Difference between revisions of "2022 AIME II Problems/Problem 7"
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(Taking diagram names from Solution 1. Also say the line that passes through O_1 and is parallel to line EF, call the points of intersection of that line and the circumference of circle O_1 points X and Y.) | (Taking diagram names from Solution 1. Also say the line that passes through O_1 and is parallel to line EF, call the points of intersection of that line and the circumference of circle O_1 points X and Y.) | ||
− | First notice that | + | First notice that DO_1 is a straight line because DXY is an isosceles triangle(or you can realize it by symmetry). That means, because DO_1 is a straight line, so angle BDO_2 = angle ADO_1, triangle ADO_1 is similar to triangle BDO_2. Also name DO_2 = x. By our similar triangles, BO_2/AO_1 = 1/4 = x/(x+30). Solving we get x = 10 = DO_2. Pythagorean Theorem on triangle DBO_2 shows BD = sqrt(10^2 - 6^2) = 8. By similar triangles, DA = 4*8 = 32 which means AB = DA - DB = 32 - 8 = 24. Because BE = CE = AE, AB = 2*BE = 24. BE = 12, which means CE = 12. CD = DO_2(its value found earlier in this solution) + CO_2(O_2 's radius) = 10 + 6 = 16. The area of DEF is 1/2 * CD * EF = CD * CE (because CE is 1/2 of EF) = 16 * 12 = 192. |
~Professor Rat's solution, added by heheman | ~Professor Rat's solution, added by heheman |
Revision as of 19:46, 29 April 2022
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
,
,
,
,
,
,
,
,
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 O_1 and is parallel to line EF, call the points of intersection of that line and the circumference of circle O_1 points X and Y.)
First notice that DO_1 is a straight line because DXY is an isosceles triangle(or you can realize it by symmetry). That means, because DO_1 is a straight line, so angle BDO_2 = angle ADO_1, triangle ADO_1 is similar to triangle BDO_2. Also name DO_2 = x. By our similar triangles, BO_2/AO_1 = 1/4 = x/(x+30). Solving we get x = 10 = DO_2. Pythagorean Theorem on triangle DBO_2 shows BD = sqrt(10^2 - 6^2) = 8. By similar triangles, DA = 4*8 = 32 which means AB = DA - DB = 32 - 8 = 24. Because BE = CE = AE, AB = 2*BE = 24. BE = 12, which means CE = 12. CD = DO_2(its value found earlier in this solution) + CO_2(O_2 's radius) = 10 + 6 = 16. The area of DEF is 1/2 * CD * EF = CD * CE (because CE is 1/2 of EF) = 16 * 12 = 192.
~Professor Rat's solution, added by heheman
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=7NGkVu0kE08
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.