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Difference between revisions of "2021 AIME II Problems/Problem 10"

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Two spheres with radii <math>36</math> and one sphere with radius <math>13</math> are each externally tangent to the other two spheres and to two different planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>. The intersection of planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> is the line <math>\ell</math>. The distance from line <math>\ell</math> to the point where the sphere with radius <math>13</math> is tangent to plane <math>\mathcal{P}</math> is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
 
Two spheres with radii <math>36</math> and one sphere with radius <math>13</math> are each externally tangent to the other two spheres and to two different planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>. The intersection of planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> is the line <math>\ell</math>. The distance from line <math>\ell</math> to the point where the sphere with radius <math>13</math> is tangent to plane <math>\mathcal{P}</math> is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
−
==Reference Diagram==
+
==Diagram==
−
[[File:2021 AIME II Problem 10 Diagram.png|center]]
+
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(275);
 +
import graph3;
 +
import solids;
  
−
<u><b>Remarks</b></u>
+
currentprojection=orthographic((1,1/2,0));
 +
triple O1, O2, O3, T1, T2, T3, A, L1, L2;
 +
O1 = (0,-36,0);
 +
O2 = (0,36,0);
 +
O3 = (0,0,-sqrt(1105));
 +
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
 +
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105);
 +
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
 +
A = (0,0,-36*sqrt(1105)/23);
 +
L1 = shift(0,-80,0)*A;
 +
L2 = shift(0,80,0)*A;
 +
 
 +
draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink);
 +
draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unitsphere,yellow,light=White);
 +
draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unitsphere,yellow,light=White);
 +
draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unitsphere,red,light=White);
 +
draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen);
 +
draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen);
 +
draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink);
 +
draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red);
 +
 
 +
label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen);
 +
label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta);
 +
 
 +
dot(O1,linewidth(4.5));
 +
dot(O2,linewidth(4.5));
 +
dot(O3,linewidth(4.5));
 +
dot(T1,heavygreen+linewidth(4.5));
 +
dot(T2,heavygreen+linewidth(4.5));
 +
dot(T3,heavygreen+linewidth(4.5));
 +
dot(A,red+linewidth(4.5));
 +
</asy>
 +
~MRENTHUSIASM
 +
 
 +
==Solution 1 (Similar Triangles and Pythagorean Theorem)==
 +
This solution refers to the <b>Diagram</b> section.
 +
 
 +
As shown below, let <math>O_1,O_2,O_3</math> be the centers of the spheres (where sphere <math>O_3</math> has radius <math>13</math>) and <math>T_1,T_2,T_3</math> be their respective points of tangency to plane <math>\mathcal{P}.</math> Let <math>\mathcal{R}</math> be the plane that is determined by <math>O_1,O_2,</math> and <math>O_3.</math> Suppose <math>A</math> is the foot of the perpendicular from <math>O_3</math> to line <math>\ell,</math> so <math>\overleftrightarrow{O_3A}</math> is the perpendicular bisector of <math>\overline{O_1O_2}.</math> We wish to find <math>T_3A.</math>
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
 +
import graph3;
 +
import solids;
 +
 
 +
currentprojection=orthographic((1,1/2,0));
 +
triple O1, O2, O3, T1, T2, T3, A, L1, L2;
 +
O1 = (0,-36,0);
 +
O2 = (0,36,0);
 +
O3 = (0,0,-sqrt(1105));
 +
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
 +
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105);
 +
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
 +
A = (0,0,-36*sqrt(1105)/23);
 +
L1 = shift(0,-80,0)*A;
 +
L2 = shift(0,80,0)*A;
 +
 
 +
draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink);
 +
draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray);
 +
draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
 +
draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
 +
draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White);
 +
draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen);
 +
draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen);
 +
draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray);
 +
draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink);
 +
draw(O1--O2--O3--cycle^^O3--A,dashed);
 +
draw(T1--T2--T3--cycle^^T3--A,heavygreen);
 +
draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed);
 +
draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red);
 +
 
 +
label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen);
 +
label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta);
 +
label("$\mathcal{R}$",O1,(0,-24,0));
 +
 
