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Difference between revisions of "2017 UNCO Math Contest II Problems/Problem 9"

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triangles whose sides lie along the segments, including
 
triangles whose sides lie along the segments, including
 
triangles that overlap with other triangles. For example,
 
triangles that overlap with other triangles. For example,
−
for n = 3 there is one triangle and for n = 4 (shown
+
for n = 3, there is one triangle and for n = 4 (shown
−
in the diagram) there are 8 triangles.
+
in the diagram), there are 8 triangles.
  
 
== Solution ==
 
== Solution ==
−
<math>{n}\choose{6}</math> + 5<math>{n}\choose{5}</math> + 4<math>{n}\choose{4}</math> + <math>{n}\choose{3}</math>
+
Since any <math>3</math> pairwise intersecting [[line]]s make a triangle, so we can split the possibilities into <math>4</math> cases:
 +
 
 +
'''Case 1: All vertices of the triangle are points on the circle'''
 +
 
 +
Any <math>3</math> points will work, so it's <math>{n}\choose{3}</math>
 +
 
 +
<asy>
 +
 
 +
pair A=dir(0),B=dir(80),C=dir(160);
 +
draw(unitcircle);
 +
draw(A--B--C--A);
 +
 
 +
</asy>
 +
 
 +
 
 +
'''Case 2: 2 vertices of the triangle are points on the circle'''
 +
 
 +
There will be <math>4</math> points to choose from, but you also have to multiply by the number of orientations (<math>4</math>), so <math>4</math><math>{n}\choose{4}</math>
 +
 
 +
<asy>
 +
 
 +
pair A=dir(0),B=dir(80),C=dir(160),D=dir(310);
 +
draw(unitcircle);
 +
draw(A--C--B--D);
 +
 
 +
</asy>
 +
 
 +
'''Case 3: 1 vertex of the triangle is a point on the circle'''
 +
 
 +
There will be <math>5</math> points to choose from, but you also have to multiply by the number of orientations (<math>5</math>), so <math>5</math><math>{n}\choose{5}</math>
 +
 
 +
<asy>
 +
 
 +
pair A=dir(0),B=dir(80),C=dir(160),D=dir(210),E=dir(300);
 +
draw(unitcircle);
 +
draw(A--C--E);
 +
draw(B--D);
 +
 
 +
</asy>
 +
 
 +
'''Case 4: All vertices of the triangle are points on the circle'''
 +
 
 +
Any <math>6</math> points will work, so it's <math>{n}\choose{6}</math>
 +
 
 +
<asy>
 +
 
 +
pair A=dir(0),B=dir(130),C=dir(160),D=dir(210),E=dir(270),F=dir(320);
 +
draw(unitcircle);
 +
draw(A--D);
 +
draw(B--E);
 +
draw(C--F);
 +
 
 +
</asy>
 +
 
 +
Thus there are a total of <math>\boxed{\binom{n}{6}+5\binom{n}{5}+4\binom{n}{4}+\binom{n}{3}}</math> different triangles.
  
 
== See also ==
 
== See also ==
−
{{UNCO Math Contest box|year=2017|n=II|num-b=8|num-a=9}}
+
{{UNCO Math Contest box|year=2017|n=II|num-b=8|num-a=10}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 23:45, 18 January 2023

Problem

[asy] pair A=dir(0),B=dir(80),C=dir(160),D=dir(310),O=(0,0); draw(circle(O,1),black); draw(A--B--C--D--A--C--B--D); [/asy]

Suppose n points on the circumference of a circle are joined by straight line segments in all possible ways and that no point that is not one of the original n points is contained in more than two of the segments. How many triangles are formed by the segments? Count all triangles whose sides lie along the segments, including triangles that overlap with other triangles. For example, for n = 3, there is one triangle and for n = 4 (shown in the diagram), there are 8 triangles.

Solution

Since any $3$ pairwise intersecting lines make a triangle, so we can split the possibilities into $4$ cases:

Case 1: All vertices of the triangle are points on the circle

Any $3$ points will work, so it's ${n}\choose{3}$

[asy] pair A=dir(0),B=dir(80),C=dir(160); draw(unitcircle); draw(A--B--C--A); [/asy]


Case 2: 2 vertices of the triangle are points on the circle

There will be $4$ points to choose from, but you also have to multiply by the number of orientations ($4$), so $4$${n}\choose{4}$

[asy] pair A=dir(0),B=dir(80),C=dir(160),D=dir(310); draw(unitcircle); draw(A--C--B--D); [/asy]

Case 3: 1 vertex of the triangle is a point on the circle

There will be $5$ points to choose from, but you also have to multiply by the number of orientations ($5$), so $5$${n}\choose{5}$

[asy] pair A=dir(0),B=dir(80),C=dir(160),D=dir(210),E=dir(300); draw(unitcircle); draw(A--C--E); draw(B--D); [/asy]

Case 4: All vertices of the triangle are points on the circle

Any $6$ points will work, so it's ${n}\choose{6}$

[asy] pair A=dir(0),B=dir(130),C=dir(160),D=dir(210),E=dir(270),F=dir(320); draw(unitcircle); draw(A--D); draw(B--E); draw(C--F); [/asy]

Thus there are a total of $\boxed{\binom{n}{6}+5\binom{n}{5}+4\binom{n}{4}+\binom{n}{3}}$ different triangles.

See also

2017 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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