AoPS Wiki talk:Problem of the Day/June 11, 2011

Problem

AoPSWiki:Problem of the Day/June 11, 2011

Solution

$[asy] //size(280); pointpen=black; pathpen = black; linewidth(1); pair A=(0,8), B=(8,8), C=(8,0), D=(0,0), E, F, G, H, J, K, L, M, X; E=(A+B)/2; F=(B+C)/2; G=(C+D)/2; H=(A+D)/2; X=(E+G)/2; J=(X+E)/2; K=(X+F)/2; L=(X+G)/2; M=(X+H)/2; D(A--B--K--L--D--M--J--B--C--L--M--A--J--K--C--D--A);D(H--F);D(E--G); MP("A",A,(-1, 1));MP("B",B,( 1, 1));MP("C",C,( 1,-1));MP("D",D,(-1,-1)); MP("E",E,( 0, 1));MP("F",F,( 1, 0));MP("G",G,( 0,-1));MP("H",H,(-1, 0)); MP("J",J,( 3, 0));MP("K",K,(0,3));MP("L",L,( 3, 0));MP("M",M,(0,3)); MP("X",X,(-1,-1)); [/asy]$

If one were to find the volume of the pyramid, you would need to look for the area of the base and the height.

The area of the base is square $JKLM$ .

$(\dfrac{1}{2})^3=\frac{1}{8}$

To find the height, you would have to apply the Pythagorean Theorem on a triangle with vertices that are the apex, point X, and a vertex of square JKLM. This triangle however is identical to triangle $AEJ$ . Thus, the height of the pyramid is equal to $\dfrac{1}{2}$ .

Therefore, the volume of the pyramid is :

$\dfrac{1}{8}\times\dfrac{1}{2}\times\dfrac{1}{3}=\boxed{\dfrac{1}{48}}$

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AoPS Wiki talk:Problem of the Day/June 11, 2011

Problem

Solution