Difference between revisions of "2011 AIME I Problems/Problem 8"
Sugar rush (talk | contribs) |
Sugar rush (talk | contribs) |
||
Line 57: | Line 57: | ||
==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>. | ||
Line 65: | Line 94: | ||
== 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 .
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.