Difference between revisions of "2021 Fall AMC 10A Problems/Problem 22"
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~ Leo.Euler | ~ Leo.Euler | ||
+ | |||
+ | == Solution 2 (Cross section & angle bisector) == | ||
+ | |||
+ | We can take half of a cross section of the sphere, as such: | ||
+ | <asy> | ||
+ | unitsize(0.5cm); | ||
+ | real r = (90-40*sqrt(3))/11; | ||
+ | pair C = (0,0); | ||
+ | pair A = (-5,0); | ||
+ | pair B = (0,12); | ||
+ | pair O = (-((2*sqrt(3))/3) * r, r); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(circle(O,r)); | ||
+ | pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r); | ||
+ | pair E = (-2.2, 0); | ||
+ | draw(O--E); | ||
+ | draw(D--O); | ||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, N); | ||
+ | label("$C$", C, SE); | ||
+ | label("$O$", O, N); | ||
+ | label("$D$", D, NW); | ||
+ | label("$E$", E, SW); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | </asy> | ||
+ | Notice that we chose a cross section where one of the spheres was tangent to the lateral surface of the cone at <math>D</math>. | ||
+ | |||
+ | To evaluate <math>r</math>, we will find <math>AE</math> and <math>EC</math> in terms of <math>r</math>; we also know that <math>AE+EC = 5</math>, so with this, we can solve <math>r</math>. Firstly, to find <math>EC</math>, we can take a bird's eye view of the cone: | ||
+ | <asy> | ||
+ | unitsize(0.8cm); | ||
+ | pair C = (0,0); | ||
+ | draw(circle(C,5)); | ||
+ | label("$C$", C, N); | ||
+ | dot(C); | ||
+ | |||
+ | |||
+ | |||
+ | real r = (90-40*sqrt(3))/11; | ||
+ | real raise = r*(2/3*sqrt(3)); | ||
+ | pair E = (-r,raise/-2); | ||
+ | pair X = (0,raise); | ||
+ | pair Y = (r,raise/-2); | ||
+ | label("$E$", E, SW); | ||
+ | dot(E); | ||
+ | label("$X$", X, NW); | ||
+ | dot(X); | ||
+ | label("$Y$", Y, SW); | ||
+ | dot(Y); | ||
+ | |||
+ | draw(circle(X,r),dashed); | ||
+ | draw(circle(E,r),dashed); | ||
+ | draw(circle(Y,r),dashed); | ||
+ | draw(E--X,dashed); | ||
+ | draw(X--Y,dashed); | ||
+ | draw(E--Y,dashed); | ||
+ | </asy> | ||
+ | <math>C</math> is the centroid of equilateral triangle <math>EXY</math>. Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from <math>E</math> to <math>XY</math>; this is because medians cut each other into a <math>2</math> to <math>1</math> ratio. This equilateral triangle has a side length of <math>2r</math>, therefore it has an altitude of length <math>r \sqrt{3}</math>; two thirds of this is <math>\frac{2r \sqrt{3}}{3}</math>, so | ||
+ | <cmath>EC = \frac{2r \sqrt{3}}{3}.</cmath> | ||
+ | <asy> | ||
+ | unitsize(0.5cm); | ||
+ | real r = (90-40*sqrt(3))/11; | ||
+ | pair C = (0,0); | ||
+ | pair A = (-5,0); | ||
+ | pair B = (0,12); | ||
+ | pair O = (-((2*sqrt(3))/3) * r, r); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(circle(O,r)); | ||
+ | pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r); | ||
+ | pair E = (-2.2, 0); | ||
+ | pair F = (-2.2, 6.72); | ||
+ | draw(E--F); | ||
+ | draw(D--O); | ||
+ | draw(A--O, dotted); | ||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, N); | ||
+ | label("$C$", C, SE); | ||
+ | label("$O$", O, NE); | ||
+ | label("$D$", D, NW); | ||
+ | label("$E$", E, SW); | ||
+ | label("$F$", F, NW); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | </asy> | ||
+ | To evaluate <math>AE</math> in terms of <math>r</math>, we will extend <math>\overline{OE}</math> past point <math>O</math> to <math>\overline{AB}</math> at point <math>F</math>.<math>\triangle AEF</math> is similar to <math>\triangle ACB</math>. Also, <math>AO</math> is the angle bisector of <math>\angle EAB</math>. Therefore, by the angle bisector theorem, <math>\frac{OE}{OF} = \frac{AE}{AF} = \frac{5}{13}</math>. Also, <math>OE = r</math>, so <math>\frac{r}{OF} = \frac{5}{13}</math>, so <math>OF = \frac{13r}{5}</math>. This means that<cmath>AE = \frac{5 \cdot EF}{12} = \frac{5 \cdot (OE + OF)}{12} = \frac{5 \cdot (r + \frac{13r}{5})}{12} = \frac{18r}{12} = \frac{3r}{2}.</cmath> | ||
+ | We have that <math>EC = \frac{2r \sqrt{3}}{3}</math> and that <math>AE = \frac{3r}{2}</math>, so <math>AC = EC + AE = \frac{2r \sqrt{3}}{3} + \frac{3r}{2} = \frac{4r \sqrt{3} + 9r}{6}</math>. We also were given that <math>AC = 5</math>. Therefore, we have | ||
+ | <cmath>\frac{4r \sqrt{3} + 9r}{6} = 5.</cmath> | ||
+ | This is a simple linear equation in terms of <math>r</math>. We can solve for <math>r</math> to get <math>r = \boxed{\textbf{(B) } \frac{90 - 40 \sqrt{3}}{11}}.</math> | ||
+ | |||
+ | ~ ihatemath123 |
Revision as of 23:50, 22 November 2021
Problem
Inside a right circular cone with base radius and height are three congruent spheres with radius . Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is ?
Solution 1 (Coordinates)
We will use coordinates. WLOG, let the coordinates of the center of the base of the cone be the origin. Then, let the center of one of the spheres be . Note that the distance between this point and the plane given by is . Thus, by the point-to-plane distance formula, we have
Solving for yields .
~ Leo.Euler
Solution 2 (Cross section & angle bisector)
We can take half of a cross section of the sphere, as such: Notice that we chose a cross section where one of the spheres was tangent to the lateral surface of the cone at .
To evaluate , we will find and in terms of ; we also know that , so with this, we can solve . Firstly, to find , we can take a bird's eye view of the cone: is the centroid of equilateral triangle . Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from to ; this is because medians cut each other into a to ratio. This equilateral triangle has a side length of , therefore it has an altitude of length ; two thirds of this is , so To evaluate in terms of , we will extend past point to at point . is similar to . Also, is the angle bisector of . Therefore, by the angle bisector theorem, . Also, , so , so . This means that We have that and that , so . We also were given that . Therefore, we have This is a simple linear equation in terms of . We can solve for to get
~ ihatemath123