Difference between revisions of "2011 AIME I Problems/Problem 8"
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==Solution 1== | ==Solution 1== | ||
+ | Note that triangles <math>\triangle AUV, \triangle BYZ</math> and <math>\triangle CWX</math> all have the same height because when they are folded up to create the legs of the table, the top needs to be parallel to the floor. We want to find the maximum possible value of this height, given that no two of <math>\overline{UV}, \overline{WX}</math> and <math>\overline{YZ}</math> do not intersect inside <math>\triangle ABC</math>. Let <math>h_{A}</math> denote the length of the altitude dropped from vertex <math>A,</math> and define <math>h_{B}</math> and <math>h_{C}</math> similarly. Also let <math>\{u, v, w, x, y, z\} = \{AU, AV, CW, CX, BY, BZ\}</math>. Then by similar triangles | ||
+ | <cmath>\begin{align} | ||
+ | \frac{u}{AB}=\frac{v}{AC}=\frac{h}{h_{A}}, \\ | ||
+ | \frac{w}{CA}=\frac{x}{CB}=\frac{h}{h_{C}}, \\ | ||
+ | \frac{y}{BC}=\frac{z}{BA}=\frac{h}{h_{B}}. | ||
+ | \end{align}</cmath> | ||
+ | Since <math>h_{A}=\frac{2K}{23}</math> and similarly for <math>27</math> and <math>30,</math> where <math>K</math> is the area of <math>\triangle ABC,</math> we can write | ||
+ | <cmath>\begin{align} | ||
+ | \frac{u}{30}=\frac{v}{27}=\frac{h}{\tfrac{2K}{23}}, \\ | ||
+ | \frac{w}{27}=\frac{x}{23}=\frac{h}{\tfrac{2K}{30}}, \\ | ||
+ | \frac{y}{23}=\frac{z}{30}=\frac{h}{\tfrac{2K}{27}}. | ||
+ | \end{align}</cmath> | ||
+ | and simplifying gives <math>u=x=\frac{690h}{2K}, v=y=\frac{621h}{2K}, w=z=\frac{810h}{2K}</math>. Because no two segments can intersect inside the triangle, we can form the inequalities <math>v+w\leq 27, x+y\leq 23,</math> and <math>z+u\leq 30</math>. That is, all three of the inequalities | ||
+ | <cmath>\begin{align} | ||
+ | \frac{621h+810h}{2K}\leq 27, \\ | ||
+ | \frac{690h+621h}{2K}\leq 23, \\ | ||
+ | \frac{810h+690h}{2K}\leq 30. | ||
+ | \end{align}</cmath> | ||
+ | must hold. Dividing both sides of each equation by the RHS, we have | ||
+ | <cmath>\begin{align} | ||
+ | \frac{53h}{2K}\leq 1\, \text{since}\, \frac{1431}{27}=53, \\ | ||
+ | \frac{57h}{2K}\leq 1\, \text{since}\, \frac{1311}{23}=57, \\ | ||
+ | \frac{50h}{2K}\leq 1\, \text{since}\, \frac{1500}{30}=50. | ||
+ | \end{align}</cmath> | ||
+ | It is relatively easy to see that <math>\frac{57h}{2K}\leq 1</math> restricts us the most since it cannot hold if the other two do not hold. The largest possible value of <math>h</math> is thus <math>\frac{2K}{57},</math> and note that by Heron's formula the area of <math>\triangle ABC</math> is <math>20\sqrt{221}</math>. Then <math>\frac{2K}{57}=\frac{40\sqrt{221}}{57},</math> and the answer is <math>40+221+57=261+57=\boxed{318}</math> | ||
+ | |||
+ | ~sugar_rush | ||
+ | |||
+ | ==Solution 2== | ||
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 | + | ==Solution 3== |
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 B be <math>x</math>, making the distance from C <math>23 - x</math>. Let <math>h</math> be the height of the table. From similar triangles, we have <math>\frac{x}{23} = \frac{h}{h_b} = \frac{27h}{2A}</math> where A is the area of ABC. Similarly, <math>\frac{23-x}{23}=\frac{h}{h_c}=\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 B be <math>x</math>, making the distance from C <math>23 - x</math>. Let <math>h</math> be the height of the table. From similar triangles, we have <math>\frac{x}{23} = \frac{h}{h_b} = \frac{27h}{2A}</math> where A is the area of ABC. Similarly, <math>\frac{23-x}{23}=\frac{h}{h_c}=\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>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=7|num-a=9}} | {{AIME box|year=2011|n=I|num-b=7|num-a=9}} | ||
− | + | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
[[Category:3D Geometry Problems]] | [[Category:3D Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 02:34, 5 March 2021
Contents
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)
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=Asvy1s-6Rx0
Solution 1
Note that triangles and
all have the same height because when they are folded up to create the legs of the table, the top needs to be parallel to the floor. We want to find the maximum possible value of this height, given that no two of
and
do not intersect inside
. Let
denote the length of the altitude dropped from vertex
and define
and
similarly. Also let
. Then by similar triangles
Since
and similarly for
and
where
is the area of
we can write
and simplifying gives
. Because no two segments can intersect inside the triangle, we can form the inequalities
and
. That is, all three of the inequalities
must hold. Dividing both sides of each equation by the RHS, we have
It is relatively easy to see that
restricts us the most since it cannot hold if the other two do not hold. The largest possible value of
is thus
and note that by Heron's formula the area of
is
. Then
and the answer is
~sugar_rush
Solution 2
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 3
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 B be , making the distance from C
. Let
be the height of the table. From similar triangles, we have
where A is the area of 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.