Difference between revisions of "2022 AIME I Problems/Problem 8"
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Let bottom left point as the origin, the radius of each circle is <math>36/3=12</math>, note that three centers for circles are <math>(9\sqrt{3},3),(12\sqrt{3},12),(6\sqrt{3},12)</math> | Let bottom left point as the origin, the radius of each circle is <math>36/3=12</math>, note that three centers for circles are <math>(9\sqrt{3},3),(12\sqrt{3},12),(6\sqrt{3},12)</math> | ||
| − | It is not hard to find that one intersection point lies on <math>\frac{\sqrt{3}x}{3}</math>, plug it into equation | + | It is not hard to find that one intersection point lies on <math>\frac{\sqrt{3}x}{3}</math> since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation |
<math>(x-9\sqrt{3})^2+(\frac{\sqrt{3}x}{3}-3)^2=12^2</math>, getting that <math>x=\frac{15\sqrt{3}+3\sqrt{39}}{2}</math>, the length is <math>2*(\frac{15\sqrt{3}+3\sqrt{39}-18\sqrt{3}}{2}=3\sqrt{39}-3\sqrt{3}</math>, leads to the answer <math>378</math> | <math>(x-9\sqrt{3})^2+(\frac{\sqrt{3}x}{3}-3)^2=12^2</math>, getting that <math>x=\frac{15\sqrt{3}+3\sqrt{39}}{2}</math>, the length is <math>2*(\frac{15\sqrt{3}+3\sqrt{39}-18\sqrt{3}}{2}=3\sqrt{39}-3\sqrt{3}</math>, leads to the answer <math>378</math> | ||
~bluesoul | ~bluesoul | ||
Revision as of 20:04, 17 February 2022
.==solution 1==
Let bottom left point as the origin, the radius of each circle is 
, note that three centers for circles are 
It is not hard to find that one intersection point lies on 
since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation
, getting that 
, the length is 
, leads to the answer 
~bluesoul
