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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 22"
Line 13: Line 13:
  
 
~ Leo.Euler
 
~ Leo.Euler
 +
 +
== Solution 2 (Cross section & angle bisector) ==
 +
 +
We can take half of a cross section of the sphere, as such:
 +
<asy>
 +
unitsize(0.5cm);
 +
real r = (90-40*sqrt(3))/11;
 +
pair C = (0,0);
 +
pair A = (-5,0);
 +
pair B = (0,12);
 +
pair O = (-((2*sqrt(3))/3) * r, r);
 +
draw(A--B--C--cycle);
 +
draw(circle(O,r));
 +
pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r);
 +
pair E = (-2.2, 0);
 +
draw(O--E);
 +
draw(D--O);
 +
label("$A$", A, SW);
 +
label("$B$", B, N);
 +
label("$C$", C, SE);
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label("$O$", O, N);
 +
label("$D$", D, NW);
 +
label("$E$", E, SW);
 +
dot(D);
 +
dot(E);
 +
</asy>
 +
Notice that we chose a cross section where one of the spheres was tangent to the lateral surface of the cone at <math>D</math>.
 +
 +
To evaluate <math>r</math>, we will find <math>AE</math> and <math>EC</math> in terms of <math>r</math>; we also know that <math>AE+EC = 5</math>, so with this, we can solve <math>r</math>. Firstly, to find <math>EC</math>, we can take a bird's eye view of the cone:
 +
<asy>
 +
unitsize(0.8cm);
 +
pair C = (0,0);
 +
draw(circle(C,5));
 +
label("$C$", C, N);
 +
dot(C);
 +
 +
 +
 +
real r = (90-40*sqrt(3))/11;
 +
real raise = r*(2/3*sqrt(3));
 +
pair E = (-r,raise/-2);
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pair X = (0,raise);
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pair Y = (r,raise/-2);
 +
label("$E$", E, SW);
 +
dot(E);
 +
label("$X$", X, NW);
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dot(X);
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label("$Y$", Y, SW);
 +
dot(Y);
 +
 +
draw(circle(X,r),dashed);
 +
draw(circle(E,r),dashed);
 +
draw(circle(Y,r),dashed);
 +
draw(E--X,dashed);
 +
draw(X--Y,dashed);
 +
draw(E--Y,dashed);
 +
</asy>
 +
<math>C</math> is the centroid of equilateral triangle <math>EXY</math>. Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from <math>E</math> to <math>XY</math>; this is because medians cut each other into a <math>2</math> to <math>1</math> ratio. This equilateral triangle has a side length of <math>2r</math>, therefore it has an altitude of length <math>r \sqrt{3}</math>; two thirds of this is <math>\frac{2r \sqrt{3}}{3}</math>, so
 +
<cmath>EC = \frac{2r \sqrt{3}}{3}.</cmath>
 +
<asy>
 +
unitsize(0.5cm);
 +
real r = (90-40*sqrt(3))/11;
 +
pair C = (0,0);
 +
pair A = (-5,0);
 +
pair B = (0,12);
 +
pair O = (-((2*sqrt(3))/3) * r, r);
 +
draw(A--B--C--cycle);
 +
draw(circle(O,r));
 +
pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r);
 +
pair E = (-2.2, 0);
 +
pair F = (-2.2, 6.72);
 +
draw(E--F);
 +
draw(D--O);
 +
draw(A--O, dotted);
 +
label("$A$", A, SW);
 +
label("$B$", B, N);
 +
label("$C$", C, SE);
 +
label("$O$", O, NE);
 +
label("$D$", D, NW);
 +
label("$E$", E, SW);
 +
label("$F$", F, NW);
 +
dot(D);
 +
dot(E);
 +
</asy>
 +
To evaluate <math>AE</math> in terms of <math>r</math>, we will extend <math>\overline{OE}</math> past point <math>O</math> to <math>\overline{AB}</math> at point <math>F</math>.<math>\triangle AEF</math> is similar to <math>\triangle ACB</math>. Also, <math>AO</math> is the angle bisector of <math>\angle EAB</math>. Therefore, by the angle bisector theorem, <math>\frac{OE}{OF} = \frac{AE}{AF} = \frac{5}{13}</math>. Also, <math>OE = r</math>, so <math>\frac{r}{OF} = \frac{5}{13}</math>, so <math>OF = \frac{13r}{5}</math>. This means that<cmath>AE = \frac{5 \cdot EF}{12} = \frac{5 \cdot (OE + OF)}{12} = \frac{5 \cdot (r + \frac{13r}{5})}{12} = \frac{18r}{12} = \frac{3r}{2}.</cmath>
 +
We have that <math>EC = \frac{2r \sqrt{3}}{3}</math> and that <math>AE = \frac{3r}{2}</math>, so <math>AC = EC + AE = \frac{2r \sqrt{3}}{3} + \frac{3r}{2} = \frac{4r \sqrt{3} + 9r}{6}</math>. We also were given that <math>AC = 5</math>. Therefore, we have
 +
<cmath>\frac{4r \sqrt{3} + 9r}{6} = 5.</cmath>
 +
This is a simple linear equation in terms of <math>r</math>. We can solve for <math>r</math> to get <math>r = \boxed{\textbf{(B) } \frac{90 - 40 \sqrt{3}}{11}}.</math>
 +
 +
~ ihatemath123

Revision as of 00:50, 23 November 2021

Problem

Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?

$\textbf{(A)}\ \frac{3}{2} \qquad\textbf{(B)}\ \frac{90-40\sqrt{3}}{11} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{144-25\sqrt{3}}{44} \qquad\textbf{(E)}\ \frac{5}{2}$

Solution 1 (Coordinates)

We will use coordinates. WLOG, let the coordinates of the center of the base of the cone be the origin. Then, let the center of one of the spheres be $(0, 2r/\sqrt{3}, r)$. Note that the distance between this point and the plane given by $12y+5z=60$ is $r$. Thus, by the point-to-plane distance formula, we have

\[\frac{|12 \cdot 2r/\sqrt{3} + 5r - 60|}{\sqrt{0^2+5^2+12^2}}=r.\]

Solving for $r$ yields $r = \boxed{\textbf{(B) }\dfrac{90 - 40\sqrt3}{11}}$.

~ Leo.Euler

Solution 2 (Cross section & angle bisector)

We can take half of a cross section of the sphere, as such: [asy] unitsize(0.5cm); real r = (90-40*sqrt(3))/11; pair C = (0,0); pair A = (-5,0); pair B = (0,12); pair O = (-((2*sqrt(3))/3) * r, r); draw(A--B--C--cycle); draw(circle(O,r)); pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r); pair E = (-2.2, 0); draw(O--E); draw(D--O); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$O$", O, N); label("$D$", D, NW); label("$E$", E, SW); dot(D); dot(E); [/asy] Notice that we chose a cross section where one of the spheres was tangent to the lateral surface of the cone at $D$.

To evaluate $r$, we will find $AE$ and $EC$ in terms of $r$; we also know that $AE+EC = 5$, so with this, we can solve $r$. Firstly, to find $EC$, we can take a bird's eye view of the cone: [asy] unitsize(0.8cm); pair C = (0,0); draw(circle(C,5)); label("$C$", C, N); dot(C); real r = (90-40*sqrt(3))/11; real raise = r*(2/3*sqrt(3)); pair E = (-r,raise/-2); pair X = (0,raise); pair Y = (r,raise/-2); label("$E$", E, SW); dot(E); label("$X$", X, NW); dot(X); label("$Y$", Y, SW); dot(Y); draw(circle(X,r),dashed); draw(circle(E,r),dashed); draw(circle(Y,r),dashed); draw(E--X,dashed); draw(X--Y,dashed); draw(E--Y,dashed); [/asy] $C$ is the centroid of equilateral triangle $EXY$. Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from $E$ to $XY$; this is because medians cut each other into a $2$ to $1$ ratio. This equilateral triangle has a side length of $2r$, therefore it has an altitude of length $r \sqrt{3}$; two thirds of this is $\frac{2r \sqrt{3}}{3}$, so \[EC = \frac{2r \sqrt{3}}{3}.\] [asy] unitsize(0.5cm); real r = (90-40*sqrt(3))/11; pair C = (0,0); pair A = (-5,0); pair B = (0,12); pair O = (-((2*sqrt(3))/3) * r, r); draw(A--B--C--cycle); draw(circle(O,r)); pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r); pair E = (-2.2, 0); pair F = (-2.2, 6.72); draw(E--F); draw(D--O); draw(A--O, dotted); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$O$", O, NE); label("$D$", D, NW); label("$E$", E, SW); label("$F$", F, NW); dot(D); dot(E); [/asy] To evaluate $AE$ in terms of $r$, we will extend $\overline{OE}$ past point $O$ to $\overline{AB}$ at point $F$.$\triangle AEF$ is similar to $\triangle ACB$. Also, $AO$ is the angle bisector of $\angle EAB$. Therefore, by the angle bisector theorem, $\frac{OE}{OF} = \frac{AE}{AF} = \frac{5}{13}$. Also, $OE = r$, so $\frac{r}{OF} = \frac{5}{13}$, so $OF = \frac{13r}{5}$. This means that\[AE = \frac{5 \cdot EF}{12} = \frac{5 \cdot (OE + OF)}{12} = \frac{5 \cdot (r + \frac{13r}{5})}{12} = \frac{18r}{12} = \frac{3r}{2}.\] We have that $EC = \frac{2r \sqrt{3}}{3}$ and that $AE = \frac{3r}{2}$, so $AC = EC + AE = \frac{2r \sqrt{3}}{3} + \frac{3r}{2} = \frac{4r \sqrt{3} + 9r}{6}$. We also were given that $AC = 5$. Therefore, we have \[\frac{4r \sqrt{3} + 9r}{6} = 5.\] This is a simple linear equation in terms of $r$. We can solve for $r$ to get $r = \boxed{\textbf{(B) } \frac{90 - 40 \sqrt{3}}{11}}.$

~ ihatemath123

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