Difference between revisions of "2021 JMPSC Sprint Problems/Problem 7"

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==Solution==
 
==Solution==
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Consider the equilateral triangle that the circles make. There are <math>3</math> ways to pick the circles so that the row they make is parallel to each side of the equilateral triangle. Thus, there are <math>\boxed{9}</math> ways total. [diagram needed]
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~tigerzhang
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==Solution 2 (kind of adds on to solution 1)==
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Since there are 2 ways to place 3 circles in a row in a row of 4 circles, and 1 way to put three circles in a row in a row of three circles. There are 3 ways for each side of the equilateral triangle. Therefore, <math>3</math> * <math>(2+1)</math> = <math>\boxed{9}</math> ways total.
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-Chenz1 (aka doge 34)
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==See also==
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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 09:53, 16 July 2021

Problem

How many sets of three circles are in the diagram below such that their centers lie on the same line and each circle in the set is adjacent to another circle in the set? (Assume that if the centers of three circles appear to be on the same line, then they are on the same line. For example, the three red shaded circles count as one such set.)

Sprint9.png

Solution

Consider the equilateral triangle that the circles make. There are $3$ ways to pick the circles so that the row they make is parallel to each side of the equilateral triangle. Thus, there are $\boxed{9}$ ways total. [diagram needed]

~tigerzhang

Solution 2 (kind of adds on to solution 1)

Since there are 2 ways to place 3 circles in a row in a row of 4 circles, and 1 way to put three circles in a row in a row of three circles. There are 3 ways for each side of the equilateral triangle. Therefore, $3$ * $(2+1)$ = $\boxed{9}$ ways total.

-Chenz1 (aka doge 34)

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png