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

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==Solution 2==
 
==Solution 2==
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Let the center of the circle with radius <math>6</math> be labeled <math>A</math> and the center of the circle with radius <math>24</math> be labeled <math>B</math>. Drop perpendiculars on the same side of line <math>AB</math> from <math>A</math> and <math>B</math> to each of the tangents at points <math>C</math> and <math>D</math>, respectively. Then, let line <math>AB</math> intersect the two diagonal tangents at point <math>P</math>. Since <math>\triangle{APC} \sim \triangle{BPD}</math>, we have <cmath>\frac{AP}{AP+30}=\frac14 \implies AP=10.</cmath> Next, throw everything on a coordinate plane with <math>A=(0, 0)</math> and <math>B = (30, 0)</math>. Then, <math>P = (-10, 0)</math>, and if <math>C = (x, y)</math>, we have <cmath>(x+10)^2+y^2=64,</cmath> <cmath>x^2+y^2=36.</cmath> Combining these and solving, we get <math>(x, y)=(-\frac{18}5, \frac{24}5)</math>. Notice now that <math>P</math>, <math>C</math>, and the intersections of the lines <math>x=6</math> (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 <math>\frac{-\frac{18}5+10}{\frac{24}5}=\frac34</math>. Thus, the other two vertices of the desired triangle are <math>(6, 12)</math> and <math>(6, -12)</math>. By the Shoelace Formula, the area of a triangle with coordinates <math>(-10, 0)</math>, <math>(6, 12)</math>, and <math>(6, -12)</math> is <cmath>\frac12|-120-0-72-72+0-120|=\boxed{\textbf{192}}.</cmath>  
+
Let the center of the circle with radius <math>6</math> be labeled <math>A</math> and the center of the circle with radius <math>24</math> be labeled <math>B</math>. Drop perpendiculars on the same side of line <math>AB</math> from <math>A</math> and <math>B</math> to each of the tangents at points <math>C</math> and <math>D</math>, respectively. Then, let line <math>AB</math> intersect the two diagonal tangents at point <math>P</math>. Since <math>\triangle{APC} \sim \triangle{BPD}</math>, we have <cmath>\frac{AP}{AP+30}=\frac14 \implies AP=10.</cmath> Next, throw everything on a coordinate plane with <math>A=(0, 0)</math> and <math>B = (30, 0)</math>. Then, <math>P = (-10, 0)</math>, and if <math>C = (x, y)</math>, we have <cmath>(x+10)^2+y^2=64,</cmath> <cmath>x^2+y^2=36.</cmath> Combining these and solving, we get <math>(x, y)=\left(-\frac{18}5, \frac{24}5\right)</math>. Notice now that <math>P</math>, <math>C</math>, and the intersections of the lines <math>x=6</math> (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 <math>\frac{-\frac{18}5+10}{\frac{24}5}=\frac34</math>. Thus, the other two vertices of the desired triangle are <math>(6, 12)</math> and <math>(6, -12)</math>. By the Shoelace Formula, the area of a triangle with coordinates <math>(-10, 0)</math>, <math>(6, 12)</math>, and <math>(6, -12)</math> is <cmath>\frac12|-120-0-72-72+0-120|=\boxed{\textbf{192}}.</cmath>  
  
 
~A1001
 
~A1001

Revision as of 20:33, 3 April 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

Solution 2

Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\triangle{APC} \sim \triangle{BPD}$, we have \[\frac{AP}{AP+30}=\frac14 \implies AP=10.\] Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$. Then, $P = (-10, 0)$, and if $C = (x, y)$, we have \[(x+10)^2+y^2=64,\] \[x^2+y^2=36.\] Combining these and solving, we get $(x, y)=\left(-\frac{18}5, \frac{24}5\right)$. Notice now that $P$, $C$, and the intersections of the lines $x=6$ (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 $\frac{-\frac{18}5+10}{\frac{24}5}=\frac34$. Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$. By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$, $(6, 12)$, and $(6, -12)$ is \[\frac12|-120-0-72-72+0-120|=\boxed{\textbf{192}}.\]

~A1001

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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