor arc <math>\widehat{A_6A_7}</math> of the circle is equal to <math>\tfrac{1}{8}-\tfrac{\sqrt2}{n}.</math> Find <math>n.</math> | ||
| + | ==Solution== | ||
| + | |||
| + | ==See Also== | ||
| + | {{AIME box|year=2019|n=II|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:13, 22 March 2019
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
See Also
| 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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
