# 2002 AIME I Problems/Problem 15

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## 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