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

(Solution: Fixed diagram.)
(Solution 2 (Official MAA): As to avoid copyrighted materials, I decide to delete this solution. Plus, this solution is repetitive from Solution 1.)
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~molocyxu
 
~molocyxu
 
==Solution 2 (Official MAA)==
 
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>
 
unitsize(1.5 cm);
 
 
pair A, B, C, D, E, F, G, P, Q;
 
 
C = dir(175);
 
D = dir(175 + 180);
 
P = (-2,-0.8);
 
Q = (2,-0.8);
 
A = (C + reflect(P,Q)*(C))/2;
 
B = (D + reflect(P,Q)*(D))/2;
 
E = intersectionpoint(P--A, Circle(C,1));
 
F = intersectionpoint(B--Q, Circle(D,1));
 
G = (D + reflect(A,C)*(D))/2;
 
 
draw(Circle(C,1));
 
draw(Circle(D,1));
 
draw(P--Q);
 
draw(A--(C + (0,1)));
 
draw(B--(D + (0,1)));
 
draw(E--C--D--F);
 
draw(D--G);
 
 
dot("$A$", A, NE);
 
dot("$B$", B, NW);
 
dot("$C$", C, NW);
 
dot("$D$", D, NE);
 
dot("$E$", E, SW);
 
dot("$F$", F, SE);
 
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 (With Animation)==
 
==Video solution (With Animation)==

Revision as of 19:47, 13 December 2021

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 $m+n.$

Solution

[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

Video solution (With Animation)

https://youtu.be/cOf9uTJ9J40

Video Solution

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

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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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