Difference between revisions of "2017 AMC 12A Problems/Problem 16"
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<math> \textbf{(A)}\ \frac{3}{4} | <math> \textbf{(A)}\ \frac{3}{4} | ||
\qquad \textbf{(B)}\ \frac{6}{7} | \qquad \textbf{(B)}\ \frac{6}{7} | ||
− | \qquad\textbf{(C)}\ \frac{ | + | \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} |
\qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} | \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} | ||
\qquad\textbf{(E)}\ \frac{11}{12} </math> | \qquad\textbf{(E)}\ \frac{11}{12} </math> | ||
Line 30: | Line 30: | ||
draw((-3,0)--(3,0)); | draw((-3,0)--(3,0)); | ||
pair P = (9/7,12/7); | pair P = (9/7,12/7); | ||
+ | pair A = (-1,0); | ||
+ | pair C = (0,0); | ||
+ | pair B = (2,0); | ||
draw(circle(P,6/7)); | draw(circle(P,6/7)); | ||
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | ||
Line 35: | Line 38: | ||
draw((2,0)--P); | draw((2,0)--P); | ||
draw((0,0)--(9/5,12/5)); | draw((0,0)--(9/5,12/5)); | ||
+ | label("$A$",A,SW); | ||
+ | label("$C$",C,S); | ||
+ | label("$B$",B,SE); | ||
+ | label("$P$",P,N); | ||
</asy> | </asy> | ||
Line 43: | Line 50: | ||
draw((-1,0)--(2,0)); | draw((-1,0)--(2,0)); | ||
pair P = (9/7,12/7); | pair P = (9/7,12/7); | ||
+ | pair A = (-1,0); | ||
+ | pair C = (0,0); | ||
+ | pair B = (2,0); | ||
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | ||
draw((-1,0)--P); | draw((-1,0)--P); | ||
draw((2,0)--P); | draw((2,0)--P); | ||
draw((0,0)--P); | draw((0,0)--P); | ||
+ | label("$A$",A,SW); | ||
+ | label("$C$",C,S); | ||
+ | label("$B$",B,SE); | ||
+ | label("$P$",P,N); | ||
</asy> | </asy> | ||
Line 62: | Line 76: | ||
<cmath>28r = 24</cmath> | <cmath>28r = 24</cmath> | ||
<cmath>r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}</cmath> | <cmath>r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}</cmath> | ||
− | |||
− | |||
==Solution 2== | ==Solution 2== | ||
Line 81: | Line 93: | ||
</asy> | </asy> | ||
− | Like the solution above, connecting the centers of the circles results in triangle <math>\Delta ABP</math> with cevian <math>PC</math>. The two triangles <math>\Delta APC</math> and <math>\Delta ABP</math> share angle <math>A</math>, which means we can use Law of Cosines to set up a system of 2 equations that solve for <math>r</math> respectively: | + | Like the solution above, connecting the centers of the circles results in triangle <math>\Delta ABP</math> with cevian <math>PC</math>. The two triangles <math>\Delta APC</math> and <math>\Delta ABP</math> share angle <math>A</math>, which means we can use [[Law of Cosines]] to set up a system of 2 equations that solve for <math>r</math> respectively: |
− | <math>(2 + r)^2 + 1^2 - 2(2 + r)(1) | + | <math>(2 + r)^2 + 1^2 - 2(2 + r)(1)\cos A = (3 - r)^2</math> (notice that the diameter of the largest semicircle is 6, so its radius is 3 and <math>PC</math> is 3 - r) |
− | <math>(2 + r)^2 + 3^2 - 2(2 + r)(3) | + | <math>(2 + r)^2 + 3^2 - 2(2 + r)(3)\cos A = (r+1)^2</math> |
We can eliminate the extra variable of angle <math>A</math> by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find <math>r</math>: | We can eliminate the extra variable of angle <math>A</math> by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find <math>r</math>: | ||
<math>2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26</math> | <math>2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26</math> | ||
+ | |||
<math>8r + 2 = -20r + 26</math> | <math>8r + 2 = -20r + 26</math> | ||
+ | |||
