Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AIME I Problems/Problem 15"

(→‎Solution: added the solution)
m (3d asy doesn't seem to work // temp save)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 +
<asy>
 +
void drawF(path3 p){draw(p); return;}
 +
import three; import graph; size(300); defaultpen(linewidth(0.7)); currentprojection=orthographic(50,-50,50);
 +
triple A=(-6,6,0), B = (-6,-6,0), C = (6,-6,0), D = (6,6,0), E = (2,0,12), F=(-6 + 19^.5, 3, 6), G=(-6 + 19^.5, -3, 6);
 +
drawF(A--B--C--D--cycle); drawF(A--E);</asy>
 +
  
 
Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: <math>A(-6,6,0)</math>, <math>B(-6,-6,0)</math>, <math>C(6,-6,0)</math> and <math>D(6,6,0)</math>. Since <math>ABFG</math> is an isosceles trapezoid and <math>CDE</math> is an isosceles triangle, we have symmetry about the <math>xz</math>-plane.
 
Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: <math>A(-6,6,0)</math>, <math>B(-6,-6,0)</math>, <math>C(6,-6,0)</math> and <math>D(6,6,0)</math>. Since <math>ABFG</math> is an isosceles trapezoid and <math>CDE</math> is an isosceles triangle, we have symmetry about the <math>xz</math>-plane.
Line 8: Line 14:
 
Therefore, the <math>y</math>-component of <math>E</math> is 0. We are given that the <math>z</math> component is 12, and it lies over the square, so we must have <math>E(2,0,12)</math> so <math>CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14</math> (the other solution, <math>E(10,0,12)</math> does not lie over the square). Now let <math>F(a,-3,b)</math> and <math>G(a,3,b)</math>, so <math>FG=6</math> is parallel to <math>\overline{AB}</math>. We must have <math>BF=8</math>, so <math>(a+6)^2+b^2=8^2-3^2=55</math>.
 
Therefore, the <math>y</math>-component of <math>E</math> is 0. We are given that the <math>z</math> component is 12, and it lies over the square, so we must have <math>E(2,0,12)</math> so <math>CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14</math> (the other solution, <math>E(10,0,12)</math> does not lie over the square). Now let <math>F(a,-3,b)</math> and <math>G(a,3,b)</math>, so <math>FG=6</math> is parallel to <math>\overline{AB}</math>. We must have <math>BF=8</math>, so <math>(a+6)^2+b^2=8^2-3^2=55</math>.
  
−
The last piece of information we have is that <math>ADEG</math> (and its reflection, <math>BCEF</math>) are faces of the polyhedron, so they must all lie in the same plane. Since we have <math>A</math>, <math>D</math>, and <math>E</math>, we can derive this plane. First, <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=12</math>. Since <math>G</math> lies on this plane, we must have <math>2\cdot3+b=12</math>, so <math>b=6</math>. Therefore, <math>a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}</math>. So <math>G(-6\pm\sqrt{19},-3,6)</math>.
+
The last piece of information we have is that <math>ADEG</math> (and its reflection, <math>BCEF</math>) are faces of the polyhedron, so they must all lie in the same plane. Since we have <math>A</math>, <math>D</math>, and <math>E</math>, we can derive this plane.* Let <math>H</math> be the extension of the intersection of the lines containing <math>\overline{AG}, \overline{BF}</math>. It follows that the projection of <math>\triangle AHB</math> onto the plane <math>x = 6</math> must coincide with the <math>\triangle CDE'</math>, where <math>E'</math> is the projection of <math>E</math> onto the plane <math>x = 6</math>. <math>\triangle GHF \sim \triangle AHB</math> by a ratio of <math>1/2</math>, so the distance from <math>H</math> to the plane <math>x = -6</math> is <cmath>\sqrt{\left(\sqrt{(2 \times 8)^2 - 6^2}\right)^2 - 12^2} = 2\sqrt{19};</cmath> and by the similarity, the distance from <math>G</math> to the plane <math>x = -6</math> is <math>\sqrt{19}</math>. The altitude from <math>G</math> to <math>ABCD</math> has height <math>12/2 = 6</math>. By similarity, the x-coordinate of <math>G</math> is <math>-6/2 = -3</math>. Then <math>G = (-6 \pm \sqrt{19}, -3, 6)</math>.
  
