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

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*an isosceles trapezoid that is not a parallelogram
 
*an isosceles trapezoid that is not a parallelogram
  
<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5</math>
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<math>\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5</math>
  
==Solution==
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==Solution 1==
This question is simply asking how many of the listed quadrilaterals are cyclic. A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is <math>3 \implies \boxed{C}.</math>
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This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is <math>\boxed{\textbf{(C) } 3}</math>.
  
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>.
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==Solution 2==
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We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is <math>\boxed{\textbf{(C) } 3}</math>.
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==Solution 3==
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The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is <math>\boxed{\textbf{(C) } 3}</math>.
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==Video Solution==
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https://youtu.be/_BaA86QR5ws
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Education, The Study of Everything
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==Video Solution==
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https://youtu.be/8LldR2lCmV8
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 21:58, 1 February 2021

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

$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution 1

This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is $\boxed{\textbf{(C) } 3}$.

Solution 2

We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is $\boxed{\textbf{(C) } 3}$.

Solution 3

The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is $\boxed{\textbf{(C) } 3}$.

Video Solution

https://youtu.be/_BaA86QR5ws

Education, The Study of Everything



Video Solution

https://youtu.be/8LldR2lCmV8

~savannahsolver

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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