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2021 AIME II Problems/Problem 10

Revision as of 11:47, 3 February 2022 by Puffer13 (talk | contribs) (→‎Diagram)

Problem

Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution 1 (Similar Triangles and Pythagorean Theorem)

This solution refers to the Diagram section.

As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell,$ so $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$ [asy] size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,yellow,light=White); draw(surface((-T1.x,L1.y,L1.z-abs(T1.z))--(-T2.x,L2.y,L2.z-abs(T2.z))--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(O1--O2--O3--cycle^^O3--A,dashed); draw(T1--T2--T3--cycle^^T3--A,heavygreen); draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed); draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red); label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); label("$\mathcal{R}$",O1,(0,-24,0)); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O_3$",O3,(0,-1.5,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); [/asy] Note that:

  1. In $\triangle O_1O_2O_3,$ we get $O_1O_2=72$ and $O_1O_3=O_2O_3=49.$

  2. Both $\triangle O_1O_2O_3$ and $\overline{O_3A}$ lie in plane $\mathcal{R}.$ Both $\triangle T_1T_2T_3$ and $\overline{T_3A}$ lie in plane $\mathcal{P}.$

  3. By symmetry, since planes $\mathcal{P}$ and $\mathcal{Q}$ are reflections of each other about plane $\mathcal{R},$ the three planes are concurrent to line $\ell.$

  4. Since $\overline{O_1T_1}\perp\mathcal{P}$ and $\overline{O_3T_3}\perp\mathcal{P},$ it follows that $\overline{O_1T_1}\parallel\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar.

Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$

  1. In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line.

    Clearly, cross-section $O_1O_3T_3T_1$ intersects line $\ell$ at exactly one point. Furthermore, as the intersection of planes $\mathcal{R}$ and $\mathcal{P}$ is line $\ell,$ we conclude that $\overrightarrow{O_1O_3}$ and $\overrightarrow{T_1T_3}$ must intersect line $\ell$ at the same point. Let $B$ be the point of concurrency of $\overrightarrow{O_1O_3},\overrightarrow{T_1T_3},$ and line $\ell.$

  2. In cross-section $\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\overline{O_1C}.$

We have the following diagram: [asy] size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O3, T1, T3, A, B, C, D; O1 = (0,-36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1)); C = (0,-36,-36*sqrt(1105)/23); D = (0,-36,-sqrt(1105)); draw(C--O1--O3--A^^D--O3--B,dashed); draw(T1--T3--A^^T3--B,heavygreen); draw(O1--T1^^O3--T3,mediumblue+dashed); draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label("$\ell$",position=EndPoint,align=3*E),red); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_3$",O3,(0,1,1),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); dot("$B$",B,(0,0,-2),red+linewidth(4.5)); dot("$C$",C,(0,0,-2),red+linewidth(4.5)); dot("$D$",D,(0,-2,0),linewidth(4.5)); [/asy] In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ or $O_3B=\frac{637}{23}.$

In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ or $DC=\frac{13\sqrt{1105}}{23}.$

Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$

~MRENTHUSIASM

Solution 2 (Pythagorean Theorem)

The centers of the three spheres form a $49$-$49$-$72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$, which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints.

Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$. Extend $\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$, the center of the small sphere $C$, we want to find $BD$.

By Pythagoras, $AC=\sqrt{49^2-36^2}=\sqrt{1105}$, and we know that $MA=36$ and $BC=13$. We know that $\overline{MA}$ and $\overline{BC}$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$. Apply the Pythagorean theorem to $\triangle BCD$, $BD=\frac{312}{23}$, so $312 + 23 = \boxed{335}$.

-Ross Gao

Solution 3 (Coordinate Geometry)

Let's try to see some symmetry. We can use an $xyz$-plane to plot where the circles are. The two large spheres are externally tangent, so we'll make them at $(0,-36,0)$ and $(0,36,0)$. The center of the little sphere would be $(x,0,-23)$ since we don't know how much the little sphere will be "pushed" down. We use the 3D distance formula to find that $x=-24$ (since $x=24$ wouldn't make sense). Now, we draw a line through the little sphere and the origin. It also intersects $\ell$ because of the symmetry we created.

$\ell$ lies on the plane too, so these two lines must intersect. The point at where it intersects is $(-24a,0,23a)$. We can use the distance formula again to find that $a=\dfrac{36}{23}$. Therefore, they intersect at $\left(-\dfrac{864}{23},0,-36\right)$. Since the little circle's $x$-coordinate is $-24$ and the intersection point's $x$-coordinate is $\dfrac{864}{23}$, we get $\dfrac{864}{23} - 24 = \dfrac{312}{23}$. Therefore, our answer to this problem is $312 + 23 = \boxed{335}$.

~Arcticturn

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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