Difference between revisions of "2022 AIME I Problems/Problem 8"

Revision as of 20:03, 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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-.
+.==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>
+It is not hard to find that one intersection point lies on <math>\frac{\sqrt{3}x}{3}</math>, 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>
+~bluesoul

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Difference between revisions of "2022 AIME I Problems/Problem 8"

Revision as of 20:03, 17 February 2022