AIME 2012 II · 第 5 题

AIME 2012 II — Problem 5

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem 5

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.

$AIME diagram$

解析

Solution

The volume of this pyramid can be found by the equation $V=\frac{1}{3}bh$ , where $b$ is the base and $h$ is the height. The base is easy, since it is a square and has area $15^2=225$ .

To find the height of the pyramid, the height of the four triangles is needed, which will be called $h^\prime$ . By drawing a line through the middle of the larger square, we see that its length is equal to the length of the smaller rectangle and two of the triangle's heights. Then $40=2h^\prime +15$ , which means that $h^\prime=12.5$ .

When the pyramid is made, you see that the height is the one of the legs of a right triangle, with the hypotenuse equal to $h^\prime$ and the other leg having length equal to half of the side length of the smaller square, or $7.5$ . So, the Pythagorean Theorem can be used to find the height.

h=\sqrt{12.5^2-7.5^2}=\sqrt{100}=10

Finally, $V=\frac{1}{3}bh=\frac{1}{3}(225)(10)=\boxed{750}$ .