 +
dot("$O_1$",O1,(0,-1,1),linewidth(4.5));
 +
dot("$O_2$",O2,(0,1,1),linewidth(4.5));
 +
dot("$O_3$",O3,(0,-1.5,0),linewidth(4.5));
 +
dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5));
 +
dot("$T_2$",T2,(0,1,-1),heavygreen+linewidth(4.5));
 +
dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5));
 +
dot("$A$",A,(0,0,-2),red+linewidth(4.5));
 +
</asy>
 +
Note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>In <math>\triangle O_1O_2O_3,</math> we get <math>O_1O_2=72</math> and <math>O_1O_3=O_2O_3=49.</math></li><p>
 +
  <li>Both <math>\triangle O_1O_2O_3</math> and <math>\overline{O_3A}</math> lie in plane <math>\mathcal{R}.</math> Both <math>\triangle T_1T_2T_3</math> and <math>\overline{T_3A}</math> lie in plane <math>\mathcal{P}.</math></li><p>
 +
  <li>By symmetry, since planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> are reflections of each other about plane <math>\mathcal{R},</math> the three planes are concurrent to line <math>\ell.</math></li><p>
 +
  <li>Since <math>\overline{O_1T_1}\perp\mathcal{P}</math> and <math>\overline{O_3T_3}\perp\mathcal{P},</math> it follows that <math>\overline{O_1T_1}\parallel\overline{O_3T_3},</math> from which <math>O_1,O_3,T_1,</math> and <math>T_3</math> are coplanar.</li><p>
 +
</ol>
 +
Now, we focus on cross-sections <math>O_1O_3T_3T_1</math> and <math>\mathcal{R}:</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
−
   <li>Let <math>\mathcal{R}</math> be the plane that is determined by the centers of the spheres, as shown in the black dashed triangle. Clearly, the side-lengths of this triangle are <math>49,49,</math> and <math>72.</math> Note that all black dashed line segments are in plane <math>\mathcal{R}.</math></li><p>
+
   <li><i><b>In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.</b></i><p>
−
  <li>Plane <math>\mathcal{P}</math> is determined by the points of tangency of the spheres, as shown in the green solid triangle. Therefore, the blue dashed line segments are the radii of the spheres. Note that all green solid line segments are in plane <math>\mathcal{P}.</math></li><p>
+
Clearly, cross-section <math>O_1O_3T_3T_1</math> intersects line <math>\ell</math> at exactly one point. Furthermore, as the intersection of planes <math>\mathcal{R}</math> and <math>\mathcal{P}</math> is line <math>\ell,</math> we conclude that <math>\overrightarrow{O_1O_3}</math> and <math>\overrightarrow{T_1T_3}</math> must intersect line <math>\ell</math> at the same point. Let <math>B</math> be the point of concurrency of <math>\overrightarrow{O_1O_3},\overrightarrow{T_1T_3},</math> and line <math>\ell.</math></li>
−
   <li>Since planes <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> are reflections of each other about plane <math>\mathcal{R},</math> it follows that the three planes are concurrent to line <math>\ell.</math></li><p>
+
   <li>In cross-section <math>\mathcal{R},</math> let <math>C</math> be the foot of the perpendicular from <math>O_1</math> to line <math>\ell,</math> and <math>D</math> be the foot of the perpendicular from <math>O_3</math> to <math>\overline{O_1C}.</math></li><p>
 
