Difference between revisions of "2018 AMC 12B Problems/Problem 25"
(Added another solution, which I think is a lot simpler and easy to think of. Might be multiple errors in formatting, this is my first time adding a solution!) |
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== Solution 1 == | == Solution 1 == | ||
+ | Let <math>O_1</math> and <math>O_2</math> be the centers of <math>\omega_1</math> and <math>\omega_2</math> respectively and draw <math>O_1O_2</math>, <math>O_1P_1</math>, and <math>O_2P_2</math>. Note that <math>\angle{O_1P_1P_2}</math> and <math>\angle{O_2P_2P_3}</math> are both right. Furthermore, since <math>\triangle{P_1P_2P_3}</math> is equilateral, <math>m\angle{P_1P_2P_3} = 60^\circ</math> and <math>m\angle{O_2P_2P_1} = 30^\circ</math>. Mark <math>M</math> as the base of the altitude from <math>O_2</math> to <math>P_1P_2.</math> Since <math>\triangle P_2O_2M</math> is a 30-60-90 triangle, <math>O_2M = 2</math> and <math>P_2M = 2\sqrt{3}</math>. Also, since <math>O_1O_2 = 8</math> and <math>O_1P_1 = 4</math>, we can find <math>P_1M = \sqrt{8^2 - (4 + 2)^2} = 2\sqrt{7}</math>. Thus, <math>P_1P_2 = P_1M + P_2M = 2\sqrt{3} + 2\sqrt{7}</math>. This makes <cmath>\left[P_1P_2P_3\right] = \frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}. </cmath> So, our answer is <math>252 + 300 = \boxed{\textbf{D) }552}</math>. | ||
− | + | Note by diyarv: | |
+ | |||
+ | A way to see that <math>P_1M = 2{\sqrt 7}</math> is through similar triangles. Call the intersection between <math>P_1M</math> and <math>O_1O_2</math> <math>S</math>. Since angles <math>\angle{O_1P_1S}</math> and <math>\angle{O_2MS}</math> are both right angles, and <math>\angle{P_1SO_1}</math> and <math>\angle{MSO_2}</math> are congruent, triangles <math>\triangle{P_1SO_1}</math> and <math>\triangle{MSO_2}</math> are similar by AA similarity. And, since <math>O_1P_1</math> is <math>4</math> and <math>O_2M</math> is 2, the common ratio is <math>2</math>. Using the fact that <math>O_1S + SO_2 = 8</math>, we see that these lengths are <math>\frac{16}{3}</math> and <math>\frac{8}{3}</math> respectively. Using the Pythagorean theorem, we see that <math>P_1S</math> and <math>SP_2</math> are <math>\frac{2\sqrt{7}}{3}</math> and <math>\frac{4\sqrt{7}}{3}</math> respectively, giving us a sum of <math>2\sqrt{7}</math>. | ||
== Solution 2 == | == Solution 2 == | ||
− | Let <math> | + | Let <math>O_i</math> be the center of circle <math>\omega_i</math> for <math>i=1,2,3</math>, and let <math>K</math> be the intersection of lines <math>O_1P_1</math> and <math>O_2P_2</math>. Because <math>\angle P_1P_2P_3 = 60^\circ</math>, it follows that <math>\triangle P_2KP_1</math> is a <math>30-60-90</math> triangle. Let <math>x=P_1K</math>; then <math>P_2K = 2x</math> and <math>P_1P_2 = x \sqrt 3</math>. The Law of Cosines in <math>\triangle O_1KO_2</math> gives <cmath>8^2 = (x+4)^2 + (2x-4)^2 - 2(x+4)(2x-4)\cos 60^\circ,</cmath>which simplifies to <math>3x^2 - 12x - 16 = 0</math>. The positive solution is <math>x = 2 + \tfrac23\sqrt{21}</math>. Then <math>P_1P_2 = x\sqrt 3 = 2\sqrt 3 + 2\sqrt 7</math>, and so the area of <math>\triangle P_1P_2P_3</math> is <cmath>\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}. </cmath>The requested sum is <math>300 + 252 = \boxed{552}</math>. |
