Difference between revisions of "2020 AIME II Problems/Problem 7"

m (Problem)
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==Solution==
 
==Solution==
 
Take the cross-section of the plane of symmetry formed by the two cones. Let the point where the bases intersect be the origin, <math>O</math>, and the bases form the positive <math>x</math> and <math>y</math> axes. Then label the vertices of the region enclosed by the two triangles as <math>O,A,B,C</math> in a clockwise manner. We want to find the radius of the inscribed circle of <math>OABC</math>. By symmetry, the center of this circle must be <math>(3,3)</math>. <math>\overline{OA}</math> can be represented as <math>8x-3y=0</math> Using the point-line distance formula, <cmath>r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\implies225+73=\boxed{298}</cmath>~mn28407
 
Take the cross-section of the plane of symmetry formed by the two cones. Let the point where the bases intersect be the origin, <math>O</math>, and the bases form the positive <math>x</math> and <math>y</math> axes. Then label the vertices of the region enclosed by the two triangles as <math>O,A,B,C</math> in a clockwise manner. We want to find the radius of the inscribed circle of <math>OABC</math>. By symmetry, the center of this circle must be <math>(3,3)</math>. <math>\overline{OA}</math> can be represented as <math>8x-3y=0</math> Using the point-line distance formula, <cmath>r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\implies225+73=\boxed{298}</cmath>~mn28407
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==Solution 2 (Official MAA)==
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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>.
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<asy>
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unitsize(0.6cm);
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pair A = (0,0);
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pair TriangleOneLeft = (-6,0);
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pair TriangleOneDown = (-3,-8);
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pair TriangleOneMid = (-3,0);
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pair D = (0,-3);
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pair TriangleTwoDown = (0,-6);
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pair B = (-8,-3);
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pair C = IP(TriangleOneMid -- TriangleOneDown, B--D);
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pair EE = foot(C, A, B);
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real radius = arclength(C--EE);
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path circ = Circle(C, radius);
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 +
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draw(A--B--TriangleTwoDown--cycle);
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draw(B--D);
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draw(A--TriangleOneLeft--TriangleOneDown--cycle);
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draw(circ);
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draw(C--EE);
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draw(TriangleOneMid -- TriangleOneDown, gray);
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dot("$B$", B, W);
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dot("$E$", EE, NW);
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dot("$A$", A, NE);
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dot("$D$", D, E);
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dot("$C$", C, SE);
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</asy>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:44, 9 June 2020

Problem

Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies withing both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Take the cross-section of the plane of symmetry formed by the two cones. Let the point where the bases intersect be the origin, $O$, and the bases form the positive $x$ and $y$ axes. Then label the vertices of the region enclosed by the two triangles as $O,A,B,C$ in a clockwise manner. We want to find the radius of the inscribed circle of $OABC$. By symmetry, the center of this circle must be $(3,3)$. $\overline{OA}$ can be represented as $8x-3y=0$ Using the point-line distance formula, \[r^2=\frac{15^2}{(\sqrt{8^2+3^2})^2}=\frac{225}{73}\implies225+73=\boxed{298}\]~mn28407

Solution 2 (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 $A$ be the point in the cross section where the bases of the cones meet, and let $C$ be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, $B$, to the center of its base, $D$. Let the sphere be tangent to $\overline{AB}$ at $E$. The right triangles $\triangle ABD$ and $\triangle CBE$ are similar, implying that the radius of the sphere is\[CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.\]The requested sum is $225+73=298$. [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]

Video Solution

https://youtu.be/bz5N-jI2e0U?t=44

See Also

2020 AIME II (ProblemsAnswer KeyResources)
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

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