# Problem Of The Week # 311 - May 31, 2018

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#### Ackbach

##### Indicium Physicus
Staff member
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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#### Ackbach

##### Indicium Physicus
Staff member
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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