Difference between revisions of "2021 AIME II Problems/Problem 10"
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MRENTHUSIASM (talk | contribs) (→Solution 3 (Proportion): Remade diagram by Asy.) |
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==Diagram== | ==Diagram== | ||
<asy> | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
size(275); | size(275); | ||
import graph3; | import graph3; | ||
import solids; | import solids; | ||
− | currentprojection=orthographic((1,1,0)); | + | currentprojection=orthographic((1,1/2,0)); |
triple O1, O2, O3, T1, T2, T3, A, L1, L2; | triple O1, O2, O3, T1, T2, T3, A, L1, L2; | ||
O1 = (0,-36,0); | O1 = (0,-36,0); | ||
Line 21: | Line 22: | ||
draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); | 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(O1)*rotate(90,O1,O2)*scale3(36)* | + | draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unitsphere,red,light=White); |
− | draw(shift(O2)*rotate(90,O1,O2)*scale3(36)* | + | draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); |
− | draw(shift(O3)*rotate(90,O1,O2)*scale3(13)* | + | 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( | ||
− | draw(surface(L1--L2--( | ||
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(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); | draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red); | ||
Line 47: | Line 46: | ||
This solution refers to the <b>Diagram</b> section. | 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> | + | 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> | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
size(300); | size(300); | ||
import graph3; | import graph3; | ||
Line 69: | Line 69: | ||
draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); | 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(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); | ||
− | draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere, | + | draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); |
− | draw(surface(( | + | 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--(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(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); | ||
Line 149: | Line 150: | ||
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>. | 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== | ==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 01:00, 14 January 2023
Contents
Problem
Two spheres with radii and one sphere with radius
are each externally tangent to the other two spheres and to two different planes
and
. The intersection of planes
and
is the line
. The distance from line
to the point where the sphere with radius
is tangent to plane
is
, where
and
are relatively prime positive integers. Find
.
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
This solution refers to the Diagram section.
As shown below, let be the centers of the spheres (where sphere
has radius
) and
be their respective points of tangency to plane
Let
be the plane that is determined by
and
Suppose
is the foot of the perpendicular from
to line
so
is the perpendicular bisector of
We wish to find
Note that:
- In
we get
and
- Both
and
lie in plane
Both
and
lie in plane
- By symmetry, since planes
and
are reflections of each other about plane
the three planes are concurrent to line
- Since
and
it follows that
from which
and
are coplanar.
Now, we focus on cross-sections and
- 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
intersects line
at exactly one point. Furthermore, as the intersection of planes
and
is line
we conclude that
and
must intersect line
at the same point. Let
be the point of concurrency of
and line
- In cross-section
let
be the foot of the perpendicular from
to line
and
be the foot of the perpendicular from
to
We have the following diagram:
In cross-section
since
as discussed, we obtain
by AA, with the ratio of similitude
Therefore, we get
or
In cross-section note that
and
Applying the Pythagorean Theorem to right
we have
Moreover, since
and
we obtain
so that
by AA, with the ratio of similitude
Therefore, we get
or
Finally, note that and
Since quadrilateral
is a rectangle, we have
Applying the Pythagorean Theorem to right
gives
from which the answer is
~MRENTHUSIASM
Solution 2 (Pythagorean Theorem)
The centers of the three spheres form a -
-
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
side of this triangle. Take its midpoint
, which is
away from the midpoint
of the
side, and connect these two midpoints.
Now consider the point at which the plane is tangent to the small sphere, and connect with the small sphere's tangent point
. Extend
through
until it hits the ray from
through the center of the small sphere (convince yourself that these two intersect). Call this intersection
, the center of the small sphere
, we want to find
.
By Pythagoras, , and we know that
and
. We know that
and
must be parallel, using ratios we realize that
. Apply the Pythagorean theorem to
,
, so
.
~Ross Gao
Solution 3 (Proportion)
This solution refers to the Diagram section.
The isosceles triangle of centers
(
is the center of sphere of radii
) has sides
and
Let be the midpoint
.
The isosceles triangle of points of tangency has sides
and
Let be the midpoint
The height is
The tangents of the half-angle between the planes is so
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by Interstigation
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.