Question

The length of a rectangle is $5$ inches and its width is less than $4$ inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is $\sqrt6,$ then the width is:

$\sqrt2$

$\sqrt3$

$2$

$\sqrt5$

$\sqrt{\frac{11}{2}}$

Answer: D

Concepts:paper folding perpendicular bisector coordinate geometry Pythagorean Theorem

Difficulty rating: 2290

Small Hint:

The crease is the perpendicular bisector of a rectangle diagonal

Big Hint:

If the width is $w\lt4,$ its intersections with the long sides differ horizontally by $\frac{w^2}{5}$

Solution:

Place the rectangle at $(0,0),$ $(5,0),$ $(5,w),$ and $(0,w).$ The crease sending $(0,0)$ to $(5,w)$ is $5x+wy=\frac{25+w^2}{2}.$ Because $w\lt4,$ it meets the two horizontal sides. Between those intersections the vertical change is $w$ and the horizontal change is $\frac{w^2}{5}.$ Thus $6=w^2+\frac{w^4}{25}.$ Setting $z=w^2$ gives $z^2+25z-150=0,$ so $z=5$ and $w=\sqrt5.$

Therefore, the correct answer is D.

1965 AMC 12 Problem 35

Answer: D

Difficulty rating: 2290

Small Hint:

Big Hint:

Solution:

Problem 35 in Other Years