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

(→‎Solution 2 (Coord Bash))
(→‎Solution 3 (Proportion): Remade diagram by Asy.)
 
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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>.
  
−
==Solution 1==
+
==Diagram==
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(275);
 +
import graph3;
 +
import solids;
  
−
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.
+
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;
  
−
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>.
+
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);
  
−
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>
+
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);
  
−
-Ross Gao
+
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 2 (Coord Bash)==
+
==Solution 1 (Similar Triangles and Pythagorean Theorem)==
−
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.  
+
This solution refers to the <b>Diagram</b> section.
  
−
<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 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>.
+
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;
  
−
~Arcticturn
+
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;
  
−
==See also==
+
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;">
 +
  <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>
 +
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>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>
 +
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);
 +
 
 +
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 <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>
 +
 
 +
Let <math>N</math> be the midpoint  <math>O_1 O_2 </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>
 +
 
 +
Let <math>M</math> be the midpoint <math>T_1 T_2.</math>
 +
 
 +
The height <math>TM</math>  is <math>\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.</math>
 +
 
 +
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>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/bQ3KdG4xH0A
 +
 
 +
~Interstigation
 +
 
 +
==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
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