Difference between revisions of "2019 AMC 10A Problems/Problem 6"
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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>. | + | 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. Say we inscribe trapezoid ABCD in a semicircle. This point would thus be the center, so the answer is <math>\boxed{\textbf{(C) } 3}</math>. |
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| + | ~proof for trapezoid added by yvz2900 | ||
===Solution 3=== | ===Solution 3=== | ||
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{{AMC10 box|year=2019|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2019|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category: Introductory Geometry Problems]] | ||
Latest revision as of 14:11, 20 October 2025
Contents
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
Solutions
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 
.
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. Say we inscribe trapezoid ABCD in a semicircle. This point would thus be the center, so the answer is .
~proof for trapezoid added by yvz2900
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 .
Video Solutions
Video Solution by Education, the Study of Everything
~Education, The Study of Everything
Video Solution by WhyMath
~savannahsolver
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
