Difference between revisions of "2021 Fall AMC 10B Problems/Problem 11"
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Thus, we have that <math>C-H=H-A</math>, or <math>A=2H-C</math>. | Thus, we have that <math>C-H=H-A</math>, or <math>A=2H-C</math>. | ||
By some formulas, <math>C=\pi{r}^2=\pi</math> and <math>H=6\cdot\frac12\cdot1\cdot(\frac12\cdot\sqrt3)=\frac{3\sqrt3}2</math>. | By some formulas, <math>C=\pi{r}^2=\pi</math> and <math>H=6\cdot\frac12\cdot1\cdot(\frac12\cdot\sqrt3)=\frac{3\sqrt3}2</math>. | ||
| − | Thus, <math>A=3\sqrt3-\pi</math> or <math>\boxed{(\textbf{B} | + | Thus, <math>A=3\sqrt3-\pi</math> or <math>\boxed{(\textbf{B})}</math>. |
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| + | ~Hefei417, or 陆畅 Sunny from China | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=12|num-b=10}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=12|num-b=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:57, 23 November 2021
Problem
I hope someone will enter this.
Solution
Let the hexagon described be of area 
and let the circle's area be 
.
Let the area we want to aim for be 
.
Thus, we have that 
, or 
.
By some formulas, 
and 
.
Thus, 
or 
.
~Hefei417, or 陆畅 Sunny from China
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
