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 | + | 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 17:21, 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
Solution
Notice that the lines drawn can be represented by radii of a circle, which are all equidistant from its center 
. From this, we can see that point does not necessarily need to be inside the quadrilateral. Since the quadrilateral must have all its points on Circle , it is a cyclic quadrilateral. Using the fact that opposite angles sum to 
, the only quadrilaterals that fit this description are the square, the rectangle, and the isosceles trapezoid 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
