Difference between revisions of "2017 AIME I Problems/Problem 1"

(Created page with "==Problem 1== Fifteen distinct points are designated on <math>\triangle ABC</math>: the 3 vertices <math>A</math>, <math>B</math>, and <math>C</math>; <math>3</math> other poi...")
 
 
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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==
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https://youtu.be/BiiKzctXDJg
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~Shreyas S
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==See Also==
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{{AIME box|year=2017|n=I|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 18:00, 17 June 2020

Problem 1

Fifteen distinct points are designated on $\triangle ABC$: the 3 vertices $A$, $B$, and $C$; $3$ other points on side $\overline{AB}$; $4$ other points on side $\overline{BC}$; and $5$ other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these $15$ points.

Solution

Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$, $\binom{6}{3}$ invalid cases on segment $BC$, and $\binom{7}{3}$ invalid cases on segment $CA$ for a total of $65$ invalid cases. The answer is thus $455-65=\boxed{390}$.

Video Solution

https://youtu.be/BiiKzctXDJg ~Shreyas S


See Also

2017 AIME I (ProblemsAnswer KeyResources)
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

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