A Walk on the Coordinate Plane

Geometry Level 3

Bob is walking on the coordinate plane. He walks from (-1,0) to (1,0) along path that is 3 units long. There exists a smallest closed curve which is guaranteed to contain Bob's path in its interior and boundary. What is the area enclosed by this curve?


The answer is 5.269.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ron Nissim
Jun 9, 2018

The most direct path from ( 1 , 0 ) (-1,0) to ( 1 , 0 ) (1,0) that passes through another point ( x , y ) (x,y) is a straight line path from ( 1 , 0 ) (-1,0) to ( x , y ) (x,y) followed by a straight line from ( x , y ) (x,y) to ( 1 , 0 ) (1,0) . Thus the curve described in the problem is the locus of all points who's sum of the distances from ( 1 , 0 ) (-1,0) and ( 1 , 0 ) (1,0) is 3 3 . This curve is an ellipse with foci at ( 1 , 0 ) (-1,0) and ( 1 , 0 ) (1,0) , semi-major axis 3 2 \frac{3}{2} , and semi-minor axis ( 3 2 ) 2 1 2 = 5 2 \sqrt{(\frac{3}{2})^2-1^2}=\frac{\sqrt{5}}{2} . So the area enclosed by the ellipse is π ( 3 2 ) ( 5 2 ) = 3 5 π 4 5.269 \pi(\frac{3}{2})(\frac{\sqrt{5}}{2})=\frac{3\sqrt{5}\pi}{4}\approx 5.269 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...