== Solution 3 == | == Solution 3 == | ||
Line 41: | Line 44: | ||
This makes the answer <math>252 + 300 = 552</math>. <math>\boxed{\textbf{D}.}</math> | This makes the answer <math>252 + 300 = 552</math>. <math>\boxed{\textbf{D}.}</math> | ||
+ | |||
+ | ~AlbeePach~ | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | <asy> | ||
+ | unitsize(12); | ||
+ | pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; | ||
+ | real theta = 41.5; | ||
+ | pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; | ||
+ | filldraw(P1--P2--P3--cycle, gray(0.9)); | ||
+ | |||
+ | draw(Circle(A, 4)); | ||
+ | draw(Circle(B, 4)); | ||
+ | draw(Circle(C, 4)); | ||
+ | dot(P1); | ||
+ | dot(P2); | ||
+ | dot(P3); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | label("$P_1$", P1, E*1.5); | ||
+ | label("$P_2$", P2, SW*1.5); | ||
+ | label("$P_3$", P3, N); | ||
+ | label("$\omega_1$", A, W*30); | ||
+ | label("$\omega_2$", B, E*17); | ||
+ | label("$\omega_3$", C, W*17); | ||
+ | label("$\Gamma$", A, W*15); | ||
+ | |||
+ | draw(Circle(A,2),red); | ||
+ | pair X=foot(A,P1,P3); | ||
+ | dot(X,blue); | ||
+ | draw(A--X,blue); | ||
+ | |||
+ | label("$2\sqrt{7}$", X--P3); | ||
+ | label("$2\sqrt{3}$",X--P1); | ||
+ | label("$X$",X,dir(-80),blue); | ||
+ | </asy> | ||
+ | |||
+ | First, note that because the <math>\angle P_1=\angle P_2=\angle P_3=\pi/3</math>, the arcs inside the shaded equilateral triangle are each <math>2\pi/3</math>. Also, the distances between the centers of any two of the <math>3</math> given circles are each <math>8</math>. | ||
+ | Draw the circle <math>\Gamma</math> concentric with <math>\omega_1</math> with radius <math>2</math>. Because the arc of <math>\omega_1</math> inside said triangle is <math>2\pi/3</math>, <math>\Gamma</math> touches <math>P_1P_3</math>, say at a point <math>X</math>. Thus, <math>P_1P_3</math> is a common tangent of <math>\omega_3</math> and <math>\Gamma</math>, and it can be seen from inspection of the given diagram that the line is an common internal tangent. | ||
+ | The length of the common internal tangent segment <math>XP_3</math> of <math>\Gamma</math> and <math>\omega_3</math> is then <math>\sqrt{8^2-(2+4)^2}=2\sqrt{7}</math>, and it is easily seen that <math>XP_1=4\sin \pi/3=2\sqrt{3}</math>. | ||
+ | Because <math>P_1P_3=2(\sqrt{3}+\sqrt{7})</math>, the area of the shaded equilateral triangle is <math>\sqrt{3}(\sqrt{3}+\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}</math>. We get <math>\sqrt{300}+\sqrt{252}\Rightarrow\boxed{\textbf{(D) }552}.</math> | ||
+ | |||
+ | |||
+ | ~crazyeyemoody907 | ||
+ | |||
+ | ==Solution 5 (Pure Coordinate Bash)== | ||
+ | |||
+ | This seems like a coordinate bashable problem. To do this, we notice that it is easier to graph the equilateral triangle first, then the circles, rather than the other way around. Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to <math>\triangle P_1P_2P_3</math>'s side length. | ||
+ | |||
+ | WLOG let <math>P_1P_2 = 2</math>. Place triangle <math>P_1P_2P_3</math> on the coordinate plane with <math>P_1(0,\sqrt3)</math>, <math>P_2(1,0)</math>, and <math>P_3(-1,0)</math>. Let <math>r</math> be the radius of the circles. | ||
