Difference between revisions of "2019 AMC 10A Problems/Problem 6"
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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> | ||
| + | |||
| + | ==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 <math>\boxed{\textbf{(C)}}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 21:18, 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
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 
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 
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 
.
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| 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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