Difference between revisions of "2012 AIME II Problems/Problem 5"
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<math>h=\sqrt{12.5^2-7.5^2}=\sqrt{100}=10</math> | <math>h=\sqrt{12.5^2-7.5^2}=\sqrt{100}=10</math> | ||
− | Finally, <math>V=\frac{1}{3}bh=\frac{1}{3}(225)(10)=\boxed{750 | + | Finally, <math>V=\frac{1}{3}bh=\frac{1}{3}(225)(10)=\boxed{750}</math>. |
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=4|num-a=6}} | {{AIME box|year=2012|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:34, 29 February 2020
Problem 5
In the accompanying figure, the outer square has side length . A second square of side length is constructed inside with the same center as and with sides parallel to those of . From each midpoint of a side of , segments are drawn to the two closest vertices of . The result is a four-pointed starlike figure inscribed in . The star figure is cut out and then folded to form a pyramid with base . Find the volume of this pyramid.
Solution
The volume of this pyramid can be found by the equation , where is the base and is the height. The base is easy, since it is a square and has area .
To find the height of the pyramid, the height of the four triangles is needed, which will be called . 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 , which means that .
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 and the other leg having length equal to half of the side length of the smaller square, or . So, the Pythagorean Theorem can be used to find the height.
Finally, .
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.