+ | |||
+ | Now, we find the coordinates of the centers of circles <math>\omega_1</math> and <math>\omega_2</math>. Since <math>\omega_2</math> is tangent to the x-axis at <math>(1,0)</math>, the center of <math>\omega_2</math> is <math>(1,r)</math>. Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of <math>\omega_1</math> to <math>P_1</math>. Since the slope of <math>\overline{P_1P_2}</math> is <math>-\sqrt3</math>, the slope of the hypotenuse is <math>\frac{1}{\sqrt{3}}</math>, so the right triangle is <math>30-60-90</math>. It's easy to see that the center of <math>\omega_1</math> is <math>\left(-\frac{r\sqrt{3}}{2}, \sqrt{3} - \frac{r}{2}\right)</math>. | ||
+ | |||
+ | Since <math>\omega_1</math> and <math>\omega_2</math> are tangent, the distance between the centers is <math>2r</math>, so we have | ||
+ | <cmath>\sqrt{\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2} = 2r</cmath> | ||
+ | <cmath>\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2 = 4r^2</cmath> | ||
+ | <cmath>1 + r\sqrt{3} + \frac{3r^2}{4} + \frac{9r^2}{4} - 3r\sqrt{3} + 3 = 4r^2</cmath> | ||
+ | <cmath>4 - 2r\sqrt{3} + 3r^2 = 4r^2</cmath> | ||
+ | <cmath>r^2 + 2r\sqrt{3} - 4 = 0</cmath> | ||
+ | By the Quadratic Formula, <math>r = \frac{-2\sqrt{3} \pm \sqrt{12 + 16}}{2} = -\sqrt{3}\pm\sqrt{7}</math>. | ||
+ | We take the positive value to get the radius of the circle is <math>\sqrt{7} - \sqrt{3}</math>. | ||
+ | |||
+ | Therefore, the ratio of the radius to the side length of the equilateral triangle is <math>\sqrt{7} - \sqrt{3} : 2 = 4 : \frac{8}{\sqrt{7} - \sqrt{3}} = 4 : 2(\sqrt{7} + \sqrt{3})</math>. | ||
+ | |||
+ | The side length of the equilateral triangle is <math>2(\sqrt{7} + \sqrt{3})</math>, so its area is <math>(\sqrt{7} + \sqrt{3})^2\sqrt{3} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252} \implies 300 + 252 = \boxed{\textbf{(D) }552}</math>. | ||
+ | |||
+ | ~rayfish | ||
+ | |||
+ | ==Solution 6 (Trignometry Bash)== | ||
+ | |||
+ | Let <math>O_1</math>, <math>O_2</math>, and <math>O_3</math> be the centers of <math>\omega_1</math>, <math>\omega_2</math>, and <math>\omega_3</math> | ||
+ | |||
+ | Let <math>P_1P_2 = a</math>, <math>\angle O_2O_1P_1= \alpha</math> | ||
+ | |||
+ | <cmath>[O_1P_1O_2P_2O_3P_3] = [P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = [O_1O_2O_3] + 3 \cdot [O_1P_1O_2]</cmath> | ||
+ | |||
+ | <cmath>[P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = \frac12 \cdot a \cdot \frac{ \sqrt{3} }{2} a + 3 \cdot \frac12 \cdot 4 \cdot a \cdot \sin 30^{\circ} = \frac{ \sqrt{3} }{4} a^2 + 3a</cmath> | ||
+ | |||
+ | <cmath>[O_1O_2O_3] + 3 \cdot [O_1P_1O_2] = \frac12 \cdot 8 \cdot \frac{ \sqrt{3} }{2} \cdot 8 + 3 \cdot \frac12 \cdot 4 \cdot 8 \cdot \sin \alpha = 16 \sqrt{3} + 48 \sin \alpha</cmath> | ||
+ | |||
+ | <cmath>\frac{ \sqrt{3} }{4} a^2 + 3a = 16 \sqrt{3} + 48 \sin \alpha</cmath> | ||
+ | |||
+ | <cmath>\sin \alpha = \frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3}</cmath> | ||
+ | |||
