Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AIME I Problems/Problem 6"

(→‎Solution 2 (No algebra))
m (→‎Solution 2 (Tangential Distance))
 
(16 intermediate revisions by 4 users not shown)
Line 2: Line 2:
 
A flat board has a circular hole with radius <math>1</math> and a circular hole with radius <math>2</math> such that the distance between the centers of the two holes is <math>7.</math> Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
 
A flat board has a circular hole with radius <math>1</math> and a circular hole with radius <math>2</math> such that the distance between the centers of the two holes is <math>7.</math> Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
−
== Solution ==
+
== Diagram ==
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
 +
import graph3;
 +
import solids;
 +
 
 +
currentprojection=orthographic((1,-5,1));
 +
real r = sqrt(160/13);
 +
 
 +
triple C1, C2, B1, B2;
 +
C1 = (0,0,sqrt(r^2-1^2));
 +
C2 = (7,0,sqrt(r^2-2^2));
 +
B1 = (0,0,0);
 +
B2 = (7,0,0);
 +
draw((-5,-10,0)--(-5,10,0)--(12,10,0)--(12,-10,0)--cycle);
 +
draw(shift(C1)*scale3(r)*unitsphere,yellow,light=White);
 +
draw(shift(C2)*scale3(r)*unitsphere,yellow,light=White);
 +
draw(Circle(B1,1,(0,0,1)));
 +
draw(Circle(B2,2,(0,0,1)));
 +
 
 +
dot(C1^^C2^^B1^^B2,linewidth(4.5));
 +
</asy>
 +
~MRENTHUSIASM
 +
 
 +
== Solution 1 (Pythagorean Theorem) ==
 
<asy>
 
<asy>
 
size(10cm);
 
size(10cm);
Line 55: Line 80:
 
~molocyxu
 
~molocyxu
  
−
==Solution 2  (No algebra)==
+
==Solution 2  (Tangential Distance)==
−
[[File:2020 AIME I 6a.png|500px|right]]
+
[[File:2020 AIME I 6a.png|400px|right]]
−
Let <math>A</math> and <math>B</math> be the centers of the holes, <math>C</math> is point of crossing <math>AB</math> and  radical axes of the circles. So <math>C</math> has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.)
+
Let <math>A</math> and <math>B</math> be the centers of the holes, let <math>C</math> be the point of crossing <math>AB</math> and  radical axes of the circles. So <math>C</math> has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.)
 +
[[File:2020 AIME I 6d.png|400px|right]]
 +
<cmath>CA = \frac {AB} {2} – \frac {r_A^2 – r_B^2}{2 AB} = \frac{23}{7}, CB = AB - AC =\frac{26}{7},</cmath>
 +
<cmath>CA' = CB'= \sqrt{BC^2 – r_B^2} = \frac {4}{7} \sqrt{30}.</cmath>
  
−
<cmath>CA = \frac {AB} {2} – \frac {r_A^2 – r_B^2}{2 AB} = \frac{23}{7}, CB = AB - AC =\frac{26}{7},</cmath>
+
Let <math>D</math> be the point of tangency of the spheres common radius <math>R</math> centered at <math>O</math> and <math>O'.</math> Let <math>\alpha</math> be the angle between <math>OO'</math> and flat board. In the plane, perpendicular to board <cmath>DC \perp OO', DC =  \frac {4}{7} \sqrt{30}.</cmath>
−
<cmath>CA' = CB'= BC^2 – r_A^2 = \frac {4}{7} \sqrt{30}.</cmath>
+
[[File:2020 AIME I 6b.png|400px|right]]
−
[[File:2020 AIME I 6.png|500px|right]]
+
Distance between <math>C</math> and midpoint <math>M</math> of <math>AB</math> is
−
Let <math>D</math> be the point of tangency of the spheres common radius <math>R</math> centered at <math>O</math> and <math>O'.</math> Let <math>\alpha</math> be the angle between <math>OO'</math> and flat board. In the plane, perpendicular to board <math>DC \perp OO', DC =  \frac {4}{7} \sqrt{30}.</math>
+
<cmath>d = \frac {26}{7} - \frac {7}{2} = \frac{3}{14} \implies \sin \alpha = \frac {d}{DC} = \frac {3}{8\sqrt {30}}.</cmath>
−
<math>DC \perp OO'</math> hence
 
−
<cmath>\sin \alpha = \frac {AC – BC}{DC} = \frac {3}{8\sqrt 30}.</cmath>
 
 
<cmath>\cos \alpha = \sqrt {1 – \frac {9}{64 \cdot 30}} = \sqrt{ \frac {637}{640}} = \frac {7}{2} \sqrt {\frac{13}{160}}.</cmath>
 
