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Problem Of The Week # 311 - May 31, 2018

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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
Here is this week's POTW:

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Determine the volume of a tetrahedron $ABCD$ if
$$\overline{AB}=\overline{AC}=\overline{AD}=5$$
and
$$\overline{BC}=3, \; \overline{CD}=4,\;\overline{DB}=5.$$

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
Congratulations to castor28 for his correct solution to this week's POTW, which was Problem 307 in the MAA Challenges. His solution follows:

By the converse of Pythagoras’ theorem, the triangle $BCD$ is a right triangle with hypotenuse $DB$; the area of that triangle is equal to 6.

From $A$, we draw the perpendicular to the plane $BCD$, intersecting the plane at $O$; $AO$ is the altitude of the pyramid relative to the base $BCD$.

The right triangles $AOB$, $AOC$, and $AOD$ are congruent, since they share the side $AO$ and their hypotenuses are equal. This shows that $BO=CO=DO$, and $O$ is the center of the circumcircle of the triangle $BCD$. As that triangle is a right triangle, $O$ is the midpoint of the hypotenuse $DB$, and $BO=\dfrac52$. This gives:
$$AO=\sqrt{AB^2-BO^2}=\frac{5\sqrt3}{2}$$
and the volume of the pyramid is equal to:
$$\frac13\times6\times\frac{5\sqrt3}{2}=5\sqrt3$$
 
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