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Difference between revisions of "2022 AIME II Problems/Problem 7"
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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>.
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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 \dot \left( \frac{3}{4} \dot 16 \right) = 2 \dot 12 = 24</math>. Therefore, the area is <math>\frac{1}{2} \dot 16 \dot 24 = \boxed{192}</math>.
  
 
==Video Solution (Mathematical Dexterity)==
 
==Video Solution (Mathematical Dexterity)==

Revision as of 12:01, 1 January 2024

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

Alternative Finish

Note that $\triangle{O_1 O_2 G} \sim \triangle{O_1 D A}$ by AA similarity. Setting up the ratio $\frac{18}{24}=\frac{24}{AB+8}$ and then substituting $AB$ with our known values will lead us to the same solution.

mathboy282

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

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, $\frac{BO_2}{AO_1} = \frac{1}{4} = \frac{x}{x+30}$. Solving we get $x = 10 = DO_2$. Pythagorean Theorem on triangle $DBO_2$ shows $BD = \sqrt{10^2 - 6^2} = 8$. By similar triangles, $DA = 4 \cdot 8 = 32$ which means $AB = DA - DB = 32 - 8 = 24$. Because $BE = CE = AE, AB = 2 \cdot BE = 24$. $BE = 12,$ 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 $\frac{1}{2} \cdot CD \cdot EF = CD \cdot CE$ (because $CE$ is $\tfrac{1}{2}$ of $EF$) $= 16 \cdot 12 = 192$.

~Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for $\LaTeX$.

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 $O_2D$. We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$, so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$. Solving, we get $x=10$. Therefore, $CD = 10 + 6 = 16$. Next, we find that using AA similarity, $\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \dot \left( \frac{3}{4} \dot 16 \right) = 2 \dot 12 = 24$ (Error compiling LaTeX. Unknown error_msg). Therefore, the area is $\frac{1}{2} \dot 16 \dot 24 = \boxed{192}$.

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=7NGkVu0kE08

Video Solution(The Power of Logic)

https://youtu.be/YAaiX_58Y7U

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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