<math>28r = 24</math>, so <math>r</math> = <math>6/7</math> <math>\boxed{(B)}</math> | <math>28r = 24</math>, so <math>r</math> = <math>6/7</math> <math>\boxed{(B)}</math> | ||
Line 118: | Line 132: | ||
</asy> | </asy> | ||
− | Let <math>C</math> be the center of the largest semicircle and <math>r</math> be the radius of <math>\circ P</math>. We know that <math>AC = 1</math>, <math>CB = 2</math>, <math>AP = r + 2</math>, <math>BP = r + 1</math>, and <math>CP = 3 - r</math>. Notice that <math>\Delta ACP</math> and <math>\Delta CBP</math> are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of <math>\Delta CBP</math> must be twice that of <math>\Delta ACP</math>, since the area of a triangle is <math>\frac{1}{2} Base \cdot Height</math>. | + | Let <math>C</math> be the center of the largest semicircle and <math>r</math> be the radius of <math>\circ P</math>. We know that <math>AC = 1</math>, <math>CB = 2</math>, <math>AP = r + 2</math>, <math>BP = r + 1</math>, and <math>CP = 3 - r</math>. Notice that <math>\Delta ACP</math> and <math>\Delta CBP</math> are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of <math>\Delta CBP</math> must be twice that of <math>\Delta ACP</math>, since the area of a triangle is <math>\frac{1}{2} \text{Base} \cdot \text{Height}</math>. |
Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. | Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. | ||
Line 133: | Line 147: | ||
draw((2,0)--P); | draw((2,0)--P); | ||
draw((0,0)--P); | draw((0,0)--P); | ||
− | label("$A$",A, | + | label("$A$",A,SW); |
label("$C$",C,S); | label("$C$",C,S); | ||
− | label("$B$",B, | + | label("$B$",B,SE); |
label("$P$",P,N); | label("$P$",P,N); | ||
</asy> | </asy> | ||
Line 142: | Line 156: | ||
[[Heron's Formula]] states that the area of an triangle with sides <math>a</math> <math>b</math> and <math>c</math> is | [[Heron's Formula]] states that the area of an triangle with sides <math>a</math> <math>b</math> and <math>c</math> is | ||
<cmath>\sqrt{s(s-a)(s-b)(s-c)}</cmath> | <cmath>\sqrt{s(s-a)(s-b)(s-c)}</cmath> | ||
− | where <math>s</math>, or the semiperimeter, is | + | where <math>s</math>, or the semiperimeter, is <math>\frac{a+b+c}{2}</math> |
The semiperimeter <math>s_1</math> of <math>\Delta ACP</math> is <cmath>\frac{[(r + 2) + (3 - r) + 1]}{2} = \frac{6}{2} = 3</cmath> | The semiperimeter <math>s_1</math> of <math>\Delta ACP</math> is <cmath>\frac{[(r + 2) + (3 - r) + 1]}{2} = \frac{6}{2} = 3</cmath> | ||
Line 161: | Line 175: | ||
<cmath>18r = 21r^2</cmath> | <cmath>18r = 21r^2</cmath> | ||
<cmath>r = \frac{18}{21} = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}</cmath> | <cmath>r = \frac{18}{21} = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}</cmath> | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>C</math>, the center of the large semicircle, to be at <math>(0, 0)</math>, and <math>P</math> to be at <math>(h, k)</math>. | ||
+ | |||
+ | <asy> | ||
+ | size(5cm); | ||
+ | draw(arc((0,0),3,0,180)); | ||
+ | draw(arc((2,0),1,0,180)); | ||
+ | draw(arc((-1,0),2,0,180)); | ||
+ | draw((-3,0)--(3,0)); | ||
+ | pair P = (9/7,12/7); | ||
+ | pair A = (-1,0); | ||
+ | pair C = (0,0); | ||
+ | pair B = (2,0); | ||
+ | draw(circle(P,6/7)); | ||
+ | dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); | ||
+ | draw((-1,0)--P); | ||
+ | draw((2,0)--P); | ||
+ | draw((0,0)--(9/5,12/5)); | ||
+ | label("$A$",A,SW); | ||
+ | label("$C$",C,S); | ||
+ | label("$B$",B,SE); | ||
+ | label("$P$",P,N); | ||
+ | </asy> | ||
+ | |||
