2006 AIME I Problems/Problem 14
Problem
A tripod has three legs each of length feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is
feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let
be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then
can be written in the form
where
and
are positive integers and
is not divisible by the square of any prime. Find
(The notation
denotes the greatest integer that is less than or equal to
)
Solution 1
![[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label("$T$",T,N);label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$S$",S,NW);label("$O$",O,SW);label("$M$",M,NE); label("$4$",(S+T)/2,NW);label("$1$",(S+A)/2,NW);label("$5$",(B+T)/2,NE);label("$4$",(O+T)/2,W); dot(S);dot(O); [/asy]](http://latex.artofproblemsolving.com/e/d/6/ed699e0695def4b0bfa8107f507b11631d51df89.png)
We will use to denote volume (four letters), area (three letters) or length (two letters).
Let be the top of the tripod,
are end points of three legs. Let
be the point on
such that
and
. Let
be the center of the base equilateral triangle
. Let
be the midpoint of segment
. Let
be the distance from
to the triangle
(
is what we want to find).
We have the volume ratio .
So .
We also have the area ratio .
The triangle is a
right triangle so
and
.
Applying Law of Cosines to the triangle with
,
and
, we find:
![$[SM] = \frac {\sqrt {5\cdot317}}{10}.$](http://latex.artofproblemsolving.com/5/3/e/53e5967f1c82954260339c694c9dc704fa27c539.png)
Putting it all together, we find .
![$\lfloor 144+\sqrt{5 \cdot 317}\rfloor =144+ \lfloor \sqrt{5 \cdot 317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}$](http://latex.artofproblemsolving.com/1/2/8/12813e6cb448d8a5e57c89905023d195536276cf.png)
Solution 2
We note that . From this we can derive that the side length of the equilateral is
. We now use 3D coordinate geometry.
We know three points of plane
hence we can write out the equation for the plane.
Plane
can be expressed as
Applying the distance between a point and a plane formula.
Solution by SimonSun
Solution 3
Diagram borrowed from Solution 1
![[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label("$T$",T,N);label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NE);label("$S$",S,NW);label("$O$",O,SW);label("$M$",M,NE); label("$4$",(S+T)/2,NW);label("$1$",(S+A)/2,NW);label("$5$",(B+T)/2,NE);label("$4$",(O+T)/2,W); dot(S);dot(O); [/asy]](http://latex.artofproblemsolving.com/e/d/6/ed699e0695def4b0bfa8107f507b11631d51df89.png)
See also
2006 AIME I (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.