1990 AIME Problems/Problem 14
Problem
The rectangle below has dimensions
and
. Diagonals
and
intersect at
. If triangle
is cut out and removed, edges
and
are joined, and the figure is then creased along segments
and
, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
Solution
Solution 1
import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0),A=(0,399^.5,0),D=(108^.5,0,0),C=(-108^.5,0,0); pair CENTER=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(CENTER.x,CENTER.y,99/133^.5); /*, Pa=(P.x,P.y,0); D(P--Pa--A);D(C--Pa--D); */ D((C+D)/2--A--C--D--P--C--P--A--D); MP("A",A,NE);MP("P",P,N);MP("C",C);MP("D",D); MP("13\sqrt{3}",(A+D)/2,E,small);MP("13\sqrt{3}",(A+C)/2,N,small);MP("12\sqrt{3}",(C+D)/2,SW,small); (Error making remote request. Unknown error_msg)
Our triangular pyramid has base . The area of this isosceles triangle is easy to find by
, where we can find
to be
by the Pythagorean Theorem. Thus
.
<center> size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0),A=(0,399^.5,0),D=(108^.5,0,0),C=(-108^.5,0,0); pair CENTER=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(CENTER.x,CENTER.y,99/133^.5), Pa=(P.x,P.y,0); D(A--C--D--P--C--P--A--D); D(P--Pa--A);D(C--Pa--D);D(circle(Pa,h)); MP("A",A,NE);MP("C",C,NW);MP("D",D);MP("P",P,N);MP("P'",Pa,SW); MP("13\sqrt{3}",(A+D)/2,E,small);MP("13\sqrt{3}",(A+C)/2,NW,small);MP("12\sqrt{3}",(C+D)/2,SW,small); MP("h",(P--Pa)/2,W);MP("\frac{\sqrt{939}}2",(C+P)/2,NW); (Error making remote request. Unknown error_msg)
To find the volume, we want to use the equation , so we need to find the height of the tetrahedron. By the Pythagorean Theorem,
. If we let
be the center of a sphere with radius
, then
lie on the sphere. The cross section of the sphere that contains
is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of
.
From here we just need to perform some brutish calculations. Using the formula (where
is the circumradius), we find
(there are slightly simpler ways to calculate
since we have an isosceles triangle). By the Pythagorean Theorem,
&= \left(\frac{\sqrt{939}}{2}\right)^2 - \left(\frac{13^2\sqrt{3}}{2\sqrt{133}}\right)^2\\ &= \frac{939 \cdot 133 - 13^4 \cdot 3}{4 \cdot 133} = \frac{13068 \cdot 3}{4 \cdot 133} = \frac{99^2}{133}\\ h &= \frac{99}{\sqrt{133}}
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Finally, we substitute into the volume equation to find
.
Solution 2
Let (or the triangle with sides
,
,
) be the base of our tetrahedron. We set points
and
as
and
, respectively. Using Pythagoras, we find
as
. We know that the vertex of the tetrahedron (
) has to be of the form
, where
is the altitude of the tetrahedron. Since the distance from
to points
,
, and
is
, we can write three equations using the distance formula:
x^{2} + (y - \sqrt {399})^{2} + z^{2} &=& \frac {939}{4}\\ (x - 6\sqrt {3})^{2} + y^{2} + z^{2} &=& \frac {939}{4}\\ (x + 6\sqrt {3})^{2} + y^{2} + z^{2} &=& \frac {939}{4}
\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)Subtracting the last two equations, we get . Solving for
with a bit of effort, we eventually get
,
,
.
Since the area of a triangle is
, we have the base area as
. Thus, the volume is
.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |