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

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== Problem ==
 
== Problem ==
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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 <math> h </math> be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then <math> h </math> can be written in the form <math> \frac m{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the greatest integer that is less than or equal to <math> x. </math>)
+
A tripod has three legs each of length <math>5</math> 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 <math>4</math> feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let <math> h </math> be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then <math> h </math> can be written in the form <math> \frac m{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the greatest integer that is less than or equal to <math> x. </math>)
  
−
== Solution ==
+
== Solution 1 ==
−
{{solution}}
+
<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;
 +
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></center>
 +
 
 +
We will use <math>[...]</math> to denote volume (four letters), area (three letters) or length (two letters).
 +
 
 +
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>.
 +
 
 +
So <math>\frac {h\cdot [SBC]}{[TO]\cdot [ABC]} = \frac {4}{5}</math>.
 +
 
 +
We also have the [[area ratios|area ratio]] <math>\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}</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:
 +
 
 +
<center><math>[SM] = \frac {\sqrt {5\cdot317}}{10}.</math></center>
 +
 
 +
Putting it all together, we find <math>h = \frac {144}{\sqrt {5\cdot317}}</math>.
 +
 
 +
<center><math> \lfloor 144+\sqrt{5 \cdot 317}\rfloor =144+ \lfloor \sqrt{5 \cdot 317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}</math>.</center>
 +
 
 +
== Solution 2 ==
 +
 
 +
We note that <math>AO=3</math>. From this we can derive that the side length of the equilateral is <math>3\sqrt{3}</math>. We now use 3D coordinate geometry.
 +
<cmath>A = (0,0,0)</cmath>
 +
<cmath>B = (3\sqrt{3},0,0)</cmath>
 +
<cmath>C = (\frac{3\sqrt{3}}{2}, \frac{9}{2}, 0)</cmath>
 +
<cmath>T = (\frac{3\sqrt{3}}{2}, \frac{3}{2}, 4)</cmath>
 +
<cmath>S= (\frac{3\sqrt{3}}{10}, \frac{3}{10}, \frac{4}{5})</cmath>
 +
We know three points of plane <math>SCB</math> hence we can write out the equation for the plane.
 +
Plane <math>SCB</math> can be expressed as <cmath>4\sqrt{3}x+4y+39z-36=0.</cmath>
 +
 
 +
Applying the distance between a point and a plane formula.
 +
 
 +
<cmath>\frac{ax+by+cz+d}{\sqrt{a^{2}+b^{2}+c^{2}}} = \frac{4\sqrt{3} \cdot \frac{3\sqrt{3}}{2} + 4\cdot \frac{3}{2} + 39 \cdot 4 -36}{\sqrt{(4\sqrt{3})^2+4^2+39^2}} = \frac{144}{\sqrt{1585}}</cmath>
 +
 
 +
<cmath>\lfloor m+\sqrt{n}\rfloor = \lfloor 144+\sqrt{1585}\rfloor = 183 </cmath>
 +
 
 +
Solution by SimonSun
 +
 
 +
 
 +
== Solution 3 (Cosine Law & Pythagorean Bash) ==
 +
 
 +
Diagram borrowed from Solution 1
 +
<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;
 +
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></center>
 +
 
