Difference between revisions of "2022 AIME II Problems/Problem 7"
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<math>\triangle O_2BD \sim \triangle O_1GO_2</math>, <math>\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}</math>, <math>\frac{O_2D}{30} = \frac{6}{18}</math>, <math>O_2D = 10</math> | <math>\triangle O_2BD \sim \triangle O_1GO_2</math>, <math>\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}</math>, <math>\frac{O_2D}{30} = \frac{6}{18}</math>, <math>O_2D = 10</math> | ||
− | <math>CD = O_2D + | + | <math>CD = O_2D + r_2 = 10 + 6 = 16</math>, |
<math>EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24</math> | <math>EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24</math> | ||
Line 52: | Line 52: | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ===Alternative Finish=== | ||
+ | Note that <math>\triangle{O_1 O_2 G} \sim \triangle{O_1 D A}</math> by AA similarity. Setting up the ratio <math>\frac{18}{24}=\frac{24}{AB+8}</math> and then substituting <math>AB</math> with our known values will lead us to the same solution. | ||
+ | |||
+ | '''mathboy282''' | ||
==Solution 2== | ==Solution 2== | ||
Line 59: | Line 64: | ||
==Solution 3== | ==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.) | + | (Taking diagram names from Solution 1. Also say the line that passes through <math>O_1</math> and is parallel to line EF, call the points of intersection of that line and the circumference of circle <math>O_1</math> points <math>X</math> and <math>Y</math>.) |
+ | |||
+ | First notice that <math>DO_1</math> is a straight line because <math>DXY</math> is an isosceles triangle(or you can realize it by symmetry). That means, because <math>DO_1</math> is a straight line, so angle <math>BDO_2</math> = angle <math>ADO_1,</math> triangle <math>ADO_1</math> is similar to triangle <math>BDO_2</math>. Also name <math>DO_2 = x</math>. By our similar triangles, <math>\frac{BO_2}{AO_1} = \frac{1}{4} = \frac{x}{x+30}</math>. Solving we get <math>x = 10 = DO_2</math>. Pythagorean Theorem on triangle <math>DBO_2</math> shows <math>BD = \sqrt{10^2 - 6^2} = 8</math>. By similar triangles, <math>DA = 4 \cdot 8 = 32</math> which means <math>AB = DA - DB = 32 - 8 = 24</math>. Because <math>BE = CE = AE, AB = 2 \cdot BE = 24</math>. <math>BE = 12,</math> which means <math>CE = 12</math>. <math>CD = DO_2</math>(its value found earlier in this solution) + <math>CO_2</math> (<math>O_2</math> 's radius) <math>= 10 + 6 = 16</math>. The area of <math>DEF</math> is <math>\frac{1}{2} \cdot CD \cdot EF = CD \cdot CE</math> (because <math>CE</math> is <math>\tfrac{1}{2}</math> of <math>EF</math>) <math>= 16 \cdot 12 = 192</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>. Next, we find that using AA similarity, <math>\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD</math> and they are 3-4-5 triangles. We can quickly compute <math>EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24</math>. Therefore, the area is <math>\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}</math>. | ||
+ | |||
+ | ~[https://aops.com/wiki/index.php/User:Afly afly] | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== | ||
https://www.youtube.com/watch?v=7NGkVu0kE08 | https://www.youtube.com/watch?v=7NGkVu0kE08 | ||
+ | |||
+ | ==Video Solution(The Power of Logic)== | ||
+ | https://youtu.be/YAaiX_58Y7U | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=6|num-a=8}} | {{AIME box|year=2022|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:43, 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,
. Next, we find that using AA similarity,
and they are 3-4-5 triangles. We can quickly compute
. Therefore, the area is
.
~afly
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.