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Difference between revisions of "2011 AIME I Problems/Problem 8"
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In triangle <math>ABC</math>, <math>BC = 23</math>, <math>CA = 27</math>, and <math>AB = 30</math>. Points <math>V</math> and <math>W</math> are on <math>\overline{AC}</math> with <math>V</math> on <math> \overline{AW} </math>, points <math>X</math> and <math>Y</math> are on <math> \overline{BC} </math> with <math>X</math> on <math> \overline{CY} </math>, and points <math>Z</math> and <math>U</math> are on <math> \overline{AB} </math> with <math>Z</math> on <math> \overline{BU} </math>. In addition, the points are positioned so that <math> \overline{UV}\parallel\overline{BC} </math>, <math> \overline{WX}\parallel\overline{AB} </math>, and <math> \overline{YZ}\parallel\overline{CA} </math>. Right angle folds are then made along <math> \overline{UV} </math>, <math> \overline{WX} </math>, and <math> \overline{YZ} </math>. The resulting figure is placed on a level floor to make a table with triangular legs. Let <math>h</math> be the maximum possible height of a table constructed from triangle <math>ABC</math> whose top is parallel to the floor. Then <math>h</math> can be written in the form <math> \frac{k\sqrt{m}}{n} </math>, where <math>k</math> and <math>n</math> are relatively prime positive integers and <math>m</math> is a positive integer that is not divisible by the square of any prime. Find <math>k+m+n</math>.
 
In triangle <math>ABC</math>, <math>BC = 23</math>, <math>CA = 27</math>, and <math>AB = 30</math>. Points <math>V</math> and <math>W</math> are on <math>\overline{AC}</math> with <math>V</math> on <math> \overline{AW} </math>, points <math>X</math> and <math>Y</math> are on <math> \overline{BC} </math> with <math>X</math> on <math> \overline{CY} </math>, and points <math>Z</math> and <math>U</math> are on <math> \overline{AB} </math> with <math>Z</math> on <math> \overline{BU} </math>. In addition, the points are positioned so that <math> \overline{UV}\parallel\overline{BC} </math>, <math> \overline{WX}\parallel\overline{AB} </math>, and <math> \overline{YZ}\parallel\overline{CA} </math>. Right angle folds are then made along <math> \overline{UV} </math>, <math> \overline{WX} </math>, and <math> \overline{YZ} </math>. The resulting figure is placed on a level floor to make a table with triangular legs. Let <math>h</math> be the maximum possible height of a table constructed from triangle <math>ABC</math> whose top is parallel to the floor. Then <math>h</math> can be written in the form <math> \frac{k\sqrt{m}}{n} </math>, where <math>k</math> and <math>n</math> are relatively prime positive integers and <math>m</math> is a positive integer that is not divisible by the square of any prime. Find <math>k+m+n</math>.
 +
 +
<math>[asy]
 +
unitsize(1 cm);
 +
pair translate;
 +
pair[] A, B, C, U, V, W, X, Y, Z;
 +
A[0] = (1.5,2.8);
 +
B[0] = (3.2,0);
 +
C[0] = (0,0);
 +
U[0] = (0.69*A[0] + 0.31*B[0]);
 +
V[0] = (0.69*A[0] + 0.31*C[0]);
 +
W[0] = (0.69*C[0] + 0.31*A[0]);
 +
X[0] = (0.69*C[0] + 0.31*B[0]);
 +
Y[0] = (0.69*B[0] + 0.31*C[0]);
 +
Z[0] = (0.69*B[0] + 0.31*A[0]);
 +
translate = (7,0);
 +
A[1] = (1.3,1.1) + translate;
 +
B[1] = (2.4,-0.7) + translate;
 +
C[1] = (0.6,-0.7) + translate;
 +
U[1] = U[0] + translate;
 +
V[1] = V[0] + translate;
 +
W[1] = W[0] + translate;
 +
X[1] = X[0] + translate;
 +
Y[1] = Y[0] + translate;
 +
Z[1] = Z[0] + translate;
 +
draw (A[0]--B[0]--C[0]--cycle);
 +
draw (U[0]--V[0],dashed);
 +
draw (W[0]--X[0],dashed);
 +
draw (Y[0]--Z[0],dashed);
 +
draw (U[1]--V[1]--W[1]--X[1]--Y[1]--Z[1]--cycle);
 +
draw (U[1]--A[1]--V[1],dashed);
 +
draw (W[1]--C[1]--X[1]);
 +
draw (Y[1]--B[1]--Z[1]);
 +
dot("</math>A<math>",A[0],N);
 +
dot("</math>B<math>",B[0],SE);
 +
dot("</math>C<math>",C[0],SW);
 +
dot("</math>U<math>",U[0],NE);
 +
dot("</math>V<math>",V[0],NW);
 +
dot("</math>W<math>",W[0],NW);
 +
dot("</math>X<math>",X[0],S);
 +
dot("</math>Y<math>",Y[0],S);
 +
dot("</math>Z<math>",Z[0],NE);
 +
dot(A[1]);
 +
dot(B[1]);
 +
dot(C[1]);
 +
dot("</math>U<math>",U[1],NE);
 +
dot("</math>V<math>",V[1],NW);
 +
dot("</math>W<math>",W[1],NW);
 +
dot("</math>X<math>",X[1],dir(-70));
 +
dot("</math>Y<math>",Y[1],dir(250));
 +
dot("</math>Z<math>",Z[1],NE);[/asy]</math>

