Difference between revisions of "2011 AIME I Problems/Problem 13"
JoetheFixer (talk | contribs) (→Solution 2) |
Sugar rush (talk | contribs) |
||
(18 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane. The distance from vertex <math>A</math> to the plane can be expressed as <math> \frac{r-\sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers | + | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane. The distance from vertex <math>A</math> to the plane can be expressed as <math> \frac{r-\sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers. Find <math>r+s+t</math>. |
− | ==Solution== | + | ==Solution 1== |
− | Set the cube at the origin with the three vertices along the axes and the plane equal to <math>ax+by+cz+d=0</math>, where <math>a^2+b^2+c^2=1</math>. | + | Set the cube at the origin with the three vertices along the axes and the plane equal to <math>ax+by+cz+d=0</math>, where <math>a^2+b^2+c^2=1</math>. The distance from a point <math>(X,Y,Z)</math> to a plane with equation <math>Ax+By+Cz+D=0</math> is |
+ | <cmath>\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},</cmath> | ||
+ | so the (directed) distance from any point <math>(x,y,z)</math> to the plane is <math>ax+by+cz+d</math>. So, by looking at the three vertices, we have <math>10a+d=10, 10b+d=11, 10c+d=12</math>, and by rearranging and summing, <cmath>(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100.</cmath> | ||
− | Solving the equation is easier if we substitute <math>11-d=y</math>, to get <math>3y^2+2=100</math>, or <math>y=\sqrt {98/3}</math>. The distance from the origin to the plane is simply d, which is equal to <math>11-\sqrt{98/3} =(33-\sqrt{294})/3</math>, so <math>33+294+3=330</math> | + | Solving the equation is easier if we substitute <math>11-d=y</math>, to get <math>3y^2+2=100</math>, or <math>y=\sqrt {98/3}</math>. The distance from the origin to the plane is simply <math>d</math>, which is equal to <math>11-\sqrt{98/3} =(33-\sqrt{294})/3</math>, so <math>33+294+3=\boxed{330}</math>. |
==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)}{\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> Squaring each equation and then adding yields <math>100=(10-d)^2+(11-d)^2+(12-d)^2</math>, and we can proceed as 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 <math>ax + by + cz = 0</math>. 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> Squaring each equation and then adding yields <math>100=(10-d)^2+(11-d)^2+(12-d)^2</math>, and we can proceed as in the first solution. |
+ | |||
+ | ==Solution 3== | ||
+ | Let the vertices with distance <math>10,11,12</math> be <math>B,C,D</math>, respectively. An equilateral triangle <math>\triangle BCD</math> is formed with side length <math>10\sqrt{2}</math>. We care only about the <math>z</math> coordinate: <math>B=10,C=11,D=12</math>. It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so <math>\text{centroid}=(10+11+12)/3=11</math>. Designate the midpoint of <math>BD</math> as <math>M</math>. Notice that median <math>CM</math> is parallel to the plane because the <math>\text{centroid}</math> and vertex <math>C</math> have the same <math>z</math> coordinate, <math>11</math>, and the median contains <math>C</math> and the <math>\text{centroid}</math>. We seek the angle <math>\theta</math> of the line:<math>(1)</math> through the centroid <math>(2)</math> perpendicular to the plane formed by <math>\triangle BCD</math>, <math>(3)</math> with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular <math>\textit{in slope}</math> to <math>BD</math>. Since <math>BD</math> makes a <math>2-14-10\sqrt{2}</math> right triangle, the orthogonal line makes the same right triangle rotated <math>90^\circ</math>. Therefore, <math>\sin\theta=\frac{14}{10\sqrt{2}}=\frac{7\sqrt{2}}{10}</math>. | ||
+ | |||
+ | It is also known that the centroid of <math>\triangle BCD</math> is a third of the way between vertex <math>A</math> and <math>H</math>, the vertex farthest from the plane. Since <math>AH</math> is a diagonal of the cube, <math>AH=10\sqrt{3}</math>. So the distance from the <math>\text{centroid}</math> to <math>A</math> is <math>10/\sqrt{3}</math>. So, the <math>\Delta z</math> from <math>A</math> to the centroid is <math>\frac{10}{\sqrt{3}}\sin\theta=\frac{10}{\sqrt{3}}\left(\frac{7\sqrt{2}}{10}\right)=\frac{7\sqrt{6}}{3}</math>. | ||
+ | |||
+ | Thus the distance from <math>A</math> to the plane is <math>11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}</math>, and <math>33+294+3=\boxed{330}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtube.com/watch?v=Wi-aqv8Ron0 | ||
== 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}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 21:37, 25 February 2021
Problem
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled . The three vertices adjacent to vertex are at heights 10, 11, and 12 above the plane. The distance from vertex to the plane can be expressed as , where , , and are positive integers. Find .
Solution 1
Set the cube at the origin with the three vertices along the axes and the plane equal to , where . The distance from a point to a plane with equation is so the (directed) distance from any point to the plane is . So, by looking at the three vertices, we have , and by rearranging and summing,
Solving the equation is easier if we substitute , to get , or . The distance from the origin to the plane is simply , which is equal to , so .
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 . 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 and Squaring each equation and then adding yields , and we can proceed as in the first solution.
Solution 3
Let the vertices with distance be , respectively. An equilateral triangle is formed with side length . We care only about the coordinate: . It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so . Designate the midpoint of as . Notice that median is parallel to the plane because the and vertex have the same coordinate, , and the median contains and the . We seek the angle of the line: through the centroid perpendicular to the plane formed by , with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular to . Since makes a right triangle, the orthogonal line makes the same right triangle rotated . Therefore, .
It is also known that the centroid of is a third of the way between vertex and , the vertex farthest from the plane. Since is a diagonal of the cube, . So the distance from the to is . So, the from to the centroid is .
Thus the distance from to the plane is , and .
Video Solution
https://youtube.com/watch?v=Wi-aqv8Ron0
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.