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2022 AIME II Problems/Problem 7

Revision as of 15:34, 31 December 2024 by Vedoral (talk | contribs) (rad ax solution)

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]

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

$\triangle O_2BD \sim \triangle O_1GO_2$, $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$, $\frac{O_2D}{30} = \frac{6}{18}$, $O_2D = 10$

$CD = O_2D + r_2 = 10 + 6 = 16$,

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

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

~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 $EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24$. Therefore, the area is $\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}$.

~afly

Solution 5 (Radical Axis)

The common internal tangent of the two circles intersects the two external tangents at their midpoints as the internal tangent is the radical axis of the two circles. Therefore, the length of the internal tangent is equal to the length of the external tangents which is $24$ by pythagorean theorem. Let $x$ denote the length of the extension of the external tangents such that they intersect. By similar triangles, we get \[\frac{x}{6} = \frac{x+24}{24}\] Therefore, $x = 8$. Our triangle has side lengths $8+12$, $24$, and $8+12$ making it a $20-20-24$ triangle. Dropping the altitude to the side with length $24$, we get the altitude has length $\sqrt{20^2-12^2} = 16$. Therefore, the area of our triangle is $16\cdot 12 = \fbox{192}$.

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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