+ | By the Law of Cosine from <math>\triangle O_2P_1P_2</math>, <math>O_2P_1^2 = a^2 + 4^2 - 2 \cdot a \cdot 4 \cdot \cos 30^{\circ} = a^2 + 16 - 4a \sqrt{3}</math> | ||
+ | |||
+ | By the Law of Cosine from <math>\triangle O_1O_2P_1</math>, <math>O_2P_1^2 = 4^2 + 8 ^ 2 - 2 \cdot 4 \cdot 8 \cdot \cos \alpha = 80 - 64 \cos \alpha</math> | ||
+ | |||
+ | <cmath>a^2 + 16 - 4a \sqrt{3} = 80 - 64 \cos \alpha</cmath> | ||
+ | |||
+ | <cmath>\cos \alpha = -(\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)</cmath> | ||
+ | |||
+ | <cmath>\because \sin^2 \alpha + \cos^2 \alpha = 1</cmath> | ||
+ | |||
+ | <cmath>\therefore (\frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3})^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1</cmath> | ||
+ | |||
+ | <cmath>[ \frac{1}{\sqrt{3}} (\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)]^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1</cmath> | ||
+ | |||
+ | <cmath> (\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)^2 + 3 \cdot (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 3</cmath> | ||
+ | |||
+ | <cmath> (a^2 + 4 \sqrt{3} a - 64)^2 + 3 \cdot (a^2 - 4 \sqrt{3} a - 64)^2 = 3 \cdot 64^2</cmath> | ||
+ | |||
+ | <cmath>(a^2 - 64)^2 + 2 (a^2 - 64) \cdot 4 \sqrt{3} a + (4 \sqrt{3} a)^2 + 3(a^2 - 64)^2 - 6 (a^2 - 64) \cdot 4 \sqrt{3} a + 3(4 \sqrt{3} a)^2 = 3 \cdot 64^2</cmath> | ||
+ | |||
+ | <cmath>4 (a^2 - 64)^2 - 4 (a^2 - 64) \cdot 4 \sqrt{3} a + 192 a^2 - 3 \cdot 64^2 = 0</cmath> | ||
+ | |||
+ | <cmath>4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a) + 192(a^2 - 64) = 0</cmath> | ||
+ | |||
+ | <cmath>4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a + 48) = 0</cmath> | ||
+ | |||
+ | <cmath>(a^2 - 64)(a^2 - 4 \sqrt{3} a - 16) = 0</cmath> | ||
+ | |||
+ | <cmath>a^2 -64 = 0 \quad or \quad a^2 - 4 \sqrt{3} a - 16 = 0</cmath> | ||
+ | |||
+ | If <math>a^2 =64</math>, <math>a= 8</math>, <math>P_1O_2^2 = 8^2 + 4^2 - 2 \cdot 8 \cdot 4 \cdot \cos 30^{\circ} = 80 - 32 \sqrt{3} < 16</math>, <math>P_1O_2 < 4</math>, meaning that <math>P_1</math> is inside the circle. | ||
+ | |||
+ | However, <math>P_1</math> is not inside the circle. | ||
+ | |||
+ | <cmath>So, \quad a^2 - 4 \sqrt{3} a - 16 = 0</cmath> | ||
+ | |||
+ | <cmath>a = \frac{ 4 \sqrt{3} + \sqrt{ (4 \sqrt{3})^2 - 4(-16) } }{2} = 2 \sqrt{3} + 2 \sqrt{7}</cmath> | ||
+ | |||
+ | <cmath>[P_1P_2P_3] = a^2 \cdot \frac{ \sqrt{3} }{4} = (2 \sqrt{3} + 2 \sqrt{7})^2 \cdot \frac{ \sqrt{3} }{4} = \sqrt{300} + \sqrt{252}</cmath> | ||
+ | |||
+ | Therefore, the answer is <math>300 + 252 = \boxed{\textbf{(D) } 552}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/wINL0J76HYw | ||
+ | |||
+ | ~r00tsOfUnity | ||
==See Also== | ==See Also== | ||
+ | note: this looks a LOT like an AIME problem | ||
{{AMC12 box|year=2018|ab=B|num-b=24|after=Last Problem}} | {{AMC12 box|year=2018|ab=B|num-b=24|after=Last Problem}} |
Latest revision as of 02:30, 10 July 2024
Contents
Problem
Circles ,
, and
each have radius
and are placed in the plane so that each circle is externally tangent to the other two. Points
,
, and
lie on
,
, and
respectively such that
and line
is tangent to
for each
, where
. See the figure below. The area of
can be written in the form
for positive integers
and
. What is
?
