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

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m (→‎Solution 2 (Tangential Distance))
 
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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 27: Line 52:
 
draw(circle(O, sqrt(160/13)));
 
draw(circle(O, sqrt(160/13)));
 
draw(circle(P, sqrt(160/13)));
 
draw(circle(P, sqrt(160/13)));
−
path b = brace(L, H);
+
path b = brace(L-(0,1), H-(0,1),0.5);
 
draw(b);
 
draw(b);
  
−
label("$R$", O -- Y, N);
+
label("$r$", O -- Y, N);
−
label("$R$", Y -- P, N);
+
label("$r$", Y -- P, N);
−
label("$R$", O -- A, NW);
+
label("$r$", O -- A, NW);
−
label("$R$", P -- D, NE);
+
label("$r$", P -- D, NE);
 
label("$1$", A -- H, N);
 
label("$1$", A -- H, N);
 
label("$2$", L -- D, N);
 
label("$2$", L -- D, N);
 
label("$7$", b, S);
 
label("$7$", b, S);
 +
 +
dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4));
 
</asy>
 
</asy>
 +
Set the common radius to <math>r</math>. First, take the cross section of the sphere sitting in the hole of radius <math>1</math>. 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 <math>r</math> and base <math>1</math>. Therefore, the height of this circle outside of the hole is <math>\sqrt{r^2-1}</math>.
  
−
Set the common radius to <math>r</math>. 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 <math>r</math> and base <math>1</math>. Therefore, the height of this circle outside of the hole is <math>\sqrt{r^2-1}</math>.
+
The other circle follows similarly for a height (outside the hole) of <math>\sqrt{r^2-4}</math>. 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 <math>7</math>, 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 <math>\sqrt{r^2-1} - \sqrt{r^2-4}</math>. Now we can set up an equation in terms of <math>r</math> with the Pythagorean theorem: <cmath>\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.</cmath> Simplifying a few times,
 +
<cmath>\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*}</cmath>
 +
Therefore, our answer is <math>\boxed{173}</math>.
  
−
The other circle follows similarly for a height (outside the hole) of <math>\sqrt{r^2-4}</math>. Now, if we take the cross section of the entire board, essentially making it 2-D, we can connect the centers of the two spheres, then form another right triangle with base <math>7</math>, 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 <math>\sqrt{r^2-1} - \sqrt{r^2-4}</math>. Now we can set up an equation in terms of <math>r</math> with the Pythagorean theorem: <cmath>(\sqrt{r^2-1} - \sqrt{r^2-4})^2 + 7^2 = (2r)^2.</cmath> Simplifying a few times, <cmath>r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 = 4r^2</cmath> <cmath>2r^2-44= -2\left(\sqrt{(r^2-1)(r^2-4)}\right)</cmath> <cmath>22-r^2=\left(\sqrt{r^4 - 5r^2 + 4}\right)</cmath> <cmath>r^4 -44r^2 + 484 = r^4 - 5r^2 + 4</cmath> <cmath>39r^2=480</cmath> <cmath>r^2=\frac{480}{39} = \frac{160}{13}.</cmath> Therefore, our answer is <math>\boxed{173}</math>.
+
~molocyxu
  
−
-molocyxu
+
==Solution 2  (Tangential Distance)==
−
==Video solution with animation==
+
[[File:2020 AIME I 6a.png|400px|right]]
 +
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>
  
−
https://youtu.be/cOf9uTJ9J40
+
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>
 +
[[File:2020 AIME I 6b.png|400px|right]]
 +
Distance between <math>C</math> and midpoint <math>M</math> of <math>AB</math> is
 +
<cmath>d = \frac {26}{7} - \frac {7}{2} = \frac{3}{14} \implies \sin \alpha = \frac {d}{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> 2R \cos \alpha = AB = 7 \implies R = \frac {\frac{7}{2} } {\frac{7}{2}\sqrt \frac{13}{160}} = \sqrt {\frac{160}{13}} .</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
−
==Solution 2 (Official MAA)==
+
==Video solution (With Animation)==
−
Consider a cross section of the board and spheres with a plane that passes through the centers of the holes and centers of the spheres as shown.
 
  
−
<asy>
+
https://youtu.be/cOf9uTJ9J40
−
unitsize(1.5 cm);
 
−
 
 
−
pair A, B, C, D, E, F, G, P, Q;
 
−
 
 
−
C = dir(175);
 
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D = dir(175 + 180);
 
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P = (-2,-0.8);
 
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Q = (2,-0.8);
 
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A = (C + reflect(P,Q)*(C))/2;
 
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B = (D + reflect(P,Q)*(D))/2;
 
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E = intersectionpoint(P--A, Circle(C,1));
 
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F = intersectionpoint(B--Q, Circle(D,1));
 
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G = (D + reflect(A,C)*(D))/2;
 
−
 
 
−
draw(Circle(C,1));
 
−
draw(Circle(D,1));
 
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draw(P--Q);
 
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draw(A--(C + (0,1)));
 
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draw(B--(D + (0,1)));
 
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draw(E--C--D--F);
 
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draw(D--G);
 
−
 
 
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dot("$A$", A, NE);
 
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dot("$B$", B, NW);
 
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dot("$C$", C, NW);
 
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dot("$D$", D, NE);
 
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dot("$E$", E, SW);
 
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dot("$F$", F, SE);
 
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dot("$G$", G, SE);
 
−
</asy>
 
  
−
Let <math>A</math>, <math>C</math>, and <math>E</math> be, respectively, the center of the hole with radius <math>1,</math> the center of the sphere resting in that hole, and a point on the edge of that hole. Let <math>B</math>, <math>D</math>, and <math>F</math> be the corresponding points for the hole with radius <math>2.</math> Let <math>G</math> be the point on <math>\overline{AC}</math> such that <math>\overline{AC} \perp \overline{GD}</math>. Let the radius of the spheres be <math>r = CE = DF</math>. Because <math>r^2 = AE^2 + AC^2 = 1 + AC^2</math> and <math>r^2 = BF^2 + BD^2 = 4 + BD^2</math>, it follows that<cmath>CG = AC - AG = AC - BD = \sqrt{r^2 - 1} - \sqrt{r^2-4}.</cmath>Because <math>DG = 7</math>, <math>CD = 2r</math>, and <math>CD^2 = CG^2 +GD^2</math>, it follows that<cmath>4r^2 = \left(\sqrt{r^2 - 1} - \sqrt{r^2-4}\right)^{\!2} + 7^2,</cmath>which simplifies to <math>r^2 = \frac{160}{13}</math>. The requested sum is <math>160+13 = 173</math>. The value of <math>r</math> is approximately <math>3.5082.</math>
 
 
==Video Solution==  
 
==Video Solution==  
  

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

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