 
Now that we have located <math>G</math>, we can calculate <math>EG^2</math>:
 
Now that we have located <math>G</math>, we can calculate <math>EG^2</math>:
 
<cmath>EG^2=(8\pm\sqrt{19})^2+3^2+6^2=64\pm16\sqrt{19}+19+9+36=128\pm16\sqrt{19}.</cmath> Taking the negative root because the answer form asks for it, we get <math>128-16\sqrt{19}</math>, and <math>128+16+19=\fbox{163}</math>.
 
<cmath>EG^2=(8\pm\sqrt{19})^2+3^2+6^2=64\pm16\sqrt{19}+19+9+36=128\pm16\sqrt{19}.</cmath> Taking the negative root because the answer form asks for it, we get <math>128-16\sqrt{19}</math>, and <math>128+16+19=\fbox{163}</math>.
 +
 +
----
 +
*One may also do this by vectors; <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=12</math>. Since <math>G</math> lies on this plane, we must have <math>2\cdot3+b=12</math>, so <math>b=6</math>. Therefore, <math>a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}</math>. So <math>G(-6\pm\sqrt{19},-3,6)</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=14|after=Last Question}}
 
{{AIME box|year=2002|n=I|num-b=14|after=Last Question}}

Revision as of 21:23, 13 March 2009

Problem

Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given that $EG^2 = p - q\sqrt {r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$

Solution

void drawF(path3 p){draw(p); return;}
import three; import graph; size(300); defaultpen(linewidth(0.7)); currentprojection=orthographic(50,-50,50);
triple A=(-6,6,0), B = (-6,-6,0), C = (6,-6,0), D = (6,6,0), E = (2,0,12), F=(-6 + 19^.5, 3, 6), G=(-6 + 19^.5, -3, 6); 
drawF(A--B--C--D--cycle); drawF(A--E); (Error making remote request. Unknown error_msg)


Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: $A(-6,6,0)$, $B(-6,-6,0)$, $C(6,-6,0)$ and $D(6,6,0)$. Since $ABFG$ is an isosceles trapezoid and $CDE$ is an isosceles triangle, we have symmetry about the $xz$-plane.

Therefore, the $y$-component of $E$ is 0. We are given that the $z$ component is 12, and it lies over the square, so we must have $E(2,0,12)$ so $CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14$ (the other solution, $E(10,0,12)$ does not lie over the square). Now let $F(a,-3,b)$ and $G(a,3,b)$, so $FG=6$ is parallel to $\overline{AB}$. We must have $BF=8$, so $(a+6)^2+b^2=8^2-3^2=55$.

The last piece of information we have is that $ADEG$ (and its reflection, $BCEF$) are faces of the polyhedron, so they must all lie in the same plane. Since we have $A$, $D$, and $E$, we can derive this plane.* Let $H$ be the extension of the intersection of the lines containing $\overline{AG}, \overline{BF}$. It follows that the projection of $\triangle AHB$ onto the plane $x = 6$ must coincide with the $\triangle CDE'$, where $E'$ is the projection of $E$ onto the plane $x = 6$. $\triangle GHF \sim \triangle AHB$ by a ratio of $1/2$, so the distance from $H$ to the plane $x = -6$ is \[\sqrt{\left(\sqrt{(2 \times 8)^2 - 6^2}\right)^2 - 12^2} = 2\sqrt{19};\] and by the similarity, the distance from $G$ to the plane $x = -6$ is $\sqrt{19}$. The altitude from $G$ to $ABCD$ has height $12/2 = 6$. By similarity, the x-coordinate of $G$ is $-6/2 = -3$. Then $G = (-6 \pm \sqrt{19}, -3, 6)$.

Now that we have located $G$, we can calculate $EG^2$: \[EG^2=(8\pm\sqrt{19})^2+3^2+6^2=64\pm16\sqrt{19}+19+9+36=128\pm16\sqrt{19}.\] Taking the negative root because the answer form asks for it, we get $128-16\sqrt{19}$, and $128+16+19=\fbox{163}$.

  • One may also do this by vectors; $\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)$, so the plane is $2y+z=2\cdot6=12$. Since $G$ lies on this plane, we must have $2\cdot3+b=12$, so $b=6$. Therefore, $a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}$. So $G(-6\pm\sqrt{19},-3,6)$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_15&oldid=30701"