+ | Therefore <math>A</math> is at <math>(-1, 0)</math> and <math>B</math> is at <math>(2, 0)</math>. | ||
+ | |||
+ | Let the radius of circle <math>P</math> be <math>r</math>. | ||
+ | |||
+ | Using Distance Formula, we get the following system of three equations: | ||
+ | |||
+ | <cmath>h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2</cmath> | ||
+ | |||
+ | By simplifying, we get | ||
+ | |||
+ | <cmath>h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1</cmath> | ||
+ | |||
+ | By subtracting the first equation from the second and third equations, we get | ||
+ | |||
+ | <cmath>8r=-4h+12, 10r=2h+6</cmath> | ||
+ | |||
+ | which simplifies to | ||
+ | |||
+ | <cmath>2r=3-h, 5r=h+3</cmath> | ||
+ | |||
+ | When we add these two equations, we get | ||
+ | |||
+ | <cmath>7r=6</cmath> | ||
+ | |||
+ | So <cmath>r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}</cmath> <math>w^5</math> | ||
+ | |||
+ | ==Solution 5(Inversion from [[Circular Inversion]])== | ||
+ | |||
+ | Let <math>\Omega</math> be a circle with radius of <math>6</math> and centered at the left corner of the semi-circle (O) with radius <math>3</math>. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram: | ||
+ | |||
+ | <asy> | ||
+ | size(5cm); | ||
+ | path circle1 = Circle((3, 0), 3); | ||
+ | path circle2 = Circle((2, 0), 2); | ||
+ | path circle3 = Circle((5, 0), 1); | ||
+ | pair P = (2,0)+(2+6/7)*dir(36.86989); | ||
+ | path circle4 = Circle(P, 6/7); | ||
+ | draw(circle1); | ||
+ | draw(circle2); | ||
+ | draw(circle3); | ||
+ | draw(circle4); | ||
+ | draw((0, 0)--(6, 0)); | ||
+ | dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); | ||
+ | path inversion = arc((0,0), 6, -30, 30); | ||
+ | draw(inversion, dashed); | ||
+ | label("$\Omega$", (6, 0) * dir(30), NE); | ||
+ | label("$C_1$", (3, 0), N); | ||
+ | label("$C_2$", (2, 0), N); | ||
+ | label("$C_3$", (5, 0), N); | ||
+ | label("$C_4$", P, N); | ||
+ | label("$O$", origin, W); | ||
+ | </asy> | ||
+ | |||
+ | <math>C_1</math> has radius <math>3</math>, <math>C_2</math> has radius <math>2</math>, and <math>C_3</math> has radius <math>1</math>. We want to find the radius of <math>C_4</math>. | ||
+ | |||
+ | We now invert the four circles. <math>C_1</math> inverts to a line. Given that one point is on <math>\Omega</math>, and all points on <math>\Omega</math> invert to themselves, we know that the resulting line must intersect that intersection point. <math>C_2</math> also inverts to a line. <math>C_2</math> has radius <math>4</math>, and since <math>\Omega</math> has radius of <math>6</math>, the resulting line must be <math>\frac{36}{4} = 9</math> units away from <math>O</math>. <math>C_3</math> inverts to a circle. By observing the diagram, we note that <math>C_3'</math>'s center must be on <math>\overline{OC_3}</math> and be between the two inverted lines, because <math>C_3</math> is tangent to <math>C_1</math> and <math>C_2</math> (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius <math>\frac{3}{2}</math> that is <math>\frac{15}{2}</math> units from <math>O</math>. | ||
+ | |||
+ | <asy> | ||
+ | size(10cm); | ||
+ | path circle1 = Circle((3, 0), 3); | ||
+ | path circle2 = Circle((2, 0), 2); | ||
+ | path circle3 = Circle((5, 0), 1); | ||
+ | pair P = (2,0)+(2+6/7)*dir(36.86989); | ||
+ | path circle4 = Circle(P, 6/7); | ||
+ | draw(circle1); | ||
+ | draw(circle2); | ||
+ | draw(circle3); | ||
+ | draw(circle4); | ||
+ | draw((0, 0)--(9, 0)); | ||
+ | dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); | ||
