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

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Polyhedron <math>ABCDEFG</math> has six faces.  Face <math>ABCD</math> is a square with <math>AB = 12;</math> face <math>ABFG</math> is a trapezoid with <math>\overline{AB}</math> parallel to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math>  The other three faces are <math>ADEG, BCEF,</math> and <math>EFG.</math>  The distance from <math>E</math> to face <math>ABCD</math> is 12.  Given that <math>EG^2 = p - q\sqrt {r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math>
 
Polyhedron <math>ABCDEFG</math> has six faces.  Face <math>ABCD</math> is a square with <math>AB = 12;</math> face <math>ABFG</math> is a trapezoid with <math>\overline{AB}</math> parallel to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math>  The other three faces are <math>ADEG, BCEF,</math> and <math>EFG.</math>  The distance from <math>E</math> to face <math>ABCD</math> is 12.  Given that <math>EG^2 = p - q\sqrt {r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math>
  
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== Solution ==
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== Solution 1 ==
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{{image}}
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<center><asy>
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<asy>size(120);
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size(200);
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void drawF(path3 p){draw(p); return;}
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import three;
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import three; import graph; size(300); defaultpen(linewidth(0.7)); currentprojection=orthographic(50,-50,50);
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import graph;
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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);  
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defaultpen(linewidth(0.7)+fontsize(8));
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drawF(A--B--C--D--cycle); drawF(A--E);</asy>
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currentprojection=orthographic(-30,50,40);
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triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12);
 +
F = 1/2*H+1/2*B;
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G = 1/2*H+1/2*A;
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draw((A--B--C--D--A)^^(D--E--C)^^(A--G--F--B)^^(G--E--F));draw((G--H--F)^^(H--E1),gray(0.6));
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dot(H1^^H,linewidth(2));
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label("$A$",A,( 0,-1, 0));
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label("$B$",B,( 0, 1, 0));
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label("$C$",C,( 0, 1, 0));
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label("$D$",D,( 0,-1, 0));
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label("$E$",E,(-1,-1, 1));
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label("$F$",F,( 0, 1, 0));
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label("$G$",G,(-1,-1, 1));
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label("$H$",H,( 1,-1, 1));
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label("$H'$",H1,(-1,-1, 1));
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</asy></center>
  
 
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.
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*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>.
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*One may also do this by vectors; <math>\overrightarrow{AD}\times\overrightarrow{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>.
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 +
==Solution 2==
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 +
We let <math>A</math> be the origin, or <math>(0,0,0)</math>, <math>B = (0,0,12)</math>, and <math>D = (12,0,0)</math>. Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, <math>FX = GY = \frac{12-6}2 = 3</math>, so <math>G = (a,b,3)</math>, where a and b are variables.
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We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have <math>CZ = DZ = 6</math> and <math>EZ = 4\sqrt{10}</math>. Therefore, the x coordinate of <math>E</math> is <math>12-\sqrt{(4\sqrt{10})^2-12^2}=12-\sqrt{16}=12-4=8</math>, so <math>E = (8,12,6)</math>.
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We also know that <math>A,D,E,</math> and <math>G</math> are coplanar, so they all lie on the plane <math>z = Ax+By+C</math>. Since <math>(0,0,0)</math> is on it, then <math>C = 0</math>. Also, since <math>(12,0,0)</math> is contained, then <math>A = 0</math>. Finally, since <math>(8,12,6)</math> is on the plane, then <math>B = \frac 12</math>. Therefore, <math>b = 6</math>. Since <math>GA = 8</math>, then <math>a^2+6^2+3^2=8^2</math>, or <math>a = \pm \sqrt{19}</math>. Therefore, the two permissible values of <math>EG^2</math> are <math>(8 \pm \sqrt{19})^2+6^2+3^2 = 128 \pm 16\sqrt{19}</math>. The only one that satisfies the conditions of the problem is <math>128 - 16\sqrt{19}</math>, from which the answer is <math>128+16+19=\boxed{163}</math>.
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==Solution 3 (minimal coordinates, Apollonius) ==
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Denote the foot of the altitude from <math>E</math> to <math>ABCD</math> be <math>X</math>. Let the projection of <math>X</math> onto <math>AD</math> be <math>Y</math>.
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We seek <math>YD=a</math>.
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Let <math>E=(0, 0, 0)</math>. Then we get <math>X=(0, 0, -12)</math>. Because the diagram is symmetrical, <math>Y=(a, -6, -12)</math>.
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So, <math>a^2+6^2+12^2=14^2 \rightarrow a=4</math>.
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We find <math>EA=2\sqrt{61}</math>.
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Extend <math>EG</math> and <math>FE</math> to meet the plane <math>z=0</math>. Since <math>EGAD</math> is a quadrilateral and all on a plane, then the extension of <math>EG</math> and <math>FE</math> will meet the lines <math>AD</math> and <math>BC</math>, respectively.
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Call these intersections <math>A'</math> and <math>B'</math>. Let <math>EA'=a, AA'=b</math>.
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Using the Law of Cosines on <math>\triangle EAD</math> gives <math>\cos(\angle EAD)=\frac{4}{\sqrt{61}}</math>.
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Using Law of Cosines on <math>\triangle EA'D</math> gives the equation <math>a^2=b^2+244+16b</math>.
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Now, using Apollonius' Theorem on the same triangle gives <math>a^2=2b^2+232</math>.
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Equating the two gives <math>b^2-16b-12=0</math>. Solving gives us <math>b=8-2\sqrt{19}, b^2=140-32\sqrt{19}</math>.
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Finally, plugging into either expression for <math>a</math> gives <math>a^2=512-64\sqrt{19}</math>.
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Since <math>FG=\frac{1}{2}A'B'</math> and is parallel to <math>A'B'</math>, by the midpoint theorem, <cmath>EG=\frac{1}{2} A'B' \rightarrow EG^2=\frac{1}{4}A'B'^2 \rightarrow = 128-16\sqrt{19}</cmath>.
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Then <math>128+16+19=163</math>.
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 +
-RyanZhu817
  