</ol>
 
</ol>
 +
We have the following diagram:
 +
<asy>
 +
size(300);
 +
import graph3;
 +
import solids;
 +
 +
currentprojection=orthographic((1,1/2,0));
 +
triple O1, O3, T1, T3, A, B, C, D;
 +
O1 = (0,-36,0);
 +
O3 = (0,0,-sqrt(1105));
 +
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
 +
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
 +
A = (0,0,-36*sqrt(1105)/23);
 +
B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1));
 +
C = (0,-36,-36*sqrt(1105)/23);
 +
D = (0,-36,-sqrt(1105));
 +
 +
draw(C--O1--O3--A^^D--O3--B,dashed);
 +
draw(T1--T3--A^^T3--B,heavygreen);
 +
draw(O1--T1^^O3--T3,mediumblue+dashed);
 +
draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label("$\ell$",position=EndPoint,align=3*E),red);
 +
dot("$O_1$",O1,(0,-1,1),linewidth(4.5));
 +
dot("$O_3$",O3,(0,1,1),linewidth(4.5));
 +
dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5));
 +
dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5));
 +
dot("$A$",A,(0,0,-2),red+linewidth(4.5));
 +
dot("$B$",B,(0,0,-2),red+linewidth(4.5));
 +
dot("$C$",C,(0,0,-2),red+linewidth(4.5));
 +
dot("$D$",D,(0,-2,0),linewidth(4.5));
 +
</asy>
 +
In cross-section <math>O_1O_3T_3T_1,</math> since <math>\overline{O_1T_1}\parallel\overline{O_3T_3}</math> as discussed, we obtain <math>\triangle O_1T_1B\sim\triangle O_3T_3B</math> by AA, with the ratio of similitude <math>\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.</math> Therefore, we get <math>\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},</math> or <math>O_3B=\frac{637}{23}.</math>
 +
 +
In cross-section <math>\mathcal{R},</math> note that <math>O_1O_3=49</math> and <math>DO_3=\frac{O_1O_2}{2}=36.</math> Applying the Pythagorean Theorem to right <math>\triangle O_1DO_3,</math> we have <math>O_1D=\sqrt{1105}.</math> Moreover, since <math>\ell\perp\overline{O_1C}</math> and <math>\overline{DO_3}\perp\overline{O_1C},</math> we obtain <math>\ell\parallel\overline{DO_3}</math> so that <math>\triangle O_1CB\sim\triangle O_1DO_3</math> by AA, with the ratio of similitude <math>\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.</math> Therefore, we get <math>\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},</math> or <math>DC=\frac{13\sqrt{1105}}{23}.</math>
 +
 +
Finally, note that <math>\overline{O_3T_3}\perp\overline{T_3A}</math> and <math>O_3T_3=13.</math> Since quadrilateral <math>DCAO_3</math> is a rectangle, we have <math>O_3A=DC=\frac{13\sqrt{1105}}{23}.</math> Applying the Pythagorean Theorem to right <math>\triangle O_3T_3A</math> gives <math>T_3A=\frac{312}{23},</math> from which the answer is <math>312+23=\boxed{335}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 2 (Pythagorean Theorem)==
 +
 +
The centers of the three spheres form a <math>49</math>-<math>49</math>-<math>72</math> triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the <math>72</math> side of this triangle. Take its midpoint <math>M</math>, which is <math>36</math> away from the midpoint <math>A</math> of the <math>72</math> side, and connect these two midpoints.
 +
 +
Now consider the point at which the plane is tangent to the small sphere, and connect <math>M</math> with the small sphere's tangent point <math>B</math>. Extend <math>\overline{MB}</math> through <math>B</math> until it hits the ray from <math>A</math> through the center of the small sphere (convince yourself that these two intersect). Call this intersection <math>D</math>, the center of the small sphere <math>C</math>, we want to find <math>BD</math>.
 +
 +
By Pythagoras, <math>AC=\sqrt{49^2-36^2}=\sqrt{1105}</math>, and we know that <math>MA=36</math> and <math>BC=13</math>. We know that <math>\overline{MA}</math> and <math>\overline{BC}</math> must be parallel, using ratios we realize that <math>CD=\frac{13}{23}\sqrt{1105}</math>. Apply the Pythagorean theorem to <math>\triangle BCD</math>, <math>BD=\frac{312}{23}</math>, so <math>312 + 23 = \boxed{335}</math>.
 +
 +
~Ross Gao
 +
 +
==Solution 3 (Proportion) ==
 +
This solution refers to the <b>Diagram</b> section.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
 +
import graph3;
 +
import solids;
 +
 +
currentprojection=orthographic((10,-3,-40));
 +
 +
triple O1, O2, O3, T1, T2, T3, A, L1, L2, M;
 +
O1 = (0,-36,0);
 +
O2 = (0,36,0);
 +
O3 = (0,0,-sqrt(1105));
 +
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
 +
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105);
 +
T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85);
 +
A = (0,0,-36*sqrt(1105)/23);
 +
L1 = shift(0,-80,0)*A;
 +
L2 = shift(0,80,0)*A;
 +
M = midpoint(T1--T2);
  
−
~MRENTHUSIASM (by Geometry Expressions)
+
draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
 +
draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
 +
draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White);
 +
draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray);
 +
draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
 +
draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White);
 +
draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White);
 +
draw(surface(T2--T1--T3--A--cycle),cyan);
 +
draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray);
 +
draw(T1--T2--T3--cycle^^M--A--T2,blue);
  
