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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>. | ||
− | == | + | ==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 <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;"> | ||
+ | <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 | ~MRENTHUSIASM | ||
− | ==Solution 1== | + | ==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 | + | ==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
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.