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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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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 points on side <math>\overline{AB}</math>; <math>4</math> other points on side <math>\overline{BC}</math>; and <math>5</math> other points on side <math>\overline{CA}</math>. Find the number of triangles with positive area whose vertices are among these <math>15</math> points.
 
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 points on side <math>\overline{AB}</math>; <math>4</math> other points on side <math>\overline{BC}</math>; and <math>5</math> other points on side <math>\overline{CA}</math>. Find the number of triangles with positive area whose vertices are among these <math>15</math> points.
  
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==Solution==
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==Solution 1==
 
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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==Solution 2==
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We simply choose <math>3</math> points from <math>15</math> to begin with since a triangle consists of <math>3</math> points and there are <math>15</math> points total. That gives us <math>\binom{15}{3}</math>. Now, we need to subtract of the degenerate triangles. These are just the triangles that are collinear. We can count the number of degenerate triangles on each side of <math>ABC</math>, then subtract is from <math>\binom{15}{3}</math>. For <math>\overline{AB}</math>, we have <math>\binom{5}{3}</math> degenerate triangles (dont forget <math>A</math> and <math>B</math>), for <math>\overline{BC}</math>, we have <math>\binom{6}{3}</math> degenerate triangles, and for <math>\overline{AC}</math>, we have <math>\binom{7}{3}</math> degenerate triangles. So, our final answer is <math>\binom{15}{3} - \binom{7}{3} - \binom{6}{3} - \binom{5}{3} = \boxed{390}</math>
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-jb2015007
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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 01:18, 10 July 2025

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 1

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 $455-65=\boxed{390}$.

Solution 2

We simply choose $3$ points from $15$ to begin with since a triangle consists of $3$ points and there are $15$ points total. That gives us $\binom{15}{3}$. Now, we need to subtract of the degenerate triangles. These are just the triangles that are collinear. We can count the number of degenerate triangles on each side of $ABC$, then subtract is from $\binom{15}{3}$. For $\overline{AB}$, we have $\binom{5}{3}$ degenerate triangles (dont forget $A$ and $B$), for $\overline{BC}$, we have $\binom{6}{3}$ degenerate triangles, and for $\overline{AC}$, we have $\binom{7}{3}$ degenerate triangles. So, our final answer is $\binom{15}{3} - \binom{7}{3} - \binom{6}{3} - \binom{5}{3} = \boxed{390}$

-jb2015007

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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