Difference between revisions of "2002 AMC 12P Problems/Problem 16"
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== Problem == | == Problem == | ||
− | + | The altitudes of a triangle are <math>12, 15,</math> and <math>20.</math> The largest angle in this triangle is | |
− | <math> \ | + | <math> |
+ | \text{(A) }72^\circ | ||
+ | \qquad | ||
+ | \text{(B) }75^\circ | ||
+ | \qquad | ||
+ | \text{(C) }90^\circ | ||
+ | \qquad | ||
+ | \text{(D) }108^\circ | ||
+ | \qquad | ||
+ | \text{(E) }120^\circ | ||
+ | </math> | ||
== Solution == | == Solution == |
Revision as of 23:51, 29 December 2023
Problem
The altitudes of a triangle are and The largest angle in this triangle is
Solution
If , then . Since , must be to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of .
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.