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Difference between revisions of "2006 AIME I Problems/Problem 14"

(Gotta show the answer.)
(+3d asymptote, How to draw perpendicular T to SCM?)
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== Solution ==
 
== Solution ==
−
{{image}}
+
<center><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;
 +
D(T--S--B--T--C--B--S--C);D(B--A--C--A--S,ddash);D(T--O--M,ddash);
 +
MP("T",T,N);MP("A",A);MP("B",B);MP("C",C,NE);dot(MP("S",S,NW));dot(MP("O",O));dot(MP("M",M,NE));
 +
MP("4",(S+T)/2,NW);MP("1",(S+A)/2,NW);MP("5",(B+T)/2,NE);MP("4",(O+T)/2,W);
 +
</asy></center>
  
 
We will use <math>[...]</math> to denote volume (four letters), area (three letters) or length (two letters).
 
We will use <math>[...]</math> to denote volume (four letters), area (three letters) or length (two letters).
Line 9: Line 18:
 
Let <math>T</math> be the top of the tripod, <math>A,B,C</math> are end points of three legs. Let <math>S</math> be the point on <math>TA</math> such that <math>[TS] = 4</math> and <math>[SA] = 1</math>. Let <math>O</math> be the center of the base [[equilateral triangle]] <math>ABC</math>. Let <math>M</math> be the [[midpoint]] of segment <math>BC</math>. Let <math>h</math> be the distance from <math>T</math> to the triangle <math>SBC</math> (<math>h</math> is what we want to find).
 
Let <math>T</math> be the top of the tripod, <math>A,B,C</math> are end points of three legs. Let <math>S</math> be the point on <math>TA</math> such that <math>[TS] = 4</math> and <math>[SA] = 1</math>. Let <math>O</math> be the center of the base [[equilateral triangle]] <math>ABC</math>. Let <math>M</math> be the [[midpoint]] of segment <math>BC</math>. Let <math>h</math> be the distance from <math>T</math> to the triangle <math>SBC</math> (<math>h</math> is what we want to find).
  
−
We have the volume ratio: <math>\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}</math>
+
We have the volume ratio <math>\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}</math>.
  
−
So <math>\frac {h\cdot [SBC]}{[TO]\cdot [ABC]} = \frac {4}{5}</math>
+
So <math>\frac {h\cdot [SBC]}{[TO]\cdot [ABC]} = \frac {4}{5}</math>.
  
−
We also have the area ratio: <math>\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}</math>
+
We also have the [[area ratios|area ratio]] <math>\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}</math>.
  
−
The triangle <math>TOA</math> is a 3-4-5 [[right triangle]] so <math>[AM] = \frac {3}{2}\cdot[AO] = \frac {9}{2}</math> and <math>\cos{\angle{TAO}} = \frac {3}{5}</math>.
+
The triangle <math>TOA</math> is a <math>3-4-5</math> [[right triangle]] so <math>[AM] = \frac {3}{2}\cdot[AO] = \frac {9}{2}</math> and <math>\cos{\angle{TAO}} = \frac {3}{5}</math>.
  
 
Applying [[Law of Cosines]] to the triangle <math>SAM</math> with <math>[SA] = 1</math>, <math>[AM] = \frac {9}{2}</math> and <math>\cos{\angle{SAM}} = \frac {3}{5}</math>, we find:
 
Applying [[Law of Cosines]] to the triangle <math>SAM</math> with <math>[SA] = 1</math>, <math>[AM] = \frac {9}{2}</math> and <math>\cos{\angle{SAM}} = \frac {3}{5}</math>, we find:
  
−
<math>[SM] = \frac {\sqrt {5\cdot317}}{10}</math>.
+
<center><math>[SM] = \frac {\sqrt {5\cdot317}}{10}.</math></center>
  
 
Putting it all together, we find <math>h = \frac {144}{\sqrt {5\cdot317}}</math>.
 
Putting it all together, we find <math>h = \frac {144}{\sqrt {5\cdot317}}</math>.
  
−
<math> \lfloor 144+\sqrt{5*317}\rfloor =144+ \lfloor \sqrt{5*317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}</math>
+
<center><math> \lfloor 144+\sqrt{5*317}\rfloor =144+ \lfloor \sqrt{5*317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}</math>.</center>
  
 
== See also ==
 
== See also ==

Revision as of 20:31, 25 April 2008

Problem

A tripod has three legs each of length $5$ 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 $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$)

Solution

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;
D(T--S--B--T--C--B--S--C);D(B--A--C--A--S,ddash);D(T--O--M,ddash);
MP("T",T,N);MP("A",A);MP("B",B);MP("C",C,NE);dot(MP("S",S,NW));dot(MP("O",O));dot(MP("M",M,NE));
MP("4",(S+T)/2,NW);MP("1",(S+A)/2,NW);MP("5",(B+T)/2,NE);MP("4",(O+T)/2,W);
 (Error making remote request. Unknown error_msg)

We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters).

Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS] = 4$ and $[SA] = 1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let $M$ be the midpoint of segment $BC$. Let $h$ be the distance from $T$ to the triangle $SBC$ ($h$ is what we want to find).

We have the volume ratio $\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}$.

So $\frac {h\cdot [SBC]}{[TO]\cdot [ABC]} = \frac {4}{5}$.

We also have the area ratio $\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}$.

The triangle $TOA$ is a $3-4-5$ right triangle so $[AM] = \frac {3}{2}\cdot[AO] = \frac {9}{2}$ and $\cos{\angle{TAO}} = \frac {3}{5}$.

Applying Law of Cosines to the triangle $SAM$ with $[SA] = 1$, $[AM] = \frac {9}{2}$ and $\cos{\angle{SAM}} = \frac {3}{5}$, we find:

$[SM] = \frac {\sqrt {5\cdot317}}{10}.$

Putting it all together, we find $h = \frac {144}{\sqrt {5\cdot317}}$.

$\lfloor 144+\sqrt{5*317}\rfloor =144+ \lfloor \sqrt{5*317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}$.

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
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