Difference between revisions of "2022 AIME II Problems/Problem 3"

Revision as of 09:12, 11 June 2022

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 $m+n$ .

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: $l^2=(\frac{9}{2}-l)^2+(3\sqrt{2})^2$ according to pythagorean theorem.

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

~DSAERF-CALMIT (https://binaryphi.site)

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

The height of the pyramid is $9/2$ , so let the pyramid's vertices be at $A = (0,0,0)$ , $B = (6, 0, 0)$ , $C= (0,6,0)$ , $D = (6,6,0)$ , and $E = (3,3, 9/2)$ . The center of the sphere is clearly directly above the center of $ABCD$ , so let $O = (3,3, z)$ be the center of the sphere. Then, $OE = OA\implies 9/2 - z = \sqrt{3^2 + 3^2 + z^2}\implies z = 1/4$ . Then, $OA = \sqrt{3^2 + 3^2 + (1/4)^2} = 17/4$ , so the answer is $\boxed{21}$ .

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=UJAYW6YNFVU

@@ Line 18: / Line 18: @@
 ~A1001
+==Solution 3==
+The height of the pyramid is <math>9/2</math>, so let the pyramid's vertices be at <math>A = (0,0,0)</math>, <math>B = (6, 0, 0)</math>, <math>C= (0,6,0)</math>, <math>D = (6,6,0)</math>, and <math>E = (3,3, 9/2)</math>. The center of the sphere is clearly directly above the center of <math>ABCD</math>, so let <math>O = (3,3, z)</math> be the center of the sphere. Then, <math>OE = OA\implies 9/2 - z = \sqrt{3^2 + 3^2 + z^2}\implies z = 1/4</math>. Then, <math>OA = \sqrt{3^2 + 3^2 + (1/4)^2} = 17/4</math>, so the answer is <math>\boxed{21}</math>.
 ==Video Solution (Mathematical Dexterity)==

2022 AIME II (Problems • Answer Key • Resources)
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