Solution 1
Let and
be the centers of
and
respectively and draw
,
, and
. Note that
and
are both right. Furthermore, since
is equilateral,
and
. Mark
as the base of the altitude from
to
Since
is a 30-60-90 triangle,
and
. Also, since
and
, we can find
. Thus,
. This makes
So, our answer is
.
Note by diyarv:
A way to see that is through similar triangles. Call the intersection between
and
. Since angles
and
are both right angles, and
and
are congruent, triangles
and
are similar by AA similarity. And, since
is
and
is 2, the common ratio is
. Using the fact that
, we see that these lengths are
and
respectively. Using the Pythagorean theorem, we see that
and
are
and
respectively, giving us a sum of
.
Solution 2
Let be the center of circle
for
, and let
be the intersection of lines
and
. Because
, it follows that
is a
triangle. Let
; then
and
. The Law of Cosines in
gives
which simplifies to
. The positive solution is
. Then
, and so the area of
is
The requested sum is
.
Solution 3
Let be the center of circle
for
. Let
be the centroid of
, which also happens to be the centroid of
. Because
and
,
.
is
the height of
, thus
is
.
Applying cosine law on , one finds that
. Multiplying by
to solve for the height of
, one gets
. Simply multiplying by
and then calculating the equilateral triangle's area, one would get the final result of
.
This makes the answer .
~AlbeePach~
Solution 4
First, note that because the , the arcs inside the shaded equilateral triangle are each
. Also, the distances between the centers of any two of the
given circles are each
.
Draw the circle
concentric with
with radius
. Because the arc of
inside said triangle is
,
touches
, say at a point
. Thus,
is a common tangent of
and
, and it can be seen from inspection of the given diagram that the line is an common internal tangent.
The length of the common internal tangent segment
of
and
is then
, and it is easily seen that
.
Because
, the area of the shaded equilateral triangle is
. We get
~crazyeyemoody907
Solution 5 (Pure Coordinate Bash)
This seems like a coordinate bashable problem. To do this, we notice that it is easier to graph the equilateral triangle first, then the circles, rather than the other way around. Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to 's side length.
WLOG let . Place triangle
on the coordinate plane with
,
, and
. Let
be the radius of the circles.
Now, we find the coordinates of the centers of circles and
. Since
is tangent to the x-axis at
, the center of
is
. Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of
to
. Since the slope of
is
, the slope of the hypotenuse is
, so the right triangle is
. It's easy to see that the center of
is
.
Since and
are tangent, the distance between the centers is
, so we have
By the Quadratic Formula,
.
We take the positive value to get the radius of the circle is
.
Therefore, the ratio of the radius to the side length of the equilateral triangle is .
The side length of the equilateral triangle is , so its area is
.
~rayfish
Solution 6 (Trignometry Bash)
Let ,
, and
be the centers of
,
, and
Let ,
By the Law of Cosine from ,
By the Law of Cosine from ,
If ,
,
,
, meaning that
is inside the circle.
However, is not inside the circle.
Therefore, the answer is .
Video Solution by MOP 2024
~r00tsOfUnity
See Also
note: this looks a LOT like an AIME problem
2018 AMC 12B (Problems • Answer Key • Resources) | |
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