−
==Solution 1==
+
dot("$O_1$",O1,(0,-1,1),linewidth(4.5));
 +
dot("$O_2$",O2,(0,1,1),linewidth(4.5));
 +
dot("$O$",O3,(0.5,-1,0),linewidth(4.5));
 +
dot("$T_1$",T1,(0,-1,-1),blue+linewidth(4.5));
 +
dot("$T_2$",T2,(0,1,-1),blue+linewidth(4.5));
 +
dot("$T$",T3,(1,1,2),blue+linewidth(4.5));
 +
dot("$M$",M,(0,0,5),blue+linewidth(4.5));
 +
dot("$A$",A,(-0.5,-1.5,0),red+linewidth(4.5));
 +
</asy>
 +
The isosceles triangle of centers <math>O_1 O_2 O</math> (<math>O</math> is the center of sphere of radii <math>13</math>) has sides  <math> O_1 O = O_2 O = 36 + 13 = 49,</math> and  <math>O_1 O_2 = 36 + 36 = 72.</math>
  
−
The centers of the three spheres form a 49-49-72 triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the 72 side of this triangle. Take its midpoint <math>M</math>, which is 36 away from the midpoint of the 72 side <math>A</math>, and connect these two midpoints.
+
Let <math>N</math> be the midpoint <math>O_1 O_2 </math>.
  
−
Now consider the point at which the plane is tangent to the small sphere, and connect <math>M</math> with the small sphere's tangent point <math>B</math>. Extend <math>MB</math> through B until it hits the ray from <math>A</math> through the center of the small sphere (convince yourself that these two intersect). Call this intersection <math>D</math>, the center of the small sphere <math>C</math>, we want to find <math>BD</math>.
+
The isosceles triangle of points of tangency <math>T_1 T_2 T</math> has sides <math>T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13} </math> and <math>T_1 T_2 = 72.</math>  
  
−
By Pythagorus AC= <math>\sqrt{49^2-36^2}=\sqrt{1105}</math>, and we know <math>MB=36,BC=13</math>. We know that <math>MB,BC</math> must be parallel, using ratios we realize that <math>CD=\frac{13}{23}\sqrt{1105}</math>. Apply Pythagorean theorem on triangle BCD; <math>BD=\frac{312}{23}</math>, so 312 + 23 = <math>\boxed{335}</math>
+
Let <math>M</math> be the midpoint <math>T_1 T_2.</math>
  
−
-Ross Gao
+
The height <math>TM</math>  is <math>\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.</math>
  
−
==Solution 2 (Coord Bash)==
+
The tangents of the half-angle between the planes is <math>\frac {TO}{AT} = \frac {MN - TO}{TM},</math> so <math>\frac {13}{AT} = \frac {36 - 13}{24},</math> <cmath>AT = \frac{24\cdot 13}{23} = \frac {312}{23} \implies  312 + 23 = \boxed{335}.</cmath>
−
Let's try to see some symmetry. We can use a coordinate plane to plot where the circles are. The 2 large spheres are externally tangent, so we'll make them at 0, -36, 0 and 0, 36, 0. The center of the little sphere would be x, 0, and -23 since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that x is -24 (since 24 wouldn't make sense). Now, we draw a line through the little sphere and the origin. It also intersects <math>\ell</math> because of the symmetry we created.  
+
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
−
<math>\ell</math> lies on the plane too, so these 2 lines must intersect. The point at where it intersects is -24a, 0, and 23a. We can use the distance formula again to find that a = <math>\dfrac{36}{23}</math>. Therefore, they intersect at <math>\left(-\dfrac{864}{23},0,-36\right)</math>. Since the little circle's x coordinate is -24 and the intersection point's x coordinate is <math>\dfrac{864}{23}</math>, we get <math>\dfrac{864}{23}</math> - 24 = <math>\dfrac{312}{23}</math>. Therefore, our answer to this problem is 312 + 23 =  <math>\boxed{335}</math>.
+
==Video Solution by Interstigation==
 +
https://youtu.be/bQ3KdG4xH0A
  
−
~Arcticturn
+
~Interstigation
  
−
==See also==
+
==See Also==
 
{{AIME box|year=2021|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2021|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:00, 14 January 2023

Problem

Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Diagram

[asy] /* Made by MRENTHUSIASM */ size(275); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unitsphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unitsphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unitsphere,red,light=White); draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red); label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); dot(O1,linewidth(4.5)); dot(O2,linewidth(4.5)); dot(O3,linewidth(4.5)); dot(T1,heavygreen+linewidth(4.5)); dot(T2,heavygreen+linewidth(4.5)); dot(T3,heavygreen+linewidth(4.5)); dot(A,red+linewidth(4.5)); [/asy] ~MRENTHUSIASM

Solution 1 (Similar Triangles and Pythagorean Theorem)

This solution refers to the Diagram section.

