Question

In the accompanying figure, the outer square $S$ has side length $40.$ A second square $S'$ of side length $15$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S.$ From each midpoint of a side of $S,$ segments are drawn to the two closest vertices of $S'.$ The result is a four-pointed starlike figure inscribed in $S.$ The star figure is cut out and then folded to form a pyramid with base $S'.$ Find the volume of this pyramid.

Answer

Answer: 750

Concepts:net (3D geometry)pyramid volume Pythagorean Theorem

Difficulty rating: 2230

Solution:

Folding the star along the sides of $S'$ lifts the four triangular points so that their tips (the midpoints of the sides of $S$ ) meet at a single apex $V.$ Let $M$ be the center of $S'$ and $P$ the midpoint of one of its sides. In the flat figure, the distance from $P$ to the tip of its triangle is $20 - \frac{15}{2} = \frac{25}{2},$ and this becomes the slant $PV$ after folding.

Triangle $PMV$ has a right angle at $M,$ with $PM = \frac{15}{2},$ so the height is $\begin{aligned} VM &= \sqrt{\left(\tfrac{25}{2}\right)^2 - \left(\tfrac{15}{2}\right)^2} \\ &= \sqrt{100} = 10. \end{aligned}$ The volume is $\frac{1}{3} \cdot 15^2 \cdot 10 = 750.$

2012 AIME II Problem 5

Answer: 750

Difficulty rating: 2230

Solution:

Problem 5 in Other Years