Difference between revisions of "2002 AIME I Problems/Problem 15"
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== Problem == | == Problem == | ||
+ | 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> | ||
− | == Solution == | + | == Solution 1 == |
+ | <center><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></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. | ||
+ | |||
+ | 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.* 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>: | ||
+ | <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>\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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 3 (minimal coordinates, Apollonius) == | ||
+ | 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>. | ||
+ | We seek <math>YD=a</math>. | ||
+ | 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>. | ||
+ | So, <math>a^2+6^2+12^2=14^2 \rightarrow a=4</math>. | ||
+ | We find <math>EA=2\sqrt{61}</math>. | ||
+ | |||
+ | 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. | ||
+ | Call these intersections <math>A'</math> and <math>B'</math>. Let <math>EA'=a, AA'=b</math>. | ||
+ | |||
+ | Using the Law of Cosines on <math>\triangle EAD</math> gives <math>\cos(\angle EAD)=\frac{4}{\sqrt{61}}</math>. | ||
+ | Using Law of Cosines on <math>\triangle EA'D</math> gives the equation <math>a^2=b^2+244+16b</math>. | ||
+ | Now, using Apollonius' Theorem on the same triangle gives <math>a^2=2b^2+232</math>. | ||
+ | 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>. | ||
+ | |||
+ | Finally, plugging into either expression for <math>a</math> gives <math>a^2=512-64\sqrt{19}</math>. | ||
+ | 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>. | ||
+ | |||
+ | Then <math>128+16+19=163</math>. | ||
+ | |||
+ | -RyanZhu817 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2002|n=I|num-b=14|after=Last Question}} | |
+ | {{MAA Notice}} |
Latest revision as of 15:56, 1 July 2024
Contents
Problem
Polyhedron has six faces. Face
is a square with
face
is a trapezoid with
parallel to
and
and face
has
The other three faces are
and
The distance from
to face
is 12. Given that
where
and
are positive integers and
is not divisible by the square of any prime, find
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]](http://latex.artofproblemsolving.com/1/6/7/167e69f38f11e4b219976293223c73b2bdec1e72.png)
Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: ,
,
and
. Since
is an isosceles trapezoid and
is an isosceles triangle, we have symmetry about the
-plane.
Therefore, the -component of
is 0. We are given that the
component is 12, and it lies over the square, so we must have
so
(the other solution,
does not lie over the square). Now let
and
, so
is parallel to
. We must have
, so
.
The last piece of information we have is that (and its reflection,
) are faces of the polyhedron, so they must all lie in the same plane. Since we have
,
, and
, we can derive this plane.* Let
be the extension of the intersection of the lines containing
. It follows that the projection of
onto the plane
must coincide with the
, where
is the projection of
onto the plane
.
by a ratio of
, so the distance from
to the plane
is
and by the similarity, the distance from
to the plane
is
. The altitude from
to
has height
. By similarity, the x-coordinate of
is
. Then
.
Now that we have located , we can calculate
:
Taking the negative root because the answer form asks for it, we get
, and
.
- One may also do this by vectors;
, so the plane is
. Since
lies on this plane, we must have
, so
. Therefore,
. So
.
Solution 2
We let be the origin, or
,
, and
. Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry,
, so
, 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 and
. Therefore, the x coordinate of
is
, so
.
We also know that and
are coplanar, so they all lie on the plane
. Since
is on it, then
. Also, since
is contained, then
. Finally, since
is on the plane, then
. Therefore,
. Since
, then
, or
. Therefore, the two permissible values of
are
. The only one that satisfies the conditions of the problem is
, from which the answer is
.
Solution 3 (minimal coordinates, Apollonius)
Denote the foot of the altitude from to
be
. Let the projection of
onto
be
.
We seek
.
Let
. Then we get
. Because the diagram is symmetrical,
.
So,
.
We find
.
Extend and
to meet the plane
. Since
is a quadrilateral and all on a plane, then the extension of
and
will meet the lines
and
, respectively.
Call these intersections
and
. Let
.
Using the Law of Cosines on gives
.
Using Law of Cosines on
gives the equation
.
Now, using Apollonius' Theorem on the same triangle gives
.
Equating the two gives
. Solving gives us
.
Finally, plugging into either expression for gives
.
Since
and is parallel to
, by the midpoint theorem,
.
Then .
-RyanZhu817
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.