+ | path inversion = arc((0,0), 6, -30, 30); | ||
+ | draw(inversion, dashed); | ||
+ | label("$\Omega$", (6, 0) * dir(30), NW); | ||
+ | label("$C_1$", (3, 0), N); | ||
+ | label("$C_2$", (2, 0), N); | ||
+ | label("$C_3$", (5, 0), N); | ||
+ | label("$C_4$", P, N); | ||
+ | label("$O$", origin, W); | ||
+ | draw((6, 3)--(6, -3)); | ||
+ | draw((9, 3)--(9, -3)); | ||
+ | draw(Circle((15/2, 0), 3/2)); | ||
+ | label("$C_1'$", (6, 3), N); | ||
+ | label("$C_2'$", (9, 3), N); | ||
+ | label("$C_3'$", (15/2, 0), N); | ||
+ | dot((15/2, 0)); | ||
+ | </asy> | ||
+ | |||
+ | Now, we invert <math>C_4</math>. Note that <math>C_4</math> is tangent to the three other original circles. So, in the inversion, <math>C_4'</math> must be tangent to the two lines ''and'' <math>C_3'</math>. It is then quickly seen that <math>C_3'</math> and <math>C_4'</math> have the same radius: <math>\frac{3}{2}</math>. | ||
+ | |||
+ | Now, we can determine the radius of <math>C_4</math> using the formula <math>r = \frac{k^2 \cdot r'}{\overline{OC_2}^2 - r'^2}</math>. <math>k^2 = 36</math>, and <math>r' = \frac{3}{2}</math>. <math>\overline{OC}</math> is just the distance from the center of the inverted circle to the center of inversion. The center of <math>C_4'</math> is <math>3</math> units above the center of <math>C_3'</math>. Since <math>\overline{OC_3'} = \frac{15}{2}</math>, we use Pythagoras to learn that <math>\overline{OC_4'}^2 = \left(\frac{15}{2}\right)^2 + 9</math>. We do not take the square root because our relationship formula takes <math>\overline{OC_4'}^2</math>. | ||
+ | |||
+ | Therefore, we have: <cmath>r = \frac{36 \cdot \frac{3}{2}}{\left(\frac{15}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 9} = \frac{54}{9 \cdot 6 + 9} = \frac{6}{7} = \boxed{B}.</cmath> | ||
+ | |||
+ | Here is the diagram with <math>C_4'</math>. | ||
+ | |||
+ | <asy> | ||
+ | size(10cm); | ||
+ | path circle1 = Circle((3, 0), 3); | ||
+ | path circle2 = Circle((2, 0), 2); | ||
+ | path circle3 = Circle((5, 0), 1); | ||
+ | pair P = (2,0)+(2+6/7)*dir(36.86989); | ||
+ | path circle4 = Circle(P, 6/7); | ||
+ | draw(circle1); | ||
+ | draw(circle2); | ||
+ | draw(circle3); | ||
+ | draw(circle4); | ||
+ | draw((0, 0)--(9, 0)); | ||
+ | dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); | ||
+ | path inversion = arc((0,0), 6, -45, 45); | ||
+ | draw(inversion, dashed); | ||
+ | label("$\Omega$", (6, 0) * dir(45), NW); | ||
+ | label("$C_1$", (3, 0), N); | ||
+ | label("$C_2$", (2, 0), N); | ||
+ | label("$C_3$", (5, 0), N); | ||
+ | label("$C_4$", P, N); | ||
+ | label("$O$", origin, W); | ||
+ | draw((6, 4.5)--(6, -3)); | ||
+ | draw((9, 4.5)--(9, -3)); | ||
+ | draw(Circle((15/2, 0), 3/2)); | ||
+ | label("$C_1'$", (6, -3), SW); | ||
+ | label("$C_2'$", (9, -3), SE); | ||
+ | label("$C_3'$", (15/2, 0), NE); | ||
+ | dot((15/2, 0)); | ||
+ | draw(Circle((15/2, 3), 3/2)); | ||
+ | dot((15/2, 3)); | ||
+ | label("$C_4'$", (15/2, 3), N); | ||
+ | draw((15/2, 0)--(15/2, 3)); | ||
+ | draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 6 (Kissing Circles) == | ||
+ | Draw the other half of the largest circle and proceed with [[Descartes' Circle Formula]]. | ||
+ | |||
+ | The curvatures of circles A,B,C, and P are <math>1, \frac{1}{2}, -\frac{1}{3},</math> and <math>\frac{1}{r}</math>, respectively. | ||
+ | |||
+ | == Video Solution by Punxsutawney Phil == | ||
+ | https://www.youtube.com/watch?v=f_cdwEDlWXQ | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2017|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:24, 11 April 2021
