Difference between revisions of "2006 AIME I Problems/Problem 14"
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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>) | 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 == |
<center><asy> | <center><asy> | ||
size(200); | size(200); | ||
Line 33: | Line 33: | ||
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>. | ||
− | <center><math> \lfloor 144+\sqrt{5 | + | <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 39: | Line 100: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:45, 31 March 2020
Contents
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 (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]](http://latex.artofproblemsolving.com/e/d/6/ed699e0695def4b0bfa8107f507b11631d51df89.png)
Apply Pythagorean Theorem on yields
Since
is equilateral, we have
and
Apply Pythagorean Theorem on
yields
Apply Law of Cosines on
we have
Apply Law of Cosines on
using the fact that
we have
Apply Pythagorean Theorem on
yields
Let the perpendicular from
hits
at
. Let
and
. Apply Pythagorean Theorem on
and
we have
Cancelling out the
term and solving gets
.
Finally, by Pythagorean Theorem,
so
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.