# 2022 AIME II Problems/Problem 3

## Problem

A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Diagram

$[asy] import three; import settings; leftbutton=new string[] {"rotate","zoom","shift","pan"}; middlebutton=new string[] {""}; rightbutton=new string[] {"zoom","rotateX","rotateY","rotateZ"}; wheelup=new string[] {"zoomin"}; wheeldown=new string[] {"zoomout"}; import three; pen p = rgb("00FF80")+opacity(0.25); pen q = rgb("FF8000")+opacity(0.75); triple A = (-3,-3,0); triple B = (-3,3,0); triple C = (3,3,0); triple D = (3,-3,0); triple F = (0,0,9/2); surface sphere = shift((0,0,1/4))*scale3(17/4)*unitsphere; draw(sphere,p); draw(A--B--C--D--A--F--B--F--C--F--D--F--A,q); dot(A); dot(B); dot(C); dot(D); dot(F); [/asy]$

## Solution 1

Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$. Because of the symmetrical property of the pyramid, we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.

Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$, where $h=\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2$ according to the Pythagorean theorem.

Solve this equation will give us $l = \frac{17}{4},$ therefore $m+n=\boxed{021}.$

## Solution 2

To start, we find the height of the pyramid. By the volume of a pyramid formula, we have $$\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.$$ Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\frac12 \cdot 6\sqrt2=3\sqrt2$, we have $$x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.$$ Taking the cross section of the pyramid and transforming the problem into $2$-d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\frac{3\sqrt{17}}2$, $\frac{3\sqrt{17}}2$, $6\sqrt2$. This turns out to be easy by the formula $R=\frac{abc}{4A}$, and through computing this value (the work has been left out) we find that $R=\frac{17}4$, so our answer is $\boxed{\textbf{021}}$.

~A1001

## Solution 3 (Coord bash)

By the volume of a pyramid formula, we have that the height of the pyramid is $\frac{9}{2}$. Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\frac{9}{2})$. Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$. Symmetry gives that $a=b=0$. Plug in $(3,3,0)$ and $(0,0,\frac{9}{2})$ and you get the system of equation

$18+c^2=r^2$

$(\frac{9}{2}-c)^2=r^2$

Solving gives $c=1/4$ and $r=17/4$, so our answer is $17+4=\boxed{021}$.~Ddk001

## See Also

 2022 AIME II (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 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.