Difference between revisions of "2022 AIME II Problems/Problem 7"
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~Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for <math>\LaTeX</math>. | ~Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for <math>\LaTeX</math>. | ||
+ | |||
+ | ==Solution 4 (similar to solution 1)== | ||
+ | |||
+ | <asy> | ||
+ | //Created by isabelchen and edited by afly | ||
+ | |||
+ | size(12cm, 12cm); | ||
+ | |||
+ | draw(circle((0,0),24)); | ||
+ | draw(circle((30,0),6)); | ||
+ | draw((72/5, 96/5) -- (40,0)); | ||
+ | draw((72/5, -96/5) -- (40,0)); | ||
+ | draw((24, 12) -- (24, -12)); | ||
+ | draw((0, 0) -- (40, 0)); | ||
+ | draw((72/5, 96/5) -- (0,0)); | ||
+ | draw((168/5, 24/5) -- (30,0)); | ||
+ | draw((54/5, 72/5) -- (30,0)); | ||
+ | draw((30,0)--(30,15/2)); | ||
+ | |||
+ | dot((72/5, 96/5)); | ||
+ | label("$A$",(72/5, 96/5),NE); | ||
+ | dot((168/5, 24/5)); | ||
+ | label("$B$",(168/5, 24/5),NE); | ||
+ | dot((24,0)); | ||
+ | label("$C$",(24,0),NW); | ||
+ | dot((40, 0)); | ||
+ | label("$D$",(40, 0),NE); | ||
+ | dot((24, 12)); | ||
+ | label("$E$",(24, 12),NE); | ||
+ | dot((24, -12)); | ||
+ | label("$F$",(24, -12),SE); | ||
+ | dot((54/5, 72/5)); | ||
+ | label("$G$",(54/5, 72/5),NW); | ||
+ | dot((0, 0)); | ||
+ | label("$O_1$",(0, 0),S); | ||
+ | dot((30, 0)); | ||
+ | label("$O_2$",(30, 0),S); | ||
+ | dot((30,15/2)); | ||
+ | label("$H$",(30,15/2),N); | ||
+ | |||
+ | label("$x$",(30,0)--(40,0),N); | ||
+ | </asy> | ||
+ | |||
+ | First, we want to find <math>O_2D</math>. We know that <math>\angle O_1AD = \angle O_2BD = 90^{\circ}</math>, so by AA similarity, <math>\triangle O_1AD \sim \triangle O_2BD</math>. We want to find the length of <math>x</math>, and using the similar triangles, we write an equation: <math>\frac{30 + x}{4} = x</math>. Solving, we get <math>x=10</math>. Therefore, <math>CD = 10 + 6 = 16</math>. | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== |
Revision as of 11:56, 1 January 2024
Contents
Problem
A circle with radius is externally tangent to a circle with radius . Find the area of the triangular region bounded by the three common tangent lines of these two circles.
Solution 1
, , , ,
, , ,
,
Alternative Finish
Note that by AA similarity. Setting up the ratio and then substituting with our known values will lead us to the same solution.
mathboy282
Solution 2
Let the center of the circle with radius be labeled and the center of the circle with radius be labeled . Drop perpendiculars on the same side of line from and to each of the tangents at points and , respectively. Then, let line intersect the two diagonal tangents at point . Since , we have Next, throw everything on a coordinate plane with and . Then, , and if , we have Combining these and solving, we get . Notice now that , , and the intersections of the lines (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is . Thus, the other two vertices of the desired triangle are and . By the Shoelace Formula, the area of a triangle with coordinates , , and is
~A1001
Solution 3
(Taking diagram names from Solution 1. Also say the line that passes through and is parallel to line EF, call the points of intersection of that line and the circumference of circle points and .)
First notice that is a straight line because is an isosceles triangle(or you can realize it by symmetry). That means, because is a straight line, so angle = angle triangle is similar to triangle . Also name . By our similar triangles, . Solving we get . Pythagorean Theorem on triangle shows . By similar triangles, which means . Because . which means . (its value found earlier in this solution) + ( 's radius) . The area of is (because is of ) .
~Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for .
Solution 4 (similar to solution 1)
First, we want to find . We know that , so by AA similarity, . We want to find the length of , and using the similar triangles, we write an equation: . Solving, we get . Therefore, .
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=7NGkVu0kE08
Video Solution(The Power of Logic)
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.