Triangle inscribed in a square

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Triangle inscribed in a square

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Source: Problem 4, Polish NO 1990

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Megus

#1 h Sep 30, 2005, 5:39 PM • 1 Y

Y by Adventure10

A triangle whose all sides have length not smaller than $1$ is inscribed in a square of side length $1$ . Prove that the center of the square lies inside the triangle or on its boundary.

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#2 h Oct 1, 2005, 12:12 AM • 2 Y

Y by Adventure10, Mango247

If a triangle XYZ doesn't contain the center of the square, there is a line through the center of the square that doesn't intersect XYZ.
Proof: Suppose that every line through the center intersects XYZ. If the four vertices of a convex quadrilateral are part of a convex figure, then all points that line inside that quadrilateral are part of the convex figure. By taking two lines through the origin and finding their intersections with the triangle, we can find a convex quadrilateral containing the origin, inside of XYZ.

Draw a line perpendicular to the line through the center of the square, as shown in the diagram.

All sides of the triangle must lie in the union of the blue and yellow regions.

The quadrilaterals surrounding the blue and yellow regions are concyclic, inscribable in circles of diameter $\sqrt{2}/2$ . Any line segment of length 1 that lies in the union of the colored regions must lie in both, since $1>\sqrt{2}/2$

If X lies in the blue region, then Y lies in the yellow region, then Z lies in the blue region, then X lies in the yellow region, a contradiction.

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#3 h Feb 21, 2006, 8:15 AM • 2 Y

Y by Adventure10, Mango247

schulmannerism wrote:

By taking two lines through the origin and finding their intersections with the triangle, we can find a convex quadrilateral containing the origin, inside of XYZ.

Is it always that there must be a quadrilateral containing the origin?

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