As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell,$ so $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$ [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(O1--O2--O3--cycle^^O3--A,dashed); draw(T1--T2--T3--cycle^^T3--A,heavygreen); draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed); draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red); label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); label("$\mathcal{R}$",O1,(0,-24,0)); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O_3$",O3,(0,-1.5,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); [/asy] Note that:

  1. In $\triangle O_1O_2O_3,$ we get $O_1O_2=72$ and $O_1O_3=O_2O_3=49.$

  2. Both $\triangle O_1O_2O_3$ and $\overline{O_3A}$ lie in plane $\mathcal{R}.$ Both $\triangle T_1T_2T_3$ and $\overline{T_3A}$ lie in plane $\mathcal{P}.$

  3. By symmetry, since planes $\mathcal{P}$ and $\mathcal{Q}$ are reflections of each other about plane $\mathcal{R},$ the three planes are concurrent to line $\ell.$

  4. Since $\overline{O_1T_1}\perp\mathcal{P}$ and $\overline{O_3T_3}\perp\mathcal{P},$ it follows that $\overline{O_1T_1}\parallel\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar.

Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$

  1. In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.

    Clearly, cross-section $O_1O_3T_3T_1$ intersects line $\ell$ at exactly one point. Furthermore, as the intersection of planes $\mathcal{R}$ and $\mathcal{P}$ is line $\ell,$ we conclude that $\overrightarrow{O_1O_3}$ and $\overrightarrow{T_1T_3}$ must intersect line $\ell$ at the same point. Let $B$ be the point of concurrency of $\overrightarrow{O_1O_3},\overrightarrow{T_1T_3},$ and line $\ell.$

  2. In cross-section $\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\overline{O_1C}.$

We have the following diagram: [asy] size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O3, T1, T3, A, B, C, D; O1 = (0,-36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1)); C = (0,-36,-36*sqrt(1105)/23); D = (0,-36,-sqrt(1105)); draw(C--O1--O3--A^^D--O3--B,dashed); draw(T1--T3--A^^T3--B,heavygreen); draw(O1--T1^^O3--T3,mediumblue+dashed); draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label("$\ell$",position=EndPoint,align=3*E),red); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_3$",O3,(0,1,1),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); dot("$B$",B,(0,0,-2),red+linewidth(4.5)); dot("$C$",C,(0,0,-2),red+linewidth(4.5)); dot("$D$",D,(0,-2,0),linewidth(4.5)); [/asy] In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ or $O_3B=\frac{637}{23}.$

In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ or $DC=\frac{13\sqrt{1105}}{23}.$

Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$

~MRENTHUSIASM

Solution 2 (Pythagorean Theorem)

The centers of the three spheres form a $49$-$49$-$72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$, which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints.

Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$. Extend $\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$, the center of the small sphere $C$, we want to find $BD$.

By Pythagoras, $AC=\sqrt{49^2-36^2}=\sqrt{1105}$, and we know that $MA=36$ and $BC=13$. We know that $\overline{MA}$ and $\overline{BC}$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$. Apply the Pythagorean theorem to $\triangle BCD$, $BD=\frac{312}{23}$, so $312 + 23 = \boxed{335}$.

~Ross Gao

Solution 3 (Proportion)

This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((10,-3,-40)); triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; M = midpoint(T1--T2); draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(T2--T1--T3--A--cycle),cyan); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(T1--T2--T3--cycle^^M--A--T2,blue); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O$",O3,(0.5,-1,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),blue+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),blue+linewidth(4.5)); dot("$T$",T3,(1,1,2),blue+linewidth(4.5)); dot("$M$",M,(0,0,5),blue+linewidth(4.5)); dot("$A$",A,(-0.5,-1.5,0),red+linewidth(4.5)); [/asy] The isosceles triangle of centers $O_1 O_2 O$ ($O$ is the center of sphere of radii $13$) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$

Let $N$ be the midpoint $O_1 O_2$.

The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13}$ and $T_1 T_2 = 72.$

Let $M$ be the midpoint $T_1 T_2.$

The height $TM$ is $\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.$

The tangents of the half-angle between the planes is $\frac {TO}{AT} = \frac {MN - TO}{TM},$ so $\frac {13}{AT} = \frac {36 - 13}{24},$ \[AT = \frac{24\cdot 13}{23} = \frac {312}{23} \implies 312 + 23 = \boxed{335}.\] vladimir.shelomovskii@gmail.com, vvsss

Video Solution by Interstigation

https://youtu.be/bQ3KdG4xH0A

~Interstigation

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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