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 .

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,

$\begin{align*}h^2 &= PA^2 - R^2 \\

&= \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:

$\begin{eqnarray*}

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 .