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2011 AIME I Problems/Problem 13: Difference between revisions

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==Solution 2==
==Solution 2==
Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the other vertices to the plane is 10-d, 11-d, and 12-d respectively. The distance formula gives <cmath>\frac{a(10-d)}{\sqrt{a^2 + b^2 + c^2}} = 10,</cmath> <cmath>\frac{b(10-d)}{\sqrt{a^2 + b^2 + c^2}} = 11,</cmath> and <cmath>\frac{c(12-d)}{\sqrt{a^2 + b^2 + c^2}} = 12.</cmath> An easy algebraic manipulation gives the equation in the first solution.
Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the other vertices to the plane is 10-d, 11-d, and 12-d respectively. The distance formula gives <cmath>\frac{a(10)}{\sqrt{a^2 + b^2 + c^2}} = 10-d,</cmath> <cmath>\frac{b(10)}{\sqrt{a^2 + b^2 + c^2}} = 11-d,</cmath> and <cmath>\frac{c(10)}{\sqrt{a^2 + b^2 + c^2}} = 12-d.</cmath> An easy algebraic manipulation gives the equation in the first solution.


== See also ==
== See also ==
{{AIME box|year=2011|n=I|num-b=12|num-a=14}}
{{AIME box|year=2011|n=I|num-b=12|num-a=14}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 21:18, 16 March 2015

Problem

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t$.

Solution

Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. Then the (directed) distance from any point (x,y,z) to the plane is $ax+by+cz+d$. So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, $(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100$.

Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\sqrt {98/3}$. The distance from the origin to the plane is simply d, which is equal to $11-\sqrt{98/3} =(33-\sqrt{294})/3$, so $33+294+3=330$

Solution 2

Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the other vertices to the plane is 10-d, 11-d, and 12-d respectively. The distance formula gives \[\frac{a(10)}{\sqrt{a^2 + b^2 + c^2}} = 10-d,\] \[\frac{b(10)}{\sqrt{a^2 + b^2 + c^2}} = 11-d,\] and \[\frac{c(10)}{\sqrt{a^2 + b^2 + c^2}} = 12-d.\] An easy algebraic manipulation gives the equation in the first solution.

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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