Difference between revisions of "2017 AIME I Problems/Problem 1"
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==Solution== | ==Solution== | ||
Every triangle is uniquely determined by 3 points. There are <math>\binom{15}{3}=455</math> ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are <math>\binom{5}{3}</math> invalid cases on segment <math>AB</math>, <math>\binom{6}{3}</math> invalid cases on segment <math>BC</math>, and <math>\binom{7}{3}</math> invalid cases on segment <math>CA</math> for a total of <math>65</math> invalid cases. The answer is thus <math>455-65=\boxed{390}</math>. | Every triangle is uniquely determined by 3 points. There are <math>\binom{15}{3}=455</math> ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are <math>\binom{5}{3}</math> invalid cases on segment <math>AB</math>, <math>\binom{6}{3}</math> invalid cases on segment <math>BC</math>, and <math>\binom{7}{3}</math> invalid cases on segment <math>CA</math> for a total of <math>65</math> invalid cases. The answer is thus <math>455-65=\boxed{390}</math>. | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/BiiKzctXDJg | ||
| + | ~Shreyas S | ||
| + | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2017|n=I|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:00, 17 June 2020
Contents
Problem 1
Fifteen distinct points are designated on 
: the 3 vertices 
, 
, and 
; 
other points on side 
; 
other points on side 
; and 
other points on side 
. Find the number of triangles with positive area whose vertices are among these 
points.
Solution
Every triangle is uniquely determined by 3 points. There are 
ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are 
invalid cases on segment 
, 
invalid cases on segment 
, and 
invalid cases on segment 
for a total of 
invalid cases. The answer is thus 
.
Video Solution
https://youtu.be/BiiKzctXDJg ~Shreyas S
See Also
| 2017 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
