Difference between revisions of "2011 AIME I Problems/Problem 8"
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dot("$Z$",Z[1],NE);</asy></center> | dot("$Z$",Z[1],NE);</asy></center> | ||
− | ==Solution== | + | ==Solution 1== |
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Note that the area is given by Heron's formula and it is <math>20\sqrt{221}</math>. Let <math>h_i</math> denote the length of the altitude dropped from vertex i. It follows that <math>h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}, h_a = \frac{40\sqrt{221}}{23}</math>. From similar triangles we can see that <math>\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rightarrow h \le \frac{h_ah_c}{h_a+h_c}</math>. We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields <math>h = \frac{40\sqrt{221}}{57} \rightarrow \boxed{318}</math>. | Note that the area is given by Heron's formula and it is <math>20\sqrt{221}</math>. Let <math>h_i</math> denote the length of the altitude dropped from vertex i. It follows that <math>h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}, h_a = \frac{40\sqrt{221}}{23}</math>. From similar triangles we can see that <math>\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rightarrow h \le \frac{h_ah_c}{h_a+h_c}</math>. We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields <math>h = \frac{40\sqrt{221}}{57} \rightarrow \boxed{318}</math>. | ||
− | Solution 2 | + | ==Solution 2== |
As from above, we can see that the length of the altitude from A is the longest. Thus the highest table is formed when X and Y meet up. Let the distance of this point from C be x, then the distance from B will be 23 - x. Let h be the height of the table. From similar triangles, we have <math>\frac{x}{23} = \frac{h}{h_c} = \frac{27h}{2A}</math> where A is the area of triangle ABC. Similarly, <math>\frac{23-x}{23}=\frac{h}{h_b}=\frac{30h}{2A}</math>. Therefore, <math>1-\frac{x}{23}=\frac{30h}{2A} \rightarrow1-\frac{27h}{2A}=\frac{30h}{2A}</math> and hence <math> h = \frac{2A}{57} = \frac{40\sqrt{221}}{57}\rightarrow \boxed{318}</math>. | As from above, we can see that the length of the altitude from A is the longest. Thus the highest table is formed when X and Y meet up. Let the distance of this point from C be x, then the distance from B will be 23 - x. Let h be the height of the table. From similar triangles, we have <math>\frac{x}{23} = \frac{h}{h_c} = \frac{27h}{2A}</math> where A is the area of triangle ABC. Similarly, <math>\frac{23-x}{23}=\frac{h}{h_b}=\frac{30h}{2A}</math>. Therefore, <math>1-\frac{x}{23}=\frac{30h}{2A} \rightarrow1-\frac{27h}{2A}=\frac{30h}{2A}</math> and hence <math> h = \frac{2A}{57} = \frac{40\sqrt{221}}{57}\rightarrow \boxed{318}</math>. |
Revision as of 16:23, 2 January 2016
Contents
[hide]Problem
In triangle ,
,
, and
. Points
and
are on
with
on
, points
and
are on
with
on
, and points
and
are on
with
on
. In addition, the points are positioned so that
,
, and
. Right angle folds are then made along
,
, and
. The resulting figure is placed on a level floor to make a table with triangular legs. Let
be the maximum possible height of a table constructed from triangle
whose top is parallel to the floor. Then
can be written in the form
, where
and
are relatively prime positive integers and
is a positive integer that is not divisible by the square of any prime. Find
.
![[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]](http://latex.artofproblemsolving.com/4/4/f/44fd6e9fb5acf1d28fe1e2acd699bd5be1230685.png)
Solution 1
Note that the area is given by Heron's formula and it is . Let
denote the length of the altitude dropped from vertex i. It follows that
. From similar triangles we can see that
. We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields
.
Solution 2
As from above, we can see that the length of the altitude from A is the longest. Thus the highest table is formed when X and Y meet up. Let the distance of this point from C be x, then the distance from B will be 23 - x. Let h be the height of the table. From similar triangles, we have where A is the area of triangle ABC. Similarly,
. Therefore,
and hence
.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.