 +
Apply [[Pythagorean Theorem]] on <math>\bigtriangleup TOB</math> yields
 +
<cmath>BO=\sqrt{TB^2-TO^2}=3</cmath>
 +
Since <math>\bigtriangleup ABC</math> is equilateral, we have <math>\angle MOB=60^{\circ}</math> and
 +
<cmath>BC=2BM=2(OB\sin MOB)=3\sqrt{3}</cmath>
 +
Apply Pythagorean Theorem on <math>\bigtriangleup TMB</math> yields
 +
<cmath>TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}</cmath>
 +
Apply [[Law of Cosines]] on <math>\bigtriangleup TBC</math> we have
 +
<cmath>BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC</cmath>
 +
<cmath>(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC</cmath>
 +
<cmath>\cos BTC=\frac{23}{50}</cmath>
 +
Apply Law of Cosines on <math>\bigtriangleup STB</math> using the fact that <math>\angle STB=\angle BTC</math> we have
 +
<cmath>SB^2=ST^2+BT^2-2(ST)(BT)\cos STB</cmath>
 +
<cmath>SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}</cmath>
 +
Apply Pythagorean Theorem on <math>\bigtriangleup BSM</math> yields
 +
<cmath>SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}</cmath>
 +
Let the perpendicular from <math>T</math> hits <math>SBC</math> at <math>P</math>. Let <math>SP=x</math> and <math>PM=\frac{\sqrt{1585}}{10}-x</math>. Apply Pythagorean Theorem on <math>TSP</math> and <math>TMP</math> we have
 +
<cmath>TP^2=TS^2-SP^2=TM^2-PM^2</cmath>
 +
<cmath>4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2</cmath>
 +
Cancelling out the <math>x^2</math> term and solving gets <math>x=\frac{181}{2\sqrt{1585}}</math>.
 +
 
 +
Finally, by Pythagorean Theorem,
 +
<cmath>TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}</cmath>
 +
so <math>\lfloor m+\sqrt{n}\rfloor=\boxed{183}</math>
 +
 
 +
~ Nafer
  
 
== See also ==
 
== See also ==
Line 9: Line 100:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:45, 31 March 2020

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 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]

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 $h = \frac {144}{\sqrt {5\cdot317}}$.

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

Solution 2

We note that $AO=3$. From this we can derive that the side length of the equilateral is $3\sqrt{3}$. We now use 3D coordinate geometry. \[A = (0,0,0)\] \[B = (3\sqrt{3},0,0)\] \[C = (\frac{3\sqrt{3}}{2}, \frac{9}{2}, 0)\] \[T = (\frac{3\sqrt{3}}{2}, \frac{3}{2}, 4)\] \[S= (\frac{3\sqrt{3}}{10}, \frac{3}{10}, \frac{4}{5})\] We know three points of plane $SCB$ hence we can write out the equation for the plane. Plane $SCB$ can be expressed as \[4\sqrt{3}x+4y+39z-36=0.\]

Applying the distance between a point and a plane formula.

\[\frac{ax+by+cz+d}{\sqrt{a^{2}+b^{2}+c^{2}}} = \frac{4\sqrt{3} \cdot \frac{3\sqrt{3}}{2} + 4\cdot \frac{3}{2} + 39 \cdot 4 -36}{\sqrt{(4\sqrt{3})^2+4^2+39^2}} = \frac{144}{\sqrt{1585}}\]

\[\lfloor m+\sqrt{n}\rfloor = \lfloor 144+\sqrt{1585}\rfloor = 183\]

Solution by SimonSun


Solution 3 (Cosine Law & Pythagorean Bash)

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]

Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields \[BO=\sqrt{TB^2-TO^2}=3\] Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and \[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\] Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields \[TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}\] Apply Law of Cosines on $\bigtriangleup TBC$ we have \[BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC\] \[(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC\] \[\cos BTC=\frac{23}{50}\] Apply Law of Cosines on $\bigtriangleup STB$ using the fact that $\angle STB=\angle BTC$ we have \[SB^2=ST^2+BT^2-2(ST)(BT)\cos STB\] \[SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}\] Apply Pythagorean Theorem on $\bigtriangleup BSM$ yields \[SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}\] Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\frac{\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \[TP^2=TS^2-SP^2=TM^2-PM^2\] \[4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2\] Cancelling out the $x^2$ term and solving gets $x=\frac{181}{2\sqrt{1585}}$.

Finally, by Pythagorean Theorem, \[TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}\] so $\lfloor m+\sqrt{n}\rfloor=\boxed{183}$

~ Nafer

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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