 
== 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}}
 +
{{MAA Notice}}

Latest revision as of 14:56, 1 July 2024

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 1

[asy] size(200); import three; import graph; defaultpen(linewidth(0.7)+fontsize(8)); currentprojection=orthographic(-30,50,40); triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12); F = 1/2*H+1/2*B; G = 1/2*H+1/2*A; draw((A--B--C--D--A)^^(D--E--C)^^(A--G--F--B)^^(G--E--F));draw((G--H--F)^^(H--E1),gray(0.6)); dot(H1^^H,linewidth(2)); label("$A$",A,( 0,-1, 0)); label("$B$",B,( 0, 1, 0)); label("$C$",C,( 0, 1, 0)); label("$D$",D,( 0,-1, 0)); label("$E$",E,(-1,-1, 1)); label("$F$",F,( 0, 1, 0)); label("$G$",G,(-1,-1, 1)); label("$H$",H,( 1,-1, 1)); label("$H'$",H1,(-1,-1, 1)); [/asy]

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; $\overrightarrow{AD}\times\overrightarrow{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)$, 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 $G(-6\pm\sqrt{19},-3,6)$.

Solution 2

We let $A$ be the origin, or $(0,0,0)$, $B = (0,0,12)$, and $D = (12,0,0)$. Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \frac{12-6}2 = 3$, so $G = (a,b,3)$, where a and b are variables.

We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have $CZ = DZ = 6$ and $EZ = 4\sqrt{10}$. Therefore, the x coordinate of $E$ is $12-\sqrt{(4\sqrt{10})^2-12^2}=12-\sqrt{16}=12-4=8$, so $E = (8,12,6)$.

We also know that $A,D,E,$ and $G$ are coplanar, so they all lie on the plane $z = Ax+By+C$. Since $(0,0,0)$ is on it, then $C = 0$. Also, since $(12,0,0)$ is contained, then $A = 0$. Finally, since $(8,12,6)$ is on the plane, then $B = \frac 12$. Therefore, $b = 6$. Since $GA = 8$, then $a^2+6^2+3^2=8^2$, or $a = \pm \sqrt{19}$. Therefore, the two permissible values of $EG^2$ are $(8 \pm \sqrt{19})^2+6^2+3^2 = 128 \pm 16\sqrt{19}$. The only one that satisfies the conditions of the problem is $128 - 16\sqrt{19}$, from which the answer is $128+16+19=\boxed{163}$.

Solution 3 (minimal coordinates, Apollonius)

Denote the foot of the altitude from $E$ to $ABCD$ be $X$. Let the projection of $X$ onto $AD$ be $Y$. We seek $YD=a$. Let $E=(0, 0, 0)$. Then we get $X=(0, 0, -12)$. Because the diagram is symmetrical, $Y=(a, -6, -12)$. So, $a^2+6^2+12^2=14^2 \rightarrow a=4$. We find $EA=2\sqrt{61}$.

Extend $EG$ and $FE$ to meet the plane $z=0$. Since $EGAD$ is a quadrilateral and all on a plane, then the extension of $EG$ and $FE$ will meet the lines $AD$ and $BC$, respectively. Call these intersections $A'$ and $B'$. Let $EA'=a, AA'=b$.

Using the Law of Cosines on $\triangle EAD$ gives $\cos(\angle EAD)=\frac{4}{\sqrt{61}}$. Using Law of Cosines on $\triangle EA'D$ gives the equation $a^2=b^2+244+16b$. Now, using Apollonius' Theorem on the same triangle gives $a^2=2b^2+232$. Equating the two gives $b^2-16b-12=0$. Solving gives us $b=8-2\sqrt{19}, b^2=140-32\sqrt{19}$.

Finally, plugging into either expression for $a$ gives $a^2=512-64\sqrt{19}$. Since $FG=\frac{1}{2}A'B'$ and is parallel to $A'B'$, by the midpoint theorem, \[EG=\frac{1}{2} A'B' \rightarrow EG^2=\frac{1}{4}A'B'^2 \rightarrow = 128-16\sqrt{19}\].

Then $128+16+19=163$.

-RyanZhu817

See also

2002 AIME I (Problems • Answer Key • Resources)
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