Difference between revisions of "FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 3"

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==Solution==
 
==Solution==
All regular octagons can be inscribed in a circle, thus any subset of <math>4</math> vertice's from this octagon also all lie on the same circle. It is easy to see that none of the other answer choices work. Therefore, the answer is $\boxed{\textbf{(B) }\text{All of the 70 quadrilaterals are cyclic.}}
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All regular octagons can be inscribed in a circle, thus any subset of <math>4</math> vertice's from this octagon also all lie on the same circle. It is easy to see that none of the other answer choices work. Therefore, the answer is <math>\boxed{\textbf{(B) }\text{All of the 70 quadrilaterals are cyclic.}}</math>
  
 
==See also==
 
==See also==

Latest revision as of 16:04, 19 November 2020

Problem

There are $\binom{8}{4}=70$ distinct quadrilaterals that can be formed from the vertices of a regular octagon. Which of these statements must hold true for all those quadrilaterals?

$\textbf{(A) }\text{All of the 70 quadrilaterals are rectangles.}\qquad\textbf{(B) }\text{All of the 70 quadrilaterals are cyclic.}\qquad\textbf{(C) }\text{All of the 70 quadrilaterals are trapezoids.}\qquad\textbf{(D) }\text{All of the 70 quadrilaterals are kites.}\qquad\textbf{(E) }\text{None of the above statements are true.}\qquad$

Solution

All regular octagons can be inscribed in a circle, thus any subset of $4$ vertice's from this octagon also all lie on the same circle. It is easy to see that none of the other answer choices work. Therefore, the answer is $\boxed{\textbf{(B) }\text{All of the 70 quadrilaterals are cyclic.}}$

See also

FidgetBoss 4000's 2019 Mock AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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