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Difference between revisions of "2022 AIME II Problems/Problem 7"

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<math>r_1 = O_1A = 24</math>, <math>r_2 = O_2B = 6</math>, <math>AG = BO_2 = r_2 = 6</math>, <math>O_1G = r_1 - r_2 = 24 - 6 = 18</math>, <math>O_1O_2 = r_1 + r_2 = 30</math>
 
<math>r_1 = O_1A = 24</math>, <math>r_2 = O_2B = 6</math>, <math>AG = BO_2 = r_2 = 6</math>, <math>O_1G = r_1 - r_2 = 24 - 6 = 18</math>, <math>O_1O_2 = r_1 + r_2 = 30</math>
  
−
<math>\triangle O_2BD \sim \triangle O_1AD</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>\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}</math>,  <math>\frac{O_2D}{30} = \frac{6}{18}</math>

Revision as of 09:32, 19 February 2022

Problem

A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles.

Solution 1

[asy] //Created by isabelchen 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)); 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); [/asy]

$r_1 = O_1A = 24$, $r_2 = O_2B = 6$, $AG = BO_2 = r_2 = 6$, $O_1G = r_1 - r_2 = 24 - 6 = 18$, $O_1O_2 = r_1 + r_2 = 30$

$\triangle O_2BD \sim \triangle O_1GO_2$

$\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$, $\frac{O_2D}{30} = \frac{6}{18}$

$O_2D = 10$, $CD = O_2D + r_1 = 10 + 6 = 16$,

$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-O_1G^2} = \sqrt{30^2-18^2} = 24$

$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{\textbf{192}}$

~isabelchen

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

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