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Difference between revisions of "2022 AIME I Problems/Problem 8"
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.==solution 1==
 
.==solution 1==
 +
 
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>
  

Revision as of 19:03, 17 February 2022

.==solution 1==

Let bottom left point as the origin, the radius of each circle is $36/3=12$, note that three centers for circles are $(9\sqrt{3},3),(12\sqrt{3},12),(6\sqrt{3},12)$

It is not hard to find that one intersection point lies on $\frac{\sqrt{3}x}{3}$, plug it into equation $(x-9\sqrt{3})^2+(\frac{\sqrt{3}x}{3}-3)^2=12^2$, getting that $x=\frac{15\sqrt{3}+3\sqrt{39}}{2}$, the length is $2*(\frac{15\sqrt{3}+3\sqrt{39}-18\sqrt{3}}{2}=3\sqrt{39}-3\sqrt{3}$, leads to the answer $378$

~bluesoul

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