Difference between revisions of "2019 AMC 10A Problems/Problem 6"

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==Solution==
 
==Solution==
  
Notice that the lines drawn can be represented by radii of a circle, which are all equidistant from its center <math>O</math>. From this, we can see that point <math>O</math> does not necessarily need to be inside the quadrilateral. Since the quadrilateral must have all its points on Circle <math>O</math>, it is a cyclic quadrilateral. Using the fact that opposite angles sum to <math>180\circ</math>, the only quadrilaterals that fit this description are the square, the rectangle, and the isosceles trapezoid $\rightarrow \boxed{C}.
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Notice that the lines drawn can be represented by radii of a circle, which are all equidistant from its center <math>O</math>. From this, we can see that point <math>O</math> does not necessarily need to be inside the quadrilateral. Since the quadrilateral must have all its points on Circle <math>O</math>, it is a cyclic quadrilateral. Using the fact that opposite angles sum to <math>180\circ</math>, the only quadrilaterals that fit this description are the square, the rectangle, and the isosceles trapezoid <math>\rightarrow \boxed{C}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:21, 9 February 2019

Problem

For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?

  • a square
  • a rectangle that is not a square
  • a rhombus that is not a square
  • a parallelogram that is not a rectangle or a rhombus
  • an isosceles trapezoid that is not a parallelogram

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution

Notice that the lines drawn can be represented by radii of a circle, which are all equidistant from its center $O$. From this, we can see that point $O$ does not necessarily need to be inside the quadrilateral. Since the quadrilateral must have all its points on Circle $O$, it is a cyclic quadrilateral. Using the fact that opposite angles sum to $180\circ$, the only quadrilaterals that fit this description are the square, the rectangle, and the isosceles trapezoid $\rightarrow \boxed{C}$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 10 Problems and Solutions

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