<cmath>\cos \alpha = \sqrt {1 – \frac {9}{64 \cdot 30}} = \sqrt{ \frac {637}{640}} = \frac {7}{2} \sqrt {\frac{13}{160}}.</cmath>
 
<cmath> 2R \cos \alpha = AB = 7 \implies R = \frac {\frac{7}{2} } {\frac{7}{2}\sqrt \frac{13}{160}} = \sqrt {\frac{160}{13}} .</cmath>
 
<cmath> 2R \cos \alpha = AB = 7 \implies R = \frac {\frac{7}{2} } {\frac{7}{2}\sqrt \frac{13}{160}} = \sqrt {\frac{160}{13}} .</cmath>
−
~vvsss, www.deoma-cmd.ru
+
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==Video solution (With Animation)==
 
==Video solution (With Animation)==

Latest revision as of 06:54, 15 December 2023

Problem

A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((1,-5,1)); real r = sqrt(160/13); triple C1, C2, B1, B2; C1 = (0,0,sqrt(r^2-1^2)); C2 = (7,0,sqrt(r^2-2^2)); B1 = (0,0,0); B2 = (7,0,0); draw((-5,-10,0)--(-5,10,0)--(12,10,0)--(12,-10,0)--cycle); draw(shift(C1)*scale3(r)*unitsphere,yellow,light=White); draw(shift(C2)*scale3(r)*unitsphere,yellow,light=White); draw(Circle(B1,1,(0,0,1))); draw(Circle(B2,2,(0,0,1))); dot(C1^^C2^^B1^^B2,linewidth(4.5)); [/asy] ~MRENTHUSIASM

Solution 1 (Pythagorean Theorem)

[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label("$r$", O -- Y, N); label("$r$", Y -- P, N); label("$r$", O -- A, NW); label("$r$", P -- D, NE); label("$1$", A -- H, N); label("$2$", L -- D, N); label("$7$", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy] Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$. Therefore, the height of this circle outside of the hole is $\sqrt{r^2-1}$.

The other circle follows similarly for a height (outside the hole) of $\sqrt{r^2-4}$. Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\sqrt{r^2-1} - \sqrt{r^2-4}$. Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \[\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.\] Simplifying a few times, \begin{align*} r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 &= 4r^2 \\ 2r^2-44 &= -2\left(\sqrt{(r^2-1)(r^2-4)}\right) \\ 22-r^2 &= \left(\sqrt{r^4 - 5r^2 + 4}\right) \\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\ 39r^2&=480 \\ r^2&=\frac{480}{39} = \frac{160}{13}. \end{align*} Therefore, our answer is $\boxed{173}$.

~molocyxu

Solution 2 (Tangential Distance)

2020 AIME I 6a.png

Let $A$ and $B$ be the centers of the holes, let $C$ be the point of crossing $AB$ and radical axes of the circles. So $C$ has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.)

2020 AIME I 6d.png

\[CA = \frac {AB} {2} – \frac {r_A^2 – r_B^2}{2 AB} = \frac{23}{7}, CB = AB - AC =\frac{26}{7},\] \[CA' = CB'= \sqrt{BC^2 – r_B^2} = \frac {4}{7} \sqrt{30}.\]

Let $D$ be the point of tangency of the spheres common radius $R$ centered at $O$ and $O'.$ Let $\alpha$ be the angle between $OO'$ and flat board. In the plane, perpendicular to board \[DC \perp OO', DC = \frac {4}{7} \sqrt{30}.\]

2020 AIME I 6b.png

Distance between $C$ and midpoint $M$ of $AB$ is \[d = \frac {26}{7} - \frac {7}{2} = \frac{3}{14} \implies \sin \alpha = \frac {d}{DC} = \frac {3}{8\sqrt {30}}.\] \[\cos \alpha = \sqrt {1 – \frac {9}{64 \cdot 30}} = \sqrt{ \frac {637}{640}} = \frac {7}{2} \sqrt {\frac{13}{160}}.\] \[2R \cos \alpha = AB = 7 \implies R = \frac {\frac{7}{2} } {\frac{7}{2}\sqrt \frac{13}{160}} = \sqrt {\frac{160}{13}} .\] vladimir.shelomovskii@gmail.com, vvsss

Video solution (With Animation)

https://youtu.be/cOf9uTJ9J40

Video Solution

https://www.youtube.com/watch?v=qCTq8KhZfYQ

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_6&oldid=207574"