Revision as of 21:45, 24 March 2011

In triangle $ABC$, $BC = 23$, $CA = 27$, and $AB = 30$. Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$, points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$, and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$. In addition, the points are positioned so that $\overline{UV}\parallel\overline{BC}$, $\overline{WX}\parallel\overline{AB}$, and $\overline{YZ}\parallel\overline{CA}$. Right angle folds are then made along $\overline{UV}$, $\overline{WX}$, and $\overline{YZ}$. The resulting figure is placed on a level floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n}$, where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k+m+n$.

$[asy] unitsize(1 cm); pair translate; pair[] A, B, C, U, V, W, X, Y, Z; A[0] = (1.5,2.8); B[0] = (3.2,0); C[0] = (0,0); U[0] = (0.69*A[0] + 0.31*B[0]); V[0] = (0.69*A[0] + 0.31*C[0]); W[0] = (0.69*C[0] + 0.31*A[0]); X[0] = (0.69*C[0] + 0.31*B[0]); Y[0] = (0.69*B[0] + 0.31*C[0]); Z[0] = (0.69*B[0] + 0.31*A[0]); translate = (7,0); A[1] = (1.3,1.1) + translate; B[1] = (2.4,-0.7) + translate; C[1] = (0.6,-0.7) + translate; U[1] = U[0] + translate; V[1] = V[0] + translate; W[1] = W[0] + translate; X[1] = X[0] + translate; Y[1] = Y[0] + translate; Z[1] = Z[0] + translate; draw (A[0]--B[0]--C[0]--cycle); draw (U[0]--V[0],dashed); draw (W[0]--X[0],dashed); draw (Y[0]--Z[0],dashed); draw (U[1]--V[1]--W[1]--X[1]--Y[1]--Z[1]--cycle); draw (U[1]--A[1]--V[1],dashed); draw (W[1]--C[1]--X[1]); draw (Y[1]--B[1]--Z[1]); dot("$A$",A[0],N); dot("$B$",B[0],SE); dot("$C$",C[0],SW); dot("$U$",U[0],NE); dot("$V$",V[0],NW); dot("$W$",W[0],NW); dot("$X$",X[0],S); dot("$Y$",Y[0],S); dot("$Z$",Z[0],NE); dot(A[1]); dot(B[1]); dot(C[1]); dot("$U$",U[1],NE); dot("$V$",V[1],NW); dot("$W$",W[1],NW); dot("$X$",X[1],dir(-70)); dot("$Y$",Y[1],dir(250)); dot("$Z$",Z[1],NE);[/asy]$

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