Difference between revisions of "2011 AIME I Problems/Problem 8"
Jabberwock2 (talk | contribs) m |
LightningX48 (talk | contribs) (→Solution) |
||
Line 54: | Line 54: | ||
==Solution== | ==Solution== | ||
− | 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 | + | 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>. |
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:22, 1 March 2015
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 .
Solution
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 $h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}}, h_a = \frac{40\sqrt{221}{23}$ (Error compiling LaTeX. ! Extra }, or forgotten $.). 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 .
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.