Contents
Problem
In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?
Solution 1
Connect the centers of the tangent circles! (call the center of the large circle )
Notice that we don't even need the circles anymore; thus, draw triangle with cevian :
and use Stewart's Theorem:
From what we learned from the tangent circles, we have , , , , , and , where is the radius of the circle centered at that we seek.
Thus:
Solution 2
Like the solution above, connecting the centers of the circles results in triangle with cevian . The two triangles and share angle , which means we can use Law of Cosines to set up a system of 2 equations that solve for respectively:
(notice that the diameter of the largest semicircle is 6, so its radius is 3 and is 3 - r)
We can eliminate the extra variable of angle by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find :
, so =
Solution 3
Let be the center of the largest semicircle and be the radius of . We know that , , , , and . Notice that and are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of must be twice that of , since the area of a triangle is .
Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles.
Let equal to the area of and equal to the area of . Heron's Formula states that the area of an triangle with sides and is where , or the semiperimeter, is
The semiperimeter of is Use Heron's Formula to obtain
Using Heron's Formula again, find the area of with sides , , and .
Now,
Solution 4
Let , the center of the large semicircle, to be at , and to be at .
Therefore is at and is at .
Let the radius of circle be .
Using Distance Formula, we get the following system of three equations:
By simplifying, we get
By subtracting the first equation from the second and third equations, we get
which simplifies to
When we add these two equations, we get
So
Solution 5(Inversion from Circular Inversion)
Let be a circle with radius of and centered at the left corner of the semi-circle (O) with radius . Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram:
has radius , has radius , and has radius . We want to find the radius of .
We now invert the four circles. inverts to a line. Given that one point is on , and all points on invert to themselves, we know that the resulting line must intersect that intersection point. also inverts to a line. has radius , and since has radius of , the resulting line must be units away from . inverts to a circle. By observing the diagram, we note that 's center must be on and be between the two inverted lines, because is tangent to and (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius that is units from .
Now, we invert . Note that is tangent to the three other original circles. So, in the inversion, must be tangent to the two lines and . It is then quickly seen that and have the same radius: .
Now, we can determine the radius of using the formula . , and . is just the distance from the center of the inverted circle to the center of inversion. The center of is units above the center of . Since , we use Pythagoras to learn that . We do not take the square root because our relationship formula takes .
Therefore, we have:
Here is the diagram with .
Solution 6 (Kissing Circles)
Draw the other half of the largest circle and proceed with Descartes' Circle Formula.
The curvatures of circles A,B,C, and P are and , respectively.
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=f_cdwEDlWXQ
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.