Boxed In

Geometry Level 3

A square with side $1$ falls on top of an open top square box with side $1$ (pictured above). Assuming the square doesn't fall completely in the box, find the smallest value for $x$ , where $x$ is the the line that connects the vertex of the square inside the box to the box base perpendicularly.

The answer is 0.5.

1 solution

Michael Fuller
Mar 28, 2015

As the square falls on top of the box, a right angled triangle forms. The hypotenuse must be $1$ , and the legs could be a variety of lengths. As we know, angles in a semicircle are $90°$ so the vertex of the triangle could lie anywhere on the semi-circle pictured above.

To find the smallest value for $x$ , we must find the largest value for the depth of this triangle, and subtract that value from $1$ . The maximum depth of the semicircle is its radius, which is of course half of the diameter ( $1$ ).

$x=1-\frac { 1 }{ 2 } \\ x=\large \color{#20A900}{\boxed { \frac { 1 }{ 2 }} }$

1 < sqrt(2), since sqrt(2) < 2 therefore, the hypotenuse can be sqrt(2), right ?

Rony C B - 6 years, 2 months ago

The hypotenuse must equal $1$ because the vertex angle of the square that lies inside the box is always $90°$ , so the hypotenuse must be the side opposite to that angle, which is the box side that has a fixed length of $1$ .

Michael Fuller - 6 years, 2 months ago

Boxed In

The answer is 0.5.

1 solution

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