any ideas ?

Geometry Level 5

Ans to 2 decimal places.

The answer is 0.57.

1 solution

Marta Reece
Apr 19, 2017

Triangle $OAB$ is equilateral with side $2$ . Points closer to $O$ than to either of the other vertices are inside the quadrilateral $OCDE$ pictured above.

The quadrilateral $OCDE$ is composed of two right triangles.

In the lower one $OC=1, CD=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$

Area of $OCDE=2\times\frac{1}{2}\times 1\times \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}\approx0.58$

Just wondering, without so much symmetry involved, if it would have been random scalene triangle (acute, obtuse or right), then what would be the most general way to approach a problem like this?

Tapas Mazumdar - 4 years, 1 month ago

Generally you would be constructing the area containing points closer to a given vertex by making perpendicular lines at midpoints.

Marta Reece - 4 years, 1 month ago

Great visual solution! Thank you so much. However, there's one piece that needs to be put in, i.e., a good mathematical proof.

Tapas Mazumdar - 4 years, 1 month ago

any ideas ?

The answer is 0.57.

1 solution

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