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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 9"

(Created page with "==Problem== <math>P-ABCD</math> is a right pyramid with square base <math>ABCD</math> edge length 6, and <math>PA=PB=PC=PD=6\sqrt{2}.</math> The probability that a randomly sele...")
 
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==Solution==
 
==Solution==
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[I believe solution is wrong - IP' is not sqrt(6)/3 as stated in the second paragraph - The_Turtle]
  
 
Let <math>R</math> be the set of all points that are at least <math>\frac{\sqrt{6}} {3}</math> units away from each face. <math>R</math> is tetrahedron, and it is similar to <math>P-ABCD</math>. This can be proved by showing that <math>R</math> is bounded by 5 planes, each of which is parallel to a corresponding plane of <math>P-ABCD</math>. Let the vertices of <math>R</math> be <math>P'A'B'C'D'</math> such that <math>P'</math> is the closest vertex to <math>P</math> and so forth. Consider cross section <math>\Delta PDB</math>. This cross section contains two concentric, similar triangles, <math>\Delta PDB</math> and <math>\Delta P'D'B'</math>. Furthermore, these triangles are equilateral; <math>BD</math> is the diagonal of a square with a side length of <math>6</math> and so <math>BD=6\sqrt{2}=PB=PD</math>.  
 
Let <math>R</math> be the set of all points that are at least <math>\frac{\sqrt{6}} {3}</math> units away from each face. <math>R</math> is tetrahedron, and it is similar to <math>P-ABCD</math>. This can be proved by showing that <math>R</math> is bounded by 5 planes, each of which is parallel to a corresponding plane of <math>P-ABCD</math>. Let the vertices of <math>R</math> be <math>P'A'B'C'D'</math> such that <math>P'</math> is the closest vertex to <math>P</math> and so forth. Consider cross section <math>\Delta PDB</math>. This cross section contains two concentric, similar triangles, <math>\Delta PDB</math> and <math>\Delta P'D'B'</math>. Furthermore, these triangles are equilateral; <math>BD</math> is the diagonal of a square with a side length of <math>6</math> and so <math>BD=6\sqrt{2}=PB=PD</math>.  

Latest revision as of 16:54, 16 February 2018

Problem

$P-ABCD$ is a right pyramid with square base $ABCD$ edge length 6, and $PA=PB=PC=PD=6\sqrt{2}.$ The probability that a randomly selected point inside the pyramid is at least $\frac{\sqrt{6}} {3}$ units away from each face can be expressed in the form $\frac{m}{n}$ where $m,n$ are relatively prime positive integers. Find $m+n.$

Solution

[I believe solution is wrong - IP' is not sqrt(6)/3 as stated in the second paragraph - The_Turtle]

Let $R$ be the set of all points that are at least $\frac{\sqrt{6}} {3}$ units away from each face. $R$ is tetrahedron, and it is similar to $P-ABCD$. This can be proved by showing that $R$ is bounded by 5 planes, each of which is parallel to a corresponding plane of $P-ABCD$. Let the vertices of $R$ be $P'A'B'C'D'$ such that $P'$ is the closest vertex to $P$ and so forth. Consider cross section $\Delta PDB$. This cross section contains two concentric, similar triangles, $\Delta PDB$ and $\Delta P'D'B'$. Furthermore, these triangles are equilateral; $BD$ is the diagonal of a square with a side length of $6$ and so $BD=6\sqrt{2}=PB=PD$.


From symmetry it follows that $PP' \perp B'D',BD$. Let $PP'$ intersect $B'D'$ at $H'$ and $BD$ at $H$. Then $PH=PP'+P'H'+HH'$. We can calculate $PH$, it is the height of an equilateral triangle with a side length of $6\sqrt2$. Then $PH=6\sqrt2 * \frac{\sqrt3}{2}=3\sqrt{6}$. Similarly, let $s$ be the sidelenth of $\Delta P'B'D'$. Then $P'H'$ is the height of this triangle and so is equal to $\frac{s\sqrt3}{2}$. Let $I$ be the foot of the perpendicular from $P'$ to $PB$. $PP'$ bisects $\angle BPD$ by symmetry, and so $\angle IPP'=30$ and $PP'=\frac{IP'}{\sin{\angle IPP'}}= \frac{\frac{\sqrt6}{3}}{\frac{1}{2}}=\frac{2\sqrt{6}}{3}$. Also $HH' = \frac{\sqrt{6}} {3}$ as it just the distance from $B'D'$ to $BD$.


Plugging these values in yields $3\sqrt{6}=\frac{2\sqrt{6}}{3}+\frac{s\sqrt3}{2}+\frac{\sqrt{6}} {3}$. Solving yields $s=4\sqrt2$. Therefore the ratio $R$ to $P-ABCD$ is $\frac{s}{PD}=\frac{4\sqrt2}{6\sqrt2}=\frac{2}{3}$. The ratio of their volumes is then the ratio of their sides cubed, or $\frac{8}{27}$. The ratio of the volumes of $R$ to $P-ABCD$ is equivalent to the probability a point will be in $R$. Hence $\frac{m}{n}=\frac{8}{27}$ and $m+n=\boxed{035}$.

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