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>. | 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> | <asy> | ||
Line 38: | Line 35: | ||
dot("$C$", C, SE); | dot("$C$", C, SE); | ||
</asy> | </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 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 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 17:03, 11 April 2021
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 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
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.