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

(Solution: Added solution (basic POE))
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==Solution 2==
 
==Solution 2==
 
We can use the process of elimination. Going down, we can see a square obviously applies. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point where the lines are equidistant. But, isosceles trapezoids DO have a point, so the answer is <math>\boxed{3(C)}</math>
 
We can use the process of elimination. Going down, we can see a square obviously applies. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point where the lines are equidistant. But, isosceles trapezoids DO have a point, so the answer is <math>\boxed{3(C)}</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)}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 22:18, 12 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

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 $3 \implies \boxed{\textbf{(C)}}.$

Solution 2

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

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 $\boxed{\textbf{(C)}}$.

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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