Difference between revisions of "2020 AIME II Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | Two congruent right circular cones each with base radius <math>3</math> and height <math>8</math> have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance <math>3</math> from the base of each cone. A sphere with radius <math>r</math> lies | + | Two congruent right circular cones each with base radius <math>3</math> and height <math>8</math> have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance <math>3</math> from the base of each cone. A sphere with radius <math>r</math> lies within both cones. The maximum possible value of <math>r^2</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
− | ==Solution== | + | ==Solution (Official MAA)== |
− | + | Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let <math>A</math> be the point in the cross section where the bases of the cones meet, and let <math>C</math> be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, <math>B</math>, to the center of its base, <math>D</math>. Let the sphere be tangent to <math>\overline{AB}</math> at <math>E</math>. The right triangles <math>\triangle ABD</math> and <math>\triangle CBE</math> are similar, implying that the radius of the sphere is<cmath>CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.</cmath>The requested sum is <math>225+73=298</math>. | |
+ | <asy> | ||
+ | unitsize(0.6cm); | ||
+ | pair A = (0,0); | ||
+ | pair TriangleOneLeft = (-6,0); | ||
+ | pair TriangleOneDown = (-3,-8); | ||
+ | pair TriangleOneMid = (-3,0); | ||
− | == | + | pair D = (0,-3); |
− | {{ | + | pair TriangleTwoDown = (0,-6); |
− | {{ | + | pair B = (-8,-3); |
+ | |||
+ | pair C = IP(TriangleOneMid -- TriangleOneDown, B--D); | ||
+ | pair EE = foot(C, A, B); | ||
+ | real radius = arclength(C--EE); | ||
+ | path circ = Circle(C, radius); | ||
+ | |||
+ | |||
+ | |||
+ | draw(A--B--TriangleTwoDown--cycle); | ||
+ | draw(B--D); | ||
+ | draw(A--TriangleOneLeft--TriangleOneDown--cycle); | ||
+ | draw(circ); | ||
+ | draw(C--EE); | ||
+ | draw(TriangleOneMid -- TriangleOneDown, gray); | ||
+ | |||
+ | dot("$B$", B, W); | ||
+ | dot("$E$", EE, NW); | ||
+ | dot("$A$", A, NE); | ||
+ | dot("$D$", D, E); | ||
+ | dot("$C$", C, SE); | ||
+ | </asy> | ||
+ | |||
+ | Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra. | ||
+ | |||
+ | ==Solution (Incircles + Coordinate Bash)== | ||
+ | <asy> | ||
+ | unitsize(0.6cm); | ||
+ | |||
+ | // Coordinates | ||
+ | pair A = (0,0); | ||
+ | pair B = (6,0); | ||
+ | pair C = (0,6); | ||
+ | |||
+ | // Calculate point C | ||
+ | pair D = (3,8); | ||
+ | pair E = (8,3); | ||
+ | pair F = (144/73,384/73); | ||
+ | |||
+ | // Draw triangles (cones) | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(A--C--E--cycle); | ||
+ | draw(incircle(A,E,F)); | ||
+ | pair EE = foot(C, A, B); | ||
+ | real radius = arclength(C--EE); | ||
+ | path circ = Circle(C, radius); | ||
+ | |||
+ | // Label points | ||
+ | dot("$A$", A, NW); | ||
+ | dot("$B$", B, SW); | ||
+ | dot("$C$", C, NE); | ||
+ | dot("$D$", D, SE); | ||
+ | dot("$E$", E, SE); | ||
+ | dot("$F$", F, NE); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>A</math> be the origin in the above diagram. Then <math>B</math> is <math>(6,0)</math>, <math>C</math> is <math>(0,6)</math>, <math>D</math> is <math>(3,8)</math>, and <math>E</math> is <math>(8,3)</math>. Also, it is easy to see that the inscribed sphere is simply the inscribed circle of <math>AEF</math>. Then we want to find the intersection of <math>AD</math> and <math>CE</math> to determine the coordinates of point <math>F</math>. Note that <math>AD</math> is line <math>y=\frac{8}{3}x</math>, and <math>CE</math> is line <math>y=-\frac{3}{8}x+6</math>. Then, you can see that these lines are perpendicular, indicating that <math>AEF</math> is a right triangle with right angle at <math>F</math>. Finding the intersection of the lines by solving the system, we get that <math>F</math> is the point <math>(\frac{144}{73},\frac{384}{73})</math> in this plane. Then, we can find the distances <math>EF</math> and <math>AF</math> by the distance formula, which are <math>\frac{55}{\sqrt{73}}</math> and <math>\frac{48}{\sqrt{73}}</math> respectively. Also, <math>AE=\sqrt{73}</math>. Then, to find the radius of this triangle's incircle, we use the formula <math>a=rs</math> from which we get that <math>r=\frac{15}{\sqrt{73}}</math> and <math>r^2=\frac{225}{73} \implies \boxed{298}</math> is the answer. | ||
+ | |||
+ | ~SirAppel | ||
+ | |||
+ | == Solution (Clean analytic geometry) == | ||
+ | Using the diagram above, we notice that the desired length is simply the distance between the point <math>C</math> and <math>\overline{AB}</math>. We can mark <math>C</math> as <math>(3,3)</math> since it is <math>3</math> units away from each of the bases. Point <math>B</math> is <math>(8,3)</math>. Thus, line <math>\overline{AB}</math> is <math>y = \frac{3}{8}x \Rightarrow 3x + 8y = 0</math>. We can use the distance from point to line formula <math>\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}</math>, where <math>x_0</math> and <math>y_0</math> are the coordinates of the point, and A, B, and C are the coefficients of the line in form <math>Ax + By + C = 0</math>. Plugging everything in, we get <cmath>\frac{|3(3) - 8(3)|}{\sqrt{8^2 + 3^2}} = \frac{15}{\sqrt{73}} \Rightarrow \frac{225}{73} \Rightarrow \boxed{298}</cmath>. | ||
+ | |||
+ | ==Solution 1: Graph paper coordbash== | ||
+ | We graph this on graph paper, with the scale of <math>\sqrt{2}:1</math>. So, we can find <math>OT</math> then divide by <math>\sqrt{2}</math> to convert to our desired units, then square the result. With 5 minutes' worth of coordbashing, we finally arrive at <math>298</math>. | ||
+ | |||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -8.325025958411356, xmax = 8, ymin = -0.6033105644019334, ymax = 12.237120576121757; /* image dimensions */ | ||
+ | pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); | ||
+ | |||
+ | draw((5,11)--(0,0)--(-6,6)--cycle, linewidth(1) + zzttqq); | ||
+ | draw((6,6)--(-5,11)--(0,0)--cycle, linewidth(1) + qqwuqq); | ||
+ | draw((0.2328977836854361,5.767102216314564)--(0.4657955673708722,6)--(0.2328977836854361,6.232897783685436)--(0,6)--cycle, linewidth(1) + qqwuqq); | ||
+ | /* draw grid of horizontal/vertical lines */ | ||
+ | pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1, gridy = 1; /* grid intervals */ | ||
+ | for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) | ||
+ | draw((i,ymin)--(i,ymax), gridstyle); | ||
+ | for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) | ||
+ | draw((xmin,i)--(xmax,i), gridstyle); | ||
+ | /* end grid */ | ||
+ | |||
+ | Label laxis; laxis.p = fontsize(10); | ||
+ | xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); | ||
+ | yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ | ||
+ | /* draw figures */ | ||
+ | draw((5,11)--(0,0), linewidth(1) + zzttqq); | ||
+ | draw((0,0)--(-6,6), linewidth(1) + zzttqq); | ||
+ | draw((-6,6)--(5,11), linewidth(1) + zzttqq); | ||
+ | draw((6,6)--(-5,11), linewidth(1) + qqwuqq); | ||
+ | draw((-5,11)--(0,0), linewidth(1) + qqwuqq); | ||
+ | draw((0,0)--(6,6), linewidth(1) + qqwuqq); | ||
+ | draw((-3,3)--(5,11), linewidth(1)); | ||
+ | draw((-5,11)--(3,3), linewidth(1)); | ||
+ | draw(circle((0,6), 2.482817665807104), linewidth(1)); | ||
+ | draw((0,6)--(2.2602739726027394,4.972602739726027), linewidth(1)); | ||
+ | /* dots and labels */ | ||
+ | dot((0,0),linewidth(1pt) + dotstyle); | ||
+ | dot((3,3),dotstyle); | ||
+ | dot((-3,3),dotstyle); | ||
+ | dot((6,6),dotstyle); | ||
+ | dot((-6,6),dotstyle); | ||
+ | dot((5,11),dotstyle); | ||
+ | dot((-5,11),dotstyle); | ||
+ | dot((0,6),linewidth(1pt) + dotstyle); | ||
+ | label("$O$", (0.059294254264342997,6.119672124650978), NE * labelscalefactor); | ||
+ | dot((2.2602739726027394,4.972602739726027),linewidth(1pt) + dotstyle); | ||
+ | label("$T$", (2.326166015469254,5.094921876435061), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | ~samrocksnature | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Using the graph drawn above in MAA solution, we see that <math>\sqrt{AC^2-r^2}</math> + <math>\sqrt{BC^2-r^2}</math> = <math>AB</math>. <math>AB</math> = <math>\sqrt{8^2+3^2}</math> = <math>\sqrt{73}</math>, <math>AC</math> = 3<math>\sqrt{2}</math>, <math>AB</math> = 5. Plugging in and solve for <math>r^2</math> gives us <math>r^2</math> = <math>\frac{225}{73}</math>. | ||
+ | |||
+ | ~S17209 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/bz5N-jI2e0U?t=44 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://www.youtube.com/watch?v=0XJddG43pIk ~ MathEx | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/dHGXtB0FxXs | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=6|num-a=8}} | {{AIME box|year=2020|n=II|num-b=6|num-a=8}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:43, 24 June 2024
Contents
Problem
Two congruent right circular cones each with base radius and height have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance from the base of each cone. A sphere with radius lies within both cones. The maximum possible value of is , where and are relatively prime positive integers. Find .
Solution (Official MAA)
Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let be the point in the cross section where the bases of the cones meet, and let be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, , to the center of its base, . Let the sphere be tangent to at . The right triangles and are similar, implying that the radius of the sphere isThe requested sum is .
Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra.
Solution (Incircles + Coordinate Bash)
Let be the origin in the above diagram. Then is , is , is , and is . Also, it is easy to see that the inscribed sphere is simply the inscribed circle of . Then we want to find the intersection of and to determine the coordinates of point . Note that is line , and is line . Then, you can see that these lines are perpendicular, indicating that is a right triangle with right angle at . Finding the intersection of the lines by solving the system, we get that is the point in this plane. Then, we can find the distances and by the distance formula, which are and respectively. Also, . Then, to find the radius of this triangle's incircle, we use the formula from which we get that and is the answer.
~SirAppel
Solution (Clean analytic geometry)
Using the diagram above, we notice that the desired length is simply the distance between the point and . We can mark as since it is units away from each of the bases. Point is . Thus, line is . We can use the distance from point to line formula , where and are the coordinates of the point, and A, B, and C are the coefficients of the line in form . Plugging everything in, we get .
Solution 1: Graph paper coordbash
We graph this on graph paper, with the scale of . So, we can find then divide by to convert to our desired units, then square the result. With 5 minutes' worth of coordbashing, we finally arrive at .
~samrocksnature
Solution 2
Using the graph drawn above in MAA solution, we see that + = . = = , = 3, = 5. Plugging in and solve for gives us = .
~S17209
Video Solution
https://youtu.be/bz5N-jI2e0U?t=44
Video Solution 2
https://www.youtube.com/watch?v=0XJddG43pIk ~ MathEx
Video